diff --git "a/theorems_contfunc.json" "b/theorems_contfunc.json" deleted file mode 100644--- "a/theorems_contfunc.json" +++ /dev/null @@ -1,2113 +0,0 @@ -{ - "dataset": { - "theorems": [ - { - "id": 0, - "type": "proposition", - "label": "Lebl-contfunc:0", - "categories": [ - "limits", - "topology", - "characterization", - "sequences" - ], - "title": "Let ", - "contents": [ - "Let$S \\subset \\R$. Then$x \\in \\R$is a cluster point of$S$if and only if there exists a convergent sequence of numbers$\\{ x_n \\}_{n=1}^\\infty$such that$x_n \\not= x$and$x_n \\in S$for all$n$, and$\\lim\\limits_{n\\to\\infty} x_n = x$." - ], - "refs": [], - "proofs": [ - { - "contents": [ - "First suppose $x$ is a cluster point of $S$.", - "For every $n \\in \\N$, pick $x_n$ to be an arbitrary point of", - "$(x-\\nicefrac{1}{n},x+\\nicefrac{1}{n}) \\cap S \\setminus \\{x\\}$, which", - "is nonempty because $x$ is a cluster point of $S$.", - "Then", - "$x_n$ is within $\\nicefrac{1}{n}$ of $x$, that is,", - "\\begin{equation*}", - "\\abs{x-x_n} < \\nicefrac{1}{n} .", - "\\avoidbreak", - "\\end{equation*}", - "As $\\{ \\nicefrac{1}{n} \\}_{n=1}^\\infty$ converges to zero,", - "$\\{ x_n \\}_{n=1}^\\infty$ converges to $x$.", - "On the other hand, if we start with a sequence of numbers", - "$\\{ x_n \\}_{n=1}^\\infty$ in $S$", - "converging to $x$ such that $x_n \\not= x$ for all $n$, then for every", - "$\\epsilon > 0$ there is an $M$ such that, in particular, $\\abs{x_M - x} <", - "\\epsilon$. That is, $x_M \\in (x-\\epsilon,x+\\epsilon) \\cap S \\setminus \\{x\\}$." - ], - "refs": [], - "ref_ids": [] - } - ], - "ref_ids": [] - }, - { - "id": 1, - "type": "proposition", - "label": "Lebl-contfunc:1", - "categories": [ - "limits", - "topology", - "sequences" - ], - "title": "Let be a cluster point of and let be a function su...", - "contents": [ - "Let$c$be a cluster point of$S \\subset \\R$and let$f \\colon S \\to \\R$be a function such that$f(x)$converges as$x$goes to$c$. Then the limit of$f(x)$as$x$goes to$c$is unique." - ], - "refs": [], - "proofs": [ - { - "contents": [ - "Let $L_1$ and $L_2$ be two numbers that both satisfy the definition.", - "Take an $\\epsilon > 0$ and find a $\\delta_1 > 0$ such that", - "$\\abs{f(x)-L_1} < \\nicefrac{\\epsilon}{2}$", - "for all $x \\in S \\setminus \\{c\\}$ with $\\abs{x-c} < \\delta_1$.", - "Also find $\\delta_2 > 0$ such that", - "$\\abs{f(x)-L_2} < \\nicefrac{\\epsilon}{2}$", - "for all $x \\in S \\setminus \\{c\\}$ with $\\abs{x-c} < \\delta_2$.", - "Put $\\delta \\coloneqq \\min \\{ \\delta_1, \\delta_2 \\}$. Suppose $x \\in S$,", - "$\\abs{x-c} < \\delta$, and $x \\not= c$. As $\\delta > 0$ and $c$ is a cluster", - "point, such an $x$ exists. Then", - "\\begin{equation*}", - "\\abs{L_1 - L_2} =", - "\\abs{L_1 - f(x) + f(x) - L_2} \\leq", - "\\abs{L_1 - f(x)} + \\abs{f(x) - L_2} < \\frac{\\epsilon}{2} + \\frac{\\epsilon}{2}", - "= \\epsilon.", - "\\end{equation*}", - "As $\\abs{L_1-L_2} < \\epsilon$ for arbitrary $\\epsilon > 0$, then", - "$L_1 = L_2$." - ], - "refs": [], - "ref_ids": [] - } - ], - "ref_ids": [] - }, - { - "id": 2, - "type": "example", - "label": "Lebl-contfunc:2", - "categories": [ - "example" - ], - "title": "Consider defined by ", - "contents": [ - "Consider$f \\colon \\R \\to \\R$defined by$f(x) \\coloneqq x^2$. Then for any$c \\in \\R$,", - "\\begin{equation*}\n\\lim_{x\\to c} f(x) = \\lim_{x\\to c} x^2 = c^2 .\n\\end{equation*}", - "Proof: Let$c \\in \\R$be fixed, and suppose$\\epsilon > 0$is given. Write", - "\\begin{equation*}\n\\delta \\coloneqq \\min \\left\\{ 1 , \\, \\frac{\\epsilon}{2\\abs{c}+1} \\right\\} .\n\\end{equation*}", - "Take$x \\not= c$such that$\\abs{x-c} < \\delta$. In particular,$\\abs{x-c} < 1$. By reverse triangle inequality,", - "\\begin{equation*}\n\\abs{x}-\\abs{c} \\leq \\abs{x-c} < 1 .\n\\end{equation*}", - "Adding$2\\abs{c}$to both sides, we obtain$\\abs{x} + \\abs{c} < 2\\abs{c} + 1$. Estimate", - "\\begin{equation*}\n\\begin{split}\n\\abs{f(x) - c^2} &= \\abs{x^2-c^2} \\\\\n&= \\abs{(x+c)(x-c)} \\\\\n&= \\abs{x+c}\\abs{x-c} \\\\\n&\\leq (\\abs{x}+\\abs{c})\\abs{x-c} \\\\\n&< (2\\abs{c}+1)\\abs{x-c} \\\\\n&< (2\\abs{c}+1)\\frac{\\epsilon}{2\\abs{c}+1} = \\epsilon .\n\\end{split}\n\\end{equation*}" - ], - "refs": [], - "proofs": [ - { - "contents": [ - "Let$c \\in \\R$be fixed, and suppose$\\epsilon > 0$is given. Write \\begin{equation*}", - "\\delta \\coloneqq \\min \\left\\{ 1 , \\, \\frac{\\epsilon}{2\\abs{c}+1} \\right\\} .", - "\\end{equation*} Take$x \\not= c$such that$\\abs{x-c} < \\delta$. In particular,$\\abs{x-c} < 1$. By reverse triangle inequality, \\begin{equation*}", - "\\abs{x}-\\abs{c} \\leq \\abs{x-c} < 1 .", - "\\end{equation*} Adding$2\\abs{c}$to both sides, we obtain$\\abs{x} + \\abs{c} < 2\\abs{c} + 1$. Estimate \\begin{equation*}", - "\\begin{split}", - "\\abs{f(x) - c^2} &= \\abs{x^2-c^2} \\\\", - "&= \\abs{(x+c)(x-c)} \\\\", - "&= \\abs{x+c}\\abs{x-c} \\\\", - "&\\leq (\\abs{x}+\\abs{c})\\abs{x-c} \\\\", - "&< (2\\abs{c}+1)\\abs{x-c} \\\\", - "&< (2\\abs{c}+1)\\frac{\\epsilon}{2\\abs{c}+1} = \\epsilon .", - "\\end{split}", - "\\end{equation*}" - ], - "refs": [], - "ref_ids": [] - } - ], - "ref_ids": [] - }, - { - "id": 3, - "type": "example", - "label": "Lebl-contfunc:3", - "categories": [ - "limits", - "example" - ], - "title": "Define by f(x) x & x > 0 , \\\\ 1 & x = 0", - "contents": [ - "Define$f \\colon [0,1) \\to \\R$by", - "\\begin{equation*}\nf(x) \\coloneqq \n\\begin{cases}\nx & \\text{if } x > 0 , \\\\\n1 & \\text{if } x = 0 .\n\\end{cases}\n\\end{equation*}", - "Then$\\lim\\limits_{x\\to 0} f(x) = 0$, even though$f(0) = 1$. See \\figureref{fig:limvaldiff}. \\begin{myfigureht} \\includegraphics{figures/limvaldiff} \\caption{Function with a different limit and value at$0$.\\label{fig:limvaldiff}} \\end{myfigureht}", - "Proof: Let$\\epsilon > 0$be given. Let$\\delta \\coloneqq \\epsilon$. For$x \\in [0,1)$,$x \\not= 0$, and$\\abs{x-0} < \\delta$, we get", - "\\begin{equation*}\n\\abs{f(x) - 0} = \\abs{x} < \\delta = \\epsilon .\n\\end{equation*}" - ], - "refs": [], - "proofs": [ - { - "contents": [ - "Let$\\epsilon > 0$be given. Let$\\delta \\coloneqq \\epsilon$. For$x \\in [0,1)$,$x \\not= 0$, and$\\abs{x-0} < \\delta$, we get \\begin{equation*}", - "\\abs{f(x) - 0} = \\abs{x} < \\delta = \\epsilon .", - "\\end{equation*}" - ], - "refs": [], - "ref_ids": [] - } - ], - "ref_ids": [] - }, - { - "id": 4, - "type": "lemma", - "label": "Lebl-contfunc:seqflimit:lemma", - "categories": [ - "sequences", - "topology", - "auxiliary result", - "characterization" - ], - "title": "Let , let be a cluster point of , let $f S L $", - "contents": [ - "Let$S \\subset \\R$, let$c$be a cluster point of$S$, let$f \\colon S \\to \\R$be a function, and let$L \\in \\R$.", - "Then$f(x) \\to L$as$x \\to c$if and only if for every sequence$\\{ x_n \\}_{n=1}^\\infty$such that$x_n \\in S \\setminus \\{c\\}$for all$n$, and such that$\\lim_{n\\to\\infty} x_n = c$, we have that the sequence$\\bigl\\{ f(x_n) \\bigr\\}_{n=1}^\\infty$converges to$L$." - ], - "refs": [], - "proofs": [ - { - "contents": [ - "Suppose", - "$f(x) \\to L$ as $x \\to c$, and $\\{ x_n \\}_{n=1}^\\infty$ is a sequence", - "such that", - "$x_n \\in S \\setminus \\{c\\}$ and", - "$\\lim_{n\\to\\infty} x_n = c$.", - "We wish to show that $\\bigl\\{ f(x_n) \\bigr\\}_{n=1}^\\infty$ converges to $L$.", - "Let $\\epsilon > 0$ be given. Find a $\\delta > 0$ such that", - "if $x \\in S \\setminus \\{c\\}$ and $\\abs{x-c} < \\delta$, then", - "$\\abs{f(x) - L} < \\epsilon$. As", - "$\\{ x_n \\}_{n=1}^\\infty$ converges to $c$, find an $M$ such that for $n \\geq M$,", - "we have that $\\abs{x_n - c} < \\delta$. Therefore, for $n \\geq M$,", - "\\begin{equation*}", - "\\abs{f(x_n) - L} < \\epsilon .", - "\\end{equation*}", - "Thus $\\bigl\\{ f(x_n) \\bigr\\}_{n=1}^\\infty$ converges to $L$.", - "For the other direction, we use proof by contrapositive. Suppose", - "it is not true that $f(x) \\to L$ as $x \\to c$. The negation of the", - "definition is that there exists an $\\epsilon > 0$ such that for every", - "$\\delta > 0$ there exists an $x \\in S \\setminus \\{c\\}$, where", - "$\\abs{x-c} < \\delta$", - "and $\\abs{f(x)-L} \\geq \\epsilon$.", - "Let us use $\\nicefrac{1}{n}$ for $\\delta$ in the statement above to", - "construct a sequence $\\{ x_n \\}_{n=1}^\\infty$. We have", - "that there exists an $\\epsilon > 0$ such that for every $n$,", - "there exists a point $x_n \\in S \\setminus \\{c\\}$, where", - "$\\abs{x_n-c} < \\nicefrac{1}{n}$", - "and $\\abs{f(x_n)-L} \\geq \\epsilon$.", - "The sequence $\\{ x_n \\}_{n=1}^\\infty$ just constructed converges to $c$, but", - "the sequence $\\bigl\\{ f(x_n) \\bigr\\}_{n=1}^\\infty$ does not converge to $L$.", - "And we are done." - ], - "refs": [], - "ref_ids": [] - } - ], - "ref_ids": [] - }, - { - "id": 5, - "type": "example", - "label": "Lebl-contfunc:5", - "categories": [ - "limits", - "example", - "sequences" - ], - "title": " does not exist, but ", - "contents": [ - "$\\displaystyle \\lim_{x \\to 0} \\sin( \\nicefrac{1}{x} )$does not exist, but$\\displaystyle \\lim_{x \\to 0} x\\sin( \\nicefrac{1}{x} ) = 0$. See \\figureref{figsin1x}.", - "\\begin{myfigureht} %left guy also used in 10.4 \\subimport*{figures/}{sin1x_xsin1x.pdf_t} \\caption{Graphs of$\\sin(\\nicefrac{1}{x})$and$x \\sin(\\nicefrac{1}{x})$. Note that the computer cannot properly graph$\\sin(\\nicefrac{1}{x})$near zero as it oscillates too fast.\\label{figsin1x}} \\end{myfigureht}", - "Proof: We start with$\\sin(\\nicefrac{1}{x})$. Define a sequence by$x_n \\coloneqq \\frac{1}{\\pi n + \\nicefrac{\\pi}{2}}$. It is not hard to see that$\\lim_{n\\to\\infty} x_n = 0$. Furthermore,", - "\\begin{equation*}\n\\sin ( \\nicefrac{1}{x_n} )\n=\n\\sin (\\pi n + \\nicefrac{\\pi}{2})\n= {(-1)}^n .\n\\end{equation*}", - "Therefore,$\\bigl\\{ \\sin ( \\nicefrac{1}{x_n} ) \\bigr\\}_{n=1}^\\infty$does not converge. By \\lemmaref{seqflimit:lemma},$\\displaystyle \\lim_{x \\to 0} \\sin( \\nicefrac{1}{x} )$does not exist.", - "Now consider$x\\sin(\\nicefrac{1}{x})$. Let$\\{ x_n \\}_{n=1}^\\infty$be a sequence such that$x_n \\not= 0$for all$n$, and such that$\\lim_{n\\to\\infty} x_n = 0$. Notice that$\\abs{\\sin(t)} \\leq 1$for all$t \\in \\R$. Therefore,", - "\\begin{equation*}\n\\abs{x_n\\sin(\\nicefrac{1}{x_n})-0}\n=\n\\abs{x_n}\\abs{\\sin(\\nicefrac{1}{x_n})}\n\\leq\n\\abs{x_n} .