8.1: Introduction to Metric Spaces

If \(n\ge2\) and \(u_1\), \(u_2\), , \(u_n\) are arbitrary members of \(A\), then

and induction yield the inequality \[ \rho(u_1,u_n)\le \sum_{i=1}^{n-1}\rho(u_i,u_{i+1}). \nonumber \]

Motivated by this example, in an arbitrary metric space we call \(\rho(u,v)\) {}, and we call Definition~

{}.

Example~ shows that it is possible to define a metric on any nonempty set \(A\). In fact, it is possible to define infinitely many metrics on any set with more than one member (Exercise~). Therefore, to specify a metric space completely, we must specify the couple \((A,\rho)\), where \(A\) is the set and \(\rho\) is the metric. (In some cases we will not be so precise; for example, we will always refer to the real numbers with the metric \(\rho(u,v)=|u-v|\) simply as \(\R\).)

There is an important kind of metric space that arises when a definition of length is imposed on a vector space. Although we assume that you are familiar with the definition of a vector space, we restate it here for convenience. We confine the definition to vector spaces over the real numbers.

We say that \(A\) is {} if (1) is true, and that \(A\) is {} if (6) is true. It can be shown that if \(B\) is any nonempty subset of \(A\) that is closed under vector addition and scalar multiplication, then \(B\) together with these operations is itself a vector space. (See any linear algebra text for the proof.) We say that \(B\) is a {} of \(A\).

From with \(u=x-y\), \(\rho(x,y)=N(x-y)\ge0\), with equality if and only if \(x=y\). From with \(u=x-y\) and \(a=-1\),

\[ \rho(y,x)=N(y-x)=N(-(x-y))=N(x-y)=\rho(x,y). \nonumber \] From

with \(u=x-z\) and \(v=z-y\), \[ \rho(x,y)=N(x-y)\le N(x-z)+N(z-y)=\rho(x,z)+\rho(z,y). \nonumber \] -2em2em

We will say that the metric in is {} \(N\). Whenever we speak of a normed vector space \((A,N)\), it is to be understood that we are regarding it as a metric space \((A,\rho)\), where \(\rho\) is the metric induced by \(N\).

We will often write \(N(u)\) as \(\|u\|\). In this case we will denote the normed vector space as \((A,\|\cdot\|)\).

Since \[ x=y+(x-y), \nonumber \] Definition~ with \(u=y\) and \(v=x-y\) implies that

\[ N(x)\le N(y)+N(x-y), \nonumber \] or \[ N(x)-N(y)\le N(x-y). \nonumber \] Interchanging \(x\) and \(y\) yields \[ N(y)-N(x)\le N(y-x). \nonumber \] Since \(N(x-y)=N(y-x)\) (Definition~

with \(u=x-y\) and \(a=-1\)), the last two inequalities imply .

In Section~5.1 we defined the norm of a vector \(\mathbf{X}=(x_1,x_2, \dots,x_n)\) in \(\R^n\) as \[ \|\mathbf{X}\|=\left(\sum_{i=1}^nx_i^2\right)^{1/2}. \nonumber \]

The metric induced by this norm is

\[ \rho(\mathbf{X},\mathbf{Y})=\left(\sum_{i=1}^n(x_i-y_i)^2 \right)^{1/2}. \nonumber \] Whenever we write \(\R^n\) without identifying the norm or metric specifically, we are referring to \(\R^n\) with this norm and this induced metric.

The following definition provides infinitely many norms and metrics on \(\R^n\).

To justify this definition, we must verify that actually defines a norm. Since it is clear that \(\|\mathbf{X}\|_p\ge0\) with equality if and only if \(\mathbf{X}=\mathbf{0}\), and \(\|a\mathbf{X}\|_p=|a|\|\mathbf{X}\|_p\) if \(a\) is any real number and \(\mathbf{X}\in \R^n\), this reduces to showing that

\[\begin{equation} \label{eq:8.1.4} \|\mathbf{X}+\mathbf{Y}\|_p\le\|\mathbf{X}\|_p+\|\mathbf{Y}\|_p \end{equation} \]

for every \(\mathbf{X}\) and \(\mathbf{Y}\) in \(\R^n\). Since \[ |x_i+y_i|\le |x_i|+|y_i|, \nonumber \] summing both sides of this equation from \(i=1\) to \(n\) yields with \(p=1\). To handle the case where \(p>1\), we need the following lemmas. The inequality established in the first lemma is known as {}.

Let \(\alpha\) and \(\beta\) be any two positive numbers, and consider the function \[ f(\beta)=\frac{\alpha^p}{p}+\frac{\beta^q}{q}-\alpha\beta, \nonumber \]

where we regard \(\alpha\) as a constant. Since \(f'(\beta)=\beta^{q-1}-\alpha\) and \(f''(\beta)=(q-1)\beta^{q-2}>0\) for \(\beta>0\), \(f\) assumes its minimum value on \([0,\infty)\) at \(\beta=\alpha^{1/(q-1)}=\alpha^{p-1}\). But \[ f(\alpha^{p-1})=\frac{\alpha^p}{p}+\frac{\alpha^{(p-1)q}}{q}-\alpha^p =\alpha^p\left(\frac{1}{p}+\frac{1}{q}-1\right)=0. \nonumber \] Therefore, \[\begin{equation} \label{eq:8.1.7} \alpha\beta\le \frac{\alpha^p}{p}+\frac{\beta^q}{q}\mbox{\quad if \quad} \alpha, \beta\ge0. \end{equation} \] Now let \[ \alpha_i=\mu_i\left(\sum_{j=1}^n \mu_j^p\right)^{-1/p} \mbox{\quad and \quad} \beta_i=\nu_i\left(\sum_{j=1}^n \nu_j^q\right)^{-1/q}. \nonumber \] From , \[ \alpha_i\beta_i\le\frac{\mu_i^p}{p}\left(\sum_{j=1}^n \mu_j^p\right)^{-1} +\frac{\nu_i^q}{q}\left(\sum_{j=1}^n \nu_j^q\right)^{-1}. \nonumber \] From , summing this from \(i=1\) to \(n\) yields \(\sum_{i=1}^n \alpha_i\beta_i\le1\), which implies .

