William Keith (1838-1911) was a Scottish-American painter.
We shall consider an abstract space which extends the concept of Euclidean space \(\mathbb{R}^{n} \) by considering the distance.
Definition. Let \(M \) be a nonempty set, and let \(d:M\times M\rightarrow\mathbb{R} \) be a real-valued function defined on \(M\times M \). We say that \((M,d) \) is a metric space when the following conditions are satisfied: for any points \(x,y,z\in M \),
- \(d(x,x)=0\);
- \(d(x,y)>0 \) when \(x\neq y\);
- \(d(x,y)=d(y,x)\);
- \(d(x,y)\leq d(x,z)+d(z,y)\); this is called the triangle inequality.
In this case, the function \(d \) is called the metric of space \((M,d) \), and \(d(x,y) \) describes the distance between points \(x \) and \(y \) in \(M \). \(\sharp \).
Example. We provide many examples of metric spaces.
(i) Let \(M=\mathbb{R}^{n} \). Then, the triangle inequality is given by
\[\parallel {\bf x}-{\bf y}\parallel\leq\parallel {\bf x}-{\bf z}\parallel
+\parallel {\bf z}-{\bf y}\parallel .\]
Now, we take
\begin{equation}{\label{maeq424}}\tag{1}
d\left ({\bf x},{\bf y}\right )=\parallel {\bf x}-{\bf y}\parallel .
\end{equation}
Then, we can show that the four conditions in Definition 1 are satisfied. This says that the Euclidean space \(\mathbb{R}^{n} \) is a metric space with the metric \(d \) defined in (\ref{maeq424}).
(ii) Let \(M=\mathbb{R}^{n} \) and define \(d:\mathbb{R}^{n}\times\mathbb{R}^{n}\rightarrow\mathbb{R} \) by
\[d({\bf x},{\bf y})=\left |x_{1}-y_{1}\right |+\cdots +\left |x_{n}-y_{n}\right |.\]
Then, we can show that the four conditions in Definition 1 are satisfied.
(iii) Let \(M=\mathbb{R}^{n} \) and define \(d:\mathbb{R}^{n}\times\mathbb{R}^{n}\rightarrow\mathbb{R} \) by
\[d({\bf x},{\bf y})=\max\left\{\left |x_{1}-y_{1}\right |,\cdots ,\left |x_{n}-y_{n}\right |\right\}.\]
Then, we can show that the four conditions in Definition 1 are satisfied.
(iv) Let \(M \) be the unit circle in \(\mathbb{R}^{2} \), i.e.,
\[M=\left\{(x_{1},x_{2}):x_{1}^{2}+x_{2}^{2}=1\right\}.\]
We define \(d({\bf x},{\bf y}) \) to be the length of the smaller arc joining the two points \({\bf x} \) and \({\bf y} \) on the unit circle. Then, we can show that the four conditions in Definition 1 are satisfied.
(v) Let \(M \) be the set of all bounded real-valued functions defined on a subset \(S \) of \(\mathbb{R} \). Given any \(f,g\in M \), we define
\[d(f,g)=\sup_{x\in S}\left |f(x)-g(x)\right |.\]
Then \((M,d) \) is a metric space.
(vi) Let \(M \) be the set of all continuous real-valued functions defined on a compact interval \([a,b] \). Given any \(f,g\in M \), we define
\[d(f,g)=\max_{x\in [a,b]}\left |f(x)-g(x)\right |.\]
Then \((M,d) \) is a metric space.
(vii) Let \(M \) be the set of all continuous real-valued functions defined on a compact interval \([a,b] \). Given any \(f,g\in M \), we define
\[d(f,g)=\int_{a}^{b}\left |f(x)-g(x)\right |dx.\]
Then \((M,d) \) is a metric space.
(viii) Let \((M,d) \) be a metric space. We define
\[\rho (x,y)=\frac{d(x,y)}{1+d(x,y)}\mbox{ for }x,y\in M.\]
Then \((M,\rho) \) is also a metric space. \(\sharp\)
The definition of continuity given in elementary calculus can be extended to functions that are defined on one metric space to another metric space.
Definition. Let \((M_{1},d_{1}) \) and \((M_{2},d_{2}) \) be two metric spaces. The function \(f:(M_{1},d_{1})\rightarrow (M_{2},d_{2}) \) is said to be continuous at \(p\in M_{1} \) when, given any \(\epsilon >0 \), there exists \(\delta>0 \) such that \(d_{1}(x,p)<\delta \) implies \(d_{2}(f(x),f(p))<\epsilon \). We say that \(f \) is continuous on a subset \(S \) of \(M_{1} \) when \(f \) is continuous at each point \(p\in S \). \(\sharp\)
The above definition can also be described in terms of open balls. Let \((M_{1},d_{1}) \) and \((M_{2},d_{2}) \) be two metric spaces. We see that \(d_{1}(x,p)<\delta \) means \(x\in B_{M_{1}}(p;\delta ) \), and that \(d_{2}(f(x),f(p))<\epsilon \) means \(f(x)\in B_{M_{2}}(f(p);\epsilon ) \). Therefore, the function \(f:(M_{1},d_{1}) \rightarrow (M_{2},d_{2}) \) is said to be continuous at \(p\in M_{1} \) when, given any \(\epsilon >0 \), there exists \(\delta>0 \) satisfying
\[f(B_{M_{1}}(p;\delta ))\subseteq B_{M_{2}}(f(p);\epsilon ),\]
where \(B_{M_{1}}(p;\delta ) \) is an open ball in \(M_{1} \) and \(B_{M_{2}}(f(p);\epsilon ) \) is an open ball in \(M_{2} \).