\n\\end{equation*}", - "As$x_n$goes to 0, then$\\abs{x_n}$goes to zero, and hence$\\bigl\\{ x_n\\sin(\\nicefrac{1}{x_n}) \\bigr\\}_{n=1}^\\infty$converges to zero. By \\lemmaref{seqflimit:lemma},$\\displaystyle \\lim_{x \\to 0} x\\sin( \\nicefrac{1}{x} ) = 0$." - ], - "refs": [ - "seqflimit:lemma" - ], - "proofs": [ - { - "contents": [ - "We start with$\\sin(\\nicefrac{1}{x})$. Define a sequence by$x_n \\coloneqq \\frac{1}{\\pi n + \\nicefrac{\\pi}{2}}$. It is not hard to see that$\\lim_{n\\to\\infty} x_n = 0$. Furthermore, \\begin{equation*}", - "\\sin ( \\nicefrac{1}{x_n} )", - "=", - "\\sin (\\pi n + \\nicefrac{\\pi}{2})", - "= {(-1)}^n .", - "\\end{equation*} Therefore,$\\bigl\\{ \\sin ( \\nicefrac{1}{x_n} ) \\bigr\\}_{n=1}^\\infty$does not converge. By \\lemmaref{seqflimit:lemma},$\\displaystyle \\lim_{x \\to 0} \\sin( \\nicefrac{1}{x} )$does not exist. Now consider$x\\sin(\\nicefrac{1}{x})$. Let$\\{ x_n \\}_{n=1}^\\infty$be a sequence such that$x_n \\not= 0$for all$n$, and such that$\\lim_{n\\to\\infty} x_n = 0$. Notice that$\\abs{\\sin(t)} \\leq 1$for all$t \\in \\R$. Therefore, \\begin{equation*}", - "\\abs{x_n\\sin(\\nicefrac{1}{x_n})-0}", - "=", - "\\abs{x_n}\\abs{\\sin(\\nicefrac{1}{x_n})}", - "\\leq", - "\\abs{x_n} .", - "\\end{equation*} As$x_n$goes to 0, then$\\abs{x_n}$goes to zero, and hence$\\bigl\\{ x_n\\sin(\\nicefrac{1}{x_n}) \\bigr\\}_{n=1}^\\infty$converges to zero. By \\lemmaref{seqflimit:lemma},$\\displaystyle \\lim_{x \\to 0} x\\sin( \\nicefrac{1}{x} ) = 0$." - ], - "refs": [ - "seqflimit:lemma" - ], - "ref_ids": [ - 4 - ] - } - ], - "ref_ids": [ - 4 - ] - }, - { - "id": 6, - "type": "corollary", - "label": "Lebl-contfunc:6", - "categories": [ - "limits", - "topology", - "consequence" - ], - "title": "Let and let be a cluster point of ", - "contents": [ - "Let$S \\subset \\R$and let$c$be a cluster point of$S$. Suppose$f \\colon S \\to \\R$and$g \\colon S \\to \\R$are functions such that the limits of$f(x)$and$g(x)$as$x$goes to$c$both exist, and\\begin{equation*} f(x) \\leq g(x) \\qquad \\text{for all } x \\in S \\setminus \\{ c \\}. \\end{equation*}Then\\begin{equation*} \\lim_{x\\to c} f(x) \\leq \\lim_{x\\to c} g(x) . \\end{equation*}" - ], - "refs": [], - "proofs": [ - { - "contents": [ - "Take $\\{ x_n \\}_{n=1}^\\infty$ be a sequence of numbers in $S \\setminus \\{ c \\}$", - "that converges to $c$. Let", - "\\begin{equation*}", - "L_1 \\coloneqq \\lim_{x\\to c} f(x), \\qquad \\text{and} \\qquad L_2 \\coloneqq \\lim_{x\\to c} g(x) .", - "\\end{equation*}", - "\\lemmaref{seqflimit:lemma} says that $\\bigl\\{ f(x_n) \\bigr\\}_{n=1}^\\infty$ converges to", - "$L_1$ and $\\bigl\\{ g(x_n) \\bigr\\}_{n=1}^\\infty$ converges to $L_2$. We also", - "have $f(x_n) \\leq g(x_n)$ for all $n$.", - "We obtain $L_1 \\leq L_2$ using", - "\\lemmaref{limandineq:lemma}." - ], - "refs": [ - "limandineq:lemma", - "seqflimit:lemma" - ], - "ref_ids": [ - 4 - ] - } - ], - "ref_ids": [] - }, - { - "id": 7, - "type": "corollary", - "label": "Lebl-contfunc:fconstineq:cor", - "categories": [ - "limits", - "topology", - "consequence" - ], - "title": "Let and let be a cluster point of ", - "contents": [ - "Let$S \\subset \\R$and let$c$be a cluster point of$S$. Suppose$f \\colon S \\to \\R$is a function such that the limit of$f(x)$as$x$goes to$c$exists. Suppose there are two real numbers$a$and$b$such that\\begin{equation*} a \\leq f(x) \\leq b \\qquad \\text{for all } x \\in S \\setminus \\{ c \\}. \\end{equation*}Then\\begin{equation*} a \\leq \\lim_{x\\to c} f(x) \\leq b . \\end{equation*}" - ], - "refs": [], - "proofs": [], - "ref_ids": [] - }, - { - "id": 8, - "type": "corollary", - "label": "Lebl-contfunc:fsqueeze:cor", - "categories": [ - "limits", - "topology", - "consequence" - ], - "title": "Let and let be a cluster point of ", - "contents": [ - "Let$S \\subset \\R$and let$c$be a cluster point of$S$. Suppose$f \\colon S \\to \\R$,$g \\colon S \\to \\R$, and$h \\colon S \\to \\R$are functions such that\\begin{equation*} f(x) \\leq g(x) \\leq h(x) \\qquad \\text{for all } x \\in S \\setminus \\{ c \\}. \\end{equation*}Suppose the limits of$f(x)$and$h(x)$as$x$goes to$c$both exist, and\\begin{equation*} \\lim_{x\\to c} f(x) = \\lim_{x\\to c} h(x) . \\end{equation*}Then the limit of$g(x)$as$x$goes to$c$exists and\\begin{equation*} \\lim_{x\\to c} g(x) = \\lim_{x\\to c} f(x) = \\lim_{x\\to c} h(x) . \\end{equation*}" - ], - "refs": [], - "proofs": [], - "ref_ids": [] - }, - { - "id": 9, - "type": "corollary", - "label": "Lebl-contfunc:falg:cor", - "categories": [ - "limits", - "topology", - "consequence" - ], - "title": "Let and let be a cluster point of ", - "contents": [ - "Let$S \\subset \\R$and let$c$be a cluster point of$S$. Suppose$f \\colon S \\to \\R$and$g \\colon S \\to \\R$are functions such that the limits of$f(x)$and$g(x)$as$x$goes to$c$both exist. Then \\begin{enumerate}[(i)] \\item$\\displaystyle \\lim_{x\\to c} \\bigl(f(x)+g(x)\\bigr) = \\left(\\lim_{x\\to c} f(x)\\right) + \\left(\\lim_{x\\to c} g(x)\\right) . $\\item$\\displaystyle \\lim_{x\\to c} \\bigl(f(x)-g(x)\\bigr) = \\left(\\lim_{x\\to c} f(x)\\right) - \\left(\\lim_{x\\to c} g(x)\\right) . $\\item$\\displaystyle \\lim_{x\\to c} \\bigl(f(x)g(x)\\bigr) = \\left(\\lim_{x\\to c} f(x)\\right) \\left(\\lim_{x\\to c} g(x)\\right) . $\\item \\label{falg:cor:iv} If$\\displaystyle \\lim_{x\\to c} g(x) \\not= 0$and$g(x) \\not= 0$for all$x \\in S \\setminus \\{ c \\}$, then\\begin{equation*} \\lim_{x\\to c} \\frac{f(x)}{g(x)} = \\frac{\\lim_{x\\to c} f(x)}{\\lim_{x\\to c} g(x)} . \\end{equation*}\\end{enumerate}" - ], - "refs": [], - "proofs": [], - "ref_ids": [] - }, - { - "id": 10, - "type": "corollary", - "label": "Lebl-contfunc:fabs:cor", - "categories": [ - "limits", - "topology", - "consequence" - ], - "title": "Let and let be a cluster point of ", - "contents": [ - "Let$S \\subset \\R$and let$c$be a cluster point of$S$. Suppose$f \\colon S \\to \\R$is a function such that the limit of$f(x)$as$x$goes to$c$exists. Then\\begin{equation*} \\lim_{x\\to c} \\abs{f(x)} = \\abs{\\lim_{x\\to c} f(x)}. \\end{equation*}" - ], - "refs": [], - "proofs": [], - "ref_ids": [] - }, - { - "id": 11, - "type": "proposition", - "label": "Lebl-contfunc:prop:limrest", - "categories": [ - "topology", - "characterization" - ], - "title": "Let , , and let $f S $ be a function", - "contents": [ - "Let$S \\subset \\R$,$c \\in \\R$, and let$f \\colon S \\to \\R$be a function. Suppose$A \\subset S$is such that there is some$\\alpha > 0$such that$(A \\setminus \\{ c \\}) \\cap (c-\\alpha,c+\\alpha) = (S \\setminus \\{ c \\}) \\cap (c-\\alpha,c+\\alpha)$. \\begin{enumerate}[(i)] \\item The point$c$is a cluster point of$A$if and only if$c$is a cluster point of$S$. \\item Supposing$c$is a cluster point of$S$, then$f(x) \\to L$as$x \\to c$if and only if$f|_A(x) \\to L$as$x \\to c$. \\end{enumerate}" - ], - "refs": [], - "proofs": [ - { - "contents": [ - "First, let $c$ be a cluster point of $A$.", - "Since $A \\subset S$, then if $( A \\setminus \\{ c\\} ) \\cap", - "(c-\\epsilon,c+\\epsilon)$ is nonempty for every $\\epsilon > 0$,", - "then $( S \\setminus \\{ c\\} ) \\cap", - "(c-\\epsilon,c+\\epsilon)$ is nonempty for every $\\epsilon > 0$.", - "Thus $c$ is a cluster point of $S$.", - "Second, suppose $c$ is a cluster", - "point of $S$. Then for $\\epsilon > 0$ such that $\\epsilon < \\alpha$", - "we get that $( A \\setminus \\{ c\\} ) \\cap (c-\\epsilon,c+\\epsilon) =", - "( S \\setminus \\{ c\\} ) \\cap (c-\\epsilon,c+\\epsilon)$, which is nonempty. This is true for all", - "$\\epsilon < \\alpha$ and hence", - "$( A \\setminus \\{ c\\} ) \\cap (c-\\epsilon,c+\\epsilon)$ must be nonempty for all", - "$\\epsilon > 0$. Thus $c$ is a cluster point of $A$.", - "Now suppose $c$ is a cluster point of $S$ and $f(x) \\to L$ as $x \\to c$. That is, for every $\\epsilon > 0$", - "there is a $\\delta > 0$ such that if $x \\in S \\setminus \\{ c \\}$", - "and $\\abs{x-c} < \\delta$, then $\\abs{f(x)-L} < \\epsilon$. Because $A \\subset S$,", - "if $x \\in A \\setminus \\{ c \\}$, then $x \\in S \\setminus \\{ c \\}$,", - "and hence $f|_A(x) \\to L$ as $x \\to c$.", - "Finally, suppose $f|_A(x) \\to L$ as $x \\to c$ and let $\\epsilon > 0$ be", - "given.", - "There is a $\\delta' > 0$ such that if $x \\in A \\setminus \\{ c \\}$", - "and $\\abs{x-c} < \\delta'$, then $\\bigl\\lvert f|_A(x)-L \\bigr\\rvert < \\epsilon$.", - "Take $\\delta \\coloneqq \\min \\{ \\delta', \\alpha \\}$.", - "Now suppose $x \\in S \\setminus \\{ c \\}$ and", - "$\\abs{x-c} < \\delta$. As $\\abs{x-c} < \\alpha$, we find $x \\in A \\setminus \\{ c \\}$,", - "and as $\\abs{x-c} < \\delta'$,", - "we get $\\abs{f(x)-L} = \\bigl\\lvert f|_A(x)-L", - "\\bigr\\rvert < \\epsilon$." - ], - "refs": [], - "ref_ids": [] - } - ], - "ref_ids": [] - }, - { - "id": 12, - "type": "proposition", - "label": "Lebl-contfunc:prop:onesidedlimits", - "categories": [ - "topology", - "characterization" - ], - "title": "Let be such that is a cluster point of both and , ...", - "contents": [ - "Let$S \\subset \\R$be such that$c$is a cluster point of both$S \\cap (-\\infty,c)$and$S \\cap (c,\\infty)$, let$f \\colon S \\to \\R$be a function, and let$L \\in \\R$. Then$c$is a cluster point of$S$and\\begin{equation*} \\lim_{x \\to c} f(x) = L \\qquad \\text{if and only if} \\qquad \\lim_{x \\to c^-} f(x) = \\lim_{x \\to c^+} f(x) = L . \\end{equation*}" - ], - "refs": [], - "proofs": [], - "ref_ids": [] - }, - { - "id": 13, - "type": "proposition", - "label": "Lebl-contfunc:contbasic:prop", - "categories": [ - "limits", - "continuity", - "characterization", - "sequences", - "topology" - ], - "title": "Consider a function defined on a set and let ", - "contents": [ - "Consider a function$f \\colon S \\to \\R$defined on a set$S \\subset \\R$and let$c \\in S$. Then: %\\begin{enumerate}[(i),itemsep=0.5\\itemsep,parsep=0.5\\parsep,topsep=0.5\\topsep,partopsep=0.5\\partopsep] \\begin{enumerate}[(i)] \\item \\label{contbasic:prop:i} If$c$is not a cluster point of$S$, then$f$is continuous at$c$. \\item \\label{contbasic:prop:ii} If$c$is a cluster point of$S$, then$f$is continuous at$c$if and only if the limit of$f(x)$as$x \\to c$exists and\\begin{equation*} \\lim_{x\\to c} f(x) = f(c) . \\end{equation*}\\item \\label{contbasic:prop:iii} The function$f$is continuous at$c$if and only if for every sequence$\\{ x_n \\}_{n=1}^\\infty$where$x_n \\in S$and$\\lim\\limits_{n\\to\\infty} x_n = c$, the sequence$\\bigl\\{ f(x_n) \\bigr\\}_{n=1}^\\infty$converges to$f(c)$. \\end{enumerate}" - ], - "refs": [], - "proofs": [ - { - "contents": [ - "\\pagebreak[2]", - "We start with \\ref{contbasic:prop:i}. Suppose $c$ is not a cluster point of", - "$S$. Then there exists a $\\delta > 0$", - "such that $S \\cap (c-\\delta,c+\\delta) = \\{ c \\}$.", - "For any $\\epsilon > 0$, simply pick this given $\\delta$.", - "The only $x \\in S$ such that $\\abs{x-c} < \\delta$ is $x=c$. Then", - "$\\abs{f(x)-f(c)} = \\abs{f(c)-f(c)} = 0 < \\epsilon$.", - "Let us move to \\ref{contbasic:prop:ii}.", - "Suppose $c$ is a cluster point of $S$. Let us first suppose", - "that $\\lim_{x\\to c} f(x) = f(c)$. Then for every $\\epsilon > 0$,", - "there is a $\\delta > 0$ such that if $x \\in S \\setminus \\{ c \\}$", - "and $\\abs{x-c} < \\delta$, then $\\abs{f(x)-f(c)} < \\epsilon$.", - "Also $\\abs{f(c)-f(c)} = 0 < \\epsilon$, so the definition of continuity at", - "$c$ is satisfied. On the other hand, suppose $f$ is continuous", - "at $c$. For every $\\epsilon > 0$, there exists a $\\delta > 0$", - "such that for $x \\in S$ where $\\abs{x-c} < \\delta$, we have", - "$\\abs{f(x)-f(c)} < \\epsilon$. Then the statement is, of course, still true if", - "$x \\in S \\setminus \\{ c \\} \\subset S$. Therefore, $\\lim_{x\\to c} f(x) =", - "f(c)$.", - "For \\ref{contbasic:prop:iii}, first suppose $f$ is continuous at $c$.", - "Let $\\{ x_n \\}_{n=1}^\\infty$", - "be a sequence such that $x_n \\in S$ and $\\lim_{n\\to\\infty} x_n = c$. Let $\\epsilon > 0$", - "be given. Find a $\\delta > 0$ such that $\\abs{f(x)-f(c)} < \\epsilon$", - "for all $x \\in S$ where $\\abs{x-c} < \\delta$. Find an $M \\in \\N$", - "such that for $n \\geq M$, we have $\\abs{x_n-c} < \\delta$. Then for", - "$n \\geq M$, we have that $\\abs{f(x_n)-f(c)} < \\epsilon$,", - "so $\\bigl\\{ f(x_n) \\bigr\\}_{n=1}^\\infty$", - "converges to $f(c)$.", - "We prove the other direction of \\ref{contbasic:prop:iii} by contrapositive.", - "Suppose $f$ is not", - "continuous at $c$. Then there exists an $\\epsilon > 0$", - "such that for every $\\delta > 0$, there exists an $x \\in S$", - "such that $\\abs{x-c} < \\delta$ and $\\abs{f(x)-f(c)} \\geq \\epsilon$.", - "Define a sequence $\\{ x_n \\}_{n=1}^\\infty$ as follows.", - "Let $x_n \\in S$ be such that $\\abs{x_n-c} < \\nicefrac{1}{n}$", - "and $\\abs{f(x_n)-f(c)} \\geq \\epsilon$.", - "Now $\\{ x_n \\}_{n=1}^\\infty$ is", - "a sequence in $S$ such that", - "$\\lim_{n\\to\\infty} x_n = c$ and such that", - "$\\abs{f(x_n)-f(c)} \\geq \\epsilon$ for all $n \\in \\N$.", - "Thus $\\bigl\\{ f(x_n) \\bigr\\}_{n=1}^\\infty$", - "does not converge to $f(c)$. It may or may not converge, but it definitely", - "does not converge to $f(c)$." - ], - "refs": [ - "contbasic:prop:i", - "contbasic:prop:ii", - "contbasic:prop:iii" - ], - "ref_ids": [] - } - ], - "ref_ids": [] - }, - { - "id": 14, - "type": "example", - "label": "Lebl-contfunc:14", - "categories": [ - "sequences", - "continuity", - "example" - ], - "title": "The function defined by is continuous", - "contents": [ - "The function$f \\colon (0,\\infty) \\to \\R$defined by$f(x) \\coloneqq \\nicefrac{1}{x}$is continuous.", - "Proof: Fix$c \\in (0,\\infty)$. Let$\\{ x_n \\}_{n=1}^\\infty$be a sequence in$(0,\\infty)$such that$\\lim_{n\\to\\infty} x_n = c$. Then", - "\\begin{equation*}\nf(c) = \\frac{1}{c}\n=\n\\frac{1}{\\lim_{n\\to\\infty} x_n}\n=\n\\lim_{n \\to \\infty} \\frac{1}{x_n}\n=\n\\lim_{n \\to \\infty} f(x_n) .