Again, let \(q=p/(p-1)\). We write \[\begin{equation} \label{eq:8.1.9} \sum_{i=1}^n(u_i+v_i)^p=\sum_{i=1}^n u_i(u_i+v_i)^{p-1} +\sum_{i=1}^n v_i(u_i+v_i)^{p-1}. \end{equation} \] From H"older’s inequality with \(\mu_i=u_i\) and \(\nu_i=(u_i+v_i)^{p-1}\), \[\begin{equation} \label{eq:8.1.10} \sum_{i=1}^n u_i(u_i+v_i)^{p-1}\le \left(\sum_{i=1}^n u_i^p\right)^{1/p} \left(\sum_{i=1}^n(u_i+v_i)^p\right)^{1/q}, \end{equation} \] since \(q(p-1)=p\). Similarly, \[ \sum_{i=1}^n v_i(u_i+v_i)^{p-1}\le \left(\sum_{i=1}^n v_i^p\right)^{1/p} \left(\sum_{i=1}^n(u_i+v_i)^p\right)^{1/q}. \nonumber \] This, , and imply that \[ \sum_{i=1}^n(u_i+v_i)^p \le\left[\left(\sum_{i=1}^n u_i^p\right)^{1/p} +\left(\sum_{i=1}^n v_i^p\right)^{1/p}\right] \left(\sum_{i=1}^n(u_i+v_i)^p\right)^{1/q}. \nonumber \]

Since \(1-1/q=1/p\), this implies , which is known as {}.

We leave it to you to verify that Minkowski’s inequality implies if \(p>1\).

We now define the \(\infty\)-{} on \(\R^n\) by \[\begin{equation} \label{eq:8.1.11} \|\mathbf{X}\|_\infty=\max\set{|x_i|}{1\le i\le n}. \end{equation} \] We leave it to you to verify (Exercise~) that \(\|\cdot\|_\infty\) is a norm on \(\R^n\). The associated metric is \[ \rho_\infty(\mathbf{X},\mathbf{Y})=\max\set{|x_i-y_i|}{1\le i\le n}. \nonumber \]

The following theorem justifies the notation in .

Let \(u_1\), \(u_2\), , \(u_n\) be nonnegative and \(M=\max\set{u_i}{1\le i\le n}\). Define \[ \sigma(p)=\left(\sum_{i=1}^n u_i^p\right)^{1/p}. \nonumber \] Since \(u_i/\sigma(p)\le1\) and \(p_2>p_1\), \[ \left(\frac{u_i}{\sigma(p_2)}\right)^{p_1}\ge \left(\frac{u_i}{\sigma(p_2)}\right)^{p_2}; \nonumber \] therefore, \[ \frac{\sigma(p_1)}{\sigma(p_2)} =\left(\sum_{i=1}^n\left(\frac{ u_i}{\sigma(p_2)}\right)^{p_1}\right)^{1/p_1} \ge\left(\sum_{i=1}^n\left(\frac{ u_i}{\sigma(p_2)}\right)^{p_2}\right)^{1/p_1}=1, \nonumber \] so \(\sigma(p_1)\ge\sigma(p_2)\). Since \(M\le\sigma(p)\le Mn^{1/p}\), \(\lim_{p\to\infty}\sigma(p)= M\). Letting \(u_i=|x_i|\) yields and .

Since Minkowski’s inequality is false if \(p<1\) (Exercise~), is not a norm in this case. However, if \(0<p<1\), then \[ \|\mathbf{X}\|_p=\sum_{i=1}^n|x_i|^p \nonumber \] is a norm on \(\R^n\) (Exercise~).

In this section and in the exercises we will consider subsets of the vector space \(\R^\infty\) consisting of sequences \(\mathbf{X}=\{x_i\}_{i=1}^\infty\), with vector addition and scalar multiplication defined by \[ \mathbf{X}+\mathbf{Y}=\{x_i+y_i\}_{i=1}^\infty \mbox{\quad and \quad} r\mathbf{X}=\{rx_i\}_{i=1}^\infty. \nonumber \]

The metric induced by \(\|\cdot\|_p\) is \[ \rho_p(\mathbf{X},\mathbf{Y})=\left(\sum_{i=1}^\infty|x_i-y_i|^p\right)^{1/p}. \nonumber \] Henceforth, we will denote \((\ell_p,\|\cdot\|_p)\) simply by \(\ell_p\).

The metric induced by \(\|\cdot\|_\infty\) is \[ \rho_\infty(\mathbf{X},\mathbf{Y})=\sup\set{|x_i-y_i|}{i\ge 1}. \nonumber \] Henceforth, we will denote \((\ell_\infty,\|\cdot\|_\infty)\) simply by \(\ell_\infty\).

At this point you may want to review Definition~ and Exercises~ and , which apply equally well to subsets of a metric space \((A,\rho)\).

We will now state some definitions and theorems for a general metric space \((A,\rho)\) that are analogous to definitions and theorems presented in Section~1.3 for the real numbers. To avoid repetition, it is to be understood in all these definitions that we are discussing a given metric space \((A,\rho)\).

We must show that if \(u_1\in S_r(u_0)\), then there is an \(\epsilon>0\) such that \[\begin{equation} \label{eq:8.1.15} S_\epsilon(u_1)\subset S_r(u_0). \end{equation} \] If \(u_1\in S_r(u_0)\), then \(\rho(u_1,u_0)<r\). Since \[ \rho(u,u_0)\le\rho(u,u_1)+\rho(u_1,u_0) \nonumber \] for any \(u\) in \(A\), \(\rho(u,u_0)<r\) if \(\rho(u,u_1)<r-\rho(u_1,u_0)\). Therefore, holds if \(\epsilon<r-\rho(u_1,u_0)\).

The entire space \(A\) is open and therefore \(\emptyset\,(=A^c)\) is closed. However, \(\emptyset\) is also open, for to deny this is to say that it contains a point that is not an interior point, which is absurd because \(\emptyset\) contains no points. Since \(\emptyset\) is open, \(A\, (=\emptyset^c)\) is closed. If \(A=\R\), these are the only sets that are both open and closed, but this is not so in all metric spaces. For example, if \(\rho\) is the discrete metric, then every subset of \(A\) is both open and closed. (Verify!)

A {} of a point \(u_0\) is a set that contains every point of some neighborhood of \(u_0\) except \(u_0\) itself. (If \(\rho\) is the discrete metric then the empty set is a deleted neighborhood of every member of \(A\)!)

The proof of the following theorem is identical to the proof Theorem~.

Although this definition is identical to Definition~, you should not assume that conclusions valid for the real numbers are necessarily valid in all metric spaces. For example, if \(A=\R\) and \(\rho(u,v)=|u-v|\), then \[ \overline S_r(u_0)=\set{u}{\rho(u,u_0)\le r}. \nonumber \] This is not true in every metric space (Exercise~).

