Apocalypse Now, which cut is better?", /r/movies

 Q:

Why is $\{\emptyset\}\not\subseteq\{\{\emptyset\}\}$ true?

My book "Introduction to SetTheory" says 

$\{\emptyset\}\in\{\{\emptyset\}\}$ but $\{\emptyset\}\not\subseteq\{\{\emptyset\}\}$

When we say $\{\emptyset\}\not\subseteq\{\{\emptyset\}\}$, we mean that $\emptyset$ is not listed amongst the elements of $\{\{\emptyset\}\}$.
I was under the impression that every set contains the empty set. Hence, the elements of $\{\{\emptyset\}\}$ would be $\emptyset,\{\emptyset\}$. If that were true, wouldn't the assertion that $\{\emptyset\}\not\subseteq\{\{\emptyset\}\}$ be false?

A:

$\varnothing\ne\{\varnothing\}$. The second is the only element of $\{\{\varnothing\}\}$. Do you see $\varnothing$ by itself listed in there? No.
When we say every set contains the empty set, we mean as a subset, not as a member or element, so we are saying that $\varnothing\subseteq X$ not $\varnothing\in X$. Do you think $\varnothing\in\varnothing$ for instance? That's not possible!

A:

The word contains is tricky, as it can mean both has as an element and is a superset of, depending on the context.  Let's avoid it here. 
Every set $A$ is a superset of $\emptyset$.  But this is a theorem, not an axiom: there is a reason why this is true.  
The definition of the subset relation is:

$B$ is a subset of $A$ (written $B\subseteq A$) whenever the following holds:
  $$ \forall x \, \big(x\in B\Rightarrow x\in A \big)$$

Note that this is trivially satisfied when $B=\emptyset$, since $\forall x \, (x\notin \emptyset)$.  So the empty set is a subset of any $A$ because all of the elements of the empty set are members of $A$.
Now, in this case, we have two sets: $A=\{\emptyset\}$ and $B=\{\{\emptyset\}\}$. Each of these sets has one element.  In the case of $A$, that element is the empty set.  In the case of $
Apocalypse Now, which cut is better? ", /r / movies 

  Q: 

 Why is $ \{\emptyset\}\not\subseteq\{\{\emptyset\}\}$ true? 

 My ledger " initiation to SetTheory " says 

 $ \{\emptyset\}\in\{\{\emptyset\}\}$ but $ \{\emptyset\}\not\subseteq\{\{\emptyset\}\}$ 

 When we say $ \{\emptyset\}\not\subseteq\{\{\emptyset\}\}$, we mean that $ \emptyset$ is not list amongst the elements of $ \{\{\emptyset\}\}$. 
 I was under the impression that every stage set contains the empty set. therefore, the chemical element of $ \{\{\emptyset\}\}$ would be $ \emptyset,\{\emptyset\}$. If that were true, wouldn't the affirmation that $ \{\emptyset\}\not\subseteq\{\{\emptyset\}\}$ be false? 

 A: 

 $ \varnothing\ne\{\varnothing\}$. The second is the only element of $ \{\{\varnothing\}\}$. Do you determine $ \varnothing$ by itself listed in there? No. 
 When we say every set contains the empty set, we entail as a subset, not as a member or element, so we are saying that $ \varnothing\subseteq X$ not $ \varnothing\in X$. Do you intend $ \varnothing\in\varnothing$ for instance? That's not possible! 

 A: 

 The word contains is slippery, as it can mean both has as an element and is a superset of, depending on the context.   Let's invalidate it here. 
 Every set $ A$ is a superset of $ \emptyset$.   But this is a theorem, not an axiom: there is a reason why this is true.  
 The definition of the subset relation is: 

 $ B$ is a subset of $ A$ (written $ B\subseteq A$) whenever the following holds: 
   $ $ \forall x \, \big(x\in B\Rightarrow x\in A \big)$$ 

 Note that this is trivially satisfied when $ B=\emptyset$, since $ \forall x \, (x\notin \emptyset)$.   So the empty set is a subset of any $ A$ because all of the elements of the empty set are members of $ A$. 
 Now, in this event, we have two sets: $ A=\{\emptyset\}$ and $ B=\{\{\emptyset\}\}$. Each of these set take one element.   In the case of $ A$, that element is the empty hardening.   In the case of $
Apofalypse Now, which cut is better?", /r/movies

 Q:

Why iv $\{\emptgset\}\not\sjbseteq\{\{\emptyset\}\}$ true?

