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This is for homework, so please hints only.

Also I know that there are questions similar to this one, but those questions use a different definition of totally bounded.

The definition my book uses for totally bounded is:

A metric space $X$ is totally bound if $\forall \epsilon > 0$ there is a finite subset $A \subset X$ such that $X = \bigcup_{a \in A} S_{\epsilon}(a)$, where $S_r(x_0) = \{x \in X: d(x,x_0) < r\}$. The difference here is that other questions only require containment, not equality.

The question I have to answer is:

Let $X$ be a metric space, show that $A \subset X$ is totally bounded $\iff$ $\bar{A}$ is totally bounded.

I have shown that if $A$ is totally bounded then you can construct a cover in the that contains $\bar{A}$. My professor has said that I may need to show that $\forall \epsilon \exists \delta$ such that there exists a finite subset $C \subset \bar{A}$ such that $\bar{A} = \bigcup_{c \in C} S_{\delta}(c)$. But I'm not sure how to proceed.

I would really appreciate any help you can offer.

Thank you.

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  • $\begingroup$ Try using the triangle inequality for a finite $\epsilon$-cover, in order to obtain a finite $2\epsilon$-cover, where an $\epsilon$-cover is a cover as you've written above of $S_\epsilon(a)$. $\endgroup$ – Keen-ameteur Oct 16 '18 at 19:24
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Fact: $\overline{S_r(a)} \subseteq \{x: d(x,a)\le r\}$ for all $a\in X$ and $r>0$.

If$A$ is totally bounded and $\varepsilon>0$ we can find a finite subset $F$ of $X$ such that $$A \subseteq \bigcup\{ S_{\varepsilon \over 2}(x): x \in F\}$$

By an application of the above fact we see that $\overline{S_{\varepsilon \over 2}(x)} \subseteq S_\varepsilon(x)$ for all $x$ and closure distributes over finite unions, so:

$$\overline{A} \subseteq \bigcup\{ S_{\varepsilon}(x): x \in F\}$$

showing that the closure of $A$ is totally bounded. The reverse is trivial as a subset of a totally bounded subset is totally bounded.

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Let $C$ be a finite cover of $A$ by balls of radius $r/3$. Assume $x \in \bar A$. Thus there exists $y \in A \cap B(x,r/3)$ and some $B(a,r/3)$ in $C$ with $y \in B(a,r/3)$. Show $x \in B(a,r)$. Conclude $\{ B(a,r) : B(a,r/3) \in C \}$ is a finite cover of $\overline A$.

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