\n\\end{equation*}", - "Thus$f$is continuous at$c$. As$f$is continuous at all$c \\in (0,\\infty)$,$f$is continuous." - ], - "refs": [], - "proofs": [ - { - "contents": [ - "Fix$c \\in (0,\\infty)$. Let$\\{ x_n \\}_{n=1}^\\infty$be a sequence in$(0,\\infty)$such that$\\lim_{n\\to\\infty} x_n = c$. Then \\begin{equation*}", - "f(c) = \\frac{1}{c}", - "=", - "\\frac{1}{\\lim_{n\\to\\infty} x_n}", - "=", - "\\lim_{n \\to \\infty} \\frac{1}{x_n}", - "=", - "\\lim_{n \\to \\infty} f(x_n) .", - "\\end{equation*} Thus$f$is continuous at$c$. As$f$is continuous at all$c \\in (0,\\infty)$,$f$is continuous." - ], - "refs": [], - "ref_ids": [] - } - ], - "ref_ids": [] - }, - { - "id": 15, - "type": "proposition", - "label": "Lebl-contfunc:15", - "categories": [ - "continuity" - ], - "title": "Let be a }", - "contents": [ - "Let$f \\colon \\R \\to \\R$be a \\emph{\\myindex{polynomial}}. That is,\\begin{equation*} f(x) = a_d x^d + a_{d-1} x^{d-1} + \\cdots + a_1 x + a_0 , \\end{equation*}for some constants$a_0, a_1, \\ldots, a_d$. Then$f$is continuous." - ], - "refs": [], - "proofs": [ - { - "contents": [ - "Fix $c \\in \\R$.", - "Let $\\{ x_n \\}_{n=1}^\\infty$ be a sequence such that", - "$\\lim_{n\\to\\infty} x_n = c$. Then", - "\\begin{equation*}", - "\\begin{split}", - "f(c) &=", - "a_d c^d + a_{d-1} c^{d-1} + \\cdots + a_1 c + a_0", - "\\\\", - "&=", - "a_d {\\left(\\lim_{n\\to\\infty} x_n\\right)}^d", - "+ a_{d-1} {\\left(\\lim_{n\\to\\infty} x_n\\right)}^{d-1}", - "+ \\cdots", - "+ a_1 \\left(\\lim_{n\\to\\infty} x_n\\right) + a_0", - "\\\\", - "& =", - "\\lim_{n \\to \\infty}", - "\\left(", - "a_d x_n^d + a_{d-1} x_n^{d-1} + \\cdots + a_1 x_n + a_0", - "\\right)", - "=", - "\\lim_{n \\to \\infty}", - "f(x_n) .", - "\\avoidbreak", - "\\end{split}", - "\\end{equation*}", - "Thus $f$ is continuous at $c$. As $f$ is continuous at all $c \\in \\R$,", - "$f$ is continuous." - ], - "refs": [], - "ref_ids": [] - } - ], - "ref_ids": [] - }, - { - "id": 16, - "type": "proposition", - "label": "Lebl-contfunc:contalg:prop", - "categories": [ - "continuity" - ], - "title": "Let and be functions continuous at ", - "contents": [ - "Let$f \\colon S \\to \\R$and$g \\colon S \\to \\R$be functions continuous at$c \\in S$. \\begin{enumerate}[(i)] \\item The function$h \\colon S \\to \\R$defined by$h(x) \\coloneqq f(x)+g(x)$is continuous at$c$. \\item The function$h \\colon S \\to \\R$defined by$h(x) \\coloneqq f(x)-g(x)$is continuous at$c$. \\item The function$h \\colon S \\to \\R$defined by$h(x) \\coloneqq f(x)g(x)$is continuous at$c$. \\item If$g(x)\\not=0$for all$x \\in S$, the function$h \\colon S \\to \\R$given by$h(x) \\coloneqq \\frac{f(x)}{g(x)}$is continuous at$c$. \\end{enumerate}" - ], - "refs": [], - "proofs": [], - "ref_ids": [] - }, - { - "id": 17, - "type": "example", - "label": "Lebl-contfunc:sincos:example", - "categories": [ - "sequences", - "continuity", - "example" - ], - "title": "The functions and are continuous", - "contents": [ - "The functions$\\sin(x)$and$\\cos(x)$are continuous. In the following computations we use the sum-to-product trigonometric identities. We also use the simple facts that$\\abs{\\sin(x)} \\leq \\abs{x}$,$\\abs{\\cos(x)} \\leq 1$, and$\\abs{\\sin(x)} \\leq 1$.", - "\\begin{equation*}\n\\begin{split}\n\\abs{\\sin(x)-\\sin(c)} & =\n\\abs{\n2 \\sin \\left( \\frac{x-c}{2} \\right) \\cos \\left( \\frac{x+c}{2} \\right)\n}\n\\\\\n& =\n2\n\\abs{ \\sin \\left( \\frac{x-c}{2} \\right) }\n\\abs{ \\cos \\left( \\frac{x+c}{2} \\right) }\n\\\\\n& \\leq\n2\n\\abs{ \\sin \\left( \\frac{x-c}{2} \\right) }\n\\\\\n& \\leq\n2\n\\abs{ \\frac{x-c}{2} }\n= \\abs{x-c}\n\\end{split}\n\\end{equation*}", - "\\begin{equation*}\n\\begin{split}\n\\abs{\\cos(x)-\\cos(c)} & =\n\\abs{\n-2 \\sin \\left( \\frac{x-c}{2} \\right) \\sin \\left( \\frac{x+c}{2} \\right)\n}\n\\\\\n& =\n2\n\\abs{ \\sin \\left( \\frac{x-c}{2} \\right) }\n\\abs{ \\sin \\left( \\frac{x+c}{2} \\right) }\n\\\\\n& \\leq\n2\n\\abs{ \\sin \\left( \\frac{x-c}{2} \\right) }\n\\\\\n& \\leq\n2\n\\abs{ \\frac{x-c}{2} }\n= \\abs{x-c}\n\\end{split}\n\\end{equation*}", - "The claim that$\\sin$and$\\cos$are continuous follows by taking an arbitrary sequence$\\{ x_n \\}_{n=1}^\\infty$converging to$c$, or by applying the definition of continuity directly. Details are left to the reader." - ], - "refs": [], - "proofs": [], - "ref_ids": [] - }, - { - "id": 18, - "type": "proposition", - "label": "Lebl-contfunc:prop:compositioncont", - "categories": [ - "continuity" - ], - "title": "Let and and be functions", - "contents": [ - "Let$A, B \\subset \\R$and$f \\colon B \\to \\R$and$g \\colon A \\to B$be functions. If$g$is continuous at$c \\in A$and$f$is continuous at$g(c)$, then$f \\circ g \\colon A \\to \\R$is continuous at$c$." - ], - "refs": [], - "proofs": [ - { - "contents": [ - "Let $\\{ x_n \\}_{n=1}^\\infty$ be a sequence in $A$ such that", - "$\\lim_{n\\to\\infty} x_n = c$.", - "As $g$ is continuous at $c$, we have $\\bigl\\{ g(x_n) \\bigr\\}_{n=1}^\\infty$ converges to $g(c)$.", - "As $f$ is continuous at $g(c)$, we have $\\bigl\\{ f\\bigl(g(x_n)\\bigr)", - "\\bigr\\}_{n=1}^\\infty$ converges", - "to $f\\bigl(g(c)\\bigr)$.", - "Thus $f \\circ g$ is continuous at $c$." - ], - "refs": [], - "ref_ids": [] - } - ], - "ref_ids": [] - }, - { - "id": 19, - "type": "example", - "label": "Lebl-contfunc:19", - "categories": [ - "example", - "continuity", - "intervals" - ], - "title": "Claim: {x}) )}^2$ is a continuous function on ", - "contents": [ - "Claim: \\emph{${\\bigl(\\sin(\\nicefrac{1}{x})\\bigr)}^2$is a continuous function on$(0,\\infty)$.}", - "Proof: The function$\\nicefrac{1}{x}$is continuous on$(0,\\infty)$and$\\sin(x)$is continuous on$(0,\\infty)$(actually on$\\R$, but$(0,\\infty)$is the range for$\\nicefrac{1}{x}$). Hence the composition$\\sin(\\nicefrac{1}{x})$is continuous. Also,$x^2$is continuous on the interval$[-1,1]$(the range of$\\sin$). Thus the composition${\\bigl(\\sin(\\nicefrac{1}{x})\\bigr)}^2$is continuous on$(0,\\infty)$." - ], - "refs": [], - "proofs": [ - { - "contents": [ - "The function$\\nicefrac{1}{x}$is continuous on$(0,\\infty)$and$\\sin(x)$is continuous on$(0,\\infty)$(actually on$\\R$, but$(0,\\infty)$is the range for$\\nicefrac{1}{x}$). Hence the composition$\\sin(\\nicefrac{1}{x})$is continuous. Also,$x^2$is continuous on the interval$[-1,1]$(the range of$\\sin$). Thus the composition${\\bigl(\\sin(\\nicefrac{1}{x})\\bigr)}^2$is continuous on$(0,\\infty)$." - ], - "refs": [], - "ref_ids": [] - } - ], - "ref_ids": [] - }, - { - "id": 20, - "type": "proposition", - "label": "Lebl-contfunc:20", - "categories": [ - "sequences", - "continuity" - ], - "title": "Let be a function and ", - "contents": [ - "Let$f \\colon S \\to \\R$be a function and$c \\in S$. Suppose there exists a sequence$\\{ x_n \\}_{n=1}^\\infty$,$x_n \\in S$for all$n$, and$\\lim_{n\\to\\infty} x_n = c$such that$\\bigl\\{ f(x_n) \\bigr\\}_{n=1}^\\infty$does not converge to$f(c)$. Then$f$is discontinuous at$c$." - ], - "refs": [], - "proofs": [], - "ref_ids": [] - }, - { - "id": 21, - "type": "example", - "label": "Lebl-contfunc:example:jumpdiscont", - "categories": [ - "sequences", - "continuity", - "example" - ], - "title": "The function defined by f(x) -1 & x < 0, \\\\ 1 & x ...", - "contents": [ - "The function$f \\colon \\R \\to \\R$defined by", - "\\begin{equation*}\nf(x) \\coloneqq \n\\begin{cases}\n-1 & \\text{if } x < 0, \\\\\n1 & \\text{if } x \\geq 0\n\\end{cases}\n\\end{equation*}", - "is not continuous at 0.", - "Proof: Consider$\\{ \\nicefrac{-1}{n} \\}_{n=1}^\\infty$, which converges to 0. Then$f(\\nicefrac{-1}{n}) = -1$for every$n$, and so$\\lim_{n\\to\\infty} f(\\nicefrac{-1}{n}) = -1$, but$f(0) = 1$. Thus the function is not continuous at 0. See \\figureref{fig:jumpdiscont}.", - "\\begin{myfigureht} \\includegraphics{figures/jumpdiscont} \\caption{Jump discontinuity. The values of$f(\\nicefrac{-1}{n})$and$f(0)$are marked as black dots.\\label{fig:jumpdiscont}} \\end{myfigureht}", - "Notice that$f(\\nicefrac{1}{n}) = 1$for all$n \\in \\N$. Hence,$\\lim_{n\\to\\infty} f(\\nicefrac{1}{n}) = f(0) = 1$. So$\\bigl\\{ f(x_n) \\bigr\\}_{n=1}^\\infty$may converge to$f(0)$for some specific sequence$\\{ x_n \\}_{n=1}^\\infty$going to 0, despite the function being discontinuous at 0.", - "Finally, consider$f\\Bigl(\\frac{{(-1)}^n}{n}\\Bigr) = {(-1)}^n$. This sequence diverges." - ], - "refs": [], - "proofs": [ - { - "contents": [ - "Consider$\\{ \\nicefrac{-1}{n} \\}_{n=1}^\\infty$, which converges to 0. Then$f(\\nicefrac{-1}{n}) = -1$for every$n$, and so$\\lim_{n\\to\\infty} f(\\nicefrac{-1}{n}) = -1$, but$f(0) = 1$. Thus the function is not continuous at 0. See \\figureref{fig:jumpdiscont}. \\begin{myfigureht} \\includegraphics{figures/jumpdiscont} \\caption{Jump discontinuity. The values of$f(\\nicefrac{-1}{n})$and$f(0)$are marked as black dots.\\label{fig:jumpdiscont}} \\end{myfigureht} Notice that$f(\\nicefrac{1}{n}) = 1$for all$n \\in \\N$. Hence,$\\lim_{n\\to\\infty} f(\\nicefrac{1}{n}) = f(0) = 1$. So$\\bigl\\{ f(x_n) \\bigr\\}_{n=1}^\\infty$may converge to$f(0)$for some specific sequence$\\{ x_n \\}_{n=1}^\\infty$going to 0, despite the function being discontinuous at 0. Finally, consider$f\\Bigl(\\frac{{(-1)}^n}{n}\\Bigr) = {(-1)}^n$. This sequence diverges." - ], - "refs": [], - "ref_ids": [] - } - ], - "ref_ids": [] - }, - { - "id": 22, - "type": "example", - "label": "Lebl-contfunc:22", - "categories": [ - "sequences", - "continuity", - "example" - ], - "title": "For an extreme example, take the so-called } {Joha...", - "contents": [ - "For an extreme example, take the so-called \\emph{\\myindex{Dirichlet function}}\\footnote{Named after the German mathematician \\href{https://en.wikipedia.org/wiki/Peter_Gustav_Lejeune_Dirichlet}{Johann Peter Gustav Lejeune Dirichlet} (1805--1859).}.", - "\\begin{equation*}\nf(x) \\coloneqq\n\\begin{cases}\n1 & \\text{if } x \\text{ is rational,} \\\\\n0 & \\text{if } x \\text{ is irrational.