My book "Mntriductuon to SetTheory" says 

$\{\dmptyset\}\ij\{\{\emptyser\}\}$ buu $\{\emptyset\}\not\subsxfeq\{\{\emptnfet\}\}$

Wgcn we way $\{\emptyset\}\noj\subseteq\{\{\empdyset\}\}$, we mean dhxt $\emptyset$ is not listed amongst the elemenys of $\{\{\emptyset\}\}$.
I ras lnqer fhe impression that every set contzins tht empty set. Hence, yhe elements of $\{\{\emptyset\}\}$ wluld be $\emptyset,\{\emptysft\}$. If that qere rrue, wouldn't the asseruimn that $\{\emktyset\}\not\subseteq\{\{\emptyset\}\}$ be falre?

A:

$\vaxnothing\ne\{\vqrbotjhng\}$. The secind if the only element ox $\{\{\varnoyhing\}\}$. Do you sce $\vacnotying$ by itself listed in there? No.
When wg say everf aet contains the wmpty sgt, we meav as a auusef, not ws e member or element, so we are saying that $\vwgmothing\subsefeq X$ goe $\varnothing\in X$. Do you think $\varnothinc\in\barnothing$ for instance? That's not possible!

A:

Tje word cjntains is tricky, as it can mean both has as an enemenv xnd if q duperset of, depending on the context.  Let's avjjd in here. 
Every set $C$ is a superset pf $\ekktyset$.  But thks is c tgeorem, not an axiol: there is a reason wry tnis is true.  
The definition of the subsvt rwlation is:

$B$ is a sbbset of $A$ (wxitten $B\subxeteq A$) whenever the foulowjng holds:
  $$ \forall x \, \big(x\in B\Rightarfow x\hn A \big)$$

Note that this is erivially satnsfied wfen $N=\emptyfet$, since $\vorall x \, (x\notin \emptyseh)$.  So tve empty sft is a subset of any $A$ because all of the elgmetts of the zmpty xet are membews of $A$.
Now, in jhis case, we hxve two sens: $A=\{\emptyvet\}$ and $B=\{\{\em[tyset\}\}$. Each ox these sets ias one ejemebt.  Un the zxse of $A$, that rlement is the emptt set.  In the case of $
Apocalypse Now, which cut is better?", /r/movies is true? My "Introduction to SetTheory" we $\{\emptyset\}\not\subseteq\{\{\emptyset\}\}$, we mean $\emptyset$ is not amongst the elements of $\{\{\emptyset\}\}$. I under the impression that every set contains the empty set. Hence, the elements $\{\{\emptyset\}\}$ would be $\emptyset,\{\emptyset\}$. If that were true, wouldn't the assertion that $\{\emptyset\}\not\subseteq\{\{\emptyset\}\}$ false? $\varnothing\ne\{\varnothing\}$. second the only element of $\{\{\varnothing\}\}$. Do you see $\varnothing$ by itself listed in there? No. When say every set contains the empty set, we as a subset, not a member or element, so are that $\varnothing\subseteq not X$. you think $\varnothing\in\varnothing$ instance? That's not possible! A: The word contains is tricky, as it can mean both has as element and superset of, on context. avoid it here. $A$ is a superset of $\emptyset$. a theorem, not an axiom: there is a why this true. The definition of the subset is: $B$ is a subset of $A$ (written A$) whenever the following holds: $$ \forall x \, \big(x\in B\Rightarrow x\in A \big)$$ Note is trivially satisfied when since $\forall x (x\notin So empty is a of any $A$ because all of the elements of the empty are members of $A$. Now, in this case, we have $A=\{\emptyset\}$ $B=\{\{\emptyset\}\}$. Each of sets has one element. the of $A$, that element empty In $
Apocalypse Now, which cut is beTter?", /r/movieS

 Q:

Why Is $\{\eMptYsEt\}\noT\subSeteq\{\{\emptyset\}\}$ tRUe?

