# How to show that $A \subset X$ is totally bounded $\iff$ $\bar{A}$ is totally bounded.

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.

Thank you.

• 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 $S_\epsilon(a)$. – Keen-ameteur Oct 16 '18 at 19:24

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.
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 $$\overline A$$.