}\n\\end{cases}\n\\avoidbreak\n\\end{equation*}", - "The function$f$is discontinuous at all$c \\in \\R$.", - "Proof: If$c$is rational, take a sequence$\\{ x_n \\}_{n=1}^\\infty$of irrational numbers such that$\\lim_{n\\to\\infty} x_n = c$(why can we?). Then$f(x_n) = 0$and so$\\lim_{n\\to\\infty} f(x_n) = 0$, but$f(c) = 1$. If$c$is irrational, take a sequence of rational numbers$\\{ x_n \\}_{n=1}^\\infty$that converges to$c$(why can we?). Then$\\lim_{n\\to\\infty} f(x_n) = 1$, but$f(c) = 0$." - ], - "refs": [], - "proofs": [ - { - "contents": [ - "If$c$is rational, take a sequence$\\{ x_n \\}_{n=1}^\\infty$of irrational numbers such that$\\lim_{n\\to\\infty} x_n = c$(why can we?). Then$f(x_n) = 0$and so$\\lim_{n\\to\\infty} f(x_n) = 0$, but$f(c) = 1$. If$c$is irrational, take a sequence of rational numbers$\\{ x_n \\}_{n=1}^\\infty$that converges to$c$(why can we?). Then$\\lim_{n\\to\\infty} f(x_n) = 1$, but$f(c) = 0$." - ], - "refs": [], - "ref_ids": [] - } - ], - "ref_ids": [] - }, - { - "id": 23, - "type": "example", - "label": "Lebl-contfunc:popcornfunction:example", - "categories": [ - "sequences", - "continuity", - "example" - ], - "title": "Define as f(x) {k} & x= {k}, m,k \\\\ 0 & x See the ...", - "contents": [ - "Define$f \\colon (0,1) \\to \\R$as", - "\\begin{equation*}\nf(x) \\coloneqq \n\\begin{cases}\n\\nicefrac{1}{k} & \\text{if } x=\\nicefrac{m}{k}, \\text{ where } m,k \\in \\N\n\\text{ and have no common divisors (lowest terms),} \\\\\n0 & \\text{if } x \\text{ is irrational.}\n\\end{cases}\n\\end{equation*}", - "See the graph of$f$in \\figureref{popcornfig}. We claim that$f$is continuous at all irrational$c$and discontinuous at all rational$c$. \\begin{myfigureht} \\includegraphics{figures/popcornfig} \\caption{Graph of the \\myquote{popcorn function.}\\label{popcornfig}} \\end{myfigureht}", - "Proof: Let$c = \\nicefrac{m}{k}$be rational and in lowest terms. Take a sequence of irrational numbers$\\{ x_n \\}_{n=1}^\\infty$such that$\\lim_{n\\to\\infty} x_n = c$. Then$\\lim_{n\\to\\infty} f(x_n) = \\lim_{n\\to\\infty} 0 = 0$, but$f(c) = \\nicefrac{1}{k} \\not= 0$. So$f$is discontinuous at$c$.", - "Now let$c$be irrational, so$f(c) = 0$. Take a sequence$\\{ x_n \\}_{n=1}^\\infty$in$(0,1)$such that$\\lim_{n\\to\\infty} x_n = c$. Given$\\epsilon > 0$, find$K \\in \\N$such that$\\nicefrac{1}{K} < \\epsilon$by the \\hyperref[thm:arch:i]{Archimedean property}. If$\\nicefrac{m}{k} \\in (0,1)$and$m,k \\in \\N$, then$0 < m < k$. So there are only finitely many rational numbers in$(0,1)$whose denominator$k$in lowest terms is less than$K$. As$\\lim_{n\\to\\infty} x_n = c$, every number not equal to$c$can appear at most finitely many times in$\\{ x_n \\}_{n=1}^\\infty$. Hence, there is an$M$such that for$n \\geq M$, all the numbers$x_n$that are rational have a denominator larger than or equal to$K$. Thus for$n \\geq M$,", - "\\begin{equation*}\n\\abs{f(x_n) - 0} = f(x_n) \\leq \\nicefrac{1}{K} < \\epsilon .\n\\end{equation*}", - "Therefore,$f$is continuous at irrational$c$." - ], - "refs": [ - "thm:arch:i" - ], - "proofs": [ - { - "contents": [ - "Let$c = \\nicefrac{m}{k}$be rational and in lowest terms. Take a sequence of irrational numbers$\\{ x_n \\}_{n=1}^\\infty$such that$\\lim_{n\\to\\infty} x_n = c$. Then$\\lim_{n\\to\\infty} f(x_n) = \\lim_{n\\to\\infty} 0 = 0$, but$f(c) = \\nicefrac{1}{k} \\not= 0$. So$f$is discontinuous at$c$. Now let$c$be irrational, so$f(c) = 0$. Take a sequence$\\{ x_n \\}_{n=1}^\\infty$in$(0,1)$such that$\\lim_{n\\to\\infty} x_n = c$. Given$\\epsilon > 0$, find$K \\in \\N$such that$\\nicefrac{1}{K} < \\epsilon$by the \\hyperref[thm:arch:i]{Archimedean property}. If$\\nicefrac{m}{k} \\in (0,1)$and$m,k \\in \\N$, then$0 < m < k$. So there are only finitely many rational numbers in$(0,1)$whose denominator$k$in lowest terms is less than$K$. As$\\lim_{n\\to\\infty} x_n = c$, every number not equal to$c$can appear at most finitely many times in$\\{ x_n \\}_{n=1}^\\infty$. Hence, there is an$M$such that for$n \\geq M$, all the numbers$x_n$that are rational have a denominator larger than or equal to$K$. Thus for$n \\geq M$, \\begin{equation*}", - "\\abs{f(x_n) - 0} = f(x_n) \\leq \\nicefrac{1}{K} < \\epsilon .", - "\\end{equation*} Therefore,$f$is continuous at irrational$c$." - ], - "refs": [ - "thm:arch:i" - ], - "ref_ids": [] - } - ], - "ref_ids": [] - }, - { - "id": 24, - "type": "example", - "label": "Lebl-contfunc:example:removablediscont", - "categories": [ - "continuity", - "example" - ], - "title": "Define by if and ", - "contents": [ - "Define$g \\colon \\R \\to \\R$by$g(x) \\coloneqq 0$if$x \\not= 0$and$g(0) \\coloneqq 1$. Then$g$is not continuous at zero, but continuous everywhere else (why?). The point$x=0$is called a \\emph{\\myindex{removable discontinuity}}. That is because if we would change the definition of$g$, by insisting that$g(0)$be$0$, we would obtain a continuous function. On the other hand, let$f$be the function of \\exampleref{example:jumpdiscont}. Then$f$does not have a removable discontinuity at$0$. No matter how we would define$f(0)$the function would still fail to be continuous. The difference is that$\\lim_{x\\to 0} g(x)$exists while$\\lim_{x\\to 0} f(x)$does not.", - "We stay with this example to show another phenomenon. Let$A \\coloneqq \\{ 0 \\}$, then$g|_A$is continuous (why?), while$g$is not continuous on$A$. Similarly, if$B \\coloneqq \\R \\setminus \\{0 \\}$, then$g|_B$is also continuous, and$g$is in fact continuous on$B$." - ], - "refs": [], - "proofs": [], - "ref_ids": [] - }, - { - "id": 25, - "type": "lemma", - "label": "Lebl-contfunc:25", - "categories": [ - "auxiliary result", - "continuity", - "boundedness", - "intervals" - ], - "title": "A continuous function is bounded", - "contents": [ - "A continuous function$f \\colon [a,b] \\to \\R$is bounded." - ], - "refs": [], - "proofs": [ - { - "contents": [ - "We prove the claim by contrapositive. Suppose $f$ is not bounded.", - "Then for each", - "$n \\in \\N$, there is an $x_n \\in [a,b]$, such that", - "\\begin{equation*}", - "\\abs{f(x_n)} \\geq n .", - "\\end{equation*}", - "The sequence $\\{ x_n \\}_{n=1}^\\infty$ is bounded as $a \\leq x_n \\leq b$.", - "By the \\hyperref[thm:bwseq]{Bolzano--Weierstrass theorem},", - "there is a convergent subsequence $\\{ x_{n_i} \\}_{i=1}^\\infty$.", - "Let $x \\coloneqq \\lim_{i\\to\\infty} x_{n_i}$.", - "Since $a \\leq x_{n_i} \\leq b$ for all $i$, then $a \\leq x \\leq b$.", - "The sequence $\\bigl\\{ f(x_{n_i}) \\bigr\\}_{i=1}^\\infty$ is not bounded", - "as", - "$\\abs{f(x_{n_i})} \\geq n_i \\geq i$.", - "Thus $f$ is not continuous at $x$ as", - "\\begin{equation*}", - "f(x)", - "=", - "f\\Bigl( \\lim_{i\\to\\infty} x_{n_i} \\Bigr) ,", - "\\qquad \\text{but} \\qquad", - "\\lim_{i\\to\\infty} f(x_{n_i}) \\enspace \\text{does not exist.} \\qedhere", - "\\end{equation*}" - ], - "refs": [ - "named:Bolzano", - "named:Weierstrass", - "thm:bwseq" - ], - "ref_ids": [] - } - ], - "ref_ids": [] - }, - { - "id": 26, - "type": "theorem", - "label": "Lebl-contfunc:26", - "categories": [ - "continuity", - "intervals" - ], - "title": "% % % A continuous function achieves both an absol...", - "contents": [ - "\\index{minimum-maximum theorem}% \\index{maximum-minimum theorem}% \\index{extreme value theorem}% A continuous function$f \\colon [a,b] \\to \\R$achieves both an absolute minimum and an absolute maximum on$[a,b]$." - ], - "refs": [ - "named:Extreme Value" - ], - "proofs": [ - { - "contents": [ - "The lemma says that $f$ is bounded, so", - "the set $f\\bigl([a,b]\\bigr) = \\bigl\\{ f(x) : x \\in [a,b] \\bigr\\}$ has a supremum and an infimum.", - "There exist sequences", - "in the set $f\\bigl([a,b]\\bigr)$ that approach its supremum and its infimum.", - "That is, there are sequences", - "$\\bigl\\{ f(x_n) \\bigr\\}_{n=1}^\\infty$ and $\\bigl\\{ f(y_n)", - "\\bigr\\}_{n=1}^\\infty$, where $x_n$ and $y_n$ are in $[a,b]$,", - "such that", - "\\begin{equation*}", - "\\lim_{n\\to\\infty} f(x_n) = \\inf f\\bigl([a,b]\\bigr) \\qquad \\text{and} \\qquad", - "\\lim_{n\\to\\infty} f(y_n) = \\sup f\\bigl([a,b]\\bigr).", - "\\end{equation*}", - "We are not done yet; we need to find where the minima and the maxima are.", - "The problem is that the sequences $\\{ x_n \\}_{n=1}^\\infty$ and", - "$\\{ y_n \\}_{n=1}^\\infty$ need not converge.", - "We know $\\{ x_n \\}_{n=1}^\\infty$ and $\\{ y_n \\}_{n=1}^\\infty$ are bounded", - "(their elements belong to a bounded interval $[a,b]$).", - "Apply the", - "\\hyperref[thm:bwseq]{Bolzano--Weierstrass theorem}", - "to find", - "convergent subsequences", - "$\\{ x_{n_i} \\}_{i=1}^\\infty$ and", - "$\\{ y_{m_i} \\}_{i=1}^\\infty$. Let", - "\\begin{equation*}", - "x \\coloneqq \\lim_{i\\to\\infty} x_{n_i}", - "\\qquad \\text{and} \\qquad", - "y \\coloneqq \\lim_{i\\to\\infty} y_{m_i}.", - "\\end{equation*}", - "As $a \\leq x_{n_i} \\leq b$ for all $i$, we have $a \\leq x \\leq b$.", - "Similarly, $a \\leq y \\leq b$. So $x$ and $y$ are in $[a,b]$.", - "A limit of a subsequence is the same as the limit of the", - "sequence, and we can take a limit past the continuous function $f$:", - "\\begin{equation*}", - "\\inf f\\bigl([a,b]\\bigr) = \\lim_{n\\to\\infty} f(x_n)", - "= \\lim_{i\\to\\infty} f(x_{n_i}) =", - "f \\Bigl( \\lim_{i\\to\\infty} x_{n_i} \\Bigr) = f(x) .", - "\\end{equation*}", - "Similarly,", - "\\begin{equation*}", - "\\sup f\\bigl([a,b]\\bigr) = \\lim_{n\\to\\infty} f(y_n)", - "= \\lim_{i\\to\\infty} f(y_{m_i}) =", - "f \\Bigl( \\lim_{i\\to\\infty} y_{m_i} \\Bigr) = f(y) .", - "\\end{equation*}", - "Hence, $f$ achieves an absolute minimum at $x$ and", - "an absolute maximum at $y$." - ], - "refs": [ - "named:Bolzano", - "named:Weierstrass", - "thm:bwseq" - ], - "ref_ids": [] - } - ], - "ref_ids": [] - }, - { - "id": 27, - "type": "example", - "label": "Lebl-contfunc:27", - "categories": [ - "example", - "intervals" - ], - "title": "The function defined on the interval achieves a mi...", - "contents": [ - "The function$f(x) \\coloneqq x^2+1$defined on the interval$[-1,2]$achieves a minimum at$x=0$when$f(0) = 1$. It achieves a maximum at$x=2$where$f(2) = 5$. Do note that the domain of definition matters. If we instead took the domain to be$[-10,10]$, then$f$would no longer have a maximum at$x=2$. Instead, the maximum would be achieved at either$x=10$or$x=-10$." - ], - "refs": [], - "proofs": [], - "ref_ids": [] - }, - { - "id": 28, - "type": "example", - "label": "Lebl-contfunc:28", - "categories": [ - "example", - "boundedness", - "intervals" - ], - "title": "The function defined by achieves neither a minimum...", - "contents": [ - "The function$f \\colon \\R \\to \\R$defined by$f(x) \\coloneqq x$achieves neither a minimum, nor a maximum. So it is important that we are looking at a bounded interval." - ], - "refs": [], - "proofs": [], - "ref_ids": [] - }, - { - "id": 29, - "type": "example", - "label": "Lebl-contfunc:29", - "categories": [ - "continuity", - "boundedness", - "intervals", - "topology", - "example" - ], - "title": "The function defined by achieves neither a minimum...", - "contents": [ - "The function$f \\colon (0,1) \\to \\R$defined by$f(x) \\coloneqq \\nicefrac{1}{x}$achieves neither a minimum, nor a maximum. It is continuous, but$(0,1)$is not closed. The values of the function are unbounded as we approach$0$. Also as we approach$x=1$, the values of the function approach$1$, but$f(x) > 1$for all$x \\in (0,1)$. There is no$x \\in (0,1)$such that$f(x) = 1$. So it is important that we are looking at a closed interval." - ], - "refs": [], - "proofs": [], - "ref_ids": [] - }, - { - "id": 30, - "type": "example", - "label": "Lebl-contfunc:30", - "categories": [ - "topology", - "continuity", - "boundedness", - "example" - ], - "title": "Continuity is important", - "contents": [ - "Continuity is important. Define$f \\colon [0,1] \\to \\R$by$f(x) \\coloneqq \\nicefrac{1}{x}$for$x > 0$and let$f(0) \\coloneqq 0$. The function does not achieve a maximum. The domain$[0,1]$is closed and bounded, but the problem is that the function is not continuous at 0." - ], - "refs": [], - "proofs": [], - "ref_ids": [] - }, - { - "id": 31, - "type": "lemma", - "label": "Lebl-contfunc:IVT:lemma", - "categories": [ - "auxiliary result", - "continuity", - "intervals" - ], - "title": "Let be a continuous function", - "contents": [ - "Let$f \\colon [a,b] \\to \\R$be a continuous function. Suppose$f(a) < 0$and$f(b) > 0$. Then there exists a number$c \\in (a,b)$such that$f(c) = 0$." - ], - "refs": [], - "proofs": [ - { - "contents": [ - "We define two sequences $\\{ a_n \\}_{n=1}^\\infty$", - "and $\\{ b_n \\}_{n=1}^\\infty$ inductively:", - "\\begin{enumerate}[(i)]", - "\\item Let $a_1 \\coloneqq a$ and $b_1 \\coloneqq b$.", - "\\item If $f\\left(\\frac{a_n+b_n}{2}\\right) \\geq 0$, let $a_{n+1} \\coloneqq a_n$ and", - "$b_{n+1} \\coloneqq \\frac{a_n+b_n}{2}$.", - "\\item If $f\\left(\\frac{a_n+b_n}{2}\\right) < 0$, let $a_{n+1} \\coloneqq \\frac{a_n+b_n}{2}$ and", - "$b_{n+1} \\coloneqq b_n$.", - "\\end{enumerate}", - "\\begin{myfigureht}", - "\\includegraphics{figures/bisect}", - "\\caption{Finding roots (bisection method).