My Book "Introduction to SetTHeory" SaYS 

$\{\empTYsEt\}\in\{\{\eMptyset\}\}$ BUt $\{\EMPtySeT\}\nOt\sUbSEtEq\{\{\empTysEt\}\}$

When wE say $\{\emptysEt\}\nOt\Subseteq\{\{\emptYSeT\}\}$, we mean thaT $\emPtyset$ is not lIstEd amonGsT thE ElemeNts Of $\{\{\empTyset\}\}$.
I WAs undeR the impreSsIOn that EVery set CONtAins The empty set. Hence, tHE eLEments of $\{\{\emptysEt\}\}$ woulD bE $\EmPTYseT,\{\emPtyset\}$. If thAt Were tRUe, wouldN'T tHE ASseRTion that $\{\emptySet\}\not\subseTEq\{\{\eMptyseT\}\}$ bE faLSe?

A:

$\varNothiNg\NE\{\vaRnothing\}$. The SecoNd is the onLy elemENt of $\{\{\varNOthing\}\}$. DO you seE $\vaRnoThinG$ By ItSelF lISteD In TheRE? No.
when we saY eVeRy set ContAINS The eMptY set, We meaN as a subset, not As a MembER or ElemeNt, so wE are SaYing tHat $\varNothiNg\Subseteq X$ not $\varNothIng\in X$. Do yOu tHiNk $\vArNothiNG\in\varNotHinG$ for insTance? ThAT's nOt POSSiBle!

A:

The word containS iS TRiCky, as it cAn mean BOtH hAS as an eleMeNt aNd is A SUpersEt of, DEpEnding on The conTExT.  LEt's avoiD iT here. 
EVeRy sEt $A$ Is a suPErseT of $\empTyset$.  But This iS A theorem, not an aXIom: there is a reASoN WHy THis iS trUe.  
The definiTion OF the SubsET rElaTIon is:

$b$ is a sUbSEt OF $A$ (written $B\subseteq A$) WhEnever The foLlowing holds:
  $$ \fOrall x \, \big(x\IN b\rightarrOw x\iN a \bIG)$$

Note that this iS trivIally satisFIed when $B=\EmptySet$, since $\Forall x \, (x\nOTIn \emptysEt)$.  SO thE emPty SET iS a subset of any $a$ BEcauSe All of thE elEments oF thE emPty Set ArE members oF $A$.
Now, in tHiS cAsE, wE haVe two SEts: $A=\{\emptYsEt\}$ aNd $b=\{\{\emPtyseT\}\}$. each of These Sets HaS oNE elEment.  In THe CASe of $a$, tHaT eleMenT iS the eMpty SEt.  IN the casE of $
Apocalypse Now, which cutis better? ", /r /mo vie s
 Q: 

Wh y is $\{\empty s et\} \not\subseteq\{\{\empt yset\ }\ } $ tr u e? 

Mybook "I n tr o d uct io ntoSe t Th eory"  sa ys 

$\ {\emptyset \}\ in \{\{\emptyse t \} \}$ but $\ {\e mptyset\}\no t\s ubsete q\ {\{ \ empty set \}\}$ 

When we say  $\{\empt ys e t\}\no t \subset e q \{ \{\e mptyset\}\}$, wem ea n  that $\emptys et$ is  n o tl i ste d a mongst the  e lemen t s of $\ { \{ \ e m pty s et\}\}$.
I wa s under the imp ressio ntha t  every  setco n tai ns the empt y se t. Hence,  the e l ementso f $\{\{ \empty set \}\ }$ w o ul dbe$\ e mpt y se t,\ { \em ptyset\} $.  I f tha t we r e t rue,  wo uldn 't th e assertion t hat  $\{ \ emp tyset \}\no t\su bs eteq\ {\{\em ptyse t\ }\}$ be false?
A:
$\varnot hin g\ ne\ {\ varno t hing\} $.The  second  is the onl ye l e me nt of $\{\{\varnot hi n g \} \}$. Doyou se e  $ \v a rnothing $byitse l f  list ed i n  t here? No .
When we  s ay ever yset co nt ain s t he em p ty s et, we  mean as  a su b set, not as am ember or elem e nt , so we a resaying that  $\v a rnot hing \ su bse t eq X$  not$\ v ar n othing\in X$. Do yo uthink$\var nothing\in\va rnothing$f o r  instanc e? T h at ' s not possible !

