问:
About a property of stopping times
I was reading some Probability Theory notes and I saw a proof of this result :

Lemma : Let ${(\Omega,\mathcal{F},\mathbb{P})}$ be a probability space equipped with the (discrete) filtration $(\mathcal{F}_n)$. For any stopping time $\tau$ on that probability space, we define the stopped $\sigma$-algebra as ${\mathcal{F}_\tau:=\{A\in\mathcal{F}:\forall n,\{\tau=n\}\cap A\in\mathcal{F}_n\}}$.

For any random variable $X$ with finite expectation, and for any $n \in \mathbb{N}$, we have that
$$ \mathbb{E}(X|\mathcal{F}_\tau)\mathbf{1}_{\{\tau=n\}} =\mathbb{E}(X|\mathcal{F}_n)\mathbf{1}_{\{\tau=n\}}$$

What I'm having trouble with is the proof of this result, it goes like this in the notes :

Proof : $\forall n \in \mathbb{N}, \forall A \in \mathcal{F}_n$, we have by definition that $ {A\cap\{\tau=n\}\in\mathcal{F}_\tau} $, and we also have that ${\{\tau=n\}\in\mathcal{F}_\tau} $

From which it follows that
$$ \mathbb{E}(\mathbf{1}_AX\mathbf{1}_{\{\tau=n\}}) =\mathbb{E}(\mathbf{1}_{A\cap\{\tau=n\}}\mathbb{E}(X|\mathcal{F}_\tau)) =\mathbb{E}(\mathbf{1}_A\mathbb{E}(X\mathbf{1}_{\{\tau=n\}}|\mathcal{F}_\tau)) $$
From which we can conclude that
$$  {\mathbb{E}(X\mathbf{1}_{\{\tau=n\}}|\mathcal{F}_n) =\mathbb{E}(X\mathbf{1}_{\{\tau=n\}}|\mathcal{F}_\tau)} \qquad \blacksquare$$

What bugs me is the last step of the proof : it is true that for any $ A \in \mathcal{F}_n,{A\cap\{\tau=n\}\in\mathcal{F}_n} $, so I understand that what we have shown is

$$\mathbb{E}(\mathbf{1}_A\mathbb{E}(X\mathbf{1}_{\{\tau=n\}}|\mathcal{F}_n)) = \mathbb{E}(\mathbf{1}_A\mathbb{E}(X\mathbf{1}_{\{\tau=n\}}|\mathcal{F}_\tau)) \quad \forall A\in \mathcal{F}_n $$

But don't we need to prove this result for all $A \in \mathcal{F}$ to prove that the two conditional expectations are the same (and thus the lemma) ? Unfortunately, if we take $A$ in $\mathcal{F}$ instead of $\mathcal{F}_n$, the above proof does not hold ($\mathbf{1}_A$ is not $\mathcal{F}_n$ measurable).

If anyone could explain what I'm misunderstanding and/or why the proof is correct, I'd be grateful.

答:
I think you need to take $A\in\mathcal{F}_{\tau}$ s.t. $A\cap \{\tau=n\}\in \mathcal{F}_n$ (using the definition of $\mathcal{F}_{\tau}$). Then
$$
\mathsf{E}[\mathsf{E}[X1_{\{\tau=n\}}\mid \mathcal{F}_{\tau}]1_A]=\mathsf{E}[X1_{A\cap\{\tau=n\}}]=\mathsf{E}[\mathsf{E}[X1_{A\cap\{\tau=n\}}\mid \mathcal{F}_n]]=\mathsf{E}[\mathsf{E}[X\mid\mathcal{F}_n]1_{A\cap\{\tau=n\}}],
$$
where the first equality is the definition of conditional expectations, and the second equality follows from the tower property. Consequently,
$$
\mathsf{E}[X1_{\{\tau=n\}}\mid \mathcal{F}_{\tau}]=\mathsf{E}[X\mid\mathcal{F}_n]1_{\{\tau=n\}}\quad\text{a.s.}
$$

Title: About a property of stopping times


问:
$y''-2y'-4x^2y+x^2 \sqrt{8\pi}=0$
I am attempting to solve the following second-order ODE, but the integrals I'm getting in it are just getting really nasty and I'm getting lost. I believe the answer should be in the form of the $S(x)$ and $C(x)$ functions (integrals of $\sin (x^2)$ and $\cos (x^2)$ respectively), but I'm not positive on that one.

For the record, this is not a homework question. This ODE came up when attempting to solve a pretty interesting integral I came up with on my own. Thanks.

答:
$$y''-2y'-4x^2y+x^2 \sqrt{8\pi}=0$$
$$y''-2y'-4x^2(y- \frac14\sqrt{8\pi})=0$$
Change of function :$\quad u=y- \frac14\sqrt{8\pi}$
$$u''-2u'-4x^2u=0$$
Change of function :$\quad u=e^{x-x^2}v$
$$v''-4xv'-3v=0$$
This is an ODE of the Airy kind.
$$v=c_1\text{Ai}(X)+c_2\text{Bi}(X)\quad\text{with}\quad X=2^{-4/3}(4x+3)$$
Ai and Bi are the Airy functions.

ContentLicense: CC BY-SA 4.0

Title: $y''-2y'-4x^2y+x^2 \sqrt{8\pi}=0$


问:
Linear Algebra Friedberg Exercise 1.7.7 Query
I will now state the exercise:

"Prove the following statement. Let β
be a basis for a vector space V, and let S be a linearly independent
subset of V. There exists a subset S1 of β such that S ∪ S1 is a basis
for V."

My question regards a fact he uses in his solution which I cannot understand how he deduces. I will now describe his solution and explain which part confuses me:

He starts by considering the family $F$ of all linearly independent subsets of β such that their union with S is linearly independent. Then he proves by the maximal principle (as referenced in his book page 59) that $F$ has a maximal element, denoted S1.

He then goes on to state precicely that "By the maximality of $S_1$ we know that $S_1 \cup S$ generates $β$ "

I understand why that implies our desired result, but what I don't understand is why it is true.

How do we know that  $S_1 \cup S$ generates $β$ ?

Thank you in advance, any help is very much appreciated

答: So assume for contradiction that $S_1\cup S$ does not generate $\beta$. Then in particular, there is some $v\in\beta$ not generated by $S_1\cup S$. You can then check that this implies by definition that $S_1\cup S\cup\{v\}$ is also independent. Thus $S_1$ is not maximal as it is bounded above by $S_1\subsetneq S_1\cup\{v\}\subseteq \beta$.


问:
Definition of amenability by Følner sequences
I am a little bit confused with the definition of amenability using the existence of Følner sequences. It seems that some people define a Følner sequence for certain classes of groups only. For example, Prof. Terence Tao in his notes writes:

Let $G = (G,\cdot)$ be a discrete, at most countable, group.  A Følner sequence is a sequence $F_1, F_2, F_3, \ldots$ of finite subsets of $G$ with $\bigcup_{N=1}^\infty F_N = G$ with the property that $\lim_{N \to \infty} \frac{|g F_N \triangle F_N|}{|F_N|} = 0$ for all $g \in G$, where $\triangle$ denotes symmetric difference.

On the other hand, Prof. Cornelia Drutu gives a more straightforward definition, for any group $G$:

A sequence of non-empty subsets $F_n$ of $G$ is called a Følner sequence in $G$ if for every $g$ in $G$,
$$\lim\limits_{n \to \infty} \frac{|g F_n \triangle F_n|}{|F_n|}.$$

In both cases, the definition of amenability would be that $G$ (necessarily discrete and at most countable in the first case only) is amenable if it has at least one Følner sequence.

My questions are as follows.

How are these two definitions related? Is there any reason to consider the topology on $G$ and limit oneself to countability? How are these assumptions relevant?

If $G$ is countable and dicrete, are these two definitions equivalent?

Is the condition $\bigcup_{N=1}^\infty F_N = G$ in the first definition necessary?

Is the second definition equivalent to $G$ being non-paradoxical, under $\mathrm{AC}$? If so, is there a direct proof, without referring to other equivalent definitions of amenability?

Any help would be largely appreciated!

答:
The condition that $\bigcup F_n=G$ is unusual, for instance $(\{1,\dots,n\})_{n\ge 0}$ is a reasonable Følner sequence in $\mathbf{Z}$.

If $(F_n)$ is Følner, however, one can modify it to get a covering sequence (i.e. $\bigcup F_n=G$). Just fix an enumeration $(g_n)$ of $G$ and set $F'_n=F_n\cup\{g_n\}$. It's still Følner.

Indeed if $G$ is finite, necessarily $F_n=G$ for large $n$. Otherwise, the added singleton makes a negligible contribution to the ratio.

For instance, $(\{-n\}\cup\{1,\dots,n\})_{n\ge 1}$ is Følner and covers $\mathbf{Z}$. This example illustrates that the covering condition is somewhat artificial.


问:
Intersection of a family of compact sets in Hausdorff topological space is compact
Let $\{G_i\}_{i \in \Lambda}$ be a family of compact sets in a Hausdorff topological space. Show that the intersection $\bigcap_{i \in \Lambda} G_i$ is compact.

My attempt was:

Since $G_j$ for some $j \in \Lambda$ is compact, there exists an open, finite covering. And $\bigcap_{i \in \Lambda} G_i \subset G_j$. So I can take the same covering for the intersection as for $G_j.$

Is that correct?

答:
It is definitely not correct. First, you seem to have misunderstood compactness. It does not say that a compact set has a finite open cover: in any space $X$ the family $\{X\}$ of open sets is a finite open cover of every subset of $X$, compact or not. It says that every open cover of a compact set has a finite subcover.

Secondly, you have not used the hypothesis that the space is Hausdorff, which is essential: the result is not true in general for non-Hausdorff spaces. Let $X$ be the line with two origins as described here; then the sets $\{p\}\cup(0,1]$ and $\{q\}\cup(0,1]$ are compact subsets of $X$ whose intersection $(0,1]$ is not compact.

HINT: Show (if you’ve not already done so) that every compact subset of a Hausdorff space is closed. What do you know about the intersection of closed sets? What do you know about closed subsets of compact sets?


问:
Convergence of $\sum^\infty_{n=0} \frac{\cos(n + \frac{1}{n^2})}{n \cdot \ln(n^2 + 1)}$ Is my idea correct?
Convergence:

$$\sum^\infty_{n=0}\frac{\cos \left(n + \frac{1}{n^2} \right)}{n \cdot \ln \left( n^2 + 1 \right)}$$

Edit (I tried to follow the idea written by @Daniel Fischer):

$$\sum_{n=0}^{\infty} \frac{cos(n + \frac{1}{n^2})}{n\cdot ln(n^2 + 1)} = \sum_{n=0}^{\infty} \frac{cos(n)}{n \cdot ln(n^2 + 1)} + \sum_{n=0}^{\infty} \frac{cos(n + \frac{1}{n^2}) - cos(n)}{n \cdot ln(n^2 + 1)}$$

$\sum_{n=0}^{\infty} \frac{cos(n)}{n \cdot ln(n^2 + 1)}$ is convergent because of Dirichlet test and How can we sum up $\sin$ and $\cos$ series when the angles are in arithmetic progression?

$\sum_{n=0}^{\infty} \frac{cos(n + \frac{1}{n^2}) - cos(n)}{n \cdot ln(n^2 + 1)} = \sum_{n=0}^{\infty} \frac{-2 \left(sin \left( \frac{2n+\frac{1}{n^2}}{2} \right) \cdot sin \left( \frac{\frac{1}{n^2}}{2} \right) \right)}{n \cdot ln \left(n^2 + 1 \right)} = \sum_{n=0}^{\infty} \frac{-2 \left(sin \left( n+\frac{1}{2n^2} \right) \cdot sin \left( \frac{1}{2n^2} \right) \right)}{n \cdot ln \left(n^2 + 1 \right)}$

Then:
$$\sum_{n=0}^{\infty} \left| \frac{-2 \left(sin \left( n+\frac{1}{2n^2} \right) \cdot sin \left( \frac{1}{2n^2} \right) \right)}{n \cdot ln \left(n^2 + 1 \right)} \right| \leq \sum_{n=0}^{\infty} \left| 2 \cdot \frac{ \frac{1}{2n^2} }{n \cdot ln \left(n^2 + 1 \right)} \right| \leq \sum_{n=0}^{\infty} \left| \frac{ 1 }{n^3 \cdot ln \left(n^2 + 1 \right)} \right| $$

For $n \geq 3$:
$$\sum_{n=0}^{\infty} \left| \frac{ 1 }{n^3 \cdot ln \left(n^2 + 1 \right)} \right| \leq \sum_{n=0}^{\infty} \left| \frac{1}{n^2} \right|$$

Therefore we know that $\sum_{n=0}^{\infty} \frac{cos(n + \frac{1}{n^2}) - cos(n)}{n \cdot ln(n^2 + 1)}$ it is also convergent. Is that correct?

答:
Continuing from the remark made in the comments,
$$\cos\biggl(n + \frac{1}{n^2}\biggr) = \cos n + \biggl(\cos \biggl(n + \frac{1}{n^2}\biggr) - \cos n\biggr).$$
The second term on the right decays like $1/n^2$, since cosine is Lipschitz continuous (prove this using the cosine sum identity). The partial sums of the first term on the right are uniformly bounded; to see this, it is convenient to remember cosine is the real part of the complex exponential,
$$ \cos n =  \operatorname{Re} e^{in}. $$
Thus the partial sum of cosine is just the real part of a nice geometric series,
$$ \sum_{n = 0}^m \cos n = \operatorname{Re} \sum_{n = 0}^m e^{in} = \operatorname{Re} \frac{1 - e^{i(m + 1)}}{1 - e^i}. $$
The right hand side has uniformly bounded modulus in $m$, so we are done.