\\label{bisectfig}}", - "\\end{myfigureht}", - "See \\figureref{bisectfig} for an example defining the first five steps.", - "If $a_n < b_n$, then $a_n < \\frac{a_n+b_n}{2} < b_n$. So", - "$a_{n+1} < b_{n+1}$.", - "Thus by \\hyperref[induction:thm]{induction} $a_n < b_n$ for all $n$.", - "Furthermore, $a_n \\leq a_{n+1}$ and", - "$b_n \\geq b_{n+1}$ for all $n$, that is, the sequences are monotone.", - "As $a_n < b_n \\leq b_1 = b$ and", - "$b_n > a_n \\geq a_1 = a$ for all $n$,", - "the sequences are also bounded. Therefore, the", - "sequences converge.", - "Let $c \\coloneqq \\lim_{n\\to\\infty} a_n$ and $d \\coloneqq \\lim_{n\\to\\infty} b_n$,", - "where also $a \\leq c \\leq d \\leq b$. We need", - "to show that $c=d$.", - "Notice", - "\\begin{equation*}", - "b_{n+1} - a_{n+1} = \\frac{b_n-a_n}{2}.", - "\\end{equation*}", - "By \\hyperref[induction:thm]{induction},", - "\\begin{equation*}", - "b_n - a_n = \\frac{b_1-a_1}{2^{n-1}} = 2^{1-n} (b-a) .", - "\\end{equation*}", - "As $2^{1-n}(b-a)$ converges to zero, we take the limit as $n$ goes to", - "infinity to get", - "\\begin{equation*}", - "d-c = \\lim_{n\\to\\infty} (b_n - a_n) =", - "\\lim_{n\\to\\infty} 2^{1-n} (b-a) = 0.", - "\\end{equation*}", - "In other words, $d=c$.", - "By construction, for all $n$,", - "\\begin{equation*}", - "f(a_n) < 0", - "\\qquad \\text{and} \\qquad", - "f(b_n) \\geq 0 .", - "\\end{equation*}", - "Since", - "$\\lim_{n\\to\\infty} a_n = \\lim_{n\\to\\infty} b_n = c$", - "and $f$ is continuous at $c$, we may take", - "limits in those inequalities:", - "\\begin{equation*}", - "f(c) = \\lim_{n\\to\\infty} f(a_n) \\leq 0", - "\\qquad \\text{and} \\qquad", - "f(c) = \\lim_{n\\to\\infty} f(b_n) \\geq 0 .", - "\\end{equation*}", - "As $f(c) \\geq 0$ and", - "$f(c) \\leq 0$, we conclude $f(c) = 0$.", - "Thus also $c \\not=a$ and $c \\not= b$, so", - "$a < c < b$." - ], - "refs": [ - "induction:thm" - ], - "ref_ids": [] - } - ], - "ref_ids": [] - }, - { - "id": 32, - "type": "theorem", - "label": "Lebl-contfunc:IVT:thm", - "categories": [ - "continuity", - "named theorem", - "intervals" - ], - "title": "Let be a continuous function", - "contents": [ - "\\index{Bolzano's theorem} \\index{Bolzano's intermediate value theorem} \\index{intermediate value theorem} Let$f \\colon [a,b] \\to \\R$be a continuous function. Suppose$y \\in \\R$is such that$f(a) < y < f(b)$or$f(a) > y > f(b)$. Then there exists a$c \\in (a,b)$such that$f(c) = y$." - ], - "refs": [ - "named:Bolzano" - ], - "proofs": [ - { - "contents": [ - "If $f(a) < y < f(b)$, then define $g(x) \\coloneqq f(x)-y$. Then", - "$g(a) < 0$ and $g(b) > 0$, and we apply \\lemmaref{IVT:lemma}", - "to $g$ to find $c$. If $g(c) = 0$, then $f(c) = y$.", - "Similarly, if $f(a) > y > f(b)$, then define $g(x) \\coloneqq y-f(x)$.", - "Again, $g(a) < 0$ and $g(b) > 0$, and we apply \\lemmaref{IVT:lemma} to", - "find $c$.", - "As before, if $g(c) = 0$, then $f(c) = y$." - ], - "refs": [ - "IVT:lemma", - "named:Bolzano" - ], - "ref_ids": [ - 31 - ] - } - ], - "ref_ids": [] - }, - { - "id": 33, - "type": "example", - "label": "Lebl-contfunc:33", - "categories": [ - "example" - ], - "title": "% The polynomial has a real root in ", - "contents": [ - "%\\index{bisection method} The polynomial$f(x) \\coloneqq x^3-2x^2+x-1$has a real root in$(1,2)$. We simply notice that$f(1) = -1$and$f(2) = 1$. Hence there must exist a point$c \\in (1,2)$such that$f(c) = 0$. To find a better approximation of the root we follow the proof of \\lemmaref{IVT:lemma}. We look at 1.5 and find that$f(1.5) = -0.625$. Therefore, there is a root of the polynomial in$(1.5,2)$. Next we look at 1.75 and note that$f(1.75) \\approx -0.016$. Hence there is a root of$f$in$(1.75,2)$. Next we look at 1.875 and find that$f(1.875) \\approx 0.44$, thus there is a root in$(1.75,1.875)$. We follow this procedure until we gain sufficient precision. In fact, the root is at$c \\approx 1.7549$." - ], - "refs": [ - "IVT:lemma", - "named:Bolzano" - ], - "proofs": [], - "ref_ids": [ - 31 - ] - }, - { - "id": 34, - "type": "proposition", - "label": "Lebl-contfunc:34", - "categories": [], - "title": "Let be a polynomial of odd degree", - "contents": [ - "Let$f(x)$be a polynomial of odd degree. Then$f$has a real root." - ], - "refs": [], - "proofs": [ - { - "contents": [ - "Suppose $f$ is a polynomial of odd degree $d$. We write", - "\\begin{equation*}", - "f(x) = a_d x^d + a_{d-1} x^{d-1} + \\cdots + a_1 x + a_0 ,", - "\\end{equation*}", - "where $a_d \\not= 0$. We divide by $a_d$ to obtain a", - "\\emph{\\myindex{monic polynomial}}\\footnote{The word \\emph{monic} means that", - "the coefficient of $x^d$ is 1.}", - "\\begin{equation*}", - "g(x) \\coloneqq x^d + b_{d-1} x^{d-1} + \\cdots + b_1 x + b_0 ,", - "\\end{equation*}", - "where $b_k = \\nicefrac{a_k}{a_d}$.", - "Let us show that $g(n)$ is", - "positive for some large $n \\in \\N$.", - "We first compare the highest order term with the rest:", - "\\begin{equation*}", - "\\begin{split}", - "\\abs{\\frac{b_{d-1} n^{d-1} + \\cdots + b_1 n + b_0}{n^d}}", - "& =", - "\\frac{\\abs{b_{d-1} n^{d-1} + \\cdots + b_1 n + b_0}}{n^d}", - "\\\\", - "& \\leq", - "\\frac{\\abs{b_{d-1}} n^{d-1} + \\cdots + \\abs{b_1} n + \\abs{b_0}}{n^d}", - "\\\\", - "& \\leq", - "\\frac{\\abs{b_{d-1}} n^{d-1} + \\cdots + \\abs{b_1} n^{d-1} + \\abs{b_0} n^{d-1}}{n^d}", - "\\\\", - "& =", - "\\frac{n^{d-1}\\bigl(\\abs{b_{d-1}} + \\cdots + \\abs{b_1} + \\abs{b_0}\\bigr)}{n^d}", - "\\\\", - "& =", - "\\frac{1}{n}", - "\\bigl(\\abs{b_{d-1}} + \\cdots + \\abs{b_1} + \\abs{b_0}\\bigr) .", - "\\end{split}", - "\\end{equation*}", - "Therefore,", - "\\begin{equation*}", - "\\lim_{n\\to\\infty} \\frac{b_{d-1} n^{d-1} + \\cdots + b_1 n + b_0}{n^d}", - "= 0 .", - "\\end{equation*}", - "Thus there exists an $M \\in \\N$ such that", - "\\begin{equation*}", - "\\abs{\\frac{b_{d-1} M^{d-1} + \\cdots + b_1 M + b_0}{M^d}} < 1 ,", - "\\end{equation*}", - "which implies", - "\\begin{equation*}", - "-(b_{d-1} M^{d-1} + \\cdots + b_1 M + b_0) < M^d .", - "\\end{equation*}", - "Therefore, $g(M) > 0$.", - "Next, consider $g(-n)$ for $n \\in \\N$. By a similar argument,", - "there exists a $K \\in \\N$ such that", - "$b_{d-1} {(-K)}^{d-1} + \\cdots + b_1 (-K) + b_0 < K^d$", - "and therefore $g(-K) < 0$ (see \\exerciseref{exercise:odddegnegativeK}).", - "In the proof,", - "make sure you use the fact that $d$ is odd.", - "In particular, if $d$ is odd, then ${(-n)}^d = -(n^d)$.", - "We appeal to the intermediate value theorem to find a", - "$c \\in (-K,M)$, such that $g(c) = 0$. As $g(x) = \\frac{f(x)}{a_d}$,", - "then $f(c) = 0$, and the proof is done." - ], - "refs": [ - "named:Bolzano" - ], - "ref_ids": [] - } - ], - "ref_ids": [] - }, - { - "id": 35, - "type": "example", - "label": "Lebl-contfunc:35", - "categories": [ - "named theorem", - "existence", - "example" - ], - "title": "You may recall how hard we worked in to show that ...", - "contents": [ - "You may recall how hard we worked in \\exampleref{example:sqrt2} to show that$\\sqrt{2}$exists. With Bolzano's theorem, we can prove the existence$k$th root of any positive number$y > 0$without any effort. We claim that for any$k \\in \\N$and any$y > 0$, there exists a number$x > 0$such that$x^k = y$.", - "Proof: If$y=1$, then it is clear, so assume$y\\not= 1$. Let$f(x) \\coloneqq x^k - y$. We notice$f(0) = -y < 0$. If$y < 1$, then$f(1) = 1^k -y > 0$. If$y > 1$, then$f(y) = y^k-y = y(y^{k-1}-1) > 0$. In either case, apply Bolzano's theorem to find an$x > 0$such that$f(x) = 0$, or in other words$x^k = y$." - ], - "refs": [ - "named:Bolzano" - ], - "proofs": [ - { - "contents": [ - "If$y=1$, then it is clear, so assume$y\\not= 1$. Let$f(x) \\coloneqq x^k - y$. We notice$f(0) = -y < 0$. If$y < 1$, then$f(1) = 1^k -y > 0$. If$y > 1$, then$f(y) = y^k-y = y(y^{k-1}-1) > 0$. In either case, apply Bolzano's theorem to find an$x > 0$such that$f(x) = 0$, or in other words$x^k = y$." - ], - "refs": [ - "named:Bolzano" - ], - "ref_ids": [] - } - ], - "ref_ids": [] - }, - { - "id": 36, - "type": "example", - "label": "Lebl-contfunc:36", - "categories": [ - "example", - "continuity", - "intervals" - ], - "title": "Interestingly, there exist discontinuous functions...", - "contents": [ - "Interestingly, there exist discontinuous functions with the intermediate value property. The function", - "\\begin{equation*}\nf(x) \\coloneqq\n\\begin{cases}\n\\sin(\\nicefrac{1}{x}) & \\text{if } x \\not= 0, \\\\\n0 & \\text{if } x=0,\n\\end{cases}\n\\end{equation*}", - "is not continuous at$0$; however,$f$has the intermediate value property: Whenever$a < b$and$y$is such that$f(a) < y < f(b)$or$f(a) > y > f(b)$, there exists a$c \\in (a,b)$such that$f(c) = y$. See \\figureref{figsin1x} for a graph of$\\sin(\\nicefrac{1}{x})$. Proof is left as \\exerciseref{exercise:meanvaluepropsin1x}." - ], - "refs": [ - "named:Bolzano" - ], - "proofs": [], - "ref_ids": [] - }, - { - "id": 37, - "type": "corollary", - "label": "Lebl-contfunc:cor:imageofinterval", - "categories": [ - "continuity", - "boundedness", - "intervals", - "topology", - "consequence" - ], - "title": "If is continuous, then the direct image is a close...", - "contents": [ - "If$f \\colon [a,b] \\to \\R$is continuous, then the direct image$f\\bigl([a,b]\\bigr)$is a closed and bounded interval or a single number." - ], - "refs": [], - "proofs": [], - "ref_ids": [] - }, - { - "id": 38, - "type": "example", - "label": "Lebl-contfunc:38", - "categories": [ - "continuity", - "example" - ], - "title": " defined by is uniformly continuous", - "contents": [ - "$f \\colon [0,1] \\to \\R$defined by$f(x) \\coloneqq x^2$is uniformly continuous.", - "Proof: Note that$0 \\leq x,c \\leq 1$. Then", - "\\begin{equation*}\n\\sabs{x^2-c^2} = \\sabs{x+c}\\sabs{x-c}\n\\leq \\bigl(\\sabs{x}+\\sabs{c}\\bigr) \\sabs{x-c}\n\\leq (1+1)\\sabs{x-c} .