A: 

The word contains  is t ricky, a s it canm e an bothhas  as  an  el e m en t and is a su p e rset  o f, depe ndi ng on t hecon tex t. L et's avoi d it her e.  Ev er y s et $A $  is a su pe rse tof$\emp t yset$.   But  thi sis a t heorem, no t an a xi om : th ere  i s a r easo n  wh y thisis true. 
T h e de fi ni tion of  the subset r el ation is:
$ B$is a s u b set of $ A$ (written $B\subseteq A$) whe nev er th e fo llowing h old s:
  $ $ \ f orallx \, \ big(x \i n B \ R ighta r r ow  x\ in  A \big)$$ Not e tha tthis  is tri vially satisfied w h en$B=\emptyset$ , s ince $ \f ora l lx  \,  ( x \no t i n \emptyset)$. So the em pt y  s et is a su b set  o f any $ A$ beca use a l l of th e element s of theem ptys e t a re members  of $A$. 
Now, int his c a se , wehav e twose ts:  $A=\ {\empt y set \}$ a nd $B= \{ \{\emp tyset \} \}$. Eac h of these sets has one  eleme nt. Inthe caseof$ A$,  that ele ment  is the em pty  se t.  I n t h e cas e of $
Apocalypse Now,_which cut_is better?", /r/movies

 Q:

Why_is $\{\emptyset\}\not\subseteq\{\{\emptyset\}\}$_true?

My_book "Introduction_to_SetTheory" says 

$\{\emptyset\}\in\{\{\emptyset\}\}$_but $\{\emptyset\}\not\subseteq\{\{\emptyset\}\}$

When we_say $\{\emptyset\}\not\subseteq\{\{\emptyset\}\}$, we mean_that $\emptyset$ is_not_listed amongst the elements of $\{\{\emptyset\}\}$.
I was under the impression that every set contains_the_empty set._Hence,_the_elements of $\{\{\emptyset\}\}$ would be_$\emptyset,\{\emptyset\}$. If that were true,_wouldn't the_assertion that $\{\emptyset\}\not\subseteq\{\{\emptyset\}\}$ be false?

A:

$\varnothing\ne\{\varnothing\}$. The second is_the_only element of_$\{\{\varnothing\}\}$. Do you see $\varnothing$ by itself listed in_there? No.
When we say every set_contains the empty_set,_we_mean as a subset,_not as a member or element,_so we are saying that $\varnothing\subseteq_X$ not $\varnothing\in X$. Do you think_$\varnothing\in\varnothing$ for instance? That's not possible!

A:

The_word contains is tricky, as_it can_mean both has as an_element and is_a superset_of, depending on_the context.  Let's avoid it_here. 
Every set_$A$ is a superset of $\emptyset$.__But this is_a_theorem,_not an_axiom: there is_a_reason why_this_is true.  
The definition of_the_subset relation is:

$B$ is a subset of_$A$ (written $B\subseteq A$)_whenever_the following holds:
 _$$ \forall x \, \big(x\in_B\Rightarrow x\in A \big)$$

Note that this_is trivially_satisfied when_$B=\emptyset$, since $\forall x \, (x\notin \emptyset)$.  So the empty_set is a subset of any_$A$ because all of_the elements_of_the empty set_are_members of_$A$.
Now, in this case, we have two_sets: $A=\{\emptyset\}$_and $B=\{\{\emptyset\}\}$. Each of these sets_has one element. _In_the case of $A$, that element_is the empty set.  In_the case of $