For the proof of the following theorem, see the proofs of Theorem~ and Corollary~.

Since metric spaces are not ordered, concepts and results concerning the real numbers that depend on order for their definitions must be redefined and reexamined in the context of metric spaces. The first example of this kind is completeness. To discuss this concept, we begin by defining an {} (more briefly, a {}) in a metric space \((A,\rho)\) as a function defined on the integers \(n\ge k\) with values in \(A\). As we did for real sequences, we denote a sequence in \(A\) by, for example, \(\{u_n\}=\{u_n\}_{n=k}^\infty\). A subsequence of a sequence in \(A\) is defined in exactly the same way as a subsequence of a sequence of real numbers (Definition~).

We leave the proof of the following theorem to you. (See the proofs of Theorems~ and .)

-2em2em

We note that if \(\rho\) is the metric induced by a norm \(\|\cdot\|\) on \(A\), then and can be replaced by \[ \lim_{n\to\infty}\|u_n-u\|=0 \nonumber \] and \[ \|u_n-u_m\|<\epsilon\mbox{\quad and \quad}m,n>N, \nonumber \] respectively.

Suppose that \(\lim_{n\to\infty}u_n=u\). If \(\epsilon>0\), there is an integer \(N\) such that \(\rho(u_n,u)<\epsilon/2\) if \(n>N\). Therefore, if \(m\), \(n>N\), then \[ \rho(u_n,u_m)\le\rho(u_n,u)+\rho(u,u_m)<\epsilon. \nonumber \] -2em2em

This example raises a question that we should resolve before going further. In Section~1.1 we defined completeness to mean that the real numbers have the following property:

Axiom {}. Every nonempty set of real numbers that is bounded above has a supremum.

Here we are saying that the real numbers are complete because every Cauchy sequence of real numbers has a limit. We will now show that these two usages of ``complete’’ are consistent.

The proof of Theorem~ requires the existence of the (finite) limits inferior and superior of a bounded sequence of real numbers, a consequence of Axiom~{}. However, the assertion in Axiom~{} can be deduced as a theorem if Axiom~{} is replaced by the assumption that every Cauchy sequence of real numbers has a limit. To see this, let \(T\) be a nonempty set of real numbers that is bounded above. We first show that there are sequences \(\{u_i\}_{i=1}^\infty\) and \(\{v_i\}_{i=1}^\infty\) with the following properties for all \(i\ge 1\):

Since \(T\) is nonempty and bounded above, \(u_1\) and \(v_1\) can be chosen to satisfy with \(i=1\). Clearly, holds with \(i=1\). Let \(w_1=(u_1+v_1)/2\), and let \[ (u_2,v_2)= \left\{\casespace\begin{array}{ll} (w_1,v_1)&\mbox{ if $w_1\le t$ for some $t\in T$},\\ (u_1,w_1)&\mbox{ if $w_1\ge t$ for all $t\in T$}. \end{array}\right. \nonumber \] In either case, and hold with \(i=2\) and holds with \(i=1\). Now suppose that \(n>1\) and \(\{u_1, \dots,u_n\}\) and \(\{v_1, \dots, v_n\}\) have been chosen so that and hold for \(1\le i\le n\) and holds for \(1\le i\le n-1\). Let \(w_n=(u_n+v_n)/2\) and let \[ (u_{n+1},v_{n+1})= \left\{\casespace\begin{array}{ll} (w_n,v_n)&\mbox{ if $w_n\le t$ for some $t\in T$},\\ (u_n,w_n)&\mbox{ if $w_n\ge t$ for all $t\in T$}. \end{array}\right. \nonumber \] Then and hold for \(1\le i\le n+1\) and

holds for \(1\le i\le n\). This completes the induction.

Now and imply that \[ 0\le u_{i+1}-u_i\le2^{i-1}(v_1-u_1) \mbox{\quad and \quad} 0\le v_i-v_{i+1}\le2^{i-1}(v_1-u_1),\quad i\ge 1. \nonumber \] By an argument similar to the one used in Example~, this implies that \(\{u_{i}\}_{i=1}^\infty\) and \(\{v_i\}_{i=1}^\infty\) are Cauchy sequences. Therefore the sequences both converge (because of our assumption), and

implies that they have the same limit. Let \[ \lim_{i\to\infty}u_i=\lim_{i\to\infty}v_i=\beta. \nonumber \] If \(t\in T\), then \(v_i\ge t\) for all \(i\), so \(\beta=\lim_{i\to\infty}v_i\ge t\); therefore, \(\beta\) is an upper bound of \(T\). Now suppose that \(\epsilon>0\). Then there is an integer \(N\) such that \(u_N>\beta-\epsilon\). From the definition of \(u_N\), there is a \(t_N\) in \(T\) such that \(t_N\ge u_N>\beta-\epsilon\). Therefore, \(\beta=\sup T\).

-2em

We say that a sequence \(\{T_n\}\) of sets is {} if \(T_{n+1}\subset T_n\) for all \(n\).

Suppose that \((A,\rho)\) is complete and \(\{T_n\}\) is a nested sequence of nonempty closed subsets of \(A\) such that \(\lim_{n\to\infty}d(T_n)=0\). For each \(n\), choose \(t_n\in T_n\). If \(m\ge n\), then \(t_m\), \(t_n\in T_n\), so \(\rho(t_n,t_m)<d(T_n)\). Since \(\lim_{n\to\infty}d(T_n)=0\), \(\{t_n\}\) is a Cauchy sequence. Therefore, \(\lim_{n\to\infty}t_n=\overline t\) exists. Since \(\overline t\) is a limit point of \(T_n\) and \(T_n\) is closed for all \(n\), \(\overline t\in T_n\) for all \(n\). Therefore, \(\overline t\in \cap_{n=1}^\infty T_n\); in fact, \(\cap_{n=1}^\infty T_n =\{\overline t\}\). (Why?)