\n\\end{equation*}", - "Therefore, given$\\epsilon > 0$, let$\\delta \\coloneqq \\nicefrac{\\epsilon}{2}$. If$\\sabs{x-c} < \\delta$, then$\\sabs{x^2-c^2} < \\epsilon$.", - "\\medskip", - "On the other hand,$g \\colon \\R \\to \\R$defined by$g(x) \\coloneqq x^2$is not uniformly continuous.", - "Proof: Suppose it is uniformly continuous, then for every$\\epsilon > 0$, there would exist a$\\delta > 0$such that if$\\sabs{x-c} < \\delta$, then$\\sabs{x^2 -c^2} < \\epsilon$. Take$x > 0$and let$c \\coloneqq x+\\nicefrac{\\delta}{2}$. Write", - "\\begin{equation*}\n\\epsilon >\n\\sabs{x^2-c^2} = \\sabs{x+c}\\sabs{x-c}\n=\n(2x+\\nicefrac{\\delta}{2})\\nicefrac{\\delta}{2} \n\\geq \n\\delta x .\n\\end{equation*}", - "Therefore,$x < \\nicefrac{\\epsilon}{\\delta}$for all$x > 0$, which is a contradiction." - ], - "refs": [], - "proofs": [ - { - "contents": [ - "Note that$0 \\leq x,c \\leq 1$. Then \\begin{equation*}", - "\\sabs{x^2-c^2} = \\sabs{x+c}\\sabs{x-c}", - "\\leq \\bigl(\\sabs{x}+\\sabs{c}\\bigr) \\sabs{x-c}", - "\\leq (1+1)\\sabs{x-c} .", - "\\end{equation*} Therefore, given$\\epsilon > 0$, let$\\delta \\coloneqq \\nicefrac{\\epsilon}{2}$. If$\\sabs{x-c} < \\delta$, then$\\sabs{x^2-c^2} < \\epsilon$. \\medskip On the other hand,$g \\colon \\R \\to \\R$defined by$g(x) \\coloneqq x^2$is not uniformly continuous. Proof: Suppose it is uniformly continuous, then for every$\\epsilon > 0$, there would exist a$\\delta > 0$such that if$\\sabs{x-c} < \\delta$, then$\\sabs{x^2 -c^2} < \\epsilon$. Take$x > 0$and let$c \\coloneqq x+\\nicefrac{\\delta}{2}$. Write \\begin{equation*}", - "\\epsilon >", - "\\sabs{x^2-c^2} = \\sabs{x+c}\\sabs{x-c}", - "=", - "(2x+\\nicefrac{\\delta}{2})\\nicefrac{\\delta}{2}", - "\\geq", - "\\delta x .", - "\\end{equation*} Therefore,$x < \\nicefrac{\\epsilon}{\\delta}$for all$x > 0$, which is a contradiction." - ], - "refs": [], - "ref_ids": [] - } - ], - "ref_ids": [] - }, - { - "id": 39, - "type": "example", - "label": "Lebl-contfunc:39", - "categories": [ - "continuity", - "characterization", - "example" - ], - "title": "The function defined by is not uniformly continuou...", - "contents": [ - "The function$f \\colon (0,1) \\to \\R$defined by$f(x) \\coloneqq \\nicefrac{1}{x}$is not uniformly continuous.", - "Proof: Given$\\epsilon > 0$, then$\\epsilon > \\abs{\\nicefrac{1}{x}-\\nicefrac{1}{y}}$holds if and only if", - "\\begin{equation*}\n\\epsilon >\n\\abs{\\nicefrac{1}{x}-\\nicefrac{1}{y}}\n=\n\\frac{\\abs{y-x}}{\\abs{xy}} \n=\n\\frac{\\abs{y-x}}{xy} ,\n\\end{equation*}", - "or", - "\\begin{equation*}\n\\abs{x-y} < xy \\epsilon .\n\\end{equation*}", - "Suppose$\\epsilon < 1$, and we wish to see if a small$\\delta > 0$would work. If$x \\in (0,1)$and$y = x+\\nicefrac{\\delta}{2} \\in (0,1)$, then$\\abs{x-y} = \\nicefrac{\\delta}{2} < \\delta$. We plug$y$into the inequality above to get$\\nicefrac{\\delta}{2} < x \\bigl( x+\\nicefrac{\\delta}{2} \\bigr) \\epsilon < x$. If the definition of uniform continuity is satisfied, then the inequality$\\nicefrac{\\delta}{2} < x$holds for all$x > 0$. But then$\\delta \\leq 0$. Therefore, no single$\\delta > 0$works for all points." - ], - "refs": [], - "proofs": [ - { - "contents": [ - "Given$\\epsilon > 0$, then$\\epsilon > \\abs{\\nicefrac{1}{x}-\\nicefrac{1}{y}}$holds if and only if \\begin{equation*}", - "\\epsilon >", - "\\abs{\\nicefrac{1}{x}-\\nicefrac{1}{y}}", - "=", - "\\frac{\\abs{y-x}}{\\abs{xy}}", - "=", - "\\frac{\\abs{y-x}}{xy} ,", - "\\end{equation*} or \\begin{equation*}", - "\\abs{x-y} < xy \\epsilon .", - "\\end{equation*} Suppose$\\epsilon < 1$, and we wish to see if a small$\\delta > 0$would work. If$x \\in (0,1)$and$y = x+\\nicefrac{\\delta}{2} \\in (0,1)$, then$\\abs{x-y} = \\nicefrac{\\delta}{2} < \\delta$. We plug$y$into the inequality above to get$\\nicefrac{\\delta}{2} < x \\bigl( x+\\nicefrac{\\delta}{2} \\bigr) \\epsilon < x$. If the definition of uniform continuity is satisfied, then the inequality$\\nicefrac{\\delta}{2} < x$holds for all$x > 0$. But then$\\delta \\leq 0$. Therefore, no single$\\delta > 0$works for all points." - ], - "refs": [], - "ref_ids": [] - } - ], - "ref_ids": [] - }, - { - "id": 40, - "type": "theorem", - "label": "Lebl-contfunc:unifcont:thm", - "categories": [ - "continuity", - "intervals" - ], - "title": "Let be a continuous function", - "contents": [ - "Let$f \\colon [a,b] \\to \\R$be a continuous function. Then$f$is uniformly continuous." - ], - "refs": [], - "proofs": [ - { - "contents": [ - "We prove the statement by contrapositive.", - "Suppose $f$ is not uniformly continuous. We will prove", - "that there is some", - "$c \\in [a,b]$ where $f$ is not continuous. Let us negate", - "the definition of uniformly continuous.", - "There exists an $\\epsilon > 0$", - "such that for every $\\delta > 0$, there exist points $x, y$ in $[a,b]$ with", - "$\\abs{x-y} < \\delta$ and $\\abs{f(x)-f(y)} \\geq \\epsilon$.", - "So for the $\\epsilon > 0$ above,", - "we find sequences $\\{ x_n \\}_{n=1}^\\infty$ and $\\{ y_n \\}_{n=1}^\\infty$ such that", - "$\\abs{x_n-y_n} < \\nicefrac{1}{n}$ and such that $\\abs{f(x_n)-f(y_n)} \\geq", - "\\epsilon$. By", - "\\hyperref[thm:bwseq]{Bolzano--Weierstrass},", - "there exists a convergent subsequence", - "$\\{ x_{n_k} \\}_{k=1}^\\infty$. Let $c \\coloneqq \\lim_{k\\to\\infty} x_{n_k}$.", - "As $a \\leq x_{n_k} \\leq b$ for all $k$, we have $a \\leq c \\leq b$. Estimate", - "\\begin{equation*}", - "\\sabs{y_{n_k} - c} =", - "\\sabs{y_{n_k} - x_{n_k} + x_{n_k} - c} \\leq", - "\\sabs{y_{n_k} - x_{n_k}}", - "+", - "\\sabs{x_{n_k}-c}", - "<", - "\\nicefrac{1}{n_k}", - "+", - "\\sabs{x_{n_k}-c} .", - "\\end{equation*}", - "As $\\nicefrac{1}{n_k}$ and $\\abs{x_{n_k}-c}$ both go to zero when", - "$k$ goes to infinity, $\\{ y_{n_k} \\}_{k=1}^\\infty$ converges and the limit", - "is $c$. We now show that $f$ is not continuous at $c$.", - "Estimate", - "\\begin{equation*}", - "\\begin{split}", - "\\abs{f(x_{n_k}) - f(c)} & =", - "\\abs{f(x_{n_k}) - f(y_{n_k}) + f(y_{n_k}) - f(c)} \\\\", - "& \\geq", - "\\abs{f(x_{n_k}) - f(y_{n_k})} - \\abs{f(y_{n_k}) - f(c)} \\\\", - "& \\geq", - "\\epsilon - \\abs{f(y_{n_k})-f(c)} .", - "\\end{split}", - "\\end{equation*}", - "Or in other words,", - "\\begin{equation*}", - "\\abs{f(x_{n_k})-f(c)}", - "+", - "\\abs{f(y_{n_k})-f(c)} \\geq", - "\\epsilon .", - "\\end{equation*}", - "At least one of the sequences $\\bigl\\{ f(x_{n_k}) \\bigr\\}_{k=1}^\\infty$ or", - "$\\bigl\\{ f(y_{n_k}) \\bigr\\}_{k=1}^\\infty$ cannot converge to $f(c)$, otherwise the", - "left-hand side of the inequality would go to zero while the right-hand side is positive.", - "Thus $f$ cannot be continuous at $c$." - ], - "refs": [ - "named:Bolzano", - "named:Weierstrass", - "thm:bwseq" - ], - "ref_ids": [] - } - ], - "ref_ids": [] - }, - { - "id": 41, - "type": "lemma", - "label": "Lebl-contfunc:unifcauchycauchy:lemma", - "categories": [ - "sequences", - "continuity", - "auxiliary result" - ], - "title": "Let and let be a uniformly continuous function", - "contents": [ - "Let$S \\subset \\R$and let$f \\colon S \\to \\R$be a uniformly continuous function. Let$\\{ x_n \\}_{n=1}^\\infty$be a Cauchy sequence in$S$. Then$\\bigl\\{ f(x_n) \\bigr\\}_{n=1}^\\infty$is Cauchy." - ], - "refs": [], - "proofs": [ - { - "contents": [ - "Let $\\epsilon > 0$ be given. There is a $\\delta > 0$ such that", - "$\\abs{f(x)-f(y)} < \\epsilon$ whenever $x,y \\in S$ and $\\abs{x-y} < \\delta$. Find an $M", - "\\in \\N$ such that for all $n, k \\geq M$, we have $\\abs{x_n-x_k} < \\delta$.", - "Then for all $n, k \\geq M$, we have $\\abs{f(x_n)-f(x_k)} < \\epsilon$." - ], - "refs": [], - "ref_ids": [] - } - ], - "ref_ids": [] - }, - { - "id": 42, - "type": "proposition", - "label": "Lebl-contfunc:context:prop", - "categories": [ - "limits", - "continuity", - "characterization", - "intervals" - ], - "title": "A function is uniformly continuous if and only if ...", - "contents": [ - "A function$f \\colon (a,b) \\to \\R$is uniformly continuous if and only if the limits\\begin{equation*} L_a \\coloneqq \\lim_{x \\to a} f(x) \\qquad \\text{and} \\qquad L_b \\coloneqq \\lim_{x \\to b} f(x) \\end{equation*}exist and the function$\\widetilde{f} \\colon [a,b] \\to \\R$defined by \\begin{equation*} \\widetilde{f}(x) \\coloneqq \\begin{cases} f(x) & \\text{if } x \\in (a,b),", - "L_a & \\text{if } x = a,", - "L_b & \\text{if } x = b \\end{cases} \\end{equation*} is continuous." - ], - "refs": [], - "proofs": [ - { - "contents": [ - "One direction is quick. If $\\widetilde{f}$ is continuous, then", - "it is uniformly continuous by \\thmref{unifcont:thm}. As $f$ is the", - "restriction of $\\widetilde{f}$ to $(a,b)$, $f$ is also uniformly continuous", - "(exercise).", - "Now suppose $f$ is uniformly continuous. We must first show", - "that the limits $L_a$ and $L_b$ exist. Let us concentrate on $L_a$.", - "Take $\\{ x_n \\}_{n=1}^\\infty$ in $(a,b)$ such that", - "$\\lim_{n\\to\\infty} x_n = a$.", - "The sequence $\\{ x_n \\}_{n=1}^\\infty$ is Cauchy, so by", - "\\lemmaref{unifcauchycauchy:lemma}", - "the sequence $\\bigl\\{ f(x_n) \\bigr\\}_{n=1}^\\infty$ is Cauchy and thus convergent.", - "Let $L_1 \\coloneqq \\lim_{n\\to\\infty} f(x_n)$. Take another sequence", - "$\\{ y_n \\}_{n=1}^\\infty$ in $(a,b)$ such that $\\lim_{n\\to\\infty} y_n = a$. By the same reasoning", - "we get $L_2 \\coloneqq \\lim_{n\\to\\infty} f(y_n)$. If we show that $L_1 = L_2$, then", - "the limit $L_a = \\lim_{x\\to a} f(x)$ exists. Let $\\epsilon > 0$ be given.", - "Find $\\delta > 0$ such that $\\abs{x-y} < \\delta$ implies $\\abs{f(x)-f(y)} <", - "\\nicefrac{\\epsilon}{3}$. Find $M \\in \\N$ such that for", - "$n \\geq M$, we have $\\abs{a-x_n} < \\nicefrac{\\delta}{2}$,", - "$\\abs{a-y_n} < \\nicefrac{\\delta}{2}$,", - "$\\abs{f(x_n)-L_1} < \\nicefrac{\\epsilon}{3}$, and", - "$\\abs{f(y_n)-L_2} < \\nicefrac{\\epsilon}{3}$. Then for $n \\geq M$,", - "\\begin{equation*}", - "\\abs{x_n-y_n} =", - "\\abs{x_n-a+a-y_n} \\leq", - "\\abs{x_n-a}+\\abs{a-y_n} < \\nicefrac{\\delta}{2} + \\nicefrac{\\delta}{2} =", - "\\delta.", - "\\end{equation*}", - "So", - "\\begin{equation*}", - "\\begin{split}", - "\\abs{L_1-L_2} &=", - "\\abs{L_1-f(x_n)+f(x_n)-f(y_n)+f(y_n)-L_2} \\\\", - "& \\leq", - "\\abs{L_1-f(x_n)}+\\abs{f(x_n)-f(y_n)}+\\abs{f(y_n)-L_2} \\\\", - "& \\leq", - "\\nicefrac{\\epsilon}{3} + \\nicefrac{\\epsilon}{3} + \\nicefrac{\\epsilon}{3}", - "=", - "\\epsilon .", - "\\end{split}", - "\\end{equation*}", - "Therefore, $L_1 = L_2$.", - "Thus $L_a$ exists. To show that $L_b$ exists is left as an exercise.", - "If $L_a = \\lim_{x\\to a} f(x)$", - "exists, then $\\lim_{x\\to a} \\widetilde{f}(x)$ exists", - "and equals $L_a$", - "(see \\propref{prop:limrest}). Similarly for $L_b$.", - "Hence $\\widetilde{f}$ is continuous at $a$ and $b$.", - "And since $f$ is continuous at $c \\in (a,b)$, then", - "$\\widetilde{f}$ is continuous at $c \\in (a,b)$ (\\propref{prop:limrest} again)." - ], - "refs": [ - "prop:limrest", - "unifcauchycauchy:lemma" - ], - "ref_ids": [ - 11, - 41 - ] - } - ], - "ref_ids": [] - }, - { - "id": 43, - "type": "proposition", - "label": "Lebl-contfunc:43", - "categories": [ - "continuity" - ], - "title": "A Lipschitz continuous function is uniformly conti...", - "contents": [ - "A Lipschitz continuous function is uniformly continuous." - ], - "refs": [], - "proofs": [ - { - "contents": [ - "Let $f \\colon S \\to \\R$ be a function and let $K$ be a constant such that", - "$\\abs{f(x)-f(y)} \\leq K \\abs{x-y}$", - "for all $x, y$ in $S$.", - "Let $\\epsilon > 0$ be given. Take $\\delta \\coloneqq", - "\\nicefrac{\\epsilon}{K}$.", - "For all $x$ and $y$ in $S$ such that", - "$\\abs{x-y} < \\delta$,", - "\\begin{equation*}", - "\\abs{f(x)-f(y)} \\leq K \\abs{x-y} < K \\delta = K \\frac{\\epsilon}{K} =", - "\\epsilon .", - "\\end{equation*}", - "Therefore, $f$ is uniformly continuous." - ], - "refs": [], - "ref_ids": [] - } - ], - "ref_ids": [] - }, - { - "id": 44, - "type": "example", - "label": "Lebl-contfunc:44", - "categories": [ - "continuity", - "example" - ], - "title": "The functions and are Lipschitz continuous", - "contents": [ - "The functions$\\sin(x)$and$\\cos(x)$are Lipschitz continuous. In \\exampleref{sincos:example} we have seen the following two inequalities.", - "\\begin{equation*}\n\\abs{\\sin(x)-\\sin(y)} \n\\leq \\abs{x-y}\n\\qquad \\text{and} \\qquad\n\\abs{\\cos(x)-\\cos(y)}\n\\leq \\abs{x-y} .