Limits. Let \((X,\rho_X)\), \((Y,\rho_Y)\) be metric spaces. Suppose \(E \subseteq X\) and \(f:E \to Y\) and \(p\) is a limit point of \(E\). We write \[f(x) \ _{\overrightarrow{x \to p}}\ q \quad \text{ or }\quad \lim_{x \to p}f(x)=q.\] if there is a point \(q \in X\) satisfying the following \(\varepsilon\)-\(\delta\) condition:
For every \(\varepsilon>0\) there exists \(\delta>0\) such that \[\rho_Y(f(x),q)<\varepsilon\] for all points \(x \in E\) for which \(0<\rho_X(x,p)<\delta\).
If \(X=Y=\mathbb{R}\) then
\[\rho_X(x,y)=\rho_Y(x,y)=|x-y|\] and the condition reads as follows:
Limit. For every \(\varepsilon>0\) there exists \(\delta>0\) such that for all \(x \in E\) if \[0<|x-p|<\delta,\] then \[|f(x)-q|<\varepsilon.\]
Theorem (Characterizations of Continuity). Let \((X,\rho_X)\), \((Y,\rho_Y)\) be metric spaces and \(E \subseteq X\), \(f:X \to Y\), and \(p\in X\) be as in the previous definition. Then
\(\lim_{x \to p}f(x)=q\) iff
\(\lim_{n \to \infty}f(p_n)=q\) for every sequence \((p_n)_{n \in \mathbb{N}}\) in \(E\) such that \(p_n \neq p\) and \(\lim_{n \to \infty}p_n =p\).
Proof (A)\(\Longrightarrow\)(B). Suppose that (A) holds. Choose \((p_n)_{n \in \mathbb{N}}\) like in condition (B). Let \(\varepsilon>0\) be given, then there exists \(\delta>0\) such that \[\rho_Y(f(x),q)<\varepsilon \quad \text{ if }\quad x \in E \quad \text{ and }\quad 0<\rho_X(x,p)<\delta.\] Also there exists \(N \in \mathbb{N}\) such that \(n \geq N\) implies \(0<\rho_X(p_n,p)<\delta\). Thus we also have \(\rho_Y(f(p_n),q)<\varepsilon\) for \(n\ge N\) showing that (B) holds. $$\tag*{$\blacksquare$}$$
Proof (B)\(\Longrightarrow\)(A). Conversely suppose (A) is false. Then there exists some \(\varepsilon>0\) such that for every \(\delta>0\) there exists a point \(x \in E\) (depending on \(\delta\)) for which \[\rho_Y(f(x),q) \geq \varepsilon \quad \text{ but } \quad 0<\rho_X(x,p)<\delta.\] Taking \(\delta_n=\frac{1}{n}\) for each \(n \in \mathbb{N}\) we thus find a sequence \((p_n)_{n \in \mathbb{N}}\) in \(E\) satisfying \(\lim_{n \to \infty}p_n =p\) but \[\rho_Y(f(p_n),q) \geq \varepsilon.\] thus (B) is false as desired. $$\tag*{$\blacksquare$}$$
Remark. It was possible to choose the sequence \((p _n)_{n \in \mathbb{N}}\) in \(E\) in one step thanks to the Axiom of Choice. Without assuming the Axiom of Choice the previous theorem is not provable.
Theorem. Suppose that \((X,\rho_X)\) is a metric space, and \(E \subseteq X\), and \(p\) is a limit point of \(E\). Let \(f,g:E \to \mathbb{R}\) be functions such that \[\lim_{x \to p}f(x)=A\quad \text{ and }\quad \lim_{x \to p}g(x)=B.\]
Then
\(\lim_{x \to p}(f+g)(x)=A+B\),
\(\lim_{x \to p}(f\cdot g)(x)=A\cdot B\),
\(\lim_{x \to p}\left(\frac{f}{g}\right)(x)=\frac{A}{B}\) if \(B \neq 0\) and \(g(x) \neq 0\) for \(x \in E\).
Continuous at the point \(p\). Suppose that \((X,\rho_x)\) and \((Y,\rho_Y)\) are metric spaces, \(E \subseteq X\), \(p \in E\) and \(f:E \to Y\). The function \(f\) is said to be continuous at point \(p\) if for every \(\varepsilon>0\) there exists \(\delta>0\) such that
\[\rho_Y(f(x),f(p))<\varepsilon\] for all points \(x \in E\) for which \[\rho_X(x,p)<\delta.\]
Continuous function. If the function \(f:E \to Y\) is continuous at every point of \(E\) then \(f\) is said to be continuous on \(E\).