Now suppose that \((A,\rho)\) is not complete, and let \(\{t_n\}\) be a Cauchy sequence in \(A\) that does not have a limit. Choose \(n_1\) so that \(\rho(t_n,t_{n_1})<1/2\) if \(n\ge n_1\), and let \(T_1=\set{t}{\rho(t,t_{n_1})\le1}\). Now suppose that \(j>1\) and we have specified \(n_1\), \(n_2\), , \(n_{j-1}\) and \(T_1\), \(T_2\), , \(T_{j-1}\). Choose \(n_j>n_{j-1}\) so that \(\rho(t_n,t_{n_j})<2^{-j}\) if \(n\ge n_j\), and let \(T_j=\set{t}{\rho(t,t_{n_j})\le2^{-j+1}}\). Then \(T_j\) is closed and nonempty, \(T_{j+1}\subset T_j\) for all \(j\), and \(\lim_{j\to\infty}d(T_j)=0\). Moreover, \(t_n\in T_j\) if \(n\ge n_j\). Therefore, if \(\overline t\in\cap_{j=1}^\infty T_j\), then \(\rho(t_n,\overline t)<2^{-j}\), \(n\ge n_j\), so \(\lim_{n\to\infty}t_n=\overline t\), contrary to our assumption. Hence, \(\cap_{j=1}^\infty T_j=\emptyset\).

When considering more than one metric on a given set \(A\) we must be careful, for example, in saying that a set is open, or that a sequence converges, etc., since the truth or falsity of the statement will in general depend on the metric as well as the set on which it is imposed. In this situation we will alway refer to the metric space by its ``full name;" that is, \((A,\rho)\) rather than just \(A\).

Suppose that holds. Let \(S\) be an open set in \((A,\rho)\) and let \(x_0\in S\). Then there is an \(\epsilon>0\) such that \(x\in S\) if \(\rho(x,x_0)<\epsilon\), so the second inequality in implies that \(x_0\in S\) if \(\sigma(x,x_0)\le\epsilon/\beta\). Therefore, \(S\) is open in \((A,\sigma)\).

Conversely, suppose that \(S\) is open in \((A,\sigma)\) and let \(x_0\in S\). Then there is an \(\epsilon>0\) such that \(x\in S\) if \(\sigma(x,x_0)<\epsilon\), so the first inequality in implies that \(x_0\in S\) if \(\rho(x,x_0)\le\epsilon\alpha\). Therefore, \(S\) is open in \((A,\rho)\).

It suffices to show that there are positive constants \(\alpha\) and \(\beta\) such \[\begin{equation} \label{eq:8.1.19} \alpha\le\frac{N_1(\mathbf{X})}{N_2{(\bf X})}\le\beta\mbox{\quad if \quad} \mathbf{X}\ne\mathbf{0}. \end{equation} \] We will show that if \(N\) is any norm on \(\R^n\), there are positive constants \(a_N\) and \(b_N\) such that \[\begin{equation} \label{eq:8.1.20} a_N\|\mathbf{X}\|_2\le N(\mathbf{X})\le b_N\|\mathbf{X}\|_2 \mbox{\quad if \quad} \mathbf{X}\ne\mathbf{0} \end{equation} \] and leave it to you to verify that this implies with \(\alpha=a_{N_1}/b_{N_2}\) and \(\beta=b_{N_1}/a_{N_2}\).

We write \(\mathbf{X}-\mathbf{Y}=(x_1,x_2, \dots,x_n)\) as \[ \mathbf{X}-\mathbf{Y}=\sum_{i=1}^n\,(x_i-y_i)\mathbf{E}_i, \nonumber \] where \(\mathbf{E}_i\) is the vector with \(i\)th component equal to \(1\) and all other components equal to \(0\). From Definition~ , , and induction, \[ N(\mathbf{X}-\mathbf{Y})\le\sum_{i=1}^n|x_i-y_i|N(\mathbf{E_i}); \nonumber \] therefore, by Schwarz’s inequality, \[\begin{equation} \label{eq:8.1.21} N(\mathbf{X}-\mathbf{Y})\le K\|\mathbf{X}-\mathbf{Y}\|_2, \end{equation} \] where \[ K=\left(\sum_{i=1}^nN^2(\mathbf{E_i})\right)^{1/2}. \nonumber \] From and Theorem~, \[ |N(\mathbf{X})-N(\mathbf{Y})|\le K\|\mathbf{X}-\mathbf{Y}\|_2, \nonumber \] so \(N\) is continuous on \(\R_2^n=\R^n\). By Theorem~, there are vectors \(\mathbf{U}_1\) and \(\mathbf{U}_2\) such that \(\|\mathbf{U}_1\|_2= \|\mathbf{U}_2\|_2=1\), \[ N(\mathbf{U}_1)=\min\set{N(\mathbf{U})}{\|\mathbf{U}\|_2=1}, \mbox{\quad and \quad} N(\mathbf{U}_2)=\max\set{N(\mathbf{U})}{\|\mathbf{U}\|_2=1}. \nonumber \] If \(a_N=N(\mathbf{U}_1)\) and \(b_N=N(\mathbf{U}_2)\), then \(a_N\) and \(b_N\) are positive (Definition~ ), and \[ a_N\le N\left(\frac{\mathbf{X}}{\|\mathbf{X}\|_2}\right)\le b_N \mbox{\quad if \quad} \mathbf{X}\ne\mathbf{0}. \nonumber \] This and Definition~

imply .

We leave the proof of the following theorem to you.

Throughout this section it is to be understood that \((A,\rho)\) is a metric space and that the sets under consideration are subsets of \(A\).

We say that a collection \({\mathcal H}\) of open subsets of \(A\) is an {} of \(T\) if \(T\subset\cup\set{H}{H\in{\mathcal H}}\). We say that \(T\) {} if every open covering \({\mathcal H}\) of \(T\) contains a finite collection \(\widehat{\mathcal H}\) such that \[ T\subset\cup\set{H}{H\in\widehat{\mathcal H}}. \nonumber \]

From Theorem~, every nonempty closed and bounded subset of the real numbers has the Heine–Borel property. Moreover, from Exercise~, any nonempty set of reals that has the Heine–Borel property is closed and bounded. Given these results, we defined a compact set of reals to be a closed and bounded set, and we now draw the following conclusion:

{}.

The definition of boundedness of a set of real numbers is based on the ordering of the real numbers: if \(a\) and \(b\) are distinct real numbers then either \(a<b\) or \(b<a\). Since there is no such ordering in a general metric space, we introduce the following definition.

As we will see below, a closed and bounded subset of a general metric space may fail to have the Heine–Borel property. Since we want compact" andhas the Heine–Borel property" to be synonymous in connection with a general metric space, we simply make the following definition.

Suppose that \(T\) has an infinite subset \(E\) with no limit point in \(T\). Then, if \(t\in T\), there is an open set \(H_t\) such that \(t\in H_t\) and \(H_t\) contains at most one member of \(E\). Then \({\mathcal H}=\cup\set{H_t}{t\in T}\) is an open covering of \(T\), but no finite collection \(\{H_{t_1},H_{t_2}, \dots,H_{t_k}\}\) of sets from \({\mathcal H}\) can cover \(E\), since \(E\) is infinite. Therefore, no such collection can cover \(T\); that is, \(T\) is not compact.