\n\\end{equation*}", - "Hence sine and cosine are Lipschitz continuous with$K=1$." - ], - "refs": [], - "proofs": [], - "ref_ids": [] - }, - { - "id": 45, - "type": "example", - "label": "Lebl-contfunc:45", - "categories": [ - "continuity", - "example" - ], - "title": "The function defined by is Lipschitz continuous", - "contents": [ - "The function$f \\colon [1,\\infty) \\to \\R$defined by$f(x) \\coloneqq \\sqrt{x}$is Lipschitz continuous. Proof:", - "\\begin{equation*}\n\\abs{\\sqrt{x}-\\sqrt{y}} = \n\\abs{\\frac{x-y}{\\sqrt{x}+\\sqrt{y}}}\n=\n\\frac{\\abs{x-y}}{\\sqrt{x}+\\sqrt{y}} .\n\\end{equation*}", - "As$x \\geq 1$and$y \\geq 1$, we see that$\\frac{1}{\\sqrt{x}+\\sqrt{y}} \\leq \\frac{1}{2}$. Therefore,", - "\\begin{equation*}\n\\abs{\\sqrt{x}-\\sqrt{y}} = \n\\abs{\\frac{x-y}{\\sqrt{x}+\\sqrt{y}}}\n\\leq\n\\frac{1}{2}\n\\abs{x-y}.\n\\end{equation*}", - "On the other hand,$g \\colon [0,\\infty) \\to \\R$defined by$g(x) \\coloneqq \\sqrt{x}$is not Lipschitz continuous. Proof: Suppose for all$x,y \\in [0,\\infty)$,", - "\\begin{equation*}\n\\abs{\\sqrt{x}-\\sqrt{y}} \n\\leq\nK \\abs{x-y} ,\n\\end{equation*}", - "for some$K$. Set$y=0$to obtain$\\sqrt{x} \\leq K x$. If$K > 0$, then for$x > 0$we get$\\nicefrac{1}{K} \\leq \\sqrt{x}$or$\\nicefrac{1}{K^2} \\leq x$. This cannot possibly be true for all$x > 0$. Thus no such$K > 0$exists and$g$is not Lipschitz continuous. See \\figureref{fig:sqrtgraph} and note how secant lines would be more and more vertical as we get closer to$x=0$. \\begin{myfigureht} \\includegraphics{figures/sqrtgraph} \\caption{Graph of$\\sqrt{x}$and some secant lines through$(0,0)$and$(x,\\sqrt{x})$.\\label{fig:sqrtgraph}} \\end{myfigureht}", - "The last example$g$is a function that is uniformly continuous but not Lipschitz continuous. To see that$\\sqrt{x}$is uniformly continuous as a function on$[0,\\infty)$, note that it is uniformly continuous when restricted to$[0,1]$by \\thmref{unifcont:thm}. It is also Lipschitz (and so uniformly continuous) when restricted to$[1,\\infty)$. It is not hard (exercise) to show that this means that$\\sqrt{x}$is a uniformly continuous function on$[0,\\infty)$." - ], - "refs": [], - "proofs": [ - { - "contents": [ - "\\begin{equation*}", - "\\abs{\\sqrt{x}-\\sqrt{y}} =", - "\\abs{\\frac{x-y}{\\sqrt{x}+\\sqrt{y}}}", - "=", - "\\frac{\\abs{x-y}}{\\sqrt{x}+\\sqrt{y}} .", - "\\end{equation*} As$x \\geq 1$and$y \\geq 1$, we see that$\\frac{1}{\\sqrt{x}+\\sqrt{y}} \\leq \\frac{1}{2}$. Therefore, \\begin{equation*}", - "\\abs{\\sqrt{x}-\\sqrt{y}} =", - "\\abs{\\frac{x-y}{\\sqrt{x}+\\sqrt{y}}}", - "\\leq", - "\\frac{1}{2}", - "\\abs{x-y}.", - "\\end{equation*} On the other hand,$g \\colon [0,\\infty) \\to \\R$defined by$g(x) \\coloneqq \\sqrt{x}$is not Lipschitz continuous. Proof: Suppose for all$x,y \\in [0,\\infty)$, \\begin{equation*}", - "\\abs{\\sqrt{x}-\\sqrt{y}}", - "\\leq", - "K \\abs{x-y} ,", - "\\end{equation*} for some$K$. Set$y=0$to obtain$\\sqrt{x} \\leq K x$. If$K > 0$, then for$x > 0$we get$\\nicefrac{1}{K} \\leq \\sqrt{x}$or$\\nicefrac{1}{K^2} \\leq x$. This cannot possibly be true for all$x > 0$. Thus no such$K > 0$exists and$g$is not Lipschitz continuous. See \\figureref{fig:sqrtgraph} and note how secant lines would be more and more vertical as we get closer to$x=0$. \\begin{myfigureht} \\includegraphics{figures/sqrtgraph} \\caption{Graph of$\\sqrt{x}$and some secant lines through$(0,0)$and$(x,\\sqrt{x})$.\\label{fig:sqrtgraph}} \\end{myfigureht} The last example$g$is a function that is uniformly continuous but not Lipschitz continuous. To see that$\\sqrt{x}$is uniformly continuous as a function on$[0,\\infty)$, note that it is uniformly continuous when restricted to$[0,1]$by \\thmref{unifcont:thm}. It is also Lipschitz (and so uniformly continuous) when restricted to$[1,\\infty)$. It is not hard (exercise) to show that this means that$\\sqrt{x}$is a uniformly continuous function on$[0,\\infty)$." - ], - "refs": [], - "ref_ids": [] - } - ], - "ref_ids": [] - }, - { - "id": 46, - "type": "proposition", - "label": "Lebl-contfunc:liminfty:unique", - "categories": [ - "limits" - ], - "title": "The limit at or as defined above is unique if it e...", - "contents": [ - "The limit at$\\infty$or$-\\infty$as defined above is unique if it exists." - ], - "refs": [], - "proofs": [], - "ref_ids": [] - }, - { - "id": 47, - "type": "example", - "label": "Lebl-contfunc:47", - "categories": [ - "limits", - "example" - ], - "title": "Let ", - "contents": [ - "Let$f(x) \\coloneqq \\frac{1}{\\abs{x}+1}$. Then", - "\\begin{equation*}\n\\lim_{x\\to \\infty} f(x) = 0 \\qquad \\text{and} \\qquad\n\\lim_{x\\to -\\infty} f(x) = 0 .\n\\end{equation*}", - "Proof: Let$\\epsilon > 0$be given. Find$M > 0$large enough so that$\\frac{1}{M+1} < \\epsilon$. If$x \\geq M$, then$0 < \\frac{1}{\\abs{x}+1} = \\frac{1}{x+1} \\leq \\frac{1}{M+1} < \\epsilon$. The first limit follows. The proof for$-\\infty$is left to the reader." - ], - "refs": [], - "proofs": [ - { - "contents": [ - "Let$\\epsilon > 0$be given. Find$M > 0$large enough so that$\\frac{1}{M+1} < \\epsilon$. If$x \\geq M$, then$0 < \\frac{1}{\\abs{x}+1} = \\frac{1}{x+1} \\leq \\frac{1}{M+1} < \\epsilon$. The first limit follows. The proof for$-\\infty$is left to the reader." - ], - "refs": [], - "ref_ids": [] - } - ], - "ref_ids": [] - }, - { - "id": 48, - "type": "example", - "label": "Lebl-contfunc:48", - "categories": [ - "limits", - "continuity", - "example", - "sequences" - ], - "title": "Let ", - "contents": [ - "Let$f(x) \\coloneqq \\sin(\\pi x)$. Then$\\lim_{x\\to\\infty} f(x)$does not exist. To prove this fact note that if$x = 2n+\\nicefrac{1}{2}$for some$n \\in \\N$, then$f(x)=1$, while if$x = 2n+\\nicefrac{3}{2}$, then$f(x)=-1$. So they cannot both be within a small$\\epsilon$of a single real number.", - "Be careful not to confuse continuous limits with limits of sequences. We could say", - "\\begin{equation*}\n\\lim_{n \\to \\infty} \\sin(\\pi n) = 0, \\qquad \\text{but} \\qquad\n\\lim_{x \\to \\infty} \\sin(\\pi x) \\enspace \\text{does not exist}.\n\\end{equation*}", - "Of course the notation is ambiguous: Are we thinking of the sequence$\\bigl\\{ \\sin (\\pi n) \\bigr\\}_{n=1}^\\infty$or the function$\\sin(\\pi x)$of a real variable? We are simply using the convention that$n \\in \\N$, while$x \\in \\R$. When the notation is not clear, it is good to explicitly mention where the variable lives, or what kind of limit are you using. If there is possibility of confusion, one can write, for example,", - "\\begin{equation*}\n\\lim_{\\substack{n \\to \\infty\\\\n \\in \\N}} \\sin(\\pi n) .\n\\end{equation*}" - ], - "refs": [], - "proofs": [], - "ref_ids": [] - }, - { - "id": 49, - "type": "lemma", - "label": "Lebl-contfunc:seqflimitinf:lemma", - "categories": [ - "limits", - "auxiliary result", - "characterization", - "sequences" - ], - "title": "Suppose is a function, is a cluster point of , and...", - "contents": [ - "Suppose$f \\colon S \\to \\R$is a function,$\\infty$is a cluster point of$S \\subset \\R$, and$L \\in \\R$. Then\\begin{equation*} \\lim_{x\\to\\infty} f(x) = L \\qquad \\text{if and only if} \\qquad \\lim_{n\\to\\infty} f(x_n) = L \\end{equation*}for all sequences$\\{ x_n \\}_{n=1}^\\infty$in$S$such that$\\lim\\limits_{n\\to\\infty} x_n = \\infty$." - ], - "refs": [], - "proofs": [ - { - "contents": [ - "First suppose $f(x) \\to L$ as $x \\to \\infty$.", - "Given an $\\epsilon > 0$, there exists an $M$ such that for all $x \\geq M$,", - "we have $\\abs{f(x)-L} < \\epsilon$.", - "Let $\\{ x_n \\}_{n=1}^\\infty$", - "be a sequence in $S$ such that $\\lim_{n\\to\\infty} x_n = \\infty$. Then there exists an", - "$N$ such that for all $n \\geq N$, we have $x_n \\geq M$. And thus", - "$\\abs{f(x_n)-L} < \\epsilon$.", - "We prove the converse by contrapositive. Suppose $f(x)$ does", - "not go to $L$ as $x \\to \\infty$.", - "This means that there exists an $\\epsilon > 0$,", - "such that for every $n \\in \\N$, there exists an $x \\in S$, $x \\geq n$, let", - "us call it $x_n$, such that $\\abs{f(x_n)-L} \\geq \\epsilon$.", - "Consider the sequence $\\{ x_n \\}_{n=1}^\\infty$. Clearly", - "$\\bigl\\{ f(x_n) \\bigr\\}_{n=1}^\\infty$ does not converge to $L$. It remains to note", - "that $\\lim_{n\\to\\infty} x_n = \\infty$, because $x_n \\geq n$ for all $n$." - ], - "refs": [], - "ref_ids": [] - } - ], - "ref_ids": [] - }, - { - "id": 50, - "type": "example", - "label": "Lebl-contfunc:50", - "categories": [ - "limits", - "example" - ], - "title": "Let us show that ", - "contents": [ - "Let us show that$\\lim\\limits_{x \\to \\infty} \\frac{1+x^2}{1+x} = \\infty$.", - "Proof: For$x \\geq 1$, we have", - "\\begin{equation*}\n\\frac{1+x^2}{1+x} \\geq \n\\frac{x^2}{x+x} = \n\\frac{x}{2} .\n\\end{equation*}", - "Given$N \\in \\R$, take$M = \\max \\{ 2N+1 , 1 \\}$. If$x \\geq M$, then$x \\geq 1$and$\\nicefrac{x}{2} > N$. So", - "\\begin{equation*}\n\\frac{1+x^2}{1+x} \\geq \n\\frac{x}{2} > N .\n\\end{equation*}" - ], - "refs": [], - "proofs": [ - { - "contents": [ - "For$x \\geq 1$, we have \\begin{equation*}", - "\\frac{1+x^2}{1+x} \\geq", - "\\frac{x^2}{x+x} =", - "\\frac{x}{2} .", - "\\end{equation*} Given$N \\in \\R$, take$M = \\max \\{ 2N+1 , 1 \\}$. If$x \\geq M$, then$x \\geq 1$and$\\nicefrac{x}{2} > N$. So \\begin{equation*}", - "\\frac{1+x^2}{1+x} \\geq", - "\\frac{x}{2} > N .", - "\\end{equation*}" - ], - "refs": [], - "ref_ids": [] - } - ], - "ref_ids": [] - }, - { - "id": 51, - "type": "proposition", - "label": "Lebl-contfunc:prop:inflimcompositions", - "categories": [ - "topology" - ], - "title": "Suppose , , , is a cluster point of , and is a clu...", - "contents": [ - "Suppose$f \\colon A \\to B$,$g \\colon B \\to \\R$,$A, B \\subset \\R$,$a \\in \\R \\cup \\{ -\\infty, \\infty\\}$is a cluster point of$A$, and$b \\in \\R \\cup \\{ -\\infty, \\infty\\}$is a cluster point of$B$. Suppose\\begin{equation*} \\lim_{x \\to a} f(x) = b\\qquad \\text{and} \\qquad \\lim_{y \\to b} g(y) = c \\end{equation*}for some$c \\in \\R \\cup \\{ -\\infty, \\infty \\}$. If$b \\in B$, then suppose$g(b) = c$. Then\\begin{equation*} \\lim_{x \\to a} g\\bigl(f(x)\\bigr) = c . \\end{equation*}" - ], - "refs": [], - "proofs": [], - "ref_ids": [] - }, - { - "id": 52, - "type": "example", - "label": "Lebl-contfunc:52", - "categories": [ - "example" - ], - "title": "Let ", - "contents": [ - "Let$h(x) \\coloneqq e^{-x^2+x}$. Then", - "\\begin{equation*}\n\\lim_{x\\to \\infty} h(x) = 0 .