LEC ANA2 4

Completions, reprise

We have already seen a construction of the completion. We will now prove that the completion of a metric space is unique. We have also not proven that $\mathbb{R}$ is complete yet, which we will also do in this course.

Uniqueness of completion

Remark 164.1(Remark).

Recall continuous functions map convergent sequences to convergent sequences. Similarly, uniformly continuous functions map Cauchy sequences to Cauchy sequences (should be clear from the definition). Important to note here that continuous functions may not preserve Cauchyness: Consider the image of the sequence $\{1/n{\}}_{n=2}^{\infty }$ under the continuous map $f:(0,1)\to \mathbb{R}$ defined by $x\mapsto 1/x$.

Theorem 164.2(Universal property of completions).

The completion $\tilde{X}$ of a metric space $X$ satisfies the following universal property: if $Y$ is any complete metric space and $f$ is any uniformly continuous function from $X$ to $Y$, then there exists a unique uniformly continuous function $\overline{f}$ from $\tilde{X}$ to $Y$ that extends $f$:

Proof.

Let$x\in \tilde{X}$. Let $\{x_n\}$ be a Cauchy sequence in¹ $X$ converging to $x$. Since $f$ is uniformly continuous, $\{f(x_n)\}$ is Cauchy. $Y$ is complete, so $\{f(x_n)\}$ must converge. We are forced to define $\overline{f}(x)$ to be the limit of $\{f(x_n)\}$, since we need $\overline{f}$ to be continuous.

We have to show $\overline{f}$ is well defined. Suppose $\{y_n\}\subseteq X$ also converges to $x$. Consider the sequence $\{z_n\}=x_1,y_1,x_2,y_2,\dots$. It clearly converges to $x$, and hence it and its image $\{f(z_n)\}$ under $f$ is Cauchy. $\{f(z_n)\}$ has $\{f(x_n)\}$ as a subsequence, which forces it to converge to $\overline{f}(x)$. It follows that the subsequence $\{f(y_n)\}$ also converges to $\overline{f}(x)$.

It remains to show $\overline{f}$ is uniformly continuous. Let $ϵ>0$. Let $\delta$ be chosen by the uniform continuity of $f$. We will show that the same $\delta$ works for $\overline{f}$. Let $x,y\in \tilde{X}$. Let $\{x_n\},\{y_n\}\subseteq X$ such that $\{x_n\}\to x$ and $\{y_n\}\to y$. Note that $d(x_n,y_n)\to d(x,y)$. If $d(x,y)<\delta$, then there exists $N$ such that for all $n\ge N$, $d(x_n,y_n)<\delta$, so $d(f(x_n),f(y_n))<ϵ$. Now,

$$\begin{array}{rl}d(\overline{f}(x),\overline{f}(y))& \le \underset{\to 0}{\underbrace{d(\overline{f}(x),f(x_n))}}+\underset{<ϵ\text{ for }n\ge N}{\underbrace{d(f(x_n),f(y_n))}}+\underset{\to 0}{\underbrace{d(f(y_n),\overline{f}(y))}}\\ & <ϵ\end{array}$$

for sufficiently large $n$.□

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Proposition 164.3(Proposition).

Completion of a metric space is unique. That is, if $X$ is a metric space and $X_1$ and $X_2$ are complete metric spaces such that there exist isometries ${\phi }_1:X\to X_1$ and ${\phi }_2:X\to X_2$ and ${\phi }_1(X)\subseteq X_1$, ${\phi }_2(X)\subseteq X_2$ are dense, there exists a bijective isometry $\phi :X_1\to X_2$ such that $\phi ({\phi }_1(x))={\phi }_2(x)$.

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This follows immediately from the fact that initial objects in a category are isomorphic.

Vasanth's way

Vasanth proved the following lemma. It is essentially the same as Theorem 2, and the same proof works; just replace $\tilde{X}$ with $\overline{A}$.

Lemma 164.4(Lemma).

Let $X,Y$ be metric spaces. Suppose $Y$ is complete. Let $f:A(\subseteq X)\to Y$ be uniformly continuous. Then, there exists a unique uniformly continuous $\overline{f}:\overline{A}\to Y$ such that $\overline{f}|{}_A=f$.

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Here is his proof for Proposition 3: ${\phi }_1^{-1}:{\phi }_1(X)\to X$ is a bijective isometry; compose with ${\phi }_2$ to obtain another bijective isometry ${\phi }_2{\phi }_1^{-1}:{\phi }_1(X)\to {\phi }_2(X)$. By Lemma 4, there exists a uniformly continuous $\phi :\overline{{\phi }_1(X)}=X_1\to X_2$ such that $\phi |{}_{{\phi }_1(X)}={\phi }_2{\phi }_1^{-1}$. It is clear that $\phi$ is a bijective isometry.

Example 164.5(Example).

Prove that the image of a totally bounded set under a uniformly continuous map is totally bounded.

Proof.

Let$A$ be a totally bounded set in $X$. Let $f:X\to Y$ be a uniformly continuous function. Consider the induced map $g:\tilde{X}\to \tilde{Y}$, where $\tilde{X}$ and $\tilde{Y}$ are the completions of $X$ and $Y$. Then the closure $\overline{A}$ in $\tilde{X}$ is compact hence $g(\overline{A})\subseteq \tilde{Y}$ is compact. Thus $f(A)=g(A)\cap Y$ is totally bounded.□

Alternate proof of completeness of $(\tilde{X},\tilde{d})$

An alternate (and neater) proof of completeness of the completion constructed here. We assume we have already shown $\varphi (X)$ is dense in $\tilde{X}$ (our previous proof of this does not depend on $(\tilde{X},\tilde{d})$ being complete, so we can keep that).