Now suppose that every infinite subset of \(T\) has a limit point in \(T\), and let \({\mathcal H}\) be an open covering of \(T\). We first show that there is a sequence \(\{H_i\}_{i=1}^\infty\) of sets from \({\mathcal H}\) that covers \(T\).

If \(\epsilon>0\), then \(T\) can be covered by \(\epsilon\)-neighborhoods of finitely many points of \(T\). We prove this by contradiction. Let \(t_1\in T\). If \(N_\epsilon(t_1)\) does not cover \(T\), there is a \(t_2\in T\) such that \(\rho(t_1,t_2)\ge\epsilon\). Now suppose that \(n\ge 2\) and we have chosen \(t_1\), \(t_2\), , \(t_n\) such that \(\rho(t_i,t_j)\ge\epsilon\), \(1\le i<j\le n\). If \(\cup_{i=1}^nN_\epsilon(t_i)\) does not cover \(T\), there is a \(t_{n+1}\in T\) such that \(\rho(t_i,t_{n+1})\ge\epsilon\), \(1\le i\le n\). Therefore, \(\rho(t_i,t_j)\ge\epsilon\), \(1\le i<j\le n+1\). Hence, by induction, if no finite collection of \(\epsilon\)-neighborhoods of points in \(T\) covers \(T\), there is an infinite sequence \(\{t_n\}_{n=1}^\infty\) in \(T\) such that \(\rho(t_i,t_j)\ge\epsilon\), \(i\ne j\). Such a sequence could not have a limit point, contrary to our assumption.

By taking \(\epsilon\) successively equal to \(1\), \(1/2\), , \(1/n\), , we can now conclude that, for each \(n\), there are points \(t_{1n}\), \(t_{2n}\), , \(t_{k_n,n}\) such that \[ T\subset \bigcup_{i=1}^{k_n} N_{1/n}(t_{in}). \nonumber \] Denote \(B_{in}=N_{1/n}(t_{in})\), \(1\le i\le n\), \(n\ge 1\), and define \[ \{G_1,G_2,G_3,...\}=\{B_{11}, \dots, B_{k_1,1}, B_{12}, \dots, B_{k_2,2}, B_{13},\dots,B_{k_3,3}, \dots\}. \nonumber \]

If \(t\in T\), there is an \(H\) in \({\mathcal H}\) such that \(t\in H\). Since \(H\) is open, there is an \(\epsilon>0\) such that \(N_\epsilon(t)\subset H\). Since \(t\in G_j\) for infinitely many values of \(j\) and \(\lim_{j\to\infty}d(G_j)=0\), \[ G_j\subset N_\epsilon(t)\subset H \nonumber \] for some \(j\). Therefore, if \(\{G_{j_i}\}_{i=1}^\infty\) is the subsequence of \(\{G_j\}\) such that \(G_{j_i}\) is a subset of some \(H_i\) in \({\mathcal H}\) (the \(\{H_i\}\) are not necessarily distinct), then \[\begin{equation} \label{eq:8.2.1} T\subset\bigcup_{i=1}^\infty H_i. \end{equation} \]

We will now show that \[\begin{equation} \label{eq:8.2.2} T\subset \bigcup_{i=1}^N H_i. \end{equation} \] for some integer \(N\). If this is not so, there is an infinite sequence \(\{t_n\}_{n=1}^\infty\) in \(T\) such that \[\begin{equation} \label{eq:8.2.3} t_n\notin \bigcup_{i=1}^n H_i, \quad n\ge 1. \end{equation} \] From our assumption, \(\{t_n\}_{n=1}^\infty\) has a limit \(\overline t\) in \(T\). From , \(\overline t\in H_k\) for some \(k\), so \(N_\epsilon(\overline t)\subset H_k\) for some \(\epsilon>0\). Since \(\lim_{n\to\infty}t_n=\overline t\), there is an integer \(N\) such that \[ t_n\in N_\epsilon(\overline t)\subset H_k\subset \bigcup_{i=1}^nH_i,\quad n>k, \nonumber \] which contradicts . This verifies , so \(T\) is compact.

Any finite subset of a metric space obviously has the Heine–Borel property and is therefore compact. Since Theorem~ does not deal with finite sets, it is often more convenient to work with the following criterion for compactness, which is also applicable to finite sets.

Suppose that \(T\) is compact and \(\{t_n\}\subset T\). If \(\{t_n\}\) has only finitely many distinct terms, there is a \(\overline t\) in \(T\) such that \(t_n=\overline t\) for infinitely many values of \(n\); if this is so for \(n_1<n_2<\cdots\), then \(\lim_{j\to\infty}t_{n_j}=\overline t\). If \(\{t_n\}\) has infinitely many distinct terms, then \(\{t_n\}\) has a limit point \(\overline t\) in \(T\), so there are integers \(n_1<n_2<\cdots\) such that \(\rho(t_{n_j},\overline t)<1/j\); therefore, \(\lim_{j\to\infty}t_{n_j}=\overline t\).

Conversely, suppose that every sequence in \(T\) has a subsequence that converges to a limit in \(T\). If \(S\) is an infinite subset of \(T\), we can choose a sequence \(\{t_n\}\) of distinct points in \(S\). By assumption, \(\{t_n\}\) has a subsequence that converges to a member \(\overline t\) of \(T\). Since \(\overline t\) is a limit point of \(\{t_n\}\), and therefore of \(T\), \(T\) is compact.

By Theorem~, \(\{t_n\}\) has a subsequence \(\{t_{n_j}\}\) such that \[\begin{equation} \label{eq:8.2.4} \lim_{j\to\infty}t_{n_j}=\overline t\in T. \end{equation} \] We will show that \(\lim_{n\to\infty}t_n=\overline t\).

Suppose that \(\epsilon>0\). Since \(\{t_n\}\) is a Cauchy sequence, there is an integer \(N\) such that \(\rho(t_n,t_m)<\epsilon\), \(n>m\ge N\). From , there is an \(m=n_j\ge N\) such that \(\rho(t_m,\overline t)<\epsilon\). Therefore, \[ \rho(t_n,\overline t)\le \rho(t_n,t_m)+\rho(t_m,\overline t)<2\epsilon,\quad n\ge m. \nonumber \] -2em2em

Suppose that \(\overline t\) is a limit point of \(T\). For each \(n\), choose \(t_n\ne\overline t\in B_{1/n}(\overline t)\cap T\). Then \(\lim_{n\to\infty}t_n=\overline t\). Since every subsequence of \(\{t_n\}\) also converges to \(\overline t\), \(\overline t\in T\), by Theorem~. Therefore, \(T\) is closed.