\begin{equation}{\label{ma91}}\tag{2}\mbox{}\end{equation}
Remark \ref{ma91}. We have the following observations.
- Let \(p \) be an accumulation point of \(M_{1} \). Then, the definition of continuity implies \[\lim_{x\rightarrow p}f(x)=f(p).\]
- Let \(p \) is an isolated point of \(M_{1} \) (this is a point in \(M_{1} \) and is not an accumulation point of \(M_{1}\)). Then, every function \(f \) defined at \(p \) will be continuous at \(p \). The reason is that, for sufficiently small \(\delta >0 \), there is only one \(x \) satisfying \(d_{1}(x,p)<\delta \), i.e., \(x=p \), which implies \(d_{2}(f(p),f(p))=0 \). \(\sharp\)
Proposition. Let \((M_{1},d_{1}) \) and \((M_{2},d_{2}) \) be two metric spaces. We consider a function \(f:(M_{1},d_{1})\rightarrow (M_{2},d_{2}) \). Then \(f \) is continuous at \(p \) if and only if,
for every sequence \(\{x_{n}\}_{n=1}^{\infty} \) in \(M_{1} \) satisfying
\[\lim_{n\rightarrow\infty}x_{n}=p,\]
the sequence \(\{f(x_{n})\}_{n=1}^{\infty} \) in \(M_{2} \) satisfying
\[\lim_{n\rightarrow\infty}f(x_{n})=f(p),\]
In other words, we have
\[\lim_{n\rightarrow\infty}f(x_{n})=f\left (\lim_{n\rightarrow\infty}x_{n}\right ).\]
Proof. The proof is similar to that of Proposition 2 in page limits of functions. \(\blacksquare\)
Proposition. Let \((M_{1},d_{1}) \), \((M_{2},d_{2}) \) and \((M_{3},d_{3}) \) be three metric spaces. We consider the functions
\[f:(M_{1},d_{1})\rightarrow (M_{2},d_{2})\]
and
\[g:(f(M_{2}),d_{2})\rightarrow (M_{3},d_{3}).\]
We define the composition of \(f \) and \(g \) as \(h(x)=g(f(x)) \) for \(x\in M_{1} \). Suppose that \(f \) is continuous at \(p \) and \(g \) is continuous at \(f(p) \). Then \(h \) is continuous at \(p \).
Proof. Let \(b=f(p) \). Since \(g \) is continuous at \(p \), given any \(\epsilon >0 \), there exists \(\delta>0 \) such that \(d_{2}(y,b)<\delta \) implies \(d_{3}(g(y),g(b))<\epsilon \). By taking \(y=f(x) \), we have
\begin{equation}{\label{ma90}}\tag{3}
d_{2}(f(x),f(p))<\delta\mbox{ implies }d_{3}(g(f(x)),g(f(p)))<\epsilon.
\end{equation}
For this \(\delta \), since \(f \) is continuous at \(p \), there exists \(\delta_{0}>0 \) such that \(d_{1}(x,p)<\delta_{0} \) implies \(d_{2}(f(x),f(p))<\delta \). By referring to (\ref{ma90}), we obtain that \(d_{1}(x,p)<\delta_{0} \) implies \(d_{3}(h(x),h(p))<\epsilon \). This shows that \(h \) is continuous at \(p \), and the proof is complete. \(\blacksquare\)
Example. Given a function by
\[f(x)=\left\{\begin{array}{ll} 1 & \mbox{if \(x\in\mathbb{Q}\)}\\ 0 & \mbox{if \(x\not\in\mathbb{Q}\)}, \end{array}\right .\]
we are going to claim that \(f \) is discontinuous at any \(c\in\mathbb{R} \). Assume that \(f \) is continuous at \(c \). Given \(\epsilon =q/2 \), there exists \(\delta>0 \) such that \(|x-c|<\delta \) implies \(|f(x)-f(c)|<\frac{1}{2} \). Using the denseness of \(\mathbb{Q} \) in \(\mathbb{R} \), there exists \(r\in\mathbb{Q} \) satisfying \(|r-c|<\delta \). There also exist an irrational number \(r^{*} \) satisfying \(|r^{}-c|<\delta \). In this case, we have \(|f(r)-f(c)|<\frac{1}{2} \) and \(|f(r^{*})-f(c)|<\frac{1}{2}\); that is, we have \(|1-f(c)|<\frac{1}{2} \) and \(|0-f(c)|<\frac{1}{2} \). Therefore, we obtain
\begin{align*}1 & =\frac{1}{2}+\frac{1}{2}\\ & >|1-f(c)|+|0-f(c)|\\ & \geq |1-f(c)+0-f(c)|\\ & =1.\end{align*}
This contradiction says that \(f \) is discontinuous at \(c \). \(\sharp\)
Proposition. Let \(\hat{\bf f} \) and \({\bf g} \) be two vector-valued functions defined on \((M,d) \) taking values in \(\mathbb{R}^{n} \). Suppose that \(\hat{\bf f} \) and \({\bf g} \) are continuous at \(p\in M \). Then, the functions \(\hat{\bf f}+{\bf g} \), \(\lambda \hat{\bf f} \), the inner product \(\hat{\bf f}\bullet{\bf g} \) and \(\parallel \hat{\bf f}\parallel \) are continuous at \(p \).