问:
How to find the angle formed by an interior bisector and a suplementary from another bisector?
The problem is as follows:

$AD$ bisect angle $\angle BAC$. Point $E$ which lies on $BC$ is generated by the $DE$ which is a bisector of $\angle ADC$. Using this information and the figure from below, find the angle which is made by $DE$ and $BC$.

The alternatives given in my book are as follows:

$\begin{array}{ll}
1.&100^{\circ}\\
2.&90^{\circ}\\
3.&110^{\circ}\\
4.&120^{\circ}\\
\end{array}$

I'm totally lost in this problem. What exactly should be the way to approach this?.

The problem not only lies in solving but how can it be solved relying only in euclidean geometry postulates?. The thing here from what I can spot is that there might come into play notable angle in triangles. Such as the angle formed by an exterior bisector or an interior bisector.

Can someone help me in a way that I can understand without much fuss?. Please include a drawing in the answer because I'm lost here.

Suspiciously $\angle ABC = 2 \angle BCA$. Is this an implicit hint which makes us to conclude to draw a segment and build an isosceles there?. I don't know. Does it needs congruence or what?. Please guide me, I need help step by step.

答:
Hint:

The internal angles of any triangle sum to $180^\circ$.

The angles on a straight line sum to $180^\circ$.

That's all you need.


答:
You can just chase angles.

Find $\angle ECD$ (and $\angle ACD$)

Find $\angle BAC$ from $\triangle ABC$

$AD$ is a bisector, so half this for $\angle CAD$

Find $\angle ADC = 2\alpha$ from $\triangle ACD$

Calculate $\alpha$

Find $x$ from $\triangle ECD$

Title: How to find the angle formed by an interior bisector and a suplementary from another bisector?

Body:
The problem is as follows:

$AD$ bisect angle $\angle BAC$. Point $E$ which lies on $BC$ is generated by the $DE$ which is a bisector of $\angle ADC$. Using this information and the figure from below, find the angle which is made by $DE$ and $BC$.

The alternatives given in my book are as follows:

$\begin{array}{ll}
1.&100^{\circ}\\
2.&90^{\circ}\\
3.&110^{\circ}\\
4.&120^{\circ}\\
\end{array}$

I'm totally lost in this problem. What exactly should be the way to approach this?.

The problem not only lies in solving but how can it be solved relying only in euclidean geometry postulates?. The thing here from what I can spot is that there might come into play notable angle in triangles. Such as the angle formed by an exterior bisector or an interior bisector.

Can someone help me in a way that I can understand without much fuss?. Please include a drawing in the answer because I'm lost here.

Suspiciously $\angle ABC = 2 \angle BCA$. Is this an implicit hint which makes us to conclude to draw a segment and build an isosceles there?. I don't know. Does it needs congruence or what?. Please guide me, I need help step by step.


问:
Operation defined on a group uniquely determines that group
This is an exercise (2.2:3) in An Invitation to General Algebra and Universal Constructions by George M. Bergman:

If $G$ is a group, let us define an operation $\delta_{G}$ on $|G|$ by $\delta_{G}(x, y) = xy^{-1}x$. Does the pair $G' = (|G|, \delta_{G})$  determine the group $(|G|, {}\cdot{} , ^{-1}, e)$? (I.e., if $G_1$ and $G_2$ yield the same pair, $G'_1 = G'_2$, must $G_1 = G_2$?)

($|G|$ means underlying set of a group).

If $G'_1 = G'_2$ then they are equal as pairs and that means that $|G_1| = |G_2|$ and $\delta_{G_1} = \delta_{G_2}$. Let $\circ$, $^{-1}$, $e_1$ (resp. $\star$ , $^{'}$, $e_2$) be binary operation, operation of taking inverse and identity operation in group $G_1$ (resp. $G_2$). By virtue of $\delta_{G_1} = \delta_{G_2}$ we get
$$ x \circ y^{-1} \circ x = x \star y^{'} \star x \tag{1}$$
for every $x, y$ of $|G_1|$.
Plugging various $x$ and $y$ in $(1)$ we can get following relations of identities:
$$ e_1 \star e_1 = {e_2}^{-1} \tag{2}$$
$$ e_2 \circ e_2 = {e_1}^{'} \tag{3}$$
$$ {e_2}^{-1} \star e_1 = {e_2}^{-1} \circ {e_2}^{-1} \tag{4}$$
And I'm stuck with that: can't get any reasonable relation which would allow to get equalitity of identities or equality of inverses or just distributivity between two group operations (to prove that groups are equal).
On the other hand, to disprove that one must find a counterexample e.g. in groups of small order (i.e. two different group structures on the same set giving the same derived operations), but it is tedious.

答: Since the question asks for equality of operations rather than isomorphism: take $G$ to be any elementary abelian $2$-group. Then $xy^{-1} x = y$ so the operation $\delta_G$ gives you no information and, for example, there are already two such structures on the two-element set (given by choosing one or the other of the two elements to be the identity).


问:
Primes congruent to $5$ and $7$ modulo $8$
Let $p \equiv 5$ or $7$ ($\mathrm{mod}$ $8$) be a prime and let $a$, $b \in \mathbb{Z}$ be such that $p$ | $a^2 + 2b^2$.

Show that $p$ | $a$ and $p$ | $b$.

Clearly, if $p$ divides any of them, then $p$ divides the other, but I don't really see how $p$ must divide one of them.

答: Assume that $p\nmid b$. Then we can multiply $a^2+2b^2\equiv0\mod p$ with $b^{-2}$ and get $(ab^{-1})^2+2\equiv0$. Now look at your previous question...


问:
$\int x^{2}\sqrt{a^{2}+x^{2}}\,dx$. Is there another way to solve it faster?
I have to calculate this integral:

\begin{align} \int x^{2}\sqrt{a^{2}+x^{2}}\,dx \qquad\text{with}
\quad a \in \mathbb{R} \end{align}

My attempt:

Using, trigonometric substitution

\begin{align}
\tan \theta &= \frac{x}{a}\\ \Longrightarrow  \ x&=a \tan \theta\\ \Longrightarrow  \ dx&=a \sec^{2}\theta\\ \Longrightarrow \ x^{2}&=a^{2}\tan^{2}\theta
\end{align}

Thus,
\begin{align}
\int x^{2}\sqrt{a^{2}+x^{2}}\,dx&=\int a^2 \tan^{2}\theta \sqrt{a^2+a^2\tan^{2}\theta}\ a\sec^{2}\theta\, d \theta\\&=a^{3}\int \tan^{2}\theta \sqrt{a^{2}(1+\tan^{2}\theta)}\sec^{2}\theta\, d\theta\\&=a^{3}\int \tan^{2}\theta \sqrt{a^{2}(\sec^{2}\theta)}\sec^{2}\theta \, d\theta\\&=a^{4}\int (1-\sec^{2}\theta)\sec^{3}\theta \, d\theta\\&=a^{4}\underbrace{\int \sec^{3}\theta \, d\theta}_{\text{solve by parts}}-a^{4}\underbrace{\int \sec^{5}\theta \, d\theta}_{\text{solve by parts}}
\end{align}

My doubt is: Is there any other way to solve it faster? Because by parts is a large process to solve each one. I really appreciate your help

答:
HINT.-You do have a binomial integral $\int x^{2}\sqrt{a^{2}+x^{2}}\,dx $ in which the second case of Chebyshev is fulfilled, i.e. $\dfrac{2+1}{2}+\dfrac12$ is an integer for which we must have the change of variable
$$u=\left(\frac{a^2+x^2}{x^2}\right)^{\dfrac12}$$ so we have after calculations the integral $$-\frac{a^5}{2}\int\frac{d(u^2-1)}{(u^2-1)^4}=\frac{a^5}{2}(u^2-1)^{-3}+C$$
NOTE.-Being in a hurry the calculation has been wrong maybe but the methode is correct (I did $x^2=\dfrac{a^2}{u^2-1}$ so the integral becomes
$$\int\dfrac{a^2}{u^2-1}\sqrt{\frac{u^2a^2}{u^2-1}}\left(\frac{-a^2 }{(u^2-1)^2}\right)\frac{1}{\sqrt{u^2-1}}\,du$$


问:
The base of the product topology
Let $(E_i, \mathcal{T_i})$  a family of topological spaces and $E=\prod_{i \in I} E_i$.
The product topology is the smallest topology on $E$ such that all projection maps are continuous.
the base of this topology is given by :
$\mathcal B=\{ \prod_{i\in I} U_i : \text{ $U_i$ open in $E_i$  for all $i\in I$, and $U_i =E_i$ for all but finitely many $i\in I$}\}$.

The problem is why we define the base of the product topology like that and didn't define it as in the finite case  : $\mathcal B=\{ \prod_{i\in I} U_i : \text{ $U_i$ open in $E_i$  for all $i\in I$} \}$.
I know that $\mathcal B$ in  this case is the base of the box topology but I didn't understand yet.
Thank you in advance.

答:
It is the coarsest topology that makes all the projection morphisms continuous. This makes it align with the category theoretical definition of "product" (the "identity map" from the box topology space to the product topology space is continuous, but not the other way around), which is a good thing.

Also note that it is, formally, defined the same way for finite products. It's just that in those cases you don't see the distinction between box and product topologies. So for simplicity people usually quote the definition of box topology when they actually mean to talk about the product topology on finite products.

Title: The base of the product topology


答: If $\pi_i : E \to E_i$ is the projection onto the $i$-th coordinate, the smallest topology on $E$ that makes each of the functions in $\{\pi_i\}_{i \in I}$ continuous is the topology generated by $$\{\pi_i^{-1}(U_i) : i \in I,\, U_i \in \mathcal T_i\}.$$ Thus, a basis element for this topology is a finite intersection of the form $\pi_{i_1}^{-1}(U_1) \cap \cdots \cap \pi_{i_k}^{-1}(U_k)$, but note that $$\pi_{i_1}^{-1}(U_1) \cap \cdots \cap \pi_{i_k}^{-1}(U_k) = \{(x_i)_{i \in I} \in E : (\forall j \in \{1,\dots,k\}) \ x_{i_j} \in U_j\} = \prod_{i \in I} V_i$$ where $V_{i_j} = U_j$ for every $j \in \{1,\dots,k\}$, and $V_i = E_i$ if $i \neq i_1,\dots,i_k$.

Title: The base of the product topology

Body:
Let $(E_i, \mathcal{T_i})$  a family of topological spaces and $E=\prod_{i \in I} E_i$.
The product topology is the smallest topology on $E$ such that all projection maps are continuous.
the base of this topology is given by :
$\mathcal B=\{ \prod_{i\in I} U_i : \text{ $U_i$ open in $E_i$  for all $i\in I$, and $U_i =E_i$ for all but finitely many $i\in I$}\}$.

The problem is why we define the base of the product topology like that and didn't define it as in the finite case  : $\mathcal B=\{ \prod_{i\in I} U_i : \text{ $U_i$ open in $E_i$  for all $i\in I$} \}$.
I know that $\mathcal B$ in  this case is the base of the box topology but I didn't understand yet.
Thank you in advance.


答:
Hint

If you have a collection of set $\mathcal A=\{A_i\}_{i\in\mathcal I}$, the coarse topology that contains $\mathcal A$ is the topology of basis $$\mathcal B=\left\{\bigcap_{j\in J}A_j\mid |\mathcal J|<\infty ,A_j\in \mathcal A\right\}.$$

In your case, denote $\pi_k: \prod_{\ell\in \mathcal K}E_\ell\to E_k$ defined by $$\pi_k((x_\ell)_{\ell\in\mathcal K})=x_k.$$
In particular, $\pi_k$ is continuous if and only if $$\pi_{k}^{-1}(U)=E_1\times ...\times E_{k-1}\times U\times E_{k+1}\times ...$$ is open in $\prod_{\ell\in \mathcal K}E_\ell$ for all $U$ open in $E_k$. So, in you case, $\mathcal A$ is a collection of sets of the form $\prod_{\ell\in \mathcal K}V_\ell$, where $V_\ell$ are open in $E_\ell$ and $V_\ell=E_\ell$ for all $\ell\in \mathcal K$, but one. Then $\mathcal B$ is the collection of sets of the form $\prod_{\ell\in\mathcal K}V_\ell$ where $V_\ell$ are open, and $V_\ell=E_\ell$ for all $\ell$, but finitely many $\ell$'s.

Title: The base of the product topology


问:
Why the root of this tree has to be "1"?
Arrange $2^{n-1}-1$ zeroes and $2^{n-1}$ ones in a balanced full binary tree of depth $n$. If we want the number of edges that connect the same (and respectively different) digits are the same, then one claims that the root of this tree has to be a one. Why is that?

For example, if $n=2$, then we need to arrange 1 zero and 2 ones. One arrangement that makes the number of edges that connect the same digits (which in this case is only one: the edge with a "+" who connects 2 ones,) and the number of edges that connect two different digits (which in this case is also one: the edge with a "-") are the same. Note that the root of this tree is unexceptionally 1.

```<code>      1
    /   \
 + /     \ -
  /       \
 1         0
</code>```

And here is a case where $n=3$.

```<code>          1
        /   \
     + /     \ -
      /       \
     1         0
  + / \ -   + / \ -
   /   \     /   \
  1     0   0     1
</code>```

Again, we see that the root is a one. If we change the root to a zero, however, then we can never find an arrangement that makes the number of "+" edges equals to the number of "-" edges. Why is that?