\n\\end{equation*}", - "Proof: The claim follows once we know", - "\\begin{equation*}\n\\lim_{x\\to \\infty} -x^2+x = -\\infty\n\\end{equation*}", - "and", - "\\begin{equation*}\n\\lim_{y\\to -\\infty} e^y = 0 ,\n\\end{equation*}", - "which is usually proved when the exponential function is defined." - ], - "refs": [], - "proofs": [ - { - "contents": [ - "The claim follows once we know \\begin{equation*}", - "\\lim_{x\\to \\infty} -x^2+x = -\\infty", - "\\end{equation*} and \\begin{equation*}", - "\\lim_{y\\to -\\infty} e^y = 0 ,", - "\\end{equation*} which is usually proved when the exponential function is defined." - ], - "refs": [], - "ref_ids": [] - } - ], - "ref_ids": [] - }, - { - "id": 53, - "type": "proposition", - "label": "Lebl-contfunc:prop:monotlimits", - "categories": [ - "topology", - "monotonicity" - ], - "title": "Let , , be increasing, and be decreasing", - "contents": [ - "Let$S \\subset \\R$,$c \\in \\R$,$f \\colon S \\to \\R$be increasing, and$g \\colon S \\to \\R$be decreasing. If$c$is a cluster point of$S \\cap (-\\infty,c)$, then\\begin{equation*} \\lim_{x \\to c^-} f(x) = \\sup \\{ f(x) : x < c, x \\in S \\} \\qquad \\text{and} \\qquad \\lim_{x \\to c^-} g(x) = \\inf \\{ g(x) : x < c, x \\in S \\} . \\end{equation*}If$c$is a cluster point of$S \\cap (c,\\infty)$, then\\begin{equation*} \\lim_{x \\to c^+} f(x) = \\inf \\{ f(x) : x > c, x \\in S \\} \\qquad \\text{and} \\qquad \\lim_{x \\to c^+} g(x) = \\sup \\{ g(x) : x > c, x \\in S \\} . \\end{equation*}If$\\infty$is a cluster point of$S$, then\\begin{equation*} \\lim_{x \\to \\infty} f(x) = \\sup \\{ f(x) : x \\in S \\} \\qquad \\text{and} \\qquad \\lim_{x \\to \\infty} g(x) = \\inf \\{ g(x) : x \\in S \\} . \\end{equation*}If$-\\infty$is a cluster point of$S$, then\\begin{equation*} \\lim_{x \\to -\\infty} f(x) = \\inf \\{ f(x) : x \\in S \\} \\qquad \\text{and} \\qquad \\lim_{x \\to -\\infty} g(x) = \\sup \\{ g(x) : x \\in S \\} . \\end{equation*}" - ], - "refs": [], - "proofs": [ - { - "contents": [ - "Let us assume $f$ is increasing, and we will show the first", - "equality. The rest of the proof is very similar and is left as an", - "exercise.", - "Let $a \\coloneqq \\sup \\{ f(x) : x < c, x \\in S \\}$. If $a = \\infty$,", - "then given an $M \\in \\R$, there exists an $x_M \\in S$, $x_M < c$, such that $f(x_M) > M$.", - "As $f$ is increasing, $f(x) \\geq f(x_M) > M$ for all $x \\in S$ with $x > x_M$.", - "Take $\\delta \\coloneqq c-x_M > 0$ to obtain the definition of the limit going to", - "infinity.", - "Next suppose $a < \\infty$.", - "Let $\\epsilon > 0$ be given. Because $a$ is the supremum and", - "$S \\cap (-\\infty,c)$ is nonempty, $a \\in \\R$ and", - "there exists an", - "$x_\\epsilon \\in S$,", - "$x_\\epsilon < c$,", - "such that $f(x_\\epsilon) > a-\\epsilon$. As $f$ is increasing,", - "if $x \\in S$ and $x_\\epsilon < x < c$, we have", - "$a-\\epsilon < f(x_\\epsilon) \\leq f(x) \\leq a$. Let", - "$\\delta \\coloneqq c-x_\\epsilon$. Then for $x \\in S \\cap (-\\infty,c)$", - "with $\\abs{x-c} < \\delta$,", - "we have $\\abs{f(x)-a} < \\epsilon$." - ], - "refs": [], - "ref_ids": [] - } - ], - "ref_ids": [] - }, - { - "id": 54, - "type": "corollary", - "label": "Lebl-contfunc:cor:continterval", - "categories": [ - "continuity", - "characterization", - "monotonicity", - "intervals", - "consequence" - ], - "title": "If is an interval and is monotone and not constant...", - "contents": [ - "If$I \\subset \\R$is an interval and$f \\colon I \\to \\R$is monotone and not constant, then$f(I)$is an interval if and only if$f$is continuous." - ], - "refs": [], - "proofs": [ - { - "contents": [ - "Without loss of generality, suppose $f$ is increasing.", - "First suppose $f$ is continuous. Take two points", - "$f(x_1), f(x_2)$ in $f(I)$ and suppose", - "$f(x_1) < f(x_2)$.", - "As $f$ is increasing, then $x_1 < x_2$. By the", - "\\hyperref[IVT:thm]{intermediate value theorem},", - "given $y$ with $f(x_1) < y < f(x_2)$, we find", - "a $c \\in (x_1,x_2) \\subset I$ such that $f(c) = y$, so $y \\in f(I)$.", - "Hence, $f(I)$ is an interval.", - "Let us prove the reverse direction by contrapositive.", - "Suppose $f$ is not continuous at $c \\in I$,", - "and that $c$ is not an endpoint of $I$.", - "Let", - "\\begin{equation*}", - "a \\coloneqq \\lim_{x \\to c^-} f(x) = \\sup \\bigl\\{ f(x) : x \\in I, x < c \\bigr\\} ,", - "\\qquad", - "b \\coloneqq \\lim_{x \\to c^+} f(x) = \\inf \\bigl\\{ f(x) : x \\in I, x > c \\bigr\\} .", - "\\end{equation*}", - "As $c$ is a discontinuity, $a < b$.", - "If $x < c$, then $f(x) \\leq a$, and", - "if $x > c$, then $f(x) \\geq b$. Therefore", - "no point", - "in $(a,b) \\setminus \\bigl\\{ f(c) \\bigr\\}$ is in $f(I)$.", - "There exists $x_1 \\in I$ with $x_1 < c$, so", - "$f(x_1) \\leq a$, and there exists $x_2 \\in I$ with $x_2 > c$,", - "so $f(x_2) \\geq b$. Both $f(x_1)$ and $f(x_2)$ are in $f(I)$,", - "but there are points in between them that are not in $f(I)$.", - "So $f(I)$ is not an interval. See \\figureref{fig:figinccont}.", - "When $c \\in I$ is an endpoint, the proof is similar and is left as an exercise." - ], - "refs": [ - "IVT:thm", - "named:Bolzano" - ], - "ref_ids": [ - 32 - ] - } - ], - "ref_ids": [] - }, - { - "id": 55, - "type": "corollary", - "label": "Lebl-contfunc:cor:monotcountcont", - "categories": [ - "consequence", - "monotonicity", - "intervals" - ], - "title": "Let be an interval and be monotone", - "contents": [ - "Let$I \\subset \\R$be an interval and$f \\colon I \\to \\R$be monotone. Then$f$has at most countably many discontinuities." - ], - "refs": [], - "proofs": [ - { - "contents": [ - "Let $E \\subset I$ be the set of all discontinuities", - "that are not endpoints of $I$. As there are", - "only two endpoints, it is enough to show that $E$ is countable.", - "Without loss of generality, suppose $f$ is increasing.", - "We will define an injection $h \\colon E \\to \\Q$.", - "For each $c \\in E$,", - "both one-sided limits of $f$ exist as $c$ is not an endpoint.", - "Let", - "\\begin{equation*}", - "a \\coloneqq \\lim_{x \\to c^-} f(x) = \\sup \\bigl\\{ f(x) : x \\in I, x < c \\bigr\\} ,", - "\\qquad", - "b \\coloneqq \\lim_{x \\to c^+} f(x) = \\inf \\bigl\\{ f(x) : x \\in I, x > c \\bigr\\} .", - "\\end{equation*}", - "As $c$ is a discontinuity, $a < b$.", - "There exists a rational number $q \\in (a,b)$, so let $h(c) \\coloneqq q$.", - "Suppose $d \\in E$ is another discontinuity.", - "If $d > c$, there", - "exist an $x \\in I$ with $c < x < d$, and so $\\lim_{x \\to d^-} f(x) \\geq b$.", - "Hence the rational number we choose for $h(d)$ is different from $q$,", - "since $q=h(c) < b$ and $h(d) > b$.", - "Similarly if $d < c$. After making such a choice for", - "every element of $E$, we have a", - "one-to-one (injective) function into $\\Q$. Therefore, $E$ is countable." - ], - "refs": [], - "ref_ids": [] - } - ], - "ref_ids": [] - }, - { - "id": 56, - "type": "example", - "label": "Lebl-contfunc:example:countdiscont", - "categories": [ - "continuity", - "boundedness", - "monotonicity", - "intervals", - "topology", - "example" - ], - "title": "By denote the largest integer less than or equal t...", - "contents": [ - "By$\\lfloor x \\rfloor$denote the largest integer less than or equal to$x$. Define$f \\colon [0,1] \\to \\R$by", - "\\begin{equation*}\nf(x) \\coloneqq\nx +\n\\sum_{n=0}^{\\lfloor 1/(1-x) \\rfloor}\n2^{-n} ,\n\\end{equation*}", - "for$x < 1$and$f(1) \\coloneqq 3$. It is an exercise to show that$f$is strictly increasing, bounded, and has a discontinuity at all points$1-\\nicefrac{1}{k}$for$k \\in \\N$. In particular, there are countably many discontinuities, but the function is bounded and defined on a closed bounded interval. See \\figureref{fig:countdiscont}. \\begin{myfigureht} \\includegraphics{figures/increasing-discont-fig} \\caption{Strictly increasing function on$[0,1]$with countably many discontinuities.\\label{fig:countdiscont}} \\end{myfigureht}", - "Similarly, one can find an example of a monotone function discontinuous on a dense set such as the rational numbers. See the exercises." - ], - "refs": [], - "proofs": [], - "ref_ids": [] - }, - { - "id": 57, - "type": "proposition", - "label": "Lebl-contfunc:prop:invcont", - "categories": [ - "continuity", - "monotonicity", - "intervals" - ], - "title": "If is an interval and is strictly monotone, then t...", - "contents": [ - "If$I \\subset \\R$is an interval and$f \\colon I \\to \\R$is strictly monotone, then the inverse$f^{-1} \\colon f(I) \\to I$is continuous." - ], - "refs": [], - "proofs": [ - { - "contents": [ - "Let us suppose $f$ is strictly increasing. The proof is almost", - "identical for a strictly decreasing function.", - "Since $f$ is strictly increasing, so is $f^{-1}$. That is, if $f(x) <", - "f(y)$, then we must have $x < y$ and therefore", - "$f^{-1}\\bigl(f(x)\\bigr) < f^{-1}\\bigl(f(y)\\bigr)$.", - "Take $c \\in f(I)$.", - "If $c$ is not a cluster point of $f(I)$, then $f^{-1}$ is continuous at $c$", - "automatically. So let $c$ be a cluster point of $f(I)$.", - "Suppose both of the following one-sided limits exist:", - "\\begin{align*}", - "x_0 & \\coloneqq \\lim_{y \\to c^-} f^{-1}(y) =", - "\\sup \\bigl\\{ f^{-1}(y) : y < c, y \\in f(I) \\bigr\\}", - "=", - "\\sup \\bigl\\{ x \\in I : f(x) < c \\bigr\\} , \\\\", - "x_1 & \\coloneqq \\lim_{y \\to c^+} f^{-1}(y) =", - "\\inf \\bigl\\{ f^{-1}(y) : y > c, y \\in f(I) \\bigr\\}", - "=", - "\\inf \\bigl\\{ x \\in I : f(x) > c \\bigr\\} .", - "\\end{align*}", - "We have $x_0 \\leq x_1$ as $f^{-1}$ is increasing.", - "For all $x \\in I$ where $x > x_0$, we have $f(x) \\geq c$. As $f$ is strictly increasing,", - "we must have $f(x) > c$ for all $x \\in I$ where $x > x_0$. Therefore,", - "\\begin{equation*}", - "\\{ x \\in I : x > x_0 \\} \\subset \\bigl\\{ x \\in I : f(x) > c \\bigr\\}.", - "\\end{equation*}", - "The infimum of the left-hand set is $x_0$, and the infimum of the right-hand", - "set is $x_1$, so we obtain $x_0 \\geq x_1$.", - "So $x_1 = x_0$, and $f^{-1}$ is continuous at $c$.", - "If one of the one-sided limits does not exist, the argument is similar", - "and is left as an exercise." - ], - "refs": [], - "ref_ids": [] - } - ], - "ref_ids": [] - }, - { - "id": 58, - "type": "example", - "label": "Lebl-contfunc:58", - "categories": [ - "example", - "continuity", - "intervals" - ], - "title": "The proposition does not require itself to be cont...", - "contents": [ - "The proposition does not require$f$itself to be continuous. Let$f \\colon \\R \\to \\R$be defined by", - "\\begin{equation*}\nf(x) \\coloneqq\n\\begin{cases}\nx & \\text{if } x < 0, \\\\\nx+1 & \\text{if } x \\geq 0. \\\\\n\\end{cases}\n\\end{equation*}", - "The function$f$is not continuous at$0$. The image of$I = \\R$is the set$(-\\infty,0)\\cup [1,\\infty)$, not an interval. Then$f^{-1} \\colon (-\\infty,0)\\cup [1,\\infty) \\to \\R$can be written as", - "\\begin{equation*}\nf^{-1}(y) =\n\\begin{cases}\ny & \\text{if } y < 0, \\\\\ny-1 & \\text{if } y \\geq 1. \n\\end{cases}\n\\end{equation*}", - "It is not difficult to see that$f^{-1}$is a continuous function. See \\figureref{invcontfig} for the graphs. \\begin{myfigureht} \\subimport*{figures/}{invcontfigAB.pdf_t} \\caption{Graph of$f$on the left and$f^{-1}$on the right.\\label{invcontfig}} \\end{myfigureht}" - ], - "refs": [], - "proofs": [], - "ref_ids": [] - } - ] - } -} \ No newline at end of file