Proof. The proof is similar to that of Proposition A in page limits of functions. \(\blacksquare\)
Example. Let \(f \) be Riemann-integrable on \([a,b] \) satisfying
\[\int_{a}^{b}f^{2}(x)dx=0.\]
Suppose that \(f \) is continuous at \(x_{0} \). We are going to claim \(f(x_{0})=0 \). Assume that \(f(x_{0})\neq 0 \). Then, we see that \(f^{2}(x_{0})>0 \) and \(f^{2} \) is continuous at \(x_{0} \). Given \(\epsilon=f^{2}(x_{0})/2 \), there exists \(\delta>0 \) such that \(x\in (x_{0}-\delta,x+\delta) \) implies \(|f^{2}(x)-f^{2}(x_{0})|<\epsilon \), i.e., \(f^{2}(x)-f^{2}(x_{0})>-\epsilon \). Therefore, we obtain
\begin{align*} f^{2}(x) & >f^{2}(x_{0})-\epsilon\\ & =\frac{f^{2}(x_{0})}{2},\end{align*}
which implies
\begin{align*} \int_{x_{0}-\delta}^{x_{0}+\delta}f^{2}(x) & \geq\int_{x_{0}-\delta}^{x_{0}+\delta}
\frac{f^{2}(x_{0})}{2}\\ & =\delta\cdot f^{2}(x_{0})>0.\end{align*}
Therefore, we get a contradiction
\begin{align*} 0 & =\int_{a}^{b}f^{2}(x)dx\\ & \geq\int_{x_{0}-\delta}^{x_{0}+\delta}f^{2}(x)>0.\end{align*}
This says \(f(x_{0})=0 \). \(\sharp\)
Proposition. Let \(f_{1},\cdots ,f_{n} \) be real-valued functions defined on a subset \(S \) of a metric space \((M,d) \). Then, the vector-valued function \(\hat{\bf f}=(f_{1},\cdots ,f_{n}) \) is continuous at \(p\in S \) if and only if each one of the functions \(f_{1},\cdots ,f_{n} \) is continuous at \(p \).
Proof. When \(p \) is an isolated point of \(S \), the result is obvious. Suppose that \(p \) is an accumulation point. Then, using Proposition B in page limits of functions, we see that
\[\lim_{x\rightarrow p}\hat{\bf f}(x)=\hat{\bf f}(p)\]
if and only if
\[\lim_{x\rightarrow p}f_{k}(x)=f_{k}(p)\mbox{ for each }k=1,\cdots ,n.\]
This completes the proof. \(\blacksquare\)
The continuity of a function \(f \) at a point \(p \) is a local property of \(f \), since it depends on the behavior of \(f \) only in the vicinity of \(p \). We can consider the global property which concerns the subsets of domain of \(f \). We consider the function \(f:(M_{1},d_{1})\rightarrow (M_{2},d_{2}) \). Let \(T \) is a subset of \(M_{2} \). The inverse image of \(T \) under \(f \) is denoted by \(f^{-1}(T) \) and is defined by
\[f^{-1}(T)=\left\{x\in M_{1}:f(x)\in T\right\}.\]
It is easy to see that \(f^{-1}(A)\subseteq f^{-1}(B) \) if \(A\subseteq B\subseteq M_{2} \) and
\begin{equation}{\label{ma93}}\tag{4}
f^{-1}(A\cup B)=f^{-1}(A)\cup f^{-1}(B)
\end{equation}
for all subsets \(A \) and \(B \) of \(M_{2} \).
\begin{equation}{\label{ma92}}\tag{5}\mbox{}\end{equation}
Proposition \ref{ma92}. We consider the function \(f:(M_{1},d_{1})\rightarrow (M_{2},d_{2}) \). Let \(X\subseteq M_{1} \) and \(Y\subseteq M_{2} \).
(i) \(X=f^{-1}(Y) \) implies \(f(X)\subseteq Y \).
(ii) \(Y=f(X) \) implies \(X\subseteq f^{-1}(Y) \).
(iii) We have \(f(f^{-1}(Y))\subseteq Y \) and \(X\subseteq f^{-1}(f(X)) \).