答:
Consider a balanced depth-$n$ binary tree of $0$'s and $1$'s and label each edge $+$ or $-$ as you did.
If  all edges were $+$, all $2^n-1$ vertices would be the same as the root.  You can switch
an edge from $+$ to $-$ or vice versa if you flip all the vertices above that edge (i.e. those such that the edge is on the path from the vertex to the root).  This always involves
flipping an odd number of vertices, so the parity of the numbers of $1$'s and of $0$'s changes.  If you do an even number of edge-switches, you have the same parity you started with.  Now your tree has $2^n-2$ edges, and you need to do $2^{n-1}-1$ switches to get an equal number of $+$ and $-$ edges.  If $n \ge 2$, this is odd, so the parity is  different than at the start.  At the start, there were an odd number the same as the root, so after the switching you must have an even number.  Since you want the number of $0$'s to be odd,
the root must be $1$.</p>

ContentLicense: CC BY-SA 3.0


问:
find the best possible bound for $|f(1/4)|?$
Given $f$ analytic in $|z| < 2,$ bounded there by $2$, and such that $f(1) = 0,$ find the best possible bound for $|f(1/4)|?$

My attempt : I found the solution here
But this solution is  not correct because it contain some mistake

Mistake  is $h(z):=\phi(g(\phi^{—1}(z)))$ doesn't satisfy schwarz lemma  since $h(0)\neq0$

My solution : Take $g(z)= \frac{f(2z)}{2}$ and take $\phi(z)= \frac{z-1/2}{1-1/2z}$

then $\phi^{-1}(z)=\frac{z+1/2}{1+1/2z}    $

take $ h(z)= g \circ  \phi^{-1}(z)  = g(\phi^{-1}(z))$

put $z=0 $, then $ h(0)=                 g(1/2)  $

we have  $ g(1/2)= f(1)/2=0  $

Therefor $h(0)=0 $, and $h$ satisfy the schwarz lemma  i,e  $|h(z)|\le |z|$

This implies  that $| g(z)| \le |\phi(z)| $

$g(1/8)  = \frac{1/8-1/2}{1-1/2.(1/8)} =2/5$

$2.g(1/8)=f(1/4)\implies f(1/4)=\frac{4}{5}   $

so the     the best possible bound for $|f(1/4)|$ is $\frac{4}{5}$

Is my solution is correct or not ?

答:
Your proof is correct, only that it should be
$$
 |g(1/8)| \le  \left| \frac{1/8 - 1/2}{1-(1/2)(1/8)}\right| = \frac 25
$$
instead of $g(1/8) = 2/5$, and therefore $|f(1/4)| \le 4/5$, not $f(1/4) = 4/5$.

You also have to show that this is the best possible bound. This can be achieved by choosing $f$ such that $h(z) = z$ or equivalently, $g(z) = \phi(z)$, that is
$$
 f(z) = 2 \phi(\frac z2) = 2 \frac{z/2 - 1/2}{1-(1/2)(z/2)} = \frac{4(z-1)}{4z-1} \, .
$$
This function satisfies the given conditions with $f(1/4) = -4/5$, so that the bound is sharp.

In other words: If $f$ is analytic in $|z| < 2$ with $|f(z)| < 2$ and $f(1) = 0$ then
$$
 f(z) = 2 h(\phi(\frac z2))
$$
where $\phi(z) = \frac{z-1/2}{1-z/2}$ and $h$ maps the unit disk into itself with $h(0) = 0$. It follows that
$$
 |f(z)| \le 2 \left| \phi(\frac z2) \right|
$$
with equality if $h(z) = \lambda z$ with some $\lambda$ of modulus one.


问:
The advantage of defining open/closed sets using sequence.
I have learned that in a first-countable topological space, open and closed sets can be defined using sequences. Moreover, the interior or closure of a set can be defined in similar ways.

My question is, what is the advantage of using sequences in testing whether a set is open or closed? Are there any particular problems where sequences are more useful than neighborhoods?

答:
Sequences are more familiar from metric spaces and it's often easier to say which sequences converge than to describe the open or closed sets. As a simple example, the product topology on the Cantor set $\{ 0, 1 \}^{\mathbb{N}}$ is exactly the topology of pointwise convergence, meaning a sequence $f_n$ of functions $f_n : \mathbb{N} \to \{ 0, 1 \}$ converges iff each of its values $f_n(m)$ converges, for all $m \in \mathbb{N}$.

Meanwhile the product topology is the coarsest topology with respect to which all evaluation maps $f_n \mapsto f_n(m)$ are continuous; this is really saying the same thing but you might find the open sets of the product topology (generated by cylinder sets) abstract and harder to think about.

For non-sequential spaces the use of sequences can be replaced by nets or filters, but then you'd have to learn about convergence of nets and filters. This is a fine thing to do but many people have not done it (I don't know anything about nets myself) whereas lots more people have taken courses in real analysis at some point. So you can rely on more people understanding an argument involving sequences.


问:
Using the total probability formula, find $Z$'s law.
An insect lays eggs. The number of eggs laid is a random variable $X$ that follows a Poisson distribution of intensity $θ$. Each egg has a probability $p$ of hatching, independent of other eggs. Let $Z$ be the random variable that determines the number of eggs that hatched.

a) For $(k, n) ∈ \mathbb{N}^2$, calculate $P [Z = k | X = n]$.

b) Using the total probability formula, find $Z$'s law.

Hi, I've stuck in the part $b)$, I know what the formula means, but can't seem to find a way to use it for finding $Z$'s law

Can someone explain, how to use the formula in this case?

答: I think you just want $$P(Z=k) = \sum\limits_n P(Z=k|X=n)\cdot P(X=n)$$


问:
Determinant of a matrix and its basis
Is the determinant of a matrix independent of the basis in which that matrix is expressed? For example, when expressing in matrix form a quantum mechanics operator, or when comparing the same matrix but in its diagonal form.

Every case I try, I find it is independent of the basis. But why? How can I prove it, or what counter example could I find?

Thanks in advance.

答: Yes, it's a consequence of Binet's theorem: $\det(AB)=\det(A)\det (B)$. Therefore \begin{align}\det(PAP^{-1})&=\det (P)\det(A)\det(P^{-1})=\det(P)\det(P^{-1})\det(A)=\\&=\det(I_n)\det(A)=\det (A)\end{align}

Title: Determinant of a matrix and its basis


问:
Algebra to reconcile 2 alternative forms of negative binomial distribution
I am given that negative binomial, "failure version" follows
Where

X: Trials

Y: Failures

r: successes

${y+r-1  \choose r-1}p^r(1-p)^y$

However, I am ALSO told that I can use the expression

${y+r-1  \choose y}p^r(1-p)^y$

How is this algebraically possible, that is, how is $r-1=y$?

Body:
As per my comment,
$$C(m+n,n)=\frac{(m+n)!}{n!(m+n-n)!}=\frac{(m+n)!}{n!m!}$$
and
$$C(m+n,m)=\frac{(m+n)!}{m!(m+n-m)!}=\frac{(m+n)!}{m!n!}$$
QED.


答:
If $p\equiv5$ or $7\pmod8$, then $-2$ is not a quadratic residue $\pmod p$,

so $a^2+2b^2\equiv0\pmod p$ and $p\nmid b$, which imply $(ab^{-1})^2\equiv-2\pmod p$,

would be a contradiction.

Title: Primes congruent to $5$ and $7$ modulo $8$

Body:
Let $p \equiv 5$ or $7$ ($\mathrm{mod}$ $8$) be a prime and let $a$, $b \in \mathbb{Z}$ be such that $p$ | $a^2 + 2b^2$.

Show that $p$ | $a$ and $p$ | $b$.

Clearly, if $p$ divides any of them, then $p$ divides the other, but I don't really see how $p$ must divide one of them.


答:
A linear transformation from a (finite dimensional) vector space to itself has a determinant. Once you put a coordinate system on your space this linear transformation also has a matrix representation. You can use the entries of this matrix and a certain formula to calculate the determinant of the linear transformation (we call this "the determinant of the matrix", but most of the time that's a red herring).

The determinant was there before the coordinate system, and thus the determinant cannot in any conceivable way depend on the coordinate system.

Title: Determinant of a matrix and its basis

Body:
Is the determinant of a matrix independent of the basis in which that matrix is expressed? For example, when expressing in matrix form a quantum mechanics operator, or when comparing the same matrix but in its diagonal form.

Every case I try, I find it is independent of the basis. But why? How can I prove it, or what counter example could I find?

Thanks in advance.


问:
The frontier of a set
I'm trying to prove that a certain set is dense in a metric space.
I have a metric space X, an open subspace $Y\subset X$ (same metric) and an open set $U\subset Y$ s.t $U$ is open and dense in $Y$. Moreover, X is Baire. I define $B=X\setminus \overline Y$ and tried proving that $U\cup B$ is dense in X.
Is the following proof correct?

$cl(U\cup B)=cl(U_n\cup(X\setminus cl(Y))=cl(U_n)\cup cl(X\setminus cl(Y))$

Now I claim that: $cl(X\setminus cl(Y))=(X\setminus cl(Y))\cup Fr(X\setminus cl(Y))=cl(X\setminus cl(Y))=(X\setminus (Y\cup Fr(Y))\cup (Fr(cl(Y)))=(X\setminus Y)$

and therefore $cl(U_n)\cup cl(X\setminus cl(Y))=Y\cup(X\setminus Y)=X$

Is it true that $Fr(Y)=Fr(cl(Y))$ - specifically in a metric Baire space?

答: No, it is not true. Take $X=\Bbb R$, endowed with the usual metric. It is a Baire space, but $\operatorname{Fr}(\Bbb Q)=\Bbb R$, whereas $\operatorname{Fr}\left(\overline{\Bbb Q}\right)=\emptyset$.

Title: The frontier of a set


问:
Selecting a random number from $(0,1)\cup(1,\infty)$
A random number $a$ is selected from $(0,1) \cup  (1,\infty)$.

Let $A $ be the event that (a random number from $(0,1) \cup  (1,\infty))$ $ \in (0,1)$. Let $P(A) = x$

Let $B$ be the event such that (a random number from $(0,1) \cup  (1,\infty))$ $ \in (1,\infty)$.Therefore, $P(B)=1-x$

Now, let $b = \frac{1}{a}.$ Since, a is random, b is random.

If $a \in (0,1)$(Probablity is $x$)  then $b \in (0,1)$(Probablity is $1-x$)

Therefore, $x = 1-x.$
Therefore, $x = \frac12$ which is definitely not right.

What is going wrong?

答:
While the comments are correct that you cannot pick a random number "uniformly" from a set of infinite length, that is not the real problem here. Let's make a few changes to get the same situation:

Pick a point from $(0,1)\cup (1,4)$ with uniform distribution. This is possible on this set. There will be a probability of $\frac 14$ that the point is in $(0,1)$ and of $\frac 34$ that the point is in $(1,4)$.

Now define the function $$f(x) = \begin{cases}3x+1, &\quad x \in (0,1)\\\dfrac{4-x}3,&\quad x\in (1,4)\end{cases}$$

and define $b = f(a)$ similar to how you defined $b = \frac 1a$. Just as in your case $b$ will be in $(1,4)$ if and only if $a \in (0,1)$.

If we pick $a$ uniformly at random, then $P(a \in (0,1)) = \frac 14, P(a \in (1,4)) = \frac 34$.

So $P(b \in (0,1)) = \frac 34, P(b \in (1,4)) = \frac 14$.

And here is your mistake:

Since, a is random, b is random.

You assumed that since the probability of picking a number in $(0,1)$ is $\frac 14$, that $P(b \in (0,1))$ must also be $\frac 14$.

But this is false, because you did not pick $b$ at random by the same method as picking $a$. $a$ was picked per the indicated distribution. Because of the functional dependence on $a$, $b$ is picked by a different probability distribution. Its odds of being in a given range are different from the odds for $a$ to be in that same range.


问:
What exactly is the complex plane, and how is it useful?
A lot of functions are defined on the complex plane, like the Gamma function:

the Lambert W function,

etc.

But I have no idea about what the complex plane means and how it's useful, or just how they graph it. If someone has an idea, please tell me about it. Thanks to all.

答:
The complex plane is a two-dimensional generalization of the real numbers. Complex numbers (a, b) are added just like you'd expect: (a, b) + (c, d) = (a + c, b + d). They multiply a bit differently, though: (a, b)(c, d) = (ac - bd, ad + bc). You can get ordinary real numbers back by setting the second number of the pair to 0.

Usually (a, b) is written $a+bi$, with the understanding that $i$ is not a real number. In fact it acts as a square root of $-1$, which does not exist in the real numbers. You can see this as so: (0, 1)(0, 1) = (0-1, 0+0) = (-1, 0) or $-1+0i.$


问:
Does symmetry and monotonicity in each argument of a function guarantees Schur Convexity?
A Schur Convex function defined on $(x_1,x_2)$ is one if it satisfies the following inequality:
$$(x_1-x_2)  (\frac{\partial}{\partial {x_1}}f -\frac{\partial}{\partial {x_2}}f) \ge 0, \quad \quad\text{for all } (x_1,x_2) \in \mathbb{R^2}.$$

Question:

If $f$ (all partial derivatives exist) is

symmetric along line $x_1=x_2$,

and $f$ is monotonic increasing (or decreasing) in each of its argument

then does it implies that f is Schur concave (or Schur convex) ?

答:
Let $$f(x,y) = (\arctan(x)  + \pi/2) (\arctan(y) + \pi/2)$$ Then $f$ is symmetric in $x$ and $y$, and monotonically increasing in each argument since $\arctan$ is positive and monotonically increasing. Now
$$(x-y)(f_x - f_y) = (x-y) \left(\frac{\arctan(y)+\frac{\pi }{2}}{x^2+1}-\frac{\arctan(x)+\frac{\pi }{2}}{y^2+1}\right)$$
is not everywhere-nonnegative. For example, setting $x = 1$, $y = -1$ yields $-\pi/2 < 0$. Thus, $f$ is not schur convex.