If \(X=Y=\mathbb{R}\) then
\[\rho_X(x,y)=\rho_Y(x,y)=|x-y|\] and the function \(f:E\to \mathbb{R}\) is said to be continuous at point \(p\) if for every \(\varepsilon>0\) there exists \(\delta>0\) such that
\[|f(x)-f(p)|<\varepsilon\] for all points \(x \in E\) for which \[|x-p|<\delta.\]
Example. Let us define \(f:\mathbb{R} \to \mathbb{R}\) by \[f(x)=\begin{cases} 1 \text{ if }x \in \mathbb{Q},\\ 0 \text{ if }x \not\in \mathbb{Q}. \end{cases}\] Determine if \(f\) is continuous or not at the point \(0\).
Solution. Let us consider the sequence \((a_n)_{n \in \mathbb{N}}\), where \(a_n=\sqrt{2}/n\). Then \(\lim_{n \to \infty}a_n=0\) and \(a_n \not\in \mathbb{Q}\), so \(f(a_n)=0\). Then \[\lim_{n \to \infty}f(a_n)=0 \neq 1 =f(0),\] so \(f\) is not continuous at point \(0\).$$\tag*{$\blacksquare$}$$
Example. Let us define \(f:\mathbb{R}^2 \to \mathbb{R}\) by \[f(x,y)=\begin{cases} 1 \text{ if }x+y \in \mathbb{Q},\\ 0 \text{ if }x+y \not\in \mathbb{Q}. \end{cases}\] Determine if \(f\) is continuous or not at the point \((0,0)\).
Solution. Let us consider the sequence \((a_n)_{n \in \mathbb{N}}\), where \(a_n=(0,\sqrt{2}/n)\). Then \(\lim_{n \to \infty}a_n=(0,0)\) and \(0+\sqrt{2}/n \not\in \mathbb{Q}\), so \(f(a_n)=0\). Then \[\lim_{n \to \infty}f(a_n)=0 \neq 1 =f(0,0),\] so \(f\) is not continuous at point \((0,0)\).$$\tag*{$\blacksquare$}$$
Remark. If \(p\) is an isolated point of \(E\) then our definition implies that every function \(f\) which has \(E\) as its domain is continuous at \(p\). For, no matter which \(\varepsilon>0\) we choose, we can pick \(\delta>0\) so that the only point \(e \in E\) for which \[\rho_X(x,p)<\delta\] is \(x=p\), then \[\rho_Y(f(x),f(p))=0<\varepsilon.\]
Fact. In the situation of the definition of continuity assume also that \(p\) is a limit point of \(E\). Then \(f\) is continuous at \(p\) iff \(\lim_{x \to p}f(x)=f(p).\)
Proof. It is obvious if we compare two previous definitions.$$\tag*{$\blacksquare$}$$
Theorem. Suppose that \((X,\rho_X)\), \((Y,\rho_Y)\), and \((Z,\rho_Z)\) are metric spaces, let \(E \subseteq X\) and \(f:E \to Y\) and \(g:f[E] \to Z\) be given and define \(h:E \to Z\) by \[{\color{blue}h(x)=g(f(x)), \quad x \in E.}\] If \(f\) is continuous at a point \(p \in E\) and \(g\) is continuous at the point \(f(p)\), then \(h\) is continuous at \(p\). In other words \[\lim_{x\to p}h(x)=\lim_{x\to p}g(f(x))=g(f(p))=h(p).\]
Let \(\varepsilon>0\) be given.
Since \(g\) is continuous at \(f(p)\) there is \(\eta>0\) such that \[\rho_Z(g(y),g(f(p)))<\varepsilon\quad \text{ if }\quad \rho_Y(y,f(p))<\eta \quad \text{ and }\quad y \in f[E].\]
Since \(f\) is continuous at \(p\), there is \(\delta>0\) such that \[\rho_Y(f(x),f(p))<\eta \quad \text{ if }\quad \rho_X(x,p)<\delta \quad \text{ and }\quad x \in E.\]
If follows that \[\rho_Z(h(x),h(p))=\rho_Z(g(f(x)),g(f(p)))<\varepsilon\] if \(\rho_X(x,p)<\delta\) and \(x \in E\). Thus \(h\) is continuous at \(p \in E\).$$\tag*{$\blacksquare$}$$
Example. Assume that \(f:\mathbb{R}^2 \to (0, \infty)\) is continuous for all \((x,y) \in \mathbb{R}^2\). Prove that \(h(x,y)=\sqrt{f(x,y)}\) is continuous.