Proof. Parts (i) and (ii) can be realized by the definition of notations \(f^{-1}(Y) \) and \(f(X) \). Part (iii) follows from parts (i) and (ii) immediately. This completes the proof. \(\blacksquare\)
\begin{equation}{\label{mat437}}\tag{6}\mbox{}\end{equation}
Theorem \ref{mat437}. Let \((M_{1},d_{1}) \) and \((M_{2},d_{2}) \) be two metric spaces. We consider a function \(f:(M_{1},d_{1})\rightarrow (M_{2},d_{2}) \).
(i) \(f \) is continuous on \(M_{1} \) if and only if, for every open subset \(Y \) in \(M_{2} \), the inverse image \(f^{-1}(Y) \) is open in \(M_{1} \).
(ii) \(f \) is continuous on \(M_{1} \) if and only if, for every closed subset \(Y \) in \(M_{2} \), the inverse image \(f^{-1}(Y) \) is closed in \(M_{1} \).
Proof. To prove part (i), let \(f \) be continuous on \(M_{1} \), and let \(Y \) be open in \(M_{2} \). Given any point \(p\in f^{-1}(Y) \), we shall prove that \(p \) is an interior point of \(f^{-1}(Y) \). Let \(y=f(p) \). Since \(Y \) is open in \(M_{2} \), there exists an open ball \(B_{M_{2}}(y;\epsilon ) \) satisfying \(B_{M_{2}}(y;\epsilon )\subseteq Y \). Since \(f \) is continuous at \(p \), there exists
$\delta>0 $ satisfying \(f(B_{M_{1}}(p;\delta ))\subseteq B_{M_{2}}(y;\epsilon ) \) by Remark \ref{ma91}. Therefore, using Proposition \ref{ma92}, we have
\begin{align*} B_{M_{1}}(p;\delta ) & \subseteq f^{-1}(f(B_{M_{1}}(p;\delta )))\\ & \subseteq f^{-1}(B_{M_{2}}(y;\epsilon ))\\ & \subseteq f^{-1}(Y),\end{align*}
which says that \(p \) is an interior point of \(f^{-1}(Y) \). For the converse, we assume that \(f^{-1}(Y) \) is open in \(M_{2} \) for every open set \(Y \) in \(M_{1} \). Given any \(p\in M_{1} \),
let \(y=f(p) \). We shall prove that \(f \) is continuous at \(p \). Given any \(\epsilon >0 \), the open ball \(B_{M_{2}}(y;\epsilon ) \) is open in \(M_{2} \), which also says that \(f^{-1}(B_{M_{2}}(y;\epsilon )) \) is open in \(M_{1} \). Since \(p\in f^{-1}(B_{M_{2}}(y;\epsilon )) \), there exists \(\delta>0 \) satisfying \(B_{M_{1}}(p;\delta )\subseteq f^{-1}(B_{M_{2}}(y;\epsilon )) \). Therefore, we obtain
\begin{align*} f(B_{M_{1}}(p;\delta )) & \subseteq f(f^{-1}(B_{M_{2}}(y;\epsilon )))\\ & \subseteq B_{M_{2}}(y;\epsilon ),\end{align*}
which also says that \(f \) is continuous at \(p \) by Remark \ref{ma91} again.
To prove part (ii), if \(Y \) is closed in \(M_{2} \), then \(Y^{c} \) is open in \(M_{2} \) and
\begin{align*} f^{-1}(Y^{c}) & =f^{-1}(M_{2}\setminus Y)\\ & =M_{1}\setminus f^{-1}(Y).\end{align*}
The result follows from part (i) immediately, and the proof is complete. \(\blacksquare\)
\begin{equation}{\label{mat438}}\tag{7}\mbox{}\end{equation}
Theorem \ref{mat438}. Let \((M_{1},d_{1}) \) and \((M_{2},d_{2}) \) be two metric spaces. We define a function \(f:(M_{1},d_{1})\rightarrow (M_{2},d_{2}) \). Suppose that \(f \) is continuous on a compact subset \(X \) of \(M_{1} \). Then, the image \(f(X) \) is a compact subset of \(M_{2} \).
