LEC ANA2 5

Completions using distance functions

Recall $C_b(X)$ is the space of continuous bounded functions $f:X\to \mathbb{R}$. We know that it is complete.

Proposition 165.1(Proposition).

Let $(X,d)$ be a metric space. There exists an isometric embedding $\phi :X\to C_b(X)$. $\overline{\phi (X)}$ is the completion of $X$.

Proof.

Fix $a\in X$. For each $x\in X$, define $f_x(y)=d(x,y)-d(a,y)$. $f_x$ is uniformly continuous:

$$\begin{array}{rl}|f_x(y)-f_x(z)|& =|(d(x,y)-d(a,y))-(d(x,z)-d(a,z))|\\ & \le |d(x,y)-d(x,z)|+|d(a,z)-d(z,y)|\\ & \le 2d(y,z).\end{array}$$

$f_x$ is bounded, too: $|f_x(y)|\le d(x,a)$. Thus, $f_x\in C_b(X)$.

Define $\phi :X\to C_b(X)$ by $\phi (x)=f_x$. For any $x,x^{\prime }\in X$,

$$\begin{array}{r}|f_x(y)-f_{x^{\prime }}(y)|=|d(x,y)-d(x^{\prime },y)|\le d(x,x^{\prime })\forall y\in X.\end{array}$$

So,

$$\underset{y\in X}{\sup }|f_x(y)-f_{x^{\prime }}(y)|\le d(x,x^{\prime }).$$

At $y=x$, $|f_x(y)-f_{x^{\prime }}(y)|$ attains $d(x,x^{\prime })$. Thus, $\Vert f_x-f_{x^{\prime }}\Vert =d(x,x^{\prime })$, and $\phi$ is an isometry.

Since $C_b(X)$ is complete, the completion of $\phi (X)$ is given by $\overline{\phi (X)}$.□

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Separable spaces

Recall what a separable space is.

Lemma 165.2(Lemma).

The property of denseness is transitive, that is, If $X\subset Y\subset Z$, $X$ is dense in $Y$, and $Y$ is dense in $Z$, then $X$ is dense in $Z$.

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Proposition 165.3(Proposition).

$l_1$ is separable.

Proof.

Let $\mathcal{F}\subseteq l_1$ be the set of all sequences in $l_1$ which have finitely many non-zero terms. Let ${\mathcal{F}}_{\mathbb{Q}}=\mathcal{F}\cap {\mathbb{Q}}^{\mathbb{N}}$. ${\mathcal{F}}_{\mathbb{Q}}$ is countable:

$$\left|{\mathcal{F}}_{\mathbb{Q}}\right|=\left|\bigcup _{i=1}^{\infty }{\mathbb{Q}}^i\right|.$$

It is clear that ${\mathcal{F}}_{\mathbb{Q}}$ is dense in $\mathcal{F}$. If we show $\mathcal{F}$ is dense in $l_1$, we’ll be done by Lemma 2.

Let $\mathbf{x}=\{x_k\}\in l_1$. Let $\{{\mathbf{x}}_n\}\subseteq \mathcal{F}$ be a sequence, with ${\mathbf{x}}_n=\{x_n^k{\}}_{k=1}^{\infty }$ and

$$\begin{array}{r}{x}_n^k:=\{\begin{array}{ll}{x}_k& k\le n,\\ 0& k>n.\end{array}\end{array}$$

Now,

$$\begin{array}{rl}\Vert {\mathbf{x}}_n-\mathbf{x}{\Vert }_1& =\sum _{k=1}^{\infty }|x_n^k-x_k|\\ & =\sum _{k=n+1}^{\infty }|x_k|\to 0.\end{array}$$

Thus, $\{{\mathbf{x}}_n\}\to \mathbf{x}$.□

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Exercise 165.4(Exercise).

1. $l_p$ is separable.
2. $C[0,1]$ is separable.
   

Example 165.5(Example).

Define $C_0:=\{\{x_n\}|x_n\to 0\}\subset l_{\infty }$. We will show that $C_0$ is a separable subspace of $l_{\infty }$. Let $\mathcal{F}\subseteq C_0$ be as in Proposition 3.
Let $\mathbf{x}=\{x_k\}\in C_0$, and define ${\mathbf{x}}_n$ as we have previously.

$$\begin{array}{rl}\Vert {\mathbf{x}}_n-\mathbf{x}{\Vert }_{\infty }& =\underset{k>n}{\sup }|x_k|\to 0.\end{array}$$

Example 165.6(Example).

$l_{\infty }$ is not separable. Consider $S=\{\{x_n\}|x_n\in \{\pm 1\}\}$. For any $\mathbf{x},\mathbf{y}\in S$, $\Vert \mathbf{x}-\mathbf{y}{\Vert }_{\infty }=2$. Consider the disjoint uncountable collection of open balls $\{B_1(\mathbf{x})|\mathbf{x}\in S\}$. Since a dense subset of $l_{\infty }$ must intersect all open balls of $l_{\infty }$, a countable dense subset doesn’t exist.

Exercise 165.7(Exercise).

Completion of a separable metric space is separable.

Immediate from Lemma 2.

Exercise 165.8(Exercise).

If a space is totally bounded, it is separable.

Let $X$ be a totally bounded space. For ${ϵ}_n=1/n$, let $O_n$ be a finite cover of $X$ by balls of radius ${ϵ}_n$, and define $S_n=\{x|B_{ϵ}(x)\in S_n\}$. Take $D={\bigcup }_{n=1}^{\infty }{S}_n$. $D$ is countable. $D$ is dense in $X$ since for any $x\in X$, we have balls $B_n\in O_n$ such that $x\in B_n$ for all $n$, so the centers of the balls converge to $x$.

[!Exercise]
The space of all non-empty compact subsets of a separable metric space $M$ endowed with the Hausdorff metric is separable.

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Proposition 165.9(Proposition).

Let $\{A_n\}$ with $A_n\subseteq X$ with $d_1(A_n)\to 0$, and ${\bigcap }_{i=1}^{\infty }{A}_i\ne \varnothing$. Suppose $f:(X,d_1)\to (Y,d_2)$ is continuous. Then $d_2(f(A_n))\to 0$.

Proof.

It should be clear that${\bigcap }_{i=1}^{\infty }{A}_i=\{x\}$. Let $ϵ>0$. $f$ is continuous, so there exists $\delta$ such that $d_1(x,y)<\delta$ implies $d_2(f(x),f(y))<ϵ$. Choose $n_0$ such that $d_1(A_n)<\delta$ for all $n\ge n_0$. If $y\in A_n$, then $d_1(x,y)<\delta$, so $d_2(f(x),f(y))<ϵ$. Thus, $d_2(f(A_n))<2ϵ$.□

Counterexample when ${\bigcap }_{i=1}^{\infty }{A}_i=\varnothing$: Let $A_n=(0,1/n]$, $f(x)=\sin (1/x)$. $f(A_n)=[-1,1]$ for all $n$.

Reviewed the Cantor intersection theorem.