Solution. Let us note that the function \(g:(0, \infty) \to (0, \infty)\) defined by \[g(x)=\sqrt{x}\] is continuous. We have \[h=g \circ f,\] so \(h\) is continuous by the previous theorem. $$\tag*{$\blacksquare$}$$
Theorem. A mapping \(f\) of a metric space \((X,\rho_X)\) into a metric space \((Y,\rho_Y)\) is continuous on \(X\) iff \(f^{-1}[V]\) is open in \(X\) for every open set \(V\) in \(Y\).
Proof. Suppose that \(f\) is continuous on \(X\) and \(V \subseteq Y\) is open.
We have to show that \(f^{-1}[V]\) is open in \(X\). Let \(p \in f^{-1}[V]\). Since \(V\) is open \(B_{\rho_Y}(f(p),\varepsilon) \subseteq V\) for some \(\varepsilon>0\).
Since \(f\) is continuous at \(p \in X\) there is \(\delta>0\) such that \[\rho_{Y}(f(x),f(p))<\varepsilon \quad \text{ if }\quad \rho_X(x,p)<\delta.\] Thus \[B_{\rho_X}(p,\delta) \subseteq f^{-1}[V]=\{x \in X\;:\;f(x) \in V\}.\]
Conversely, suppose \(f^{-1}[V]\) is open in \(X\) for any open \(V \subseteq Y\).
Fix \(p \in X\) and \(\varepsilon>0\) and consider \[{\color{blue}V=B_{\rho_Y}(f(p),\varepsilon)}\] which is open thus \(f^{-1}[V]\) is open, hence there is \(\delta>0\) so that \(B_{\rho_X}(p,\delta) \subseteq f^{-1}[V]\).
Thus if \(\rho_X(x,p)<\delta\), then \(x \in f^{-1}[V]\), hence \[f(x) \in V=B_{\rho_Y}(f(p),\varepsilon) \quad \iff \quad \rho_Y(f(x),f(p))<\varepsilon. \qquad \tag*{$\blacksquare$}\]
Corollary. A mapping \(f:X \to Y\) between metric spaces \((X,\rho_X)\) and \((Y,\rho_Y)\) is continuous iff \(f^{-1}[C]\) is closed in \(X\) for any closed set \(C\) in \(Y\).
Proof. A set is closed iff its complement is open. We are done by invoking the previous theorem, since \(f^{-1}[E^c]=(f^{-1}[E])^c\) for every open set \(E \subseteq Y\). $$\tag*{$\blacksquare$}$$
Example. Let \(f:\mathbb{R} \to \mathbb{R}\) be continuous and \(a \in R\). Prove that the set \[A=\{x \in \mathbb{R}\;:\;f(x)>a\}\] is open.
Solution: We have \[\{x \in \mathbb{R}\;:\;f(x)>a\}=f^{-1}[(a,\infty)]\] and \((a,\infty)\) is open in \(\mathbb{R}\), so by the previous theorem, \(A\) is open.$$\tag*{$\blacksquare$}$$
Example. Prove that the set \[A=\{(x,y) \in \mathbb{R}\;:\;\sqrt{x^2+y^2}<1\}\] is open in \(\mathbb{R}^2\) with the Euclidean metric.
Solution: Let us consider a continuous function \(f:\mathbb{R}^2 \to \mathbb{R}\) defined by \[f(x,y)=\sqrt{x^2+y^2}.\] Moreover, by the previous theorem \[A=\{(x,y) \in \mathbb{R}\;:\;f(x,y)<1\}=f^{-1}[B(0,1)]\] is open since \(B(0,1)\) is an open unit ball in \(\mathbb{R}^2\).$$\tag*{$\blacksquare$}$$
Theorem. Let \(f,g:X \to \mathbb{R}\) be two continuous functions on a metric space \((X,\rho_X)\). Then \(f+g\), \(f\cdot g\), and \(\frac{f}{g}\) are continuous. In the last case we assume \(g(x) \neq 0\) for all \(x \in X\).