Proof. Let \({\cal F} \) be an open covering of \(f(X) \), i.e.,
\[f(X)\subseteq\bigcup_{A\in {\cal F}}A.\]
We shall show that a finite number of the sets \(A \) in \({\cal F} \) also cover \(f(X) \). Since \(f \) is continuous on the subspace \((X,d_{1}) \), using Theorem \ref{mat437}, we see that each inverse image \(f^{-1}(A) \) is open in \((X,d_{1}) \). Therefore, the sets \(f^{-1}(A) \) form an open covering of \(X \). Since \(X \) is compact, a finite number of them also cover \(X \), which is represented as
\[X\subseteq f^{-1}(A_{1})\cup\cdots\cup f^{-1}(A_{m}).\]
Therefore, using (\ref{ma93}) and Proposition \ref{ma92}, we obtain
\begin{align*} f(X) & \subseteq f\left (f^{-1}(A_{1})\cup\cdots\cup f^{-1}(A_{m})\right )\\ & =f(f^{-1}(A_{1}))\cup\cdots\cup f(f^{-1}(A_{m}))\\ & \subseteq A_{1}\cup\cdots\cup A_{m}, \end{align*}
which shows that \(f(X) \) is compact. This completes the proof. \(\blacksquare\)
Definition. The vector-valued function \({\bf f}:(M,d)\rightarrow\mathbb{R}^{n} \) is called bounded on \(M \) when there exist a positive number \(K \) satisfying \(\parallel {\bf f}(x)\parallel\leq K \) for all \(x\in M \). \(\sharp\)
\begin{equation}{\label{map439}}\tag{8}\mbox{}\end{equation}
Proposition \ref{map439}. We consider the vector-valued function \({\bf f}:(M,d)\rightarrow\mathbb{R}^{n} \). Suppose that \({\bf f} \) is continuous on a compact subset \(X \) of \(M \). Then \({\bf f} \) is bounded on the subspace \((X,d) \).
Proof. Since \({\bf f} \) is bounded on \(X \) if and only if \({\bf f}(X) \) is a bounded subset of \(\mathbb{R}^{n} \), the result follows from Theorem \ref{mat438} immediately. \(\blacksquare\)
Let \(f \) be a real-valued function. Suppose that \(f \) is bounded on \(X \). Then, Proposition \ref{map439} says that \(f(X) \) is a bounded subset of \(\mathbb{R} \). Therefore, we have
\begin{align*} \inf f(X) & =\inf_{x\in X}f(x)\\ & \leq f(x)\\ & \leq\sup_{x\in X}f(x)\\ & =\sup f(X)\end{align*}
for each \(x\in X \). The next result shows that a continuous function \(f \) can take values \(\inf f(X) \) and \(\sup f(X) \) when \(X \) is compact.
\begin{equation}{\label{mat9}}\tag{9}\mbox{}\end{equation}
Theorem \ref{mat9}. Let \(f \) be a real-valued function defined on a metric space \((M,d) \). Suppose that \(f \) is continuous on a compact subset \(X \) of \((M,d) \). Then, there exist \(p,q\in X \) satisfying
\[f(p)=\inf_{x\in X}f(x)=\min_{x\in X}f(x)\]
and
\[f(q)=\sup_{x\in X}f(x)=\max_{x\in X}f(x).\]
Proof. Using Theorem 5 in page point set topology in metric space and Theorem \ref{mat438}, we see that \(f(X) \) is a closed and bounded subset of \(\mathbb{R} \). Let \(m=\sup f(X) \). Then \(m \) is an adherent point of bounded set \(f(X) \). Since \(f(X) \) is closed, it follows \(m\in f(X) \), which says \(m=f(q) \) for some \(q\in X \). We can similarly show \(f(p)=\inf f(X) \) for some \(p\in X \). This completes the proof. \(\blacksquare\)
Proposition A. Let \((M_{1},d_{1}) \) and \((M_{2},d_{2}) \) be two metric spaces. We consider a function \(f:(M_{1},d_{1})\rightarrow (M_{2},d_{2}) \) such that \(f \) is one-to-one on \(M_{1} \), i.e., the inverse function \(f^{-1} \) exists. Suppose that \(M_{1} \) is compact, and that \(f \) is continuous on \(M_{1} \). Then \(f^{-1} \) is continuous on \(f(M_{1}) \).
Proof. By considering \(f^{-1} \), we see that \(f(X) \) is the inverse image of \(X \) under \(f^{-1} \). Using part (ii) of Theorem \ref{mat437}, we only need to show that \(f(X) \) is closed in \(M_{2} \) for every closed set \(X \) in \(M_{1} \). Since \(X \) is closed and \(M_{1} \) is compact, Theorem A in page point set topology in metric space says that \(X \) is compact. Therefore, \(f(X) \) is compact by Theorem \ref{mat438}. Finally, using Theorem 5 in page point set topology in metric space, we conclude that \(f(X) \) is closed. This completes the proof. \(\blacksquare\)
\begin{equation}{\label{map441}}\tag{9a}\mbox{}\end{equation}
Proposition \ref{map441}. Let \(f \) be a real-valued function defined on an interval \(I \) in \(\mathbb{R} \). Suppose that \(f \) is continuous at \(c\in I \) with \(f(c)\neq 0 \). Then, there exists a one-dimensional open ball \(B(c;\delta ) \) such that \(f(x) \) has the same sign as \(f(c) \) in \(B(c;\delta )\cap I \).