Edit: neither is this quantity everywhere-nonpositive! Setting $x = 0$, $y = -2$ yields $2 \left(\frac{2 \pi }{5}-\arctan(2)\right) \approx 0.3$.


问:
Is there a way of defining infinity so that it both meets the criterion of being a number as well as being greater than any real number?
This number would not be equivalent to the concept of infinity, but would be a number analogous to that concept. I'd also be interested to how the different ways of doing this (I'm assuming there are many), how they differ, and what their different applications are.

So this version of infinity would behave like this:

$\infty$ > X for all X $\in$ $\mathbb{R}$

$\infty$ + 2 > $\infty$ + 1

2 > 1

Rather than:

$\infty$ is the largest number

$\infty$ + 2 = $\infty$ + 1

2 = 1

答:
There are indeed lots of ways to do this, although they may not quite paint the picture you expect.

At the simplest level, assuming that you want a number system in which addition, subtraction, multiplication, and division work "as we expect," you're just looking for non-Archimedean ordered fields which contain $\mathbb{R}$. There are indeed lots of these. For example, consider the set of rational functions in a single variable $x$ with coefficients from $\mathbb{R}$, modulo equality almost everywhere (so that we identify e.g. $x$ with $x^2\over x$). This is a field, and carries a natural ordering compatible with the field structure: set $f<g$ iff $\lim_{x\rightarrow\infty}(g(x)-f(x))>0$. It's easy to check that this is in fact a non-Archimedean field. (In fact, note that any ordered field containing $\mathbb{R}$ must be non-Archimedean!)

A powerful result here is the compactness theorem from logic. This implies that, in a precise sense, whenever I have a "reasonably concrete" set of algebraic properties describing some (infinite) structure there are other structures satisfying those same properties which are "as big as I want" (e.g. cardinality $2^{2^{2^{2^{\aleph_0}}}}$ instead of $\vert\mathbb{R}\vert=2^{\aleph_0}$). That's overkill here, but it's a good thing to keep in mind for down the road. (Note that this, combined with the previous paragraph, means that Archimedean-ness is not a "reasonably concrete" property in the relevant sense!)

We can look further for non-Archimedean fields satisfying additional special properties. A hyperreal field, for example, is a non-Archimedean ordered field which shares all "reasonably-definable" properties with $\mathbb{R}$ in a precise sense, while the surreal numbers form a particular (proper-class-sized) non-Archimedean ordered field which is "maximal" in an appropriate sense.

In all such cases however there is not a single "distinguished" infinite element; instead, there are lots of infinite elements, with no one being particularly special. So this cuts against the idea of $\infty$ as a meaningful symbol denoting a single element.


问:
Number of partitions with x parts
I might be wrong, but this is how I define "parts" when talking about partitions:

The partition 1+1+1 has 3 parts. The partition 1+2+2 has 3 parts. The numbers, whether they be distinct or not, do not matter: it is the number of digits that matter.

The partitions of 5 are as follows: 1+1+1+1+1 = 1+1+1+2 = 1+2+2 = 1+1+3 = 2+3 = 1+4 = 5

Of those 7 partitions, two of them consist of 3 parts (1+2+2 and 1+1+3). Two of them consist of 2 parts (2+3 and 1+4).

The number of parts is defined as the variable $x$, and the number being partitioned is defined as the variable $n$. I want an equation that gives the number of partitions with $x$ amount of parts, when $n$ is partitioned. The in-values are $x$ and $n$, and the out-value is the number of partitions with $x$ amount of parts.

Let's say that the no. of partitions consisting of $x$ parts is defined as the variable $y$.

Here's what I know:

There are three constants: If $x = 1, n, (n-1)$ then $y = 1$, regardless of the $n$-value.

To show this, look at the partitions of 5 again. Only one partition has an $x$-value of 1, and that is the partition 5. Only one partition has an $x$-value of $n$, and that is the partition of 1+1+1+1+1. Only one partition has an $x$-value of $n-1$, and that is the partition of 1+1+1+2.

I hope this was clearer. I have edited this question drastically, because I think I worded myself unclearly in the first edition.

EDIT: My question is this: Is there an equation for determining $y$, based off of $n$ and $x$?

答:
It sounds like you are asking for the number $p_k(n)$ of partitions of a number $n$ into exactly $k$ parts. This is a standard type of restricted partition problem and can be solved as follows. Given such a partition, "rotate the Ferrers diagram," which shows that $p_k(n)$ also counts the number of partitions of a number $n$ whose largest part has size exactly $k$. Removing this part produces a partition of $n-k$ whose largest part has size at most $k$, and using this we can show that for fixed $k$ this sequence has a very nice generating function given by

$$P_k(x) = \sum_{n \ge 0} p_k(n) x^n = \frac{x^k}{\prod_{i=1}^k (1 - x^i)}$$

which in principle allows you to write down a closed form for $p_k(n)$ (again, for fixed $k$; it gets more complicated the larger $k$ is), although I won't do so. As a random example, setting $k = 5$ gives

$$\begin{align} \sum_{n \ge 0} p_5(n) x^n &= \frac{x^5}{(1 - x)(1 - x^2)(1 - x^3)(1 - x^4)(1 - x^5)} \\ &= x^5 + x^6 + 2x^7 + 3x^8 + 5x^9 + 7x^{10} + 10x^{11} + 13x^{12} + 18x^{13} + \dots \end{align}$$

which shows, for example, that the number of partitions $p_5(13)$ of $13$ into exactly $5$ parts is $18$. $p_5(n-5)$, which also counts the number $p_{\le 5}(n)$ of partitions of $n$ into at most $5$ parts, turns out to be A001401 on the OEIS.

Among other things, this generating function can be used to deduce the asymptotic growth rate of $p_k(n)$ (for fixed $k$, as $n \to \infty$), which turns out to be

$$p_k(n) \sim \frac{n^{k-1}}{(k-1)! k!}.$$

Edit: Okay, mea culpa - I said I'd write down the closed form you get from the generating function but for large $k$ I am really not sure how to do it even in a messy way. The basic strategy is to compute the partial fraction decomposition of the generating function $P_k(x)$ above. This gets very messy in general; here's what you get for small values of $k$.

For $k = 1$ there is a unique partition of $n$ with $1$ part, namely $n = n$. This corresponds to the generating function

$$P_1(x) = \frac{x}{1 - x} = x + x^2 + x^3 + \dots $$

which gives $\boxed{ p_1(n) = 1 }$ if $n \ge 1$ and $0$ if $n = 0$. Pretty straightforward.

For $k = 2$ the partial fraction decomposition is

$$P_2(x) = \frac{x^2}{(1 - x)(1 - x^2)} = \frac{1}{2(1 - x)^2} - \frac{3}{4(1 - x)} + \frac{1}{4(1 + x)}$$

(this can be done by hand) which gives the already-surprisingly-complicated

$$\boxed{ p_2(n) = \frac{n}{2} + \frac{(-1)^n - 1}{4} }.$$

This can be simplified, and proven, with a little casework as follows. The partitions into two parts are $n = i + j$ where, if $i \ge j$, we have $1 \le j \le \lfloor \frac{n}{2} \rfloor$. So we get $\boxed{ p_2(n) = \left\lfloor \frac{n}{2} \right\rfloor }$ which is equivalent to the above after squinting a bit.

For $k = 3$ the partial fraction decomposition is (according to WolframAlpha)

$$P_3(x) = \frac{-x^2 + 20x - 7}{72(1 - x)^3} - \frac{8}{1 + x} + \frac{x + 2}{9(1 + x + x^2)}$$

which gives, using that $\frac{x + 2}{1 + x + x^2} = \frac{2 - x - x^2}{1 - x^3}$ and $-x^2 + 20x - 7 = - (1 - x)^2 - 18(1 - x) + 12$, and after some simplification,

$$\boxed{ p_3(n) = \frac{6n^2 - 7}{72} - \frac{(-1)^n}{8} + \frac{\omega_3(n)}{9} }$$

where $\omega_3(n)$ is a periodic function with period $3$ that repeats $2, -1, -1, 2, -1, -1, \dots $. Already it's not entirely obvious that this is an integer, and I can't claim not to have made any computational errors, but the result has to look something like this even if I messed something up. It only gets worse from here - $p_k(n)$ has period-$d$ components for every $1 \le d \le k$. You can check out the partial fraction decomposition for $k = 4$ for yourself here.

So there's an important sense in which it's hopeless to expect a closed form for large $k$, although as above the leading-order asymptotic in $n$ for fixed $k$ is easy to calculate and the next few terms might not be so bad either. The good news is that for many purposes it doesn't matter; the generating function is powerful enough to answer many questions you might have about partitions with restricted parts.

Title: Number of partitions with x parts


问:
Does the kernel of $f:\operatorname{End}_A\rightarrow\mathbb{Z}$ contains all non-automorphisms?
Let:

$A$ is a non-trivial abelian group

$\operatorname{End}_A$ is the ring of endomorphisms of $A$

$\operatorname{Aut}_A$ is the set of automorphisms of $A$

$f:\operatorname{End}_A\rightarrow\mathbb{Z}$ is a unital ring homomorphism

$\ker(f)$ is the kernel of $f$

We can prove that $\operatorname{Aut}_A\cap \ker(f)=\emptyset$. Does the same hold for the complements, i.e. $\overline{\operatorname{Aut}_A}\cap\overline{\ker(f)}=\emptyset$? Can we at least say that $\phi\notin\ker(f)$ implies that $\phi$ has the left or right inverse?

答: Let $A = \mathbb Z$ and $\chi: \mathbb Z \to \mathbb Z, \chi(z) = 2z$. Set $f: \operatorname{End}(A) \to \mathbb Z, f(\phi) = \phi(1)$. Then $f$ is a ring homomorphism because $f(\phi \circ \psi) = \phi \circ \psi(1) = \phi(\psi(1)) = \phi(n) = mn$, where $\phi(z) = mz$ and $\psi(z) = nz$. Moreover, $\chi \notin \ker f$.


问:
Primitive Trinomial for $82589933$?
At Twelve new primitive binary trinomials, $x^{74207281}+x^{9999621}+1$ is shown to be a primitive trinomial in $GF(2)[x]$.

Note that $2^{74207281}-1$ was the largest known (Mersenne) prime before 2018, that is the prime in the exponent of the Mersenne prime gave rise to a trinomial.

In the interim, new Mersenne primes have been found.

Just announced (Dec 2018), $2^{82589933}-1$ is prime, and in Jan 2018, $2^{77232917}-1$ was found to be prime.

Do these also give rise to trinomials?

To be explicit, are there $j$ or $k$ such that $x^{82589933}+x^{j}+1$ or $x^{77232917}+x^{k}+1$ are primitive trinomials in $GF(2)[x]$?

答:
As per Swan's Theorem, reproduced in The Great Trinomial Hunt, $T_{r,s}(x) \equiv x^r + x^s + 1$ has an even number of factors over $GF(2)$ if $r$ is $\pm 3 \pmod{8}$ and $s$ is even and not a factor of $2r$. Given that both of the Mersenne exponents above (82589933 and 77232917) are 5 mod 8, if $s>2$ even then we know $T_{r,s}$ is reducible.

Likewise as stated in the paper, $x^r + x^s + 1$ has the same number of factors as $x^r + x^{r-s} + 1$, which eliminates all odd $s$ aside from $s = r-2$.

So it suffices to look at $x^r + x^2 + 1$ for the two values of $r$ in question.
This is not exactly an easy computation, even though polynomial squaring is easy in $GF(2)$.

Fortunately, the paper then suggests a useful trick -- sieving against small factors by using the fact that $T \mod x^d+1 = x^{(r \pmod{d})} + x^{(s\pmod{d})} + 1$.

With this, it is easy to see that $x^{82589933} + x^2 + 1$ and $x^{77232917} + x^2 + 1$ are both divisible by $x^3+x+1$.

So there are no primitive trinomials for these primes.

Title: Primitive Trinomial for $82589933$?


问:
A question about Rudin's proof that every positive real number has a unique positive $n$-th root
I am studying more deeply the first chapters of Rudin's Principles of Mathematical Analysis. What interests me are theorems' proofs strategies.
Theorem 1.21 states:

For every real $x > 0$ and every integer $n > 0$ there is one and only one real y such that $y^n=x$.

The one and only one part is soon proved; in order to prove the actual existance of $y$, the set

$E$ consisting of all positive real numbers $t$ such that $t^n<x$

is defined and it's then proved that $y=sup(E)$ exists in $R$. It remains to prove that $y^n=x$ and it's done in the last (and quite lenghty) part of the proof.

What I don't understand is why this last part is needed. I would say: once we defined $E$ as above we can consider the set $S$ containing all the upper bounds $f$ of $E$ such that  $f^n \geq x$ right? Among those upper bounds we can consider the smallest one which is indeed $y=sup(E)$ (we know it exists) and being the smallest it is the one such that $y^n = x$.

Am I being incorrect?

答: Yes, you are being incorrect. If $f$ is an upper bound of $E$, then, yes, $f^n\geqslant x$. But it is not obvious at all that if $f$ is the least upper bound, then $f^n=x$. That equality requires a proof.


问:
Geometric representation of product rule?
At time 1:06 of this video by minutephysics, there is a geometric representation of the product rule:

However, I don't understand how the sums of the areas of those thin strips represent $d(u\cdot v)$. The only way I can see it is that $d(u\cdot v)$ is a small change in the area of the square, and those thin strips do represent that; however, I'm not sure if this is correct and if it is, how formal of a proof is this? Thanks!

答:
I use the picture of the rectangle in my own teaching (without the differential notation) and show it to grad students who are starting their teaching careers. It is far superior to the usual tricky addition-of-$0$ argument found in most textbooks.

