# If $E$ is a totally bounded subset of a metric space $X$. Then, any subset of $E$ is totally bounded.

Proof: Let $D \subset E$, where $E \subset X$ and ($X,d$) is a metric space. Suppose that $E$ is totally bounded. That is: for all $\varepsilon > 0$, there exist finitely many points $x_1, \ldots , x_n \in X$ such that: $E \subset \bigcup_{i = 1}^{n} B(x_i, \varepsilon)$.

So: given $\varepsilon > 0$, we have $D \subset E$ and $E \subset \bigcup_{i = 1}^{n} B(x_i, \varepsilon)$ with $x_1, \ldots , x_n \in X$. But then, by the transitive property (applied to subsets), it follows that $D \subset \bigcup_{i = 1}^{n} B(x_i, \varepsilon)$, where $x_1, \ldots , x_n \in X$. But this is to say that $D$ is totally bounded, as claimed. QED.

Did I do anything wrong or am I correct? Thanks!

• This is correct. Commented Dec 9, 2016 at 3:12
• Here's a related result that you might want to try proving: the closure of a totally bounded set is totally bounded. Commented Dec 9, 2016 at 3:15
• I did that one earlier today! Thanks for the suggestion. That one was definitely trickier than this one. Commented Dec 9, 2016 at 3:18