Proof. We first note that the one-dimensional open ball \(B(c;\delta ) \) is given by
\[B(c;\delta )=\left\{x:|x-c|<\delta\right\}.\]
Suppose that \(f(c)>0 \). The continuity at \(c \) says that, given any \(\epsilon >0 \), there exists \(\delta>0 \) such that \(x\in B(c;\delta )\cap I \) implies
\[f(c)-\epsilon <f(x)<f(c)+\epsilon .\]
By taking \(\epsilon =f(c)/2>0 \), it follows that \(x\in B(c;\delta )\cap I \) implies
\[\frac{1}{2}f(c)<f(x)<\frac{3}{2}f(c),\]
which says that \(f(x) \) has the same sign as \(f(c) \) in \(B(c;\delta )\cap I \). For the case of \(f(c)<0 \), we take \(\epsilon =-f(c)/2 \). This completes the proof. \(\blacksquare\)
\begin{equation}{\label{mat442}}\tag{10}\mbox{}\end{equation}
Theorem \ref{mat442}. (Bolzano). Let \(f \) be a real-valued function defined on a closed interval \([a,b] \) in \(\mathbb{R} \). Suppose that \(f \) is continuous on \([a,b] \) and \(f(a)\cdot f(b)<0 \). Then, there exists at least one point \(c\in (a,b) \) satisfying \(f(c)=0 \).
Proof. We assume that \(f(a)>0 \) and \(f(b)<0 \). Let
\[A=\left\{x:x\in [a,b]\mbox{ and }f(x)>0\right\}.\]
Then \(A\neq\emptyset \), since \(a\in A \) with \(f(a)>0 \). We also see that \(A \) is bounded above by \(b \). Let \(c=\sup A \). We shall prove that \(f(c)=0 \). Suppose that \(f(c)\neq 0 \). From Proposition \ref{map441}, there exists an one-dimensional open ball \(B(c;\delta ) \) such that \(f(x) \) has the same sign as \(f(c) \) on \(B(c;\delta )\cap [a,b] \).
- Suppose that \(f(c)>0 \). There exists a point \(x>c \) satisfying \(f(x)>0 \), i.e., \(x\in A \), which contradicts \(c=\sup A \).
- Suppose that \(f(c)<0 \), Then, we see that \(f(c-\delta /2)<0 \), i.e., \(c-\delta /2\not\in A \). This says that \(c-\delta /2 \) is an upper bound of \(A \), which again contradicts \(c=\sup A \).
Therefore, we must have \(f(c)=0 \), and the proof is complete. \(\blacksquare\)
\begin{equation}{\label{mat8}}\tag{11}\mbox{}\end{equation}
Theorem \ref{mat8}. (Intermediate-Value Theorem). Let \(f \) be a real-valued and continuous function defined on a closed interval \(I \) in \(\mathbb{R} \). Suppose that \(\alpha <\beta \) in \(I \) satisfying \(f(\alpha )\neq f(\beta ) \). Then \(f \) takes every value between \(f(\alpha ) \) and \(f(\beta ) \) in the interval \((\alpha ,\beta ) \).
Proof. Let \(k \) be a number between \(f(\alpha ) \) and \(f(\beta ) \). We define \(g(x)=f(x)-k \) on \([\alpha ,\beta ] \). Then, we see that \(f(\alpha)\cdot f(\beta)<0 \). Using Bolzano Theorem \ref{mat442} to the function \(g \), there exists \(\gamma\in (\alpha ,\beta ) \) satisfying \(g(\gamma)=0 \), i.e., \(f(\gamma)=k \). This completes the proof. \(\blacksquare\)
Theorems \ref{mat9} and \ref{mat8} say that the continuous image of a closed interval \(I \) under a real-valued function \(f \) is also a closed interval given by
\[\left [\inf_{x\in I}f(x),\sup_{x\in I}f(x)\right ]=\left [\min_{x\in I}f(x),\max_{x\in I}f(x)\right ].\]
Given two metric spaces \((M_{1},d_{1}) \) and \((M_{2},d_{2}) \), we consider the function \(f:(M_{1},d_{1})\rightarrow (M_{2},d_{2}) \). Suppose that \(f \) is continuous on a subset \(S \) of \(M_{1} \). Given any point \(p\in S \) and any \(\epsilon >0 \), there exists \(\delta >0 \) (which depends on \(p \) and \(\epsilon\)) such that \(d_{1}(x,p)<\delta \) implies \(d_{2}(f(x),f(p) <\epsilon \). We expect that this \(\delta \) is independent of \(p \), which proposes the concept of uniform continuity.
Definition. We consider the function \(f:(M_{1},d_{1})\rightarrow (M_{2},d_{2}) \). Then \(f \) is said to be uniformly continuous on a subset \(S \) of \(M_{1} \) when, given any \(\epsilon >0 \), there exists \(\delta >0 \) (which depends only on \(\epsilon\)) such that, given any \(x,p\in S \), \(d_{1}(x,p)<\delta \) implies \(d_{2}(f(x),f(p))<\epsilon \). \(\sharp\)
Example. The difference of continuity and uniform continuity can be clear by the following examples.