Here is the argument in greater detail:

\begin{align*}
\frac{\Delta(uv)}{\Delta x} &= \frac{(u+\Delta u)(v+\Delta v) - uv}{\Delta x} \\
&= \frac{u\Delta v + v\Delta u + \Delta u\Delta v}{\Delta x} = u \frac{\Delta v}{\Delta x} +
v \frac{\Delta u}{\Delta x} + \Delta u\cdot\frac{\Delta v}{\Delta x}\,.
\end{align*}
Taking $\lim\limits_{\Delta x\to 0}$ gives the product rule.

This can all be written out with the usual $f(x+h)g(x+h)$ notation, if so desired.

By the way, this same picture can be used to give a more motivated proof of the product theorem for limits, as well.


答: If $A=\mathbb{Z}$, then $\operatorname{End}(A)=\mathbb{Z}$. You can then take $f$ to be the identity of $\mathbb{Z}$, and any element which is neither $0$ nor $\pm 1$ is neither in the kernel nor invertible.

Title: Does the kernel of $f:\operatorname{End}_A\rightarrow\mathbb{Z}$ contains all non-automorphisms?

Body:
Let:

$A$ is a non-trivial abelian group

$\operatorname{End}_A$ is the ring of endomorphisms of $A$

$\operatorname{Aut}_A$ is the set of automorphisms of $A$

$f:\operatorname{End}_A\rightarrow\mathbb{Z}$ is a unital ring homomorphism

$\ker(f)$ is the kernel of $f$

We can prove that $\operatorname{Aut}_A\cap \ker(f)=\emptyset$. Does the same hold for the complements, i.e. $\overline{\operatorname{Aut}_A}\cap\overline{\ker(f)}=\emptyset$? Can we at least say that $\phi\notin\ker(f)$ implies that $\phi$ has the left or right inverse?


答:
I assume you prescribe that a ring morphism sends the unit to the unit, so that the map is surjective. This implies in particular that $\ker(f) $ is a proper ideal: if it was all of $End_A$, the image would be $0$.

The answer to the first question is yes: the automorphisms are by definition the units of the ring, and they can't intersect a proper ideal.

The answer to the second question is no as it is shown by the following example. Let $A=\mathbb{Z}$ and $ f: End_A \to \mathbb{Z} $ be defined by
$$f(\varphi) = \varphi(1) $$
Then the endomorphisms $\mu_n(x) =nx $ for $n\ge 2$ are not automorphisms and they are mapped to $n$ by $f$ (so they don't belong to the kernel) .


问:
The base of the product topology
Let $(E_i, \mathcal{T_i})$  a family of topological spaces and $E=\prod_{i \in I} E_i$.
The product topology is the smallest topology on $E$ such that all projection maps are continuous.
the base of this topology is given by :
$\mathcal B=\{ \prod_{i\in I} U_i : \text{ $U_i$ open in $E_i$  for all $i\in I$, and $U_i =E_i$ for all but finitely many $i\in I$}\}$.

The problem is why we define the base of the product topology like that and didn't define it as in the finite case  : $\mathcal B=\{ \prod_{i\in I} U_i : \text{ $U_i$ open in $E_i$  for all $i\in I$} \}$.
I know that $\mathcal B$ in  this case is the base of the box topology but I didn't understand yet.
Thank you in advance.

答:
The answer is very simple: the product topology behaves much better than the box topology and is more useful. In particular, it has the following characteristics. Let $X=\prod_{i\in I}X_i$.

For any space $Y$ and any function $f:Y\to X$, $f$ is continuous if and only if $\pi_i\circ f:Y\to X_i$ is continuous for each $i\in I$, where $\pi_i:X\to X_i$ is the projection map. The product topology is the only topology on $X$ with this property.

$X$ is compact if and only if each of the spaces $X_i$ is compact. The box product makes the projection maps continuous, so half of this is still true for the box product: if $X$ is compact in the box topology, then each $X_i$ is compact. The other half, however, is the Tikhonov product theorem, which does not hold in general for the box topology. For instance, let $X_n$ be the discrete two-point space for each $n\in\Bbb N$; then the box topology on $\prod_{n\in\Bbb N}X_n$ is discrete, hence certainly not compact, while the product topology gives us a compact space homeomorphic to the middle-thirds Cantor set.

$X$ is connected if and only if each $X_i$ is connected. Here again half of the theorem is true for the box topology: if $X$ is connected in the box topology, so is each of the spaces $X_i$. The other direction, however, can fail: if $X_n=\Bbb R$ for each $n\in\Bbb N$, $\prod_{n\in\Bbb N}X_n$ is not connected.

These properties alone are enough to justify taking the usual product topology as the standard one, but it is nicer than the box topology in other ways, especially when $I$ is countably infinite. If $I$ is countably infinite and each $X_i$ has one of the following properties, then $X$ with the product topology also has that property: separability, metrizability, first countability, second countability. None of these is preserved by the box topology when infinitely many of the $X_i$ are non-trivial.

Note that while it’s perfectly true that the first of the properties that I listed means that the usual product topology makes the topological product a category-theoretic product, it was established as the standard product topology before category theory came along.

Title: The base of the product topology


问:
Proving existence and uniqueness of solution to differential equation
I have to prove existence and uniqueness of the solution to the following equations:

$f'_0(t)=-\lambda_0f_0(t)$

$f'_n(t)=-\lambda_n f_n(t)+\lambda_{n-1}f_{n-1}(t), n=1,2,...$

where $\lambda_n\geq0$ and $f_n(0)\geq0 $ for all n.

I prove by induction that  $f_n(t)\geq0 $ for all $t\geq0$.

I have no idea where to start. I appreciate any help.

答:
The conditions you have given are insufficient to guarantee that their solution will be unique. Suppose $\ \lambda_0\ne0\ $, but $\ \lambda_n=0\ $ for all $\ n\ge1\ $ for instance.   Then the conditions have the trivial solution
$$
f_n(t)=0
$$
for all $\ n\ $, but also the non-trivial solution
\begin{align}
f_0(t)&=e^{-\lambda_0t}\\
f_1(t)&=1-e^{-\lambda_0t}\\
f_n(t)&=0
\end{align}
for all $\ n\ge2\ $ (or any positive multiple of it).

In the general case, if you set
\begin{align}
\mathbf{g}_n(t)&=\pmatrix{f_0(t)\\f_1(t)\\\vdots\\f_n(t)}\ \text{, and}\\
\Lambda_n&=\pmatrix{-\lambda_0&0&0&\dots&0&0\\
\lambda_0&-\lambda_1&0&\dots&0&0\\
0&\lambda_1&-\lambda_2&\dots&0&0\\
\vdots&\vdots&\ddots&\ddots&\vdots&\vdots\\
0&0&0&\dots&\lambda_{n-1}&-\lambda_n}\ ,
\end{align}
then you can write the differential equations in vector form as
$$
\mathbf{g}_n'(t)=\Lambda_n\mathbf{g}_n(t)\ .
$$
This has the solution
\begin{align}
\mathbf{g}_n(t)&=e^{\Lambda_n t}\mathbf{g}_n(0)\\
&=\left(\sum_{k=0}^\infty\frac{t^k\Lambda_n^k}{k!}\right)\mathbf{g}_n(0)\ .\
\end{align}
which is unique if $\ \mathbf{g}_n(0)\ $ is uniquely specified, but not otherwise.

Perhaps what you are required to do is to show that the solution is unique for any given $\ f_1(0),$$ f_2(0), \dots,$$ f_n(0), \dots\ $, but if that is the case, the question should have been more clearly worded.


问:
If $\text{Mod}_{R}$ and $\text{Mod}_{S}$ are equivalent, then $R$ and $S$ have the same simple modules.
Let $R$ and $S$ be (unital, associative) rings. If $\text{Mod}_{R}$ and $\text{Mod}_{S}$ are equivalent categories, then is it true that $R$ and $S$ have the same amount of isomorphism classes of simple modules?

Since an equivalence of categories preserves 'all categorical properties': an equivalent question would be? Is being a simple module a categorical property? Intuitively, I would guess so because Schur's lemma allows us to say something about morphisms from and to a simple module but I'm struggling with making everything precise.

答: Yes.  A simple module is an object $A$ of $\mathrm{Mod}_R$ which is not a zero object such that every monic morphism $B\to A$ from any other object $B$ either factors through a zero object or is an isomorphism.  (In concrete terms, this says any injective homomorphism to $A$ is either $0$ or surjective, so every submodule of $A$ is either trivial or all of $A$.)  Each part of this definition is preserved by an equivalence of categories, so an equivalence of categories $\mathrm{Mod}_R\to \mathrm{Mod}_S$ sends simple modules to simple modules.


答:
The other answers are good, but I want to point out that this actually never happens.

Suppose for contradiction that $f : \operatorname{End}_A \to \mathbb{Z}$ such that $\ker(f)$ contains all non-automorphisms. Since $\ker(f)$ is a proper ideal of $\operatorname{End}_A$, it does not contain any elements of $\operatorname{Aut}_A$. This means that $\ker(f) = \operatorname{End}_A \setminus \operatorname{Aut}_A$. The existence of $f$ tells us that $\operatorname{End}_A \neq 0$, so we conclude $\operatorname{End}_A$ is a local ring with maximal ideal $\ker(f)$. Now $\operatorname{img}(f) \cong \operatorname{End}_A/\ker(f)$ must be a division ring. However, $\mathbb{Z}$ has no division subrings (these would be fields of characteristic $0$, and $\mathbb{Q}$ does not embed in $\mathbb{Z}$).


问:
Density of Pythagorean triples
We define a Pythagorean triple as a triple $$ such that $a,b,c\in \mathbb N$ and $a^2+b^2=c^2$.

In order to avoid duplicates, we say that a triple $$ is legit iff $b>a$.

Let $\mathcal P$ be the set of all legit Pythagorean triples.

We define $$L_{PT}^N=\{ |  \in  \mathcal P\wedge b\leq N\}$$

(If it's more convinient we can define it for $b^2\leq N$, $c\leq N$ or $c^2\leq N$).

What is the density of $|L_{PT}^N|$ as a function of $N$? e.g. is $|L_{PT}^N|=\Theta(N^2)?\Theta(N)?$

We say that a triple $$ is minimal if $gcd(a,b,c)=1$.
Let $\mathcal P_M$ be the set of all legit, minimal triples.

Let $$L_{MPT}^N=\{ |  \in  \mathcal P_M\wedge b\leq N\}$$

What is the density of $|L_{MPT}^N|$ as a function of $N$? e.g. is $|L_{MPT}^N|=\Theta(N)?$

答:
The are primitive Pythagorean triples for every $A=(2x+1), x\in\mathbb{N}$. This could mean that the density is $50\%$ but "multiple primitive triples" increase with the compositeness of $A$. . For example, using Euclid's formula

$$A=m^2-k^2,\quad B=2mk,\quad C=m^2+k^2$$

\begin{equation}
f(4,1)=(15,8,17)\qquad f(8,7)=(15,112,113)\\
f(5,2)=(21,20,29)\qquad f(11,10)=(21,220,221)\\
f(7,4)=(33,56,65)\qquad f(17,16)=(33,544,545)\\
f(6,1)=(35,12,37)\qquad f(18,17)=(35,612,613)\\
f(8,5)=(39,80,89)\qquad f(20,19)=(39,760,761)\\
\vdots\\
f(41,24)=(1105,1968,2257)\\
 f(49,36)=(1105,3528,3697)\\
 f(113,108)=(1105,24408,24433)\\
 f(553,552)=(1105,610512,610513
\end{equation}

The number of primitives for a given $A$-value is $2^{n-1}$ where $n$ is the number of distinct prime factors of $A$. For example
$
15=3\cdot5\rightarrow 2^1=2\qquad
1105=5\cdot13\cdot17\rightarrow2^2=4.\quad$ Some $A$-values will have more than this but the extras will not be primitive. This means that the density could increase arbitrarily as $N\rightarrow\infty$. A complete answer would include analysis of composite numbers as $N$ increases.

There are primitive triples for every $B=4x, x\in\mathbb{N}$ and there are multiples for composites the same way there are for $A$-values. These occur more often for $B$ values than with $A$-values so the number of multiples might approach the same value with larger $N$.

\begin{equation}
f(3,2)=(5,12,13)\qquad f(6,1)=(35,12,37)\\
f(5,2)=(21,20,29)\qquad f(10,1)=(99,20,101)\\
f(7,2)=(45,28,53)\qquad f(14,1)=(195,28,197)\\
f(4,4)=(0,32,32)\qquad f(16,1)=(255,32,257)\\
\vdots\\
f(6,5)=(11,60,61)\quad f(10,3)=(91,60,109)\\ f(15,2)=(221,60,229)\quad f(30,1)=(899,60,901)
\end{equation}

All $C$-values must of be of the form $(4x+1), x\in\mathbb{N}$ but not all such values are valid. The appear to be valid about $2/3$ of the time but that is just a casual observation of a limited data set. In any case, there are multiples for composite $C$-values the same way there are for $A$-values and $B$-values.

Since all of these have what appears to have the same frequency and number of multiples for composites in the long run, the density might be represented by the discussion of $A$-values alone.

Title: Density of Pythagorean triples


问:
Using Truth Tables to Solve Logic Riddles
I've been traversing the practice exercises in Kenneth Rosen's Discrete Math and Its Applications in preparation for an upcoming class that is heavily proofs-based.

Constructing truth tables from propositions has been relatively simple up to this point, but now I am having to use them to solve logic puzzles and I am struggling. I can, for the most part, reason through many of the problems with words, but I am specifically instructed to solve them using truth tables and I find this to be very difficult.