The family of unit open balls \({\mathcal H}=\set{B_1(t)}{t\in T}\) is an open covering of \(T\). Since \(T\) is compact, there are finitely many members \(t_1\), \(t_2\), , \(t_n\) of \(T\) such that \(S\subset \cup_{j=1}^nB_1(t_j)\). If \(u\) and \(v\) are arbitrary members of \(T\), then \(u\in B_1(t_r)\) and \(v\in B_1(t_s)\) for some \(r\) and \(s\) in \(\{1,2, \dots,n\}\), so \[\begin{eqnarray*} \rho(u,v)\ar\le \rho(u,t_r)+\rho(t_r,t_s)+\rho(t_s,v)\\ \ar\le 2+\rho(t_r,t_s)\le2+\max\set{\rho(t_i,t_j)}{1\le i<j\le n}. \end{eqnarray*} \nonumber \] Therefore, \(T\) is bounded.

The converse of Theorem~ is false; for example, if \(A\) is any infinite set equipped with the discrete metric (Example~.), then every subset of \(A\) is bounded and closed. However, if \(T\) is an infinite subset of \(A\), then \({\mathcal H}=\set{\{t\}}{t\in T}\) is an open covering of \(T\), but no finite subfamily of \({\mathcal H}\) covers \(T\).

We leave it to you (Exercise~) to show that every totally bounded set is bounded and that the converse is false.

We will prove that if \(T\) is not totally bounded, then \(T\) is not compact. If \(T\) is not totally bounded, there is an \(\epsilon>0\) such that there is no finite \(\epsilon\)-net for \(T\). Let \(t_1\in T\). Then there must be a \(t_2\) in \(T\) such that \(\rho(t_1,t_2)>\epsilon\). (If not, the singleton set \(\{t_1\}\) would be a finite \(\epsilon\)-net for \(T\).) Now suppose that \(n\ge 2\) and we have chosen \(t_1\), \(t_2\), , \(t_n\) such that \(\rho(t_i,t_j)\ge\epsilon\), \(1\le i<j\le n\). Then there must be a \(t_{n+1}\in T\) such that \(\rho(t_i,t_{n+1})\ge\epsilon\), \(1\le i\le n\). (If not, \(\{t_1,t_2, \dots,t_n\}\) would be a finite \(\epsilon\)-net for \(T\).) Therefore, \(\rho(t_i,t_j)\ge\epsilon\), \(1\le i<j\le n+1\). Hence, by induction, there is an infinite sequence \(\{t_n\}_{n=1}^\infty\) in \(T\) such that \(\rho(t_i,t_j)\ge\epsilon\), \(i\ne j\). Since such a sequence has no limit point, \(T\) is not compact, by Theorem~.

Let \(S\) be an infinite subset of \(T\), and let \(\{s_i\}_{i=1}^\infty\) be a sequence of distinct members of \(S\). We will show that \(\{s_i\}_{i=1}^\infty\) has a convergent subsequence. Since \(T\) is closed, the limit of this subsequence is in \(T\), which implies that \(T\) is compact, by Theorem~.

For \(n\ge1\), let \(T_{1/n}\) be a finite \(1/n\)-net for \(T\). Let \(\{s_{i0}\}_{i=1}^\infty=\{s_i\}_{i=1}^\infty\). Since \(T_1\) is finite and \(\{s_{i0}\}_{i=1}^\infty\) is infinite, there must be a member \(t_1\) of \(T_1\) such that \(\rho(s_{i0},t_1)\le1\) for infinitely many values of \(i\). Let \(\{s_{i1}\}_{i=1}^\infty\) be the subsequence of \(\{s_{i0}\}_{i=1}^\infty\) such that \(\rho(s_{i1},t_1)\le1\).

We continue by induction. Suppose that \(n>1\) and we have chosen an infinite subsequence \(\{s_{i,n-1}\}_{i=1}^\infty\) of \(\{s_{i,n-2}\}_{i=1}^\infty\). Since \(T_{1/n}\) is finite and \(\{s_{i,n-1}\}_{i=1}^\infty\) is infinite, there must be member \(t_n\) of \(T_{1/n}\) such that \(\rho(s_{i,n-1},t_n)\le1/n\) for infinitely many values of \(i\). Let \(\{s_{in}\}_{i=1}^\infty\) be the subsequence of \(\{s_{i,n-1}\}_{i=1}^\infty\) such that \(\rho(s_{in},t_n)\le1/n\). From the triangle inequality, \[\begin{equation} \label{eq:8.2.5} \rho(s_{in},s_{jn})\le2/n,\quad i,j\ge1,\quad n\ge 1. \end{equation} \] Now let \(\widehat s_i=s_{ii}\), \(i\ge 1\). Then \(\{\widehat s_i\}_{i=1}^\infty\) is an infinite sequence of members of \(T\). Moroever, if \(i,j\ge n\), then \(\widehat s_i\) and \(\widehat s_j\) are both included in \(\{s_{in}\}_{i=1}^\infty\), so implies that \(\rho(\widehat s_i,\widehat s_j)\le2/n\); that is, \(\{\widehat s_i\}_{i=1}^\infty\) is a Cauchy sequence and therefore has a limit, since \((A,\rho)\) is complete.

We will show that \(T\) is totally bounded in \(\ell_\infty\). Since \(\ell_\infty\) is complete (Exercise_{), Theorem} will then imply that \(T\) is compact.

Let \(\epsilon>0\). Choose \(N\) so that \(\mu_i\le\epsilon\) if \(i>N\). Let \(\mu=\max\set{\mu_i}{1\le i\le n}\) and let \(p\) be an integer such that \(p\epsilon>\mu\). Let \(Q_\epsilon=\set{r_i\epsilon}{r_i= \mbox{integer in}[-p,p]}\). Then the subset of \(\ell_\infty\) such that \(x_i\in Q_\epsilon\), \(1\le i\le N\), and \(x_i=0\), \(i>N\), is a finite \(\epsilon\)-net for \(T\).

In Example~ we showed that \(C[a,b]\) is a complete metric space under the metric \[ \rho(f,g)=\|f-g\|=\max\set{|f(x)-g(x)|}{a\le x\le b}. \nonumber \] We will now give necessary and sufficient conditions for a subset of \(C[a,b]\) to be compact.