(i) Let \(f(x)=1/x \) for \(x>0 \) and take \(M=(0,1] \). It is easy to see that \(f \) is continuous on \(M \). However, we are going to claim that \(f \) is not uniformly continuous on \(M \). Let \(\epsilon =10 \). Assume that there exists \(\delta>0 \) such that, given any \(x,p\in M \), \(|x-p|<\delta \) implies \(|f(x)-f(p)|<\epsilon \). Given \(\bar{\delta}<\min{1,\delta} \), we take \(x=\bar{\delta} \) and \(p=\bar{\delta}/11 \). Then, we obtain \(|x-p|<\bar{\delta}<\delta \). It follows \(|f(x)-f(p)|<\epsilon =10 \). However, we have
\begin{align*} |f(x)-f(p)| & =\frac{11}{\bar{\delta}}-\frac{1}{\bar{\delta}}\\ & =\frac{10}{\bar{\delta}}\\ & >10=\epsilon,\end{align*}
which contradicts the definition of uniform continuity.
(ii) Let \(f(x)=x^{2} \) for \(x\in\mathbb{R} \) and take \(M=(0,1] \). We are going to claim that this function \(f \) is uniformly continuous on \(M \). Since
\begin{align*} |f(x)-f(p)| & =|x^{2}-p^{2}|\\ & =|(x-p)(x+p)|<2|x-p|,\end{align*}
if \(|x-p|<\delta \), then \(|f(x)-f(p)|<2\delta \). Therefore, given any \(\epsilon >0 \), we can take \(\delta =\epsilon /2 \) to obtain \(|f(x)-f(p)|<\epsilon \) for any points \(x \) and \(p \) in \(M \) with \(|x-p|<\delta \). This shows that \(f \) is uniformly continuous on \(M \). \(\sharp\)
Proposition. Suppose that two real-valued functions \(f_{1},f_{2}:(M,d)\rightarrow\mathbb{R} \) are uniformly continuous on a subset \(S \) of \(M \). Then \(f_{1}-f_{2} \) is also uniformly continuous on \(S \).
Proof. Given any \(\epsilon>0 \), there exist \(\delta_{1},\delta_{2}>0 \) such that \(d(x_{1},x_{2})<\delta_{1} \) implies \(|f_{1}(x_{1})-f_{1}(x_{2})|<\frac{\epsilon}{2} \), and that \(d(x_{1},x_{2})<\delta_{2} \) implies \(|f_{2}(x_{1})-f_{2}(x_{2})|<\frac{\epsilon}{2} \) for \(x_{1},x_{2}\in S \). Let \(\delta=\min{\delta_{1},\delta_{2}} \), and let \(f=f_{1}-f_{2} \). Then, for \(d(x_{1},x_{2})<\delta \), we have
\begin{align*} & \left |f(x_{1})-f(x_{2})\right |\\ & \quad =\left |f_{1}(x_{1})-f_{2}(x_{1})-f_{1}(x_{2})+f_{2}(x_{2})\right |\\ & \quad\leq\left |f_{1}(x_{1})-f_{1}(x_{2})\right |+\left |f_{2}(x_{2})-f_{2}(x_{1})\right |\\ & \quad<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon. \end{align*}
This completes the proof. \(\blacksquare\)
Suppose that \(f:(M_{1},d_{1})\rightarrow (M_{2},d_{2}) \) is uniformly continuous on a subset \(S \) of \(M_{1} \). Then, it is also continuous on \(S \). We are going to study the converse.
\begin{equation}{\label{ma104}}\tag{12}\mbox{}\end{equation}
Theorem \ref{ma104}. (Heine). Suppose that the function \(f:(M_{1},d_{1})\rightarrow (M_{2},d_{2}) \) is continuous on a compact subset \(S \) of \(M_{1} \). Then \(f \) is uniformly continuous on \(S \).
Proof. Given any \(\epsilon >0 \), for each point \(a\in S \), there exists open ball \(B_{M_{1}}(a;r) \) for some \(r>0 \) such that \(x\in B_{M_{1}}(a;r)\cap S \) implies \(d_{2}\left (f(x),f(a)\right )<\frac{\epsilon}{2} \). We consider the collection \({\cal F} \) of all open balls \(B_{M_{1}}(a;r_{a}) \) for \(a\in S \). Then \({\cal F} \) covers \(S \). Since \(S \) is compact, a finite number of them also cover \(S \), i.e.,
\begin{equation}{\label{maeq446}}\tag{13}
S\subseteq\bigcup_{k=1}^{m}B_{M_{1}}\left (a_{k};r_{k}\right ).
\end{equation}
We have that \(x\in B_{M_{1}}(a_{k};r_{k})\cap S \) implies
\begin{equation}{\label{maeq447}}\tag{14}
d_{2}\left (f(x),f(a_{k})\right )<\frac{\epsilon}{2}.