Translating statements into logical expressions is particularly challenging for me. For example, given the following puzzle:

"The police have three suspects for the murder of Mr.
Cooper: Mr. Smith, Mr. Jones, and Mr. Williams. Smith,
Jones, and Williams each declare that they did not kill
Cooper. Smith also states that Cooper was a friend of
Jones and that Williams disliked him. Jones also states
that he did not know Cooper and that he was out of town
the day Cooper was killed. Williams also states that he saw both Smith and Jones with Cooper the day of the
killing and that either Smith or Jones must have killed
him.  Can you determine who the murderer was if one of the three men is guilty, the two innocent men
are telling the truth, but the statements of the guilty
man may or may not be true?"

how do I know what information to populate my truth table with? How am I supposed to represent a statement like

"Smith also states that Cooper was a friend of Jones and that
Williams disliked him."

using propositional logic? The fact that two people are friends, or that one person dislikes another, doesn't give me any concrete information that I can use to inform decisions, and I am worried that any assumptions I make (e.g. if person A likes person B, he would not kill him) could be incorrect and invalidate any further reasoning. I could construct truth table after truth table, but if the propositions I am evaluating are irrelevant, it renders the entire process useless.

Any general advice on solving these kinds of problems would be greatly appreciated.

答:
Hint:

If 2 of the 3 men contradict each other, then the 3rd man must be innocent, and therefore must be truthful in all of this statements.


问:
What does the notation P < NP exactly mean?
Over the last days I stepped two times over the
notation P < NP, but I am not sure what it exactly means.
I first saw it here:

P < NP In Our Stockings?

https://rjlipton.wordpress.com/2020/12/24/p-np-in-our-stockings/

And now here:

Toward Combinatorial Proof of P < NP Basic Approach (119-128)

https://www.cs.swansea.ac.uk/reports/yr2006/CSR7-2006.pdf

The abstract of the later paper says: "We present a plausibe
“school-algebraic” condition C0 that infers, in Peano Arithmetic,
the negative solution (abbr.: P < NP) to the familiar
open problem P =? NP."

Is P < NP a stronger statement than P ≠ NP? And infering P < NP
means here from PA+C0, and because C0 is doubtful, its not
really a proof of P ≠ NP?

答:
The $<$ sign here means "strict containment"; since we know that $P$ is contained in $NP$ it's equivalent to saying that $P \neq NP$.

And infering P < NP means here from PA+C0, and because C0 is doubtful, its not really a proof of P ≠ NP?

Yes. It's not stated as clearly as it could be, but $C_0$ is an unproven conjecture.


问:
Calculate the length of a polar curve

Calculate the length of the polar curve $$\theta (r)=\frac{1}{2}\left( r+\frac{1}{r}\right)$$ from r = 1 to r = 3.

I understand mostly how to get the length of a polar curve by:

$$\int_{a}^{b} \sqrt[]{(f(\theta ))^{2}+(f'(\theta ))^{2}} \ d\theta $$

But in this exercise i dont get how to do it. Maybe i need to write the function $\theta (r)$ in terms of $\theta$

Any ideas or hints? Thanks

答:
As OP states "polar curve", then we can imagine it as curve on polar plane, or curve on Cartesian plane in polar coordinates. So, we can do it in 2 ways:

Considering it as usual parametrical representation of curve on plane $(\boldsymbol r, \boldsymbol\theta)$ and considering $r$ as parameter we have $$\left(\frac{ds}{dr}\right)^2 =\left(\frac{dr}{dr}\right)^2+\left(\frac{d \theta}{dr}\right)^2$$
So, length can be calculated as
$$\int\limits_{1}^{3}\sqrt{1+\theta'^2(r)}dr $$

Considering on plane $(\boldsymbol x, \boldsymbol y)$ and taking polar representation of curve by $x =r \cos \theta, y = r \sin \theta$. Using in last formulas $\theta = \theta(r)$ we can calculate
$$\frac{dx}{dr} = \cos \theta - r \frac{d\theta}{dr} \sin \theta$$
and
$$\frac{dy}{dr} = \sin \theta + r \frac{d\theta}{dr} \cos \theta$$
Using obtained we have
$$\left(\frac{ds}{dr}\right)^2 =\left(\frac{dx}{dr}\right)^2+\left(\frac{dy}{dr}\right)^2 =\\
=\left(\ \cos \theta - r \frac{d\theta}{dr} \sin \theta \right)^2 + \left(\ \sin \theta + r \frac{d\theta}{dr} \cos \theta \right)^2 =\\
=1+ r^2\left(\frac{d\theta}{dr}\right)^2$$
and length can be calculate as integral
$$\int\limits_{1}^{3}\sqrt{1+r^2\theta'^2(r)}dr$$

Title: Calculate the length of a polar curve


问:
Can we define a topology on $\mathbb{R}^2$ using lines instead of circles or squares?
Does it make sense to define a topology on the coordinate grid $\mathbb{R}^2$ using the horizontal and vertical lines as "building-blocks"?

$B_{1, y} = \{ (x,y): y \in \mathbb{R} \}$ vertical lines are closed

$B_{2, x} = \{ (x,y): x \in \mathbb{R} \}$ horizontal lines are closed

The definition of basis of Topology is a collection of open sets:

the base elements cover $X = \mathbb{R}^2$

let $B_1$ and $B_2$ be the basis elements and $I = B_1 \cap B_2$ the there exists a $B_3$ such that $z \in B_3 \subseteq I = B_1 \cap B_2$.

Basically my question is that if we use a basis other than squares or circles can we change the topology of $\mathbb{R}^2$.  Here we coudl write one possible basis for the Euclidean topology:

$B_{(x_0, y_0), \epsilon} = \{ (x^2 - x_0)^2 - (y - y_0)^2 < \epsilon  \}  $

if we have two  bases $(\mathbb{R}^2, \mathcal{B}_1)$ $ (\mathbb{R}^2, \mathcal{B}_2)$ can these two copies of the plane be homeomorphic.

This matches the definition on p. 78 of Munkres.  We can define the topology generated by a basis $\mathcal{B}$.  A subset $U \subseteq X$ is open if for any $x \in U$ there is a basis element $B \in \mathcal{B}$ such that $x \in B \subset U$. On the next page, he shows that from the basis $\mathcal{B}$ the collections of sets $U$ is a topology.

Example:  check that $U_1 \cap U_2 \in \mathcal{T}$ (the intersection of open sets is open).  Given $x \in U_1 \cap U_2$ we can find $x \in B_1 \subset U_1$ and $x \in B_2 \subset U_2$ and then $x \in B_3 \subset B_1 \cap B_2 \subset U_1 \cap U_2$.  Our open sets might be irregular and yet the basis elements are quite standard.

Possible I have to change my question to define a basis of open  sets rather than closed.  We do have the result that if $U$ is open then $X \backslash U$ is closed.

Related

Why $B_3 \subset B_1 \cap B_2$ in the definition of a basis for a Topology in Munkres' Book?

答:
So we consider the coarsest topology on $ℝ^2$ such that horizontal and vertical lines are closed. We will characterize closed sets and the closure, and we will mention some properties of this topology.

Preliminaries.

$A ⊆ ℝ^2$ is thin if every two distinct points in $A$ are distinct in both coordinates.

$A_x := \{y ∈ ℝ: (x, y) ∈ A\}$ is the vertical slice through $A$,

$A^y := \{x ∈ ℝ: (x, y) ∈ A\}$ is the horizontal slice through $A$,

$C(x, y) := (\{x\} × ℝ) ∪ (ℝ × \{y\})$ is the cross at $(x, y)$.

Proposition 1.
Let $A ⊊ ℝ^2$. The following conditions are equivalent.

$A$ is closed.

$A$ is a finite union of horizontal lines, vertical lines, and points.

Every horizontal and vertical slice through $A$ is full or finite, and $A$ contains no infinite thin subset.

Proof. 2) ⇒ 1) clear. 1) ⇒ 3) since every horizontal or vertical line satisfies 3) and since sets satisfying 3) are stable under finite unions and arbitrary intersections. 3) ⇒ 2) follows from the following lemma.

Lemma.
Let $A ⊆ ℝ^2$ be a set containing no infinite thin subset.
Let $V := \{x ∈ ℝ: A_x \text{ infinite}\}$, $H := \{y ∈ ℝ: A^y \text{ infinite}\}$, $F := \{p ∈ A: C(p) ∩ A \text{ finite}\}$.
We claim that $V, H, F$ are finite sets, that $A ⊆ (ℝ × H) ∪ (V × ℝ) ∪ F =: B$, and that 3) ⇒ ($A = B$) ⇒ 2).

Proof.
If $V$ or $H$ is infinite, then we can easity find an infinite thin subset.
For every $p ∈ F$ the intersection $C(p) ∩ F$ is finite, so if $F$ is infinite, we can inductively construct an infinite thin subset by picking $p_n ∈ F \setminus ⋃_{i < n} C(p_i)$.
Clearly $A ⊆ B$.
3) ⇒ $(A = B)$ since if $x ∈ V$, then $A_x = ℝ$, and similarly for $H$.
$(A = B)$ ⇒ 2) since $B$ satisfies 2).

Proposition 2.
Let $A ⊆ ℝ^2$.
If $A$ contains an infinite thin subset, then $\overline{A} = ℝ^2$.
Otherwise, $\overline{A} = B$ (the notation from the Lemma).

Proof.
The density follows from Proposition 1.
$B$ is a closed superset of $A$ by 2) and the Lemma.
$B ⊆ \overline{A}$ by 3).

Note that this topology is $T_1$, but not $T_2$. In fact it is anti-$T_2$ or hyperconnected: every two non-empty open sets intersect.
It is easy to see that the topology is (quasi)compact and separable.
However it has weight continuum.

Also note that the whole construction can be generalized to a product $X × Y$ of any sets. If $X$ is finite, then $X × Y$ will be the sum of $X$-many copies of $Y$ with the cofinite topology. Similarly if $Y$ is finite. In particular, if both $X$ and $Y$ are finite, then $X × Y$ is discrete.
If both $X$ and $Y$ are infinite, then everything works as above (and the weight is $\max(|X|, |Y|)$).


问:
Prove $H$ is a normal subgroup of $K$
$G$ is a set.
$$G=\left\{\begin{vmatrix}1 & 0 \\ 0  & 1\end{vmatrix}, \begin{vmatrix}0 & 1 \\-1 & 0\end{vmatrix}, \begin{vmatrix}-1 & 0 \\ 0  & -1\end{vmatrix}, \begin{vmatrix}0 & -1 \\ 1  & 0\end{vmatrix}, \begin{vmatrix}i & 0 \\ 0  & i\end{vmatrix}, \begin{vmatrix}0 & i \\ -i  & 0\end{vmatrix}, \begin{vmatrix}-i & 0 \\ 0  & i\end{vmatrix}, \begin{vmatrix}0 & -i \\ i  & 0\end{vmatrix}\right\}$$ and $(G, \times)$ is a group.

If
$$H=\left\{\begin{vmatrix}1 & 0 \\ 0  & 1\end{vmatrix}, \begin{vmatrix}-1 & 0 \\ 0  & -1\end{vmatrix}\right\}$$ and
$$K=\left\{\begin{vmatrix}1 & 0 \\ 0  & 1\end{vmatrix}, \begin{vmatrix}-1 & 0 \\ 0  & -1\end{vmatrix}, \begin{vmatrix}i & 0 \\ 0  & i\end{vmatrix}, \begin{vmatrix}-i & 0 \\ 0  & -i\end{vmatrix}\right\},$$
then $H$ and $K$ is a subgroup of $G$, $H$ is a normal subgroup of $K$, and $K$ is a normal subgroup of $G$.

My questions are :

a. How do I prove that $H$ is a normal subgroup of $K$?
I knew that to prove if $H$ is a normal subgroup I have to prove that each left coset of $H$ in K is a right coset of $H$ in $K.$ But I don't know how apply that to the matrices.

b. What is every element of factor group $G/H?$

答: as for $H$ in $K$ you could notice that $H$ is of index 2 in K so it is normal.