Theorem~ implies that for each \(f\) in \(C[a,b]\) there is a constant \(M_f\) {} \(f\), such that \[ |f(x)|\le M_f \mbox{\quad if \quad} a\le x\le b, \nonumber \] and Theorem~ implies that there is a constant \(\delta_f\) {} such that \[ |f(x_1)-f(x_2)|\le \epsilon \mbox{\quad if \quad} x_1,x_2\in [a,b]\mbox{\quad and \quad} |x_1-x_2|<\delta_f. \nonumber \] The difference in Definition~ is that the {} \(M\) and \(\delta\) apply to {} \(f\) in \(T\).

For necessity, suppose that \(T\) is compact. Then \(T\) is closed (Theorem~) and totally bounded (Theorem~). Therefore, if \(\epsilon>0\), there is a finite subset \(T_\epsilon=\{g_1,g_2, \dots,g_k\}\) of \(C[a,b]\) such that if \(f\in T\), then \(\|f-g_i\|\le \epsilon\) for some \(i\) in \(\{1,2, \dots,k\}\). If we temporarily let \(\epsilon=1\), this implies that \[ \|f\|=\|(f-g_i)+g_i\|\le\|f-g_i\|+\|g_i\|\le 1+\|g_i\|, \nonumber \] which implies with \[ M=1+\max\set{\|g_i\|}{1\le i\le k}. \nonumber \]

For , we again let \(\epsilon\) be arbitary, and write \[\begin{equation} \label{eq:8.2.8} \begin{array}{rcl} |f(x_1)-f(x_2)| \ar\le |f(x_1)-g_i(x_1)|+|g_i(x_1)-g_i(x_2)|+|g_i(x_2)-f(x_2)|\\ \ar\le |g_i(x_1)-g_i(x_2)|+2\|f-g_i\|\\ \ar< |g_i(x_1)-g_i(x_2)|+2\epsilon. \end{array} \end{equation} \] Since each of the finitely many functions \(g_1\), \(g_2\), , \(g_k\) is uniformly continuous on \([a,b]\) (Theorem~), there is a \(\delta>0\) such that \[ |g_i(x_1)-g_i(x_2)|<\epsilon\mbox{\quad if \quad} |x_1-x_2|<\delta,\quad 1\le i\le k. \nonumber \] This and imply with \(\epsilon\) replaced by \(3\epsilon\). Since this replacement is of no consequence, this proves necessity.

For sufficiency, we will show that \(T\) is totally bounded. Since \(T\) is closed by assumption and \(C[a,b]\) is complete, Theorem~ will then imply that \(T\) is compact.

Let \(m\) and \(n\) be positive integers and let \[ \xi_r=a+\frac{r}{m}(b-a),\quad 0\le r\le m, \mbox{\quad and \quad} \eta_s=\frac{sM}{n},\quad -n\le s\le n; \nonumber \] that is, \(a=\xi_0<\xi_1<\cdots<\xi_m=b\) is a partition of \([a,b]\) into subintervals of length \((b-a)/m\), and \(-M=\eta_{-n}<\eta_{-n+1}<\cdots<\eta_{n-1}<\eta_n=M\) is a partition of the

segment of the \(y\)-axis between \(y=-M\) and \(y=M\) into subsegments of length \(M/n\). Let \(S_{mn}\) be the subset of \(C[a,b]\) consisting of functions \(g\) such that \[ \{g(\xi_0), g(\xi_1), \dots, g(\xi_m)\} \subset\{\eta_{-n},\eta_{-n+1} \dots,\eta_{n-1}, \eta_n\} \nonumber \] and \(g\) is linear on \([\xi_{i-1},\xi_i]\), \(1\le i\le m\). Since there are only \((m+1)(2n+1)\) points of the form \((\xi_r,\eta_s)\), \(S_{mn}\) is a finite subset of \(C[a,b]\).

Now suppose that \(\epsilon>0\), and choose \(\delta>0\) to satisfy . Choose \(m\) and \(n\) so that \((b-a)/m<\delta\) and \(2M/n<\epsilon\). If \(f\) is an arbitrary member of \(T\), there is a \(g\) in \(S_{mn}\) such that \[\begin{equation} \label{eq:8.2.9} |g(\xi_i)-f(\xi_i)|<\epsilon,\quad 0\le i\le m. \end{equation} \] If \(0\le i\le m-1\), \[\begin{equation} \label{eq:8.2.10} |g(\xi_i)-g(\xi_{i+1})|=|g(\xi_i)-f(\xi_i)|+|f(\xi_i)-f(\xi_{i+1})| +|f(\xi_{i+1})-g(\xi_{i+1})|. \end{equation} \] Since \(\xi_{i+1}-\xi_i<\delta\), , , and imply that \[ |g(\xi_i)-g(\xi_{i+1})|<3\epsilon. \nonumber \] Therefore, \[\begin{equation} \label{eq:8.2.11} |g(\xi_i)-g(x)|<3\epsilon,\quad \xi_i\le x\le \xi_{i+1}, \end{equation} \] since \(g\) is linear on \([\xi_i,\xi_{i+1}]\).

Now let \(x\) be an arbitrary point in \([a,b]\), and choose \(i\) so that \(x\in[\xi_i,\xi_{i+1}]\). Then \[ |f(x)-g(x)|\le|f(x)-f(\xi_i)|+|f(\xi_i)-g(\xi_i)|+|g(\xi_i)-g(x)|, \nonumber \] so , , and imply that \(|f(x)-g(x)|<5\epsilon\), \(a\le x\le b\). Therefore, \(S_{mn}\) is a finite \(5\epsilon\)-net for \(T\), so \(T\) is totally bounded.

Let \(T\) be the closure of \({\mathcal F}\); that is, \(f\in T\) if and only if either \(f\in T\) or \(f\) is the uniform limit of a sequence of members of \({\mathcal F}\). Then \(T\) is also uniformly bounded and equicontinuous (verify), and \(T\) is closed. Hence, \(T\) is compact, by Theorem~. Therefore, \({\mathcal F}\) has a limit point in \(T\). (In this context, the limit point is a function \(f\) in \(T\).) Since \(f\) is a limit point of \({\mathcal F}\), there is for each integer \(n\) a function \(f_n\) in \({\mathcal F}\) such that \(\|f_n-f\|<1/n\); that is \(\{f_n\}\) converges uniformly to \(f\) on \([a,b]\).

In Chapter~ we studied real-valued functions defined on subsets of \(\R^n\), and in Chapter~6.4. we studied functions defined on subsets of \(\R^n\) with values in \(\R^m\). These are examples of functions defined on one metric space with values in another metric space.(Of course, the two spaces are the same if \(n=m\).)