\end{equation}
Let
\[\delta=\min\left\{\frac{r_{1}}{2},\cdots ,\frac{r_{m}}{2}\right\}.\]
We consider any points \(x,p\in S \) satisfying \(d_{M_{1}}(x,p)<\delta \). From (\ref{maeq446}), we see that \(x\in B_{M_{1}}(a_{k};r_{k}) \) for some \(k \). Therefore, from (\ref{maeq447}), we also have
\[d_{2}\left (f(x),f(a_{k})\right )<\frac{\epsilon}{2}.\]
using the triangle inequality, we obtain
\begin{align} d_{1}(p,a_{k}) & \leq d_{1}(p,x)+d_{1}(x,a_{k})\nonumber\\ & <\delta +\frac{r_{k}}{2}\nonumber\\ & \leq\frac{r_{k}}{2}+\frac{r_{k}}{2}\nonumber\\ & =r_{k},\label{maeq448}\tag{15}\end{align}
which says \(p\in B_{M_{1}}(a_{k};r_{k})\cap S \). Therefore, we also obtain
\begin{equation}{\label{maeq449}}\tag{16}
d_{2}\left (f(p),f(a_{k})\right )<\frac{\epsilon}{2}.
\end{equation}
From (\ref{maeq448}) and (\ref{maeq449}), the triangle inequality says
\begin{align*} & d_{2}\left (f(x),f(p)\right )\\ & \quad\leq d_{2}\left (f(x),f(a_{k})\right )+
d_{2}\left (f(a_{k}),f(p)\right )\\ & \quad<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon .\end{align*}
This completes the proof. \(\blacksquare\)
Definition. Let \(f:(M,d)\rightarrow (M,d) \) be a function from a metric space \((M,d) \) into itself.
- A point \(p\in M \) is called a fixed point of \(f \) when \(f(p)=p \).
- The function \(f \) is called a contraction of \(M \) when there is a positive number \(\alpha <1 \) (which is called a contraction constant) satisfying \begin{equation}{\label{maeq450}}\tag{17} d(f(x),f(y))\leq\alpha d(x,y)\end{equation} for all \(x,y\in M \). \(\sharp\)
\begin{equation}{\label{ma96}}\tag{18}\mbox{}\end{equation}
Remark \ref{ma96}. It is clear to see that if \(f \) is a contraction of \(M \), then it is uniformly continuous on \(M \). \(\sharp\)
Theorem. (Fixed Point Theorem) Let \((M,d) \) be a complete metric space. Suppose that the function \(f:(M,d)\rightarrow (M,d) \) is a contraction of \(M \). Then \(f \) has a unique fixed point.
Proof. Given any point \(x\in M \), we consider the sequence of iterations given by
\[\{x,f(x),f(f(x)),\cdots\}.\]
This defines a sequence \(\{p_{n}\}_{n=1}^{\infty} \) inductively given by \(p_{0}=x \) and \(p_{n+1}=f(p_{n}) \) for \(n=0,1,2,\cdots \). We shall prove that this sequence \(\{p_{n}\}_{n=1}^{\infty} \) converges to a fixed point of \(f \). We first show that \(\{p_{n}\}_{n=1}^{\infty} \) is a Cauchy sequence. From (\ref{maeq450}), we have
\begin{align*} d\left (p_{n+1},p_{n}\right ) & =d\left (f(p_{n}),f(p_{n-1})\right )\\ & \leq\alpha d\left (p_{n},p_{n-1}\right ).\end{align*}
By induction on \(n \), we obtain
\begin{equation}{\label{maeq451}}\tag{19}
d\left (p_{n+1},p_{n}\right )\leq\alpha^{n}d\left (p_{1},p_{0}\right ).
\end{equation}
For \(m>n \), using the triangle inequality and (\ref{maeq451}), we have
\begin{align*} d\left (p_{m},p_{n}\right ) & \leq\sum_{k=n}^{m-1}d\left (p_{k+1},p_{k}\right )\\ & \leq d\left (p_{1},p_{0}\right )\cdot\sum_{k=n}^{m-1}\alpha^{k}\\ & =d\left (p_{1},p_{0}\right )\cdot\frac{\alpha^{n}(1-\alpha^{m-n})}{1-\alpha}\\ & <\frac{d\left (p_{1},p_{0}\right )\cdot\alpha^{n}}{1-\alpha}. \end{align*}
Since \(0<\alpha<1 \), we have
\[\lim_{n\rightarrow\infty}\alpha^{n}=0,\]
which implies
\[\lim_{n\rightarrow\infty}d(p_{m},p_{n})=0.\]
This shows that \({p_{n}}_{n=1}^{\infty} \) is a Cauchy sequence. By the completeness, there exists a point \(p\in M \) satisfying
\[\lim_{n\rightarrow\infty}p_{n}=p.\]
Remark \ref{ma96} says that \(f \) is continuous. Therefore, we obtain
\begin{align*} f(p) & =f\left (\lim_{n\rightarrow\infty}p_{n}\right )\\ & =\lim_{n\rightarrow\infty}f(p_{n})\\ & =\lim_{n\rightarrow\infty}p_{n+1}=p.\end{align*}
To prove the uniqueness, suppose that \(p \) and \(q \) are two distinct fixed points. Using (\ref{maeq450}), we have
\begin{align*} d(p,q) & =d(f(p),f(q))\\ & \leq\alpha d(p,q)\\ & <d(p,q).\end{align*}
This contradiction shows the uniqueness, and the proof is complete. \(\blacksquare\)