Title: Prove $H$ is a normal subgroup of $K$


问:
Computing the number of conjugacy-classes in $GL_{n}(\mathbb{F}_{p})$ of elementary abelian p-subgroups by GAP and Magma
I'm trying to compute the number of conjugacy-classes of elementary abelian p-subgroups of rank $2$ in $GL_{n}(\mathbb{F}_{p})$ by GAP and Magma. So I consider the following GAP function:

```<code>myfunction:= function(n,p)
local G, S, cclS, cclG;
  G := Image(IsomorphismPermGroup(GL(n,p)));
  S := SylowSubgroup(G,p);
  cclS := Filtered(ConjugacyClassesSubgroups(S),
    cl->Size(Representative(cl))=p^2
    and not IsCyclic(Representative(cl)));
  cclG := List(EquivalenceClasses(cclS,
    function(cl1,cl2)
      return IsConjugate(G,Representative(cl1),
        Representative(cl2));
  end),Representative);
  cclG := List(cclG,cl->Representative(cl)^G);
  return Sum(List(cclG,Size));
end;
</code>```

Questions:
Is there something I missed in this algorithm?

Is there a similar algorithm with  Magma which give the required computation?

Thank you in advance.

答:
This calculation should work, but you do a lot of extra work (calculating all subgroups of the Sylow subgroup) which you do not really need. This can become quite costly, in particular once the dimension $n$ becomes larger.
What will be more efficient is to find the subgroups from possible generating sets, e.g. by using:

```<code>cclG:=List(IsomorphicSubgroups(G,AbelianGroup([p,p])),Image);
</code>```

(replacing the lines from S:=... down to ...Representative);

