3
$\begingroup$

So, I'm basically being asked to prove the following Theorem: A metric space is totally bounded if and only if its completion is compact.

Here is the definition of completion: A completion of a metric space $(X,d)$ is a pair consisting of a complete metric space $(X^{*},d^{*})$ and an isometry $\phi :X\rightarrow X^{*}$ such that $\phi (X)$ is dense in $X^{*}$.

Proof. $(\Rightarrow )$ Suppose a metric space $(X,d)$ is totally bounded, then so is its completion. So, the completion is a totally bounded, complete metric space. (A metric space is totally bounded and complete if and only if it is compact.) Hence, the completion is compact.

$(\Leftarrow )$ Suppse the completion is compact. (A metric space is totally bounded and complete if and only if it is compact.) Then it is totally bounded and so are its subsets. So, the metric space is totally bounded.

Can someone explain to me further why we have that if a metric space is totally bounded, then so is its completion? I'm not quite understanding why this is true.

$\endgroup$
2
$\begingroup$

Suppose that $\langle X,d\rangle$ is totally bounded. Then for each $\epsilon>0$ there is a finite $F_\epsilon\subseteq X$ such that

$$\bigcup_{x\in F_\epsilon}B_d(x,\epsilon)=X\;.$$

We want to show that there is a finite $F_\epsilon^*\subseteq X^*$ such that

$$\bigcup_{x\in F_\epsilon^*}B_{d^*}(x,\epsilon)=X^*\;,$$

so let $\epsilon>0$. Since $\varphi[X]$ is dense in $X^*$, for each $y\in X^*$ and $\epsilon>0$ there is an $x\in X$ such that

$$d^*\big(y,\varphi(x)\big)<\frac{\epsilon}2\;.$$

There is a $z\in F_{\epsilon/2}$ such that

$$d^*\big(\varphi(x),\varphi(z)\big)=d(x,z)<\frac{\epsilon}2\;,$$

so $d^*\big(y,\varphi(z)\big)<\epsilon$ by the triangle inequality. Thus,

$$\bigcup_{z\in F_{\epsilon/2}}B_{d^*}\big(\varphi(z),\epsilon\big)=X^*\;,$$

and and we may set $F_\epsilon^*=\varphi[F_{\epsilon/2}]$.

$\endgroup$
  • 1
    $\begingroup$ @Michael: You’re welcome! More generally, if a metric space has a dense subspace that is totally bounded, so is the original metric space; it’s basically the same proof. $\endgroup$ – Brian M. Scott Aug 11 '16 at 18:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.