Equivalence of precompactness and completely boundedness.

Question

Definitions:

The set $A\subset X$ is called completely bounded if $\forall \epsilon >0 \ \exists x_1,...,x_k \in A$ s.t. $A \subset \bigcup_{i=1}^k B(x_i,\epsilon)$. The set $A$ is called precompact if its closure is compact.

Let $X$ be a complete metric space. Then $A$ is precompact iff $A$ is completely bounded.

$\implies$

Suppose $A$ is precompact. Take $\epsilon >0$. Now take a finite subcover $\bar{A} \subset \bigcup_{i=1}^k B(y_i,\epsilon)$ from the cover $\bar{A} \subset \bigcup_{y\in \bar{A}} B(y,\epsilon)$. Now we have necessary $x_1,...x_k$, where $x_i=y_i$ if $y\in A$ and $x_i$ is any element on the epsilon neighborhood of $y_i$ if $y$ is on the boundary of $A$.

$\impliedby$

Here I'm stucked. Can I say that $x_n$ has a convergent subsequence if $\forall \epsilon >0 , N>0 \ \exists n,m >N : |x_n-x_m|<\epsilon$ ? Is it correct statement?

Answer 1

You have the right general idea, but you need to work a little harder than that.

Let $\sigma=\langle x_n:n\in\Bbb Z^+\rangle$ be a sequence in $A$. Use the fact that $A$ is totally bounded to show that $\sigma$ has a Cauchy subsequence; that’s a somewhat stronger statement than the one that you wrote.

• First show that there is a $y_1\in A$ such that $\{n\in\Bbb Z^+:x_n\in B(y_1,1)\}$ is infinite, and let $N_1=\{n\in\Bbb Z^+:x_n\in B(y_1,1)\}$.

• Show that there is a $y_2\in A$ such that $\left\{n\in N_1:x_n\in B\left(y_2,\frac12\right)\right\}$ is infinite, and let $N_2=\left\{n\in N_1:x_n\in B\left(y_2,\frac12\right)\right\}$.

Continuing in this fashion, for $m\in\Bbb Z^+$ recursively define points $y_m\in A$ and infinite $N_m\subseteq\Bbb Z^+$ such that $x_n\in B\left(y_m,\frac1m\right)$ for each $n\in N_m$, and $N_{m+1}\subseteq N_m$.

Now let $n_1$ be the smallest member of $N_1$, $n_2$ the smallest member of $N_2$ that is larger than $n_1$, $n_3$ the smallest member of $N_3$ that is larger than $n_2$, and so on. That is, for each $m\in\Bbb Z^+$ let

$$n_{m+1}=\min\{n\in N_{m+1}:n>n_m\}\;.$$

Then $\langle n_m:m\in\Bbb Z^+\rangle$ is a subsequence of $\sigma$.

• Show that this subsequence is Cauchy. Conclude that it converges to some point of $\operatorname{cl}A$.

This shows that every sequence in $A$ has a convergent subsequence. The final step is to extend this to $\operatorname{cl}A$. Suppose that $\sigma=\langle x_n:n\in\Bbb Z^+\rangle$ is a sequence in $\operatorname{cl}A$. For each $n\in\Bbb Z^+$ there is a $y_n\in A$ such that $d(x_n,y_n)<\frac1n$. (Why?)

• We’ve just shown that $\langle y_n:n\in\Bbb Z^+\rangle$ has a convergent subsequence $\langle y_{n_k}:k\in\Bbb Z^+\rangle$. Show that $\langle x_{n_k}:k\in\Bbb Z^+\rangle$, the corresponding subsequence of $\sigma$, converges to the same point.

Once you’ve done that, you’ll have shown that $\operatorname{cl}A$ is sequentially compact and hence compact.