Claim 164.6(Claim).

For all Cauchy sequences in $\varphi (X)$, a limit exists in $\tilde{X}$.

Proof.

Let$\{y_n\}$ be a Cauchy sequence in $\varphi (X)$. Recall that each $y_n$ is an equivalence class of Cauchy sequences, and since each $y_n\in \varphi (X)$, we can write $y_n=[P_n]$, where $P_n$ is the constant sequence $x_n,x_n,\dots$ for some $x_n\in X$.

It is clear that $\{x_n\}$ is a Cauchy sequence in $X$. Consider $[\{x_n\}]$ in $\tilde{X}$.

$$\begin{array}{rl}\tilde{d}(y_m,[\{x_n\}])& =\left(\underset{n\to \infty }{\lim }d(x_m,x_n)\right)\to 0\text{ as }m\to \infty .\end{array}$$

Thus, $\{y_m\}\to [\{x_n\}]$ in $\tilde{X}$.□

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Now, If $\{z_n\}\subseteq \tilde{X}$ is a Cauchy sequence, using the fact that $\varphi (X)$ is dense in $\tilde{X}$, we can construct a sequence $\{y_n\}\subseteq \varphi (X)$ such that $\tilde{d}(z_n,y_n)<1/n$.

$\{y_n\}$ is a Cauchy sequence, since

$$\begin{array}{r}\tilde{d}(y_n,y_m)\le \tilde{d}(y_n,z_n)+\tilde{d}(z_n,z_m)+\tilde{d}(z_m,y_m)\to 0\text{ as }m,n\to \infty .\end{array}$$

Let $\{y_n\}$ converge to some $y$ in $\tilde{X}$, by Claim 6. It follows that $\{z_n\}$ also converges to $y$:

$$\begin{array}{rl}\tilde{d}(z_n,y)& \le \tilde{d}(z_n,y_n)+\tilde{d}(y_n,y)\to 0\text{ as }n\to \infty .\end{array}$$

Thus, every Cauchy sequence in $\tilde{X}$ converges, and $\tilde{X}$ is complete.

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Completeness of $l_1$

Recall $l_p$ is a NLS for $1\le p\le \infty$ with the $p$-norm. We have shown $l_{\infty }$ is complete.

We will now show $l_1$ is complete.

Let $\{{\mathbf{x}}_n\}\subseteq l_1$ be Cauchy, with ${\mathbf{x}}_n=\{x_n^k{\}}_{k=1}^{\infty }$. Then, ${\lim }_{n,m\to \infty }\Vert {\mathbf{x}}_n-{\mathbf{x}}_m\Vert =0$, so

$$\begin{array}{rl}& \underset{n,m\to \infty }{\lim }\sum _{k=1}^{\infty }|x_n^k-x_m^k|=0\\ ⟹& \underset{n,m\to \infty }{\lim }|x_n^k-x_m^k|=0\text{for each }k.\end{array}$$

$\{x_n^k{\}}_{n=1}^{\infty }$ is Cauchy for each $k$. Let $\{x_n^k\}\to x^k$ for each $k$.

$\{x^k\}$ is our candidate limit for $\{{\mathbf{x}}_n\}$. We have to prove $\{x^k\}\in l_1$ and $\{{\mathbf{x}}_n\}\to \{x^k\}$. Let $M$ be such that $\Vert {\mathbf{x}}_n\Vert \le M$ for all $n$ (Cauchy sequences are bounded).

$$\begin{array}{rl}\sum _{k=1}^N|x^k|& \le \sum _{k=1}^N|x^k-x_{n_0}^k|+\sum _{k=1}^N|x_{n_0}^k|\\ & =\underset{m\to \infty }{\lim }\sum _{k=1}^N|x_m^k-x_{n_0}^k|+\sum _{k=1}^N|x_{n_0}^k|\\ & \le \underset{m\to \infty }{\lim }\Vert {\mathbf{x}}_m-{\mathbf{x}}_{n_0}\Vert +\sum _{k=1}^N|x_{n_0}^k|\\ & \le M+\Vert {\mathbf{x}}_{n_0}\Vert \end{array}$$

for all $N$. Thus, ${\sum }_{k=1}^{\infty }|x^k|<\infty$ (monotone convergence!), and $\{x^k\}\in l_1$.

It remains to show $\{{\mathbf{x}}_n\}\to \{x^k\}$, that is,

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

Let $ϵ>0$. Choose $n_0$ such that for all $n,m\ge n_0$, $\Vert {\mathbf{x}}_n-{\mathbf{x}}_m\Vert <ϵ$. Then, for $n\ge n_0$, we have

$$\sum _{k=1}^N|x_n^k-x^k|=\underset{m\to \infty }{\lim }\sum _{k=1}^N|x_n^k-x_m^k|\le \underset{m\to \infty }{\lim }\Vert {\mathbf{x}}_n-{\mathbf{x}}_m\Vert \le ϵ$$

for all $N$. Then, for $n\ge n_0$,

$$\begin{array}{r}\sum _{k=1}^{\infty }|x_n^k-x^k|\le ϵ.\end{array}$$

We are done!

Exercise 164.7(Exercise).

Prove $l_2$ is complete.

Footnotes

1. Technically, in $\varphi (X)$, where $\varphi$ is the inclusion map $X\to \tilde{X}$. ↩