In this section we briefly consider functions defined on subsets of a metric space \((A,\rho)\) with values in a metric space \((B,\sigma)\). We indicate that \(f\) is such a function by writing \[ f:(A,\rho)\to(B,\sigma). \nonumber \] The {} and {} of \(f\) are the sets \[ D_f=\set{u\in A}{f(u)\mbox{ is defined}} \nonumber \] and \[ R_f=\set{v\in B}{v=f(u)\mbox{ for some }u\mbox{ in }D_f}. \nonumber \]

Note that can be written as \[ f(D_f\cap N_\delta(\widehat u))\subset N_\epsilon(f(\widehat u)). \nonumber \] Also, \(f\) is automatically continuous at every isolated point of \(D_f\). (Why?)

Suppose that is true, and let \(\{u_n\}\) be a sequence in \(D_f\) that satisfies . Let \(\epsilon>0\) and choose \(\delta>0\) to satisfy . From , there is an integer \(N\) such that \(\rho(u_n,\widehat u)<\delta\) if \(n\ge N\). Therefore, \(\sigma(f(u_n),\widehat v)<\epsilon\) if \(n\ge N\), which implies .

For the converse, suppose that is false. Then there is an \(\epsilon_0>0\) and a sequence \(\{u_n\}\) in \(D_f\) such that \(\rho(u_n,\widehat u)<1/n\) and \(\sigma(f(u_n),\widehat v)\ge\epsilon_0\), so is false.

We leave the proof of the next two theorems to you.

Let \(\{v_n\}\) be an infinite sequence in \(f(T)\). For each \(n\), \(v_n=f(u_n)\) for some \(u_n\in T\). Since \(T\) is compact, \(\{u_n\}\) has a subsequence \(\{u_{n_j}\}\) such that \(\lim_{j\to\infty}u_{n_j}=\widehat u\in T\) (Theorem~). From Theorem~, \(\lim_{j\to\infty}f(u_{n_j})=f(\widehat u)\); that is, \(\lim_{j\to\infty}v_{n_j}=f(\widehat u)\). Therefore, \(f(T)\) is compact, again by Theorem~.

If \(f\) is not uniformly continuous on \(T\), then for some \(\epsilon_0>0\) there are sequences \(\{u_n\}\) and \(\{v_n\}\) in \(T\) such that \(\rho(u_n,v_n)<1/n\) and \[\begin{equation} \label{eq:8.3.6} \sigma(f(u_n),f(v_n))\ge\epsilon_0. \end{equation} \] Since \(T\) is compact, \(\{u_n\}\) has a subsequence \(\{u_{n_k}\}\) that converges to a limit \(\widehat u\) in \(T\) (Theorem~). Since \(\rho(u_{n_k},v_{n_k})<1/n_k\), \(\lim_{k\to\infty}v_{n_k}=\widehat u\) also. Then \[ \lim_{k\to\infty}f(u_{n_k})=\dst\lim_{k\to \infty}f(v_{n_k})=f(\widehat u) \nonumber \] (Theorem~~), which contradicts .

We note that a contraction of \((A,\rho)\) is uniformly continuous on \(A\).

To see that cannot have more than one solution, suppose that \(u=f(u)\) and \(v=f(v)\). Then \[\begin{equation} \label{eq:8.3.9} \rho(u,v)=\rho(f(u),f(v)). \end{equation} \] However, implies that \[\begin{equation} \label{eq:8.3.10} \rho(f(u),f(v))\le\alpha\rho(u,v). \end{equation} \] Since and imply that \[ \rho(u,v)\le\alpha\rho(u,v) \nonumber \] and \(\alpha<1\), it follows that \(\rho(u,v)=0\). Hence \(u=v\).

We will now show that has a solution. With \(u_0\) arbitrary, define \[\begin{equation}\label{eq:8.3.11} u_n=f(u_{n-1}),\quad n\ge1. \end{equation} \] We will show that \(\{u_n\}\) converges. From and , \[\begin{equation} \label{eq:8.3.12} \rho(u_{n+1},u_n)=\rho(f(u_n),f(u_{n-1}))\le\alpha\rho(u_n,u_{n-1}). \end{equation} \]

The inequality \[\begin{equation}\label{eq:8.3.13} \rho(u_{n+1},u_n)\le \alpha^n \rho(u_1,u_0),\quad n\ge0, \end{equation} \] follows by induction from . If \(n>m\), repeated application of the triangle inequality yields \[ \rho(u_n,u_m) \le \rho(u_n,u_{n-1})+\rho(u_{n-1},u_{n-2})+\cdots+\rho(u_{m+1},u_m), \nonumber \] and yields \[ \rho(u_n,u_m)\le\rho(u_1,u_0)\alpha^m(1+\alpha+\cdots+\alpha^{n-m-1})< \frac{\alpha^m}{1-\alpha}. \nonumber \] Now it follows that \[ \rho(u_n,u_m)<\frac{\rho(u_1,u_0)}{1-\alpha}\alpha^N\mbox{\quad if\quad} n,m>N, \nonumber \] and, since \(\lim_{N\to\infty} \alpha^N=0\), \(\{u_n\}\) is a Cauchy sequence. Since \(A\) is complete, \(\{u_n\}\) has a limit \(\widehat u\). Since \(f\) is continuous at \(\widehat u\), \[ f(\widehat u)=\lim_{n\to\infty}f(u_{n-1})=\lim_{n\to\infty}u_n=\widehat u, \nonumber \] where Theorem~~ implies the first equality and implies the second.

Let \(A\) be \(C[a,b]\), which is complete. If \(u\in C[a,b]\), let \(f(u)=v\), where \[ v(x)=h(x)+\lambda\int_a^bK(x,y)u(y)\,dy,\quad a\le x\le b. \nonumber \] Since \(v\in C[a,b]\), \(f:C[a,b]\to C[a,b]\). If \(u_1\), \(u_2\in C[a,b]\), then \[ |v_1(x)-v_2(x)|\le|\lambda|\int_a^b|K(x,y)||u_1(y)-v_1(y)|\,dy, \nonumber \] so \[ \|v_1-v_2\|\le |\lambda|M(b-a)\|u_1-u_2\|. \nonumber \] Since \(|\lambda|M(b-a)<1\), \(f\) is a contraction. Hence, there is a unique \(u\) in \(C[a,b]\) such that \(f(u)=u\). This \(u\) satisfies .

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