Example 1. Every polynomial \[p(x)=a_nx^n+a_{n-1}x^{n-1}+\ldots+a_1x+a_0\] is a continuous function on \(\mathbb{R}\).
Example 2. The exponential function \(f(x)=e^x\) is continuous as we have shown that for any \((a_n)_{n \in \mathbb{N}}\) so that \(\lim_{n \to \infty}a_n=a\) one has \(\lim_{n \to \infty}e^{a_n}=e^a.\)
Example 3. \(f(x)=|x|\) is continuous on \(\mathbb{R}\) since \(|f(x)-f(y)| \leq |x-y|\).
Example 4. \(f(x)=\lfloor x\rfloor=\max\{n \in \mathbb{Z}\;:\; n \leq x\}\) is NOT continuous at any \(x \in \mathbb{Z}\).
Example 5. \(f(x)=x^{\alpha}\) for any \(\alpha \in \mathbb{R}\) is continuous on \((0,\infty)\).
Example 6. If \(f,g:X \to \mathbb{R}\) are continuous then \(\max\{f,g\}\) and \(\min\{f,g\}\) are continuous as well. Indeed,
\[\max\{f,g\}=\frac{f+g+|f-g|}{2},\qquad \min\{f,g\}=\frac{f+g-|f-g|}{2}.\]
Bounded function. A mapping \(f:E \to \mathbb{R}\) is said to be bounded if there is a number \(M>0\) such that \[|f(x)| \leq M\quad \text{ for all }\quad x \in E.\]
Theorem (4.4.1). Suppose that \(f\) is a continuous mapping of a compact metric space \((X,\rho_X)\) into a metric space \((Y,\rho_Y)\). Then \(f[X]\) is compact in \(Y\).
Let \((V_{\alpha})_{\alpha \in A}\) be an open cover of \(f[X]\), i.e. \[f[X] \subseteq \bigcup_{\alpha \in A}V_{\alpha}.\] Since \(f\) is continuous then each set \(f^{-1}[V_{\alpha}]\) is open in \(X\). Since \(X\) is compact and \[X \subseteq \bigcup_{\alpha \in A}f^{-1}[V_{\alpha}]\] thus there are \(\alpha_1,\alpha_2,\ldots,\alpha_n \in A\) so that \[X \subseteq \bigcup_{j=1}^n f^{-1}[V_{\alpha_j}].\] Since \(f[f^{-1}[E]] \subseteq E\) we have \[f[X] \subseteq f\big[\bigcup_{j=1}^{n}f^{-1}[V_{\alpha_j}]\big] \subseteq \bigcup_{j=1}^{n}V_{\alpha_j}. \qquad \tag*{$\blacksquare$}\]
Corollary. If \(f:X \to \mathbb{R}\) is continuous on a compact metric space \((X,\rho_{X})\) then \(f[X]\) is closed and bounded in \(\mathbb{R}\). Specifically, \(f\) is bounded.
Theorem. Suppose \(f:X \to \mathbb{R}\) is continuous on a compact metric space \((X,\rho_X)\) and \[M=\sup_{p \in X}f(p)\quad \text{ and }\quad m=\inf_{p \in X}f(p).\] Then there are \(p,q\) such that \[f(p)=M \quad \text{ and }\quad f(q)=m.\]
Proof. \(f[X] \subseteq \mathbb{R}\) is closed and bounded. Thus \(M\) and \(m\) are members of \(f[X]\) and we are done. $$\tag*{$\blacksquare$}$$
Theorem. Suppose \(f\) is continuous injective mapping of a compact metric space \(X\) onto a metric space \(Y\). Then the inverse mapping \(f^{-1}\) defined on \(Y\) by \[f^{-1}(f(x))=x, \ \ \ x \in X\] is a continuous mapping of \(Y\) onto \(X\).
Proof. The inverse \(f^{-1}:Y \to X\) is well defined since \(f:X \to Y\) is one-to-one and onto. It suffices to prove that \(f[V]\) is open in \(Y\) for every open set \(V\) in \(X\). Fix \(V \subseteq X\) open, \(V^c\) is closed in \(X\) thus compact, hence \(f[V^c]\) is compact subset of \(Y\) and consequently \(f[V^c]\) is closed. Since \(f:X \to Y\) is one-to-one and onto, hence \[f[V]=\left(f[V^c]\right)^c\] and, consequently, \(f[V]\) is open as desired.$$\tag*{$\blacksquare$}$$