Since your function only returns the count of the total number of subgroups in these classes, this can be done even faster: Count, for each element $g\in G$ of order $p$, the number of elements of order $p$ in $C_G(g)$ outside $\langle g\rangle$, and to divide the result by the order of $GL_2(p)$ (to make up for the different generating systems for the same subgroup):

```<code>cl:=Filtered(ConjugacyClasses(G),x->Order(Representative(x))=p);
return Sum(
  List(cl,x->Size(x)*Sum(List(Filtered(ConjugacyClasses(Centralizer(x)),
    y->Order(Representative(y))=p
    and not Representative(y) in Group(Representative(x))),Size))))/Size(GL(2,p));
</code>```

Presumably one could implement the same algorithm in Magma, but why would one want to do this?

Body:
This calculation should work, but you do a lot of extra work (calculating all subgroups of the Sylow subgroup) which you do not really need. This can become quite costly, in particular once the dimension $n$ becomes larger.
What will be more efficient is to find the subgroups from possible generating sets, e.g. by using:

cclG:=List(IsomorphicSubgroups(G,AbelianGroup([p,p])),Image);

(replacing the lines from S:=... down to ...Representative);

Since your function only returns the count of the total number of subgroups in these classes, this can be done even faster: Count, for each element $g\in G$ of order $p$, the number of elements of order $p$ in $C_G(g)$ outside $\langle g\rangle$, and to divide the result by the order of $GL_2(p)$ (to make up for the different generating systems for the same subgroup):

cl:=Filtered(ConjugacyClasses(G),x->Order(Representative(x))=p);
return Sum(
  List(cl,x->Size(x)*Sum(List(Filtered(ConjugacyClasses(Centralizer(x)),
    y->Order(Representative(y))=p
    and not Representative(y) in Group(Representative(x))),Size))))/Size(GL(2,p));

Presumably one could implement the same algorithm in Magma, but why would one want to do this?


问:
how to calculate complement of vector space in gap
I have a finite dimensional (but large dimension) vector space over GF(2), $V$.
I also have a subspace $ A \subset V$.
I have basis for $V$ and for $A$. I'd like to find a basis for the
complement of $A$ in $V$, $B : V = A \oplus B; A \cap B = 0$. (Note this is not
the orthogonal complement). What would be a good way to do this in gap?
If there are no built-in functions, what would be a good approach/algorithm?

答:
If you have the bases given as list of vectors,
BaseSteinitzVectors(basV,basA).factorspace gives a basis of a complement.
This should be reasonably effective. Make sure that your vectors are in compressed form, as this will be faster.


问:
$x_1, x_2,..., x_p$ are consecutive numbers, then at least one of them is divisible by $p$
Let $p$ be a prime number, and let $x_1, x_2,..., x_p$ be natural numbers.
Consider the statement

If $x_1, x_2,..., x_p$ are consecutive numbers, then at least one of them is divisible
by $p$

I think that this statement is true since we have that we will go up to some $p$ (prime) of our sequence so this means that the $x_p$ natural number will always be divisible by $p$. How would I write this more symbolically since I have the idea in my head of what we want to prove.

答: Consider the set $\{ r_i = x_i \mod p \}$.  How many elements does it have?  Use the pigeonhole principle to see that one of the $r_i$ must be $0$.


问:
How to find the angle in between two triangles?
The problem is indicated in the figure from below:

The alternatives given in my book are as follows:

$\begin{array}{ll}
1.&10^{\circ}\\
2.&15^{\circ}\\
3.&20^{\circ}\\
4.&18^{\circ}\\
\end{array}$

What exactly should be done here?. I'm stuck. The only thing which I can spot is that:

$\angle DAC + x = 55$

$\angle BAC + 40 = \angle BCD + 55$

But without any other further knowledge, I don't know what else can be done here?. Can someone help me here?. Can this problem be solved relying only in euclidean geometry?.

The other thing which I can spot is since it says $BD \parallel AC$

This means:

$\angle BDA = \angle DAC$

But again this information doesn't really help me much into solving this problem. Can someone guide me further?. Please include a drawing in the answer because I'm lost. Does this requires a construction?.

答:
At first (blue angles), notice how the $ABC$ triangle is isosceles (since $AB=BC$), and you get that the bottom angles are both $70$ (because $\frac{180-40}{2}=70$). Then, since $BD\| AC$, you can find that the angle next to the $40$º is also $70$, and that the big angle that contains the $x$ is $55$.

Finally (red angles), we also know $AB=BD$, so the triangle $ABD$ is also isosceles and we know one angle is $40+70=110$, so the other two must be $35$ (because $\frac{180-110}{2}=35$).

So your $x$ is
$$\boxed{x=55-35=20}$$

Ask me is there's anything you don't understand, maybe I explained it too fast.


答:
Note: in writing this answer I have not assumed that the asker knows about modular arithmetic, though the proof is the same albeit longer.

Let $x_1=ap+r$ for $a\ge 0$ and $0\le r<p$. This is possible by the division algorithm. Then $x_k=ap+r+(k-1)$ for $1\le k \le p$. If $r=0$, we are done because then $x_1=ap$ and hence $p\mid x_1$. If not, $1\le r \le p-1$ so that
$$ 1\le p-r\le p-1.$$
So, setting $m-1=p-r$ we have $2\le m \le p$. So,
$$ x_m=ap+r+(m-1)=ap+r+(p-r)=(a+1)p$$
and $p\mid x_m$.

Body:
Let $p$ be a prime number, and let $x_1, x_2,..., x_p$ be natural numbers.
Consider the statement

If $x_1, x_2,..., x_p$ are consecutive numbers, then at least one of them is divisible
by $p$

I think that this statement is true since we have that we will go up to some $p$ (prime) of our sequence so this means that the $x_p$ natural number will always be divisible by $p$. How would I write this more symbolically since I have the idea in my head of what we want to prove.


答: Longer method than Yotam's: conjugate each subgroup by each element of $K$ and $G$ and show that they are invariant. You'll have to do $4*2 + 8*4 = 40$ different conjugations, but this isn't so large as to make the computation unfeasible to do by hand.

Title: Prove $H$ is a normal subgroup of $K$

Body:
$G$ is a set.
$$G=\left\{\begin{vmatrix}1 & 0 \\ 0  & 1\end{vmatrix}, \begin{vmatrix}0 & 1 \\-1 & 0\end{vmatrix}, \begin{vmatrix}-1 & 0 \\ 0  & -1\end{vmatrix}, \begin{vmatrix}0 & -1 \\ 1  & 0\end{vmatrix}, \begin{vmatrix}i & 0 \\ 0  & i\end{vmatrix}, \begin{vmatrix}0 & i \\ -i  & 0\end{vmatrix}, \begin{vmatrix}-i & 0 \\ 0  & i\end{vmatrix}, \begin{vmatrix}0 & -i \\ i  & 0\end{vmatrix}\right\}$$ and $(G, \times)$ is a group.

If
$$H=\left\{\begin{vmatrix}1 & 0 \\ 0  & 1\end{vmatrix}, \begin{vmatrix}-1 & 0 \\ 0  & -1\end{vmatrix}\right\}$$ and
$$K=\left\{\begin{vmatrix}1 & 0 \\ 0  & 1\end{vmatrix}, \begin{vmatrix}-1 & 0 \\ 0  & -1\end{vmatrix}, \begin{vmatrix}i & 0 \\ 0  & i\end{vmatrix}, \begin{vmatrix}-i & 0 \\ 0  & -i\end{vmatrix}\right\},$$
then $H$ and $K$ is a subgroup of $G$, $H$ is a normal subgroup of $K$, and $K$ is a normal subgroup of $G$.

My questions are :

a. How do I prove that $H$ is a normal subgroup of $K$?
I knew that to prove if $H$ is a normal subgroup I have to prove that each left coset of $H$ in K is a right coset of $H$ in $K.$ But I don't know how apply that to the matrices.

b. What is every element of factor group $G/H?$


问:
Cartesian products and power sets
another discrete math question. I was watching this video about Cartesian products, power sets and cardinality when a thought occurred to me. If the set S were arbitrary, would its Cartesian product be a subset of the power set? In other words would this statement be true:

$S \times S \subseteq \mathcal{P}(S)$

I've looked around on the site to find a similar question, but any questions I found involved the Cartesian product of two or more sets, not one. I think the statement is false because with an arbitrary set, we don't know what the elements are, so there's a chance the statement is false by default. Any thoughts?

答:
Frederic Schuller in his notes on geometric anatomy of theoretical physics describes axiomatic set theory (ZFC) and comes up with that definition.

If you apply it to the example shown in the last answer, you will come up in that it is correct.

Title: Cartesian products and power sets


问:
$(f + g)_* = f_* + g_*$
Prove: for any maps $f,g : \Sigma X \rightarrow Y$ , then
$(f + g)_* = f_* + g_* : H_*(\Sigma X) \rightarrow H_*(Y )$. And prove the same for cohomology

I know how to prove the composition as here Proving $(f \circ g)_{*}= f_{*} \circ g_{*}$ , but here this map is from the homology of the suspension of the space not the space itself. could anyone help me in proving this, please?

The meaning of addition of $f,g$ in this context is:

As for adding functions:  for any space  X  and any space  Y,  $[\Sigma X, Y ]$   is a group (because the suspension is a co-$H$ space).  Therefore,  if  $f, g: \Sigma X \to Y$,  then  there is also their sum  $f + g: \Sigma X \to Y.$

Explicitly, you pinch the equator of the suspension to get a wedge of two copies of $\Sigma X$, and then apply $f$  to the top wedge summand and  $g$  to the bottom wedge summand.

答:
This answer is adapted from Lemma 4.60 of Hatcher. The original argument there relies on chasing elements through diagrams. Here I present a more categorical version of the proof.

Note that I have deviated a little bit from Hatcher's notation by using both $\times$ and $\oplus$ in places where he uses $\oplus$ only, even though the direct sum and direct product of two summands/factors are the same. This is to differentiate between maps into vs. out of the direct product/sum, so that it's clear which limit/colimit diagram we are using, and makes it obvious what changes when we dualize.

As clarified by the discussion in the comments as well as your edit, the map $f + g$ is the composition $\Sigma X \xrightarrow{c} \Sigma X \vee \Sigma X \xrightarrow{f\vee g} Y$, where $c$ is the "pinch" map that quotients out an equatorial copy of $X$ in the suspension. Applying the homology functor to this, we get the map $(f+g)_*$. We want to prove that this map is in fact equal to $f_* + g_*$.

The key idea is to think of $f_* + g_*$ as the composition of the diagonal homomorphism $\Delta: H_\bullet(\Sigma X) \to H_\bullet(\Sigma X) \times H_\bullet(\Sigma X)$ given by $x\mapsto (x, x)$ with the direct sum $f_*\oplus g_*$. Now recall that $H_i(\Sigma X\vee \Sigma X) \cong H_i(\Sigma X) \oplus H_i(\Sigma X)$. Thus if we can show that the following diagram commutes, and the two vertical arrows cancel (i.e. $j$ is the inverse of $(i_{1*}\oplus i_{2*})$), then we will have obtained the equality between the composition of the top row, which by definition is just $(f+g)_*$, and the composition of the curved arrows at the bottom, which is $f_*+g_*$, since $$f_* + g_* = (f_*\oplus g_*)\circ\Delta = (f\vee g)_* \circ (i_{1*}\oplus i_{2*}) \circ j \circ c_* = (f\vee g)_* \circ c_* = (f+g)_*\ .$$

Now we tackle this proof in three steps. First we show that the triangle on the right commutes.
The isomorphism $i_{1*}\oplus i_{2*}: H_\bullet(\Sigma X)\oplus H_\bullet(\Sigma X)\to H_\bullet(\Sigma X\vee \Sigma X)$ is induced by the inclusion maps $i_1, i_2: \Sigma X \hookrightarrow \Sigma X\vee \Sigma X$. Since $(f\vee g)\circ i_1 = f$ and $(f\vee g)\circ i_2 = g$, using the universal property of direct sums, we can show that the composition $(f\vee g)_*\circ (i_{1*}\oplus i_{2*})$ is the direct sum map $f_*\oplus g_*$. (See below for the relevant diagram. The horizontal arrows at the bottom are the canonical injections into the direct sum.)

Next we want to construct the inverse of $i_{1*}\oplus i_{2*}$. Consider the maps $q_i: \Sigma X \vee \Sigma X \to \Sigma X, i = 1, 2$ that are identity on the $i$-th component, and constant on the other component. It's clear that $q_1\circ i_1$ and $q_2\circ i_2$ are the identity maps, and $q_1\circ i_2$ and $q_2 \circ i_1$ are constant maps, hence an easy calculation shows that $q_{1*}\times q_{2*}$ is the inverse of $i_{1*}\oplus i_{2*}$.

Finally to show that the left triangle commutes, note that $q_1\circ c$ and $q_2\circ c$ are both homotopic to the identity. (Think of this homotopy as sliding the equator to one of the poles.) This means that the induced maps $q_{1*}\circ c_*$ and $q_{2*} \circ c_*$ are equal to the identity, so $(q_{1*}\times q_{2*})\circ c_*$ is the diagonal map. (We can also draw a diagram for this proof. Even though it will seem a bit trivial, when we dualize it to prove the cohomology case later, it will become more useful.)

This completes the proof.

For the cohomology version of the statement, you essentially dualize all diagrams (reverse all arrows, direct sums become direct products and vice versa), and argue that in diagram 1 the top composition, which becomes $(f+g)^*$, is equal to the bottom composition, which becomes $f^* + g^*$. The central idea is similar, which is to break up $f^* + g^*$ into the composition of $f^*\times g^*$ with the addition $(x, y)\mapsto x+y$. The former is the curved arrow on the right, now pointing from $H^\bullet(Y)$ to $H^\bullet(\Sigma X)\times H^\bullet(\Sigma(X)$, and the latter is the curved arrow on the left, from $H^\bullet(\Sigma X)\times H^\bullet(\Sigma X)$ to $H^\bullet(\Sigma X)$. Commutativity on the left and right follows from the dualization of diagrams 2 and 3, respectively.

As an addendum, I want to relate this proof back to the diagram chasing in Hatcher. The commutativity of the diagrams that I'm proving here can be phrased as maps having equal values on all elements. So saying that $(f\vee g)_*\circ (i_{1*}\oplus i_{2*}) = f_*\oplus g_*$ is the same as saying for all $(x, y) \in H_\bullet(\Sigma X)\oplus H_\bullet(\Sigma X)$, we have $(f\vee g)_* (i_{1*}(x) + i_{2*}(y)) = f_*(x) + g_*(y)$. To prove this, you need only show it for the generators, which are of the form $(x, 0)$ and $(0, y)$. Similarly, to prove that $(q_{1*}\times q_{2*}) \circ c_* = \Delta$, you show that $(q_{1*}\times q_{2*})(c_*(x)) = (x, x)$ for all $x\in H_\bullet(\Sigma X)$. This is the route that Hatcher takes.

Title: $(f + g)_* = f_* + g_*$


答: If $AB=BC=BD$, then $B$ is the centre of circle with $A,C,D$ on the circumference and the angle $\angle ADC$ is half the angle $\angle ABC$, i.e. $20^\circ$

Title: How to find the angle in between two triangles?

Body:
The problem is indicated in the figure from below:

The alternatives given in my book are as follows:

$\begin{array}{ll}
1.&10^{\circ}\\
2.&15^{\circ}\\
3.&20^{\circ}\\
4.&18^{\circ}\\
\end{array}$

What exactly should be done here?. I'm stuck. The only thing which I can spot is that:

$\angle DAC + x = 55$

$\angle BAC + 40 = \angle BCD + 55$

But without any other further knowledge, I don't know what else can be done here?. Can someone help me here?. Can this problem be solved relying only in euclidean geometry?.

The other thing which I can spot is since it says $BD \parallel AC$

This means:

$\angle BDA = \angle DAC$

But again this information doesn't really help me much into solving this problem. Can someone guide me further?. Please include a drawing in the answer because I'm lost. Does this requires a construction?.


问:
Example 3.84 in Axler: Quotient spaces
Axler defines a quotient space as follows:

Suppose $U$ is a subspace of $V$. Then the quotient space $V/U$ if the set of all affine subsets of $V$ parallel to $U$. In other words,
$$V/U = \{v + U : v \in V\}.$$

As an example of one, he gives:

If $U$ is a line in $\mathbb{R}^3$ containing the origin, then $\mathbb{R}^3/U$ is the set of all lines in $\mathbb{R}^3$ parallel to $U$.

Geometrically this should be a set of 'vertical lines' covering $\mathbb{R}^3$, but I do not know how to create such a plot nor do I understand where it came from. The best I have been able to do is work it out algebraically.

First, by a "line" in $\mathbb{R}^3$, I assume Axler means a subset of the form
$$U = \{(x,0,0) \mid x \in \mathbb{R}\}.$$
Then we consider $\mathbb{R}^3/U$. Having fixed a point $(a,b,c)$, its equivalence class is the set of all $(x,y,z)$ satisfying:
\begin{align*}
(a,b,c) \sim (x,y,z) \iff (a - c, b - y, c - z) \in U \iff b - c = 0, c - z = 0 \iff b = c, c = z.
\end{align*}
It made more sense when I picked an explicit point and an explicit line, but the punchline should be that I get a line through $(a,b,c)$ parallel to $U$. But I do not know how to plot this or what "parallel" means in multiple dimensions.

答:
You'd be better served with the "point-and-vector" description of a line (note that in $\mathbb{R}^3$, you would need two non-equivalent linear equations to determine a line; in $\mathbb{R}^n$ you would need $n-1$ linearly independent linear equations to determine a line).

A line in $\mathbb{R}^n$ can be described by a pair of $n$-tuples, a basepoint $\mathbf{b}_0$ and a "direction" $\mathbf{v}\neq\mathbf{0}$,
$$\mathbf{b}_0 = (a_1,\ldots,a_n),\qquad \mathbf{v}=(v_1,\ldots,v_n).$$
The line consists precisely of all points of the form
$$\mathbf{p} = \mathbf{b}_0 + t\mathbf{v},\quad t\in\mathbb{R}.$$
Geometrically, you are taking the vector $\mathbf{v}$, stretching/contracting/flipping it by a factor of $t$, and then placing its base at $\mathbf{b}_0$. The line consists of all the endpoints you can get this way. Algebraically, it consists of all $n$-tuples of the form
$$(a_1+tv_1, a_2+tv_2,\ldots,a_n+tv_n),\qquad t\in\mathbb{R}.$$

Two lines, $L_1=\mathbf{b}_0+t\mathbf{v}$ and $L_2=\mathbf{c}_0+t\mathbf{w}$ are parallel if and only if the two vectors $\mathbf{v}$ and $\mathbf{w}$ are parallel, if and only if they are each a scalar multiple of the other.

Title: Example 3.84 in Axler: Quotient spaces


问:
Dyadics (binary basis)
Can anyone tell me why, for every $s\in [0,1]$, there exists $(\varepsilon_j)_j$ such that $\varepsilon_j\in \{0,1\}$ and
$$s=\sum_j^{\infty}\frac{\varepsilon_j}{2^j}.$$

答:
HINT: Given $s\in[0,1]$, we can define the numbers $\epsilon_j$ recursively.

Let $a_0=0$. Given $a_j$ for some $j\ge 0$, let

$$\epsilon_{j+1}=\begin{cases}
0,&\text{if }s-a_j<\frac1{2^{j+1}}\\
1,&\text{if }s-a_j\ge\frac1{2^{j+1}}\,,
\end{cases}$$

and let $a_{j+1}=a_j+\epsilon_{j+1}$.

Verify that $\sum_{j=1}^n\frac{\epsilon_j}{2^j}=a_n$ for each $n\ge 1$.

Thus, we need to show that $\lim\limits_{n\to\infty}a_n=s$.

Show that if $s=1$, then $\epsilon_j=1$ for each $j\in\Bbb Z^+$, and verify that in this case $\lim\limits_{n\to\infty}a_n=s$.

What is $\sum\limits_{j=n+1}^\infty\frac1{2^j}$? Use this result to show that if $n\in\Bbb Z^+$, and $\epsilon_j=1$ for all $j>n$, then $\epsilon_n=1$.

Conclude that if $s<1$, then $\{j\in\Bbb Z^+:\epsilon_j=0\}$ is infinite.

Show that if $\epsilon_j=0$, then $s-a_j<\frac1{2^j}$.

Conclude that $\lim\limits_{n\to\infty}a_n=s$, i.e., that $s=\sum\limits_{j=1}^\infty\frac{\epsilon_j}{2^j}$.

Title: Dyadics (binary basis)


问:
Is a system of elementary system of complex symmetric polynomials solvable in polynomial time?
Give a generalized system of the form...

$$a_1 + a_2 + a_3 = x_1 + x_2 + x_3$$
$$a_1 a_2 + a_1 a_3 + a_2 a_3 = x_1 x_2 + x_1 x_3 + x_2 x_3$$
$$a_1 a_2 a_3 = x_1 x_2 x_3$$

... solvable in polynomial time?

By generalized I mean having $n$ of the $a_i$ and $x_i$ terms, with all the $a_i$ and $x_i$ being complex numbers.

See https://en.wikipedia.org/wiki/Elementary_symmetric_polynomial and https://en.wikipedia.org/wiki/Vieta%27s_formulas .

答: It is instant. The collection of $x_i$ is merely a permutation of the $a_j$


问:
Interpretation of a term in Lambda Calculus
Just started studying $λ$-calculus and I came across this $λ$-term : $$λx.λy.xy$$
As far as I understand this can be read as :

Apply $x$ to $y$

The result of $(1)$ is probably an expression say $E_1$

$y$ variable is a bound variable while $x$ is a free variable in the context of $E_1$

For some input $I$ apply $E_1$ to $I$ and the result is an expression say $E_{2}$ that depends on $x$

Everything emphasized is a hypothesis. Still trying to understand this formal system of writing functions so any insight would be really helpful.

答:
Note that, with  explicit full parenthesization, the term is $\lambda x. (\lambda y. (xy))$

Thus, the meaning of the term $\lambda x. \lambda y. xy$ is the following. It is a function of two arguments (this is the meaning of the initial $\lambda x.\lambda y$) that takes the first argument (represented by the parameter $x$) and apply it to the second argument (represented by the parameter $y$), since $xy$ is the application of $x$ to $y$.

Remember that, in the $\lambda$-calculus, everything is a function and so it is standard that the argument of a function is another function.

In general, a term of the form $\lambda x. N$ has to be seen as a function $x \mapsto N$, i.e. a function that associates $x$ to $N$ (which, in turn, is another function). Hence, a term of the form $\lambda x. \lambda y. M$ has to be seen as a function $x \mapsto (y \mapsto M)$, i.e. a function that associates $x$ to a function that associates $y$ to $M$. By (de-)currying, this is equivalent to see $\lambda x. \lambda y. M$ as a function that associates a couple of arguments $(x,y)$ to $M$.


问:
How can I show that state $0$ is positive recurrent?
I'm having trouble with Lawler problem 2.10:

Consider a branching process with offspring distribution given by $\{p_n\}$. We will make the process into an irreducible Markov chain by asserting that if the population ever dies out, then the next generation will have one new individual [in other words, $p(0,1) = 1$]. For which $\{p_n\}$ will this chain be positive recurrent, null recurrent, transient?

I somehow know that the chain is positive recurrent when $\mu = \sum_k kp_k < 1$, but I'm not sure why.

I know that it would suffice to show that state $0$ is positive recurrent, as the Markov chain is  irreducible.

I would greatily appreciate any help.

答:
So here's one approach: let $N_{n}$ denote the size of your population at time $n$ without the additional condition that $p(0,1)=1$ and let $(X_{i,j})_{1\leq{i,j}<\infty}$ be i.i.d. random variables distributed according to your offspring distribution. That is, $\mathbb{P}(X_{1,1}=k)=p_{k}\hspace{2pt}$ for all nonnegative integers $k$. Then:

$$\mathbb{E}N_{n+1}=\mathbb{E}(\sum_{k=1}^{N_n}X_{n+1,k})=\mathbb{E}N_{n}\mathbb{E}X_{n+1,1}=\mu\mathbb{E}(N_{n})$$

The second equality here follows from the fact that all of the $X_{n+1,k}$'s are independent of $N_{n}$. From this recurrence, we deduce that $\mathbb{E}N_{n}=N_{0}\mu^{n}$. Thus,

$$\mathbb{E}N_{n}=N_{0}\mu^{n}=\sum_{k=1}^{\infty}k\mathbb{P}(N_{n}=k)\geq{\sum_{k=1}^{\infty}\mathbb{P}(N_{n}=k)}=1-\mathbb{P}(N_{n}=0)$$

Rearranging, this implies that:

$$\mathbb{P}(N_{n}=0)\geq{1-N_{0}\mu^{n}}$$

Let $\tau_{0}$ denote the hitting time of $0$ by $N_{n}$. Then $\mathbb{P}(N_{n}=0)=\mathbb{P}(\tau_{0}\leq{n})$ since once our population dies, it stays dead. Thus, it follows that:

$$\mathbb{P}(\tau_{0}>n)\leq{N_{0}\mu^{n}}$$

From here, since $\tau_{0}$ is a nonnegative random variable, the layer- cake representation tells us that:

$$\mathbb{E}\tau_{0}=\sum_{k=1}^{\infty}\mathbb{P}(\tau_{0}\geq{k})\leq{1+\sum_{k=1}^{\infty}\mathbb{P}(\tau_{0}>k)}\leq{1+\sum_{k=1}^{\infty}N_{0}\mu^{k}}=1+\frac{N_{0}\mu}{1-\mu}<\infty$$

The summability of this last geometric series is where we made use of the fact that $\mu<1$. Notice now that since your Markov chain behaves the same way as the original branching process up until the point we hit $0$, $\tau_{0}$ is unchanged when we add on the condition that $p(0,1)=1$. In summa, we see that $\mathbb{E}\tau_{0}<\infty$ as desired and so the resulting Markov process is positive recurrent.


问:
Prove that $\pi_1(L(p,q))\cong\mathbb{Z}/p\mathbb{Z}$
Let $p, q>0$ be coprime integers and

$S^3\equiv\{(z, w)\in\mathbb{C}^2||z|^2+|w|^2=1\}$

Then, an equivalence relation $\sim$ on $S^3$ as follows.

$(z, w)\sim(z', w') \overset{def} \Leftrightarrow \exists k\in\mathbb{Z}, z=e^{\frac{2\pi i k}{p}}z', w=e^{\frac{2\pi i kq}{p}}w'$

Last,  $L(p, q)\equiv S^3/\sim$ (Lens space) be a quotient space and $f:S^3\longrightarrow L(p,q)$ be a quotient map.

(i) Prove that $f$ is a covering map.

(ii) Prove that $\pi_1(L(p,q))\cong\mathbb{Z}/p\mathbb{Z}$.

I know the equivalence relation is related to a group action of $\mathbb{Z}/p\mathbb{Z}$, but I don't still understand (ii) in particular. Can someone help solve this problem? I'm sorry but I'm a beginner in topology. Thanks.

答: There is a covering action of $\mathbb{Z}/p$ on $S^3$ (in the sense of Hatcher) given by $(e^{2 \pi i/p})\cdot(z,w)=(z \cdot e^{2\pi i /p},w \cdot e^{2 \pi i q/p})$. The quotient space of $S^3$ by this action is $L(p,q)$. Since $S^3$ is simply connected, Hatcher 1.40 (page 72) implies that $\mathbb Z/p$ is the fundamental group.