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$Y \subseteq(X,d)$

If $Y$ is totally bounded then for each $\epsilon$ there is $\{ y_1,y_2,...,y_n\}$ such that $Y \subset \cup_i^n B(y_i,\epsilon)$.

Now let's say that I want to bound the set Y. I can choose $\epsilon = M $ and there exist un point $y_0 \in Y$ such that $Y \subset$ B$(a,M)$ because $Y$ is totally bounded.

Is this right?

CrispyBacon
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The claim is true, but your argument is not valid. If you take, say, $\varepsilon =1$, then the definition of totally bounded does not give you a single $a\in X$ with $Y\subseteq \mathrm B(a,1)$. But, you are very close. Work correctly with the definition, and you'll get a bunch of finitely many balls of radius $1$ which together cover the entire space. Do something with them to get your result.

Ittay Weiss
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  • So I need to show that it's bounded by finitely many balls? – CrispyBacon Oct 13 '18 at 17:54
  • No, you need to show that it is bounded by one ball. You know it is bounded by finitely many balls due to it being totally bounded. You need to work with what you actually know toward what you want to show. – Ittay Weiss Oct 13 '18 at 18:32
  • Since we have n finitely many ball of raidus epsilon , let M = n * 2* epsilon, then the set is bounded by B(M/2, M) – CrispyBacon Oct 13 '18 at 18:38