Showing that the closure of a totally bounded set is totally bounded
I would like to show that if I have a subset $M$ of a metric space $(X,d)$ such that $M$ is totally bounded, then its closure $cl(M)$ is also totally bounded.
My general strategy would be to show that for every $\epsilon > 0$, there exist $x_1,...,x_n \in M$ such that $M = \bigcup B_{\epsilon/2}(x_i)$. And then I would use the fact that this works also for $\epsilon$ since it works for $\epsilon /2$. However, should I be having two cases? I am listing the cases as follows:
- $cl(M) = M;$
- $cl(M) \neq M$.
Solution 1:
Just as you mentioned, given $\varepsilon>0$, we can find $x_1,\ldots,x_n\in M$ such that $M\subseteq \bigcup_i B_{\varepsilon/2}(x_i)$. Let's show that $\operatorname{cl}(M)\subseteq\bigcup_i B_\varepsilon(x_i)$.
Let $z\in \operatorname{cl}(M)$. Then there exists $x\in M$ such that $d(z,x)<\varepsilon/2$, and there is some $i$ such that $d(x,x_i)<\varepsilon/2$, so $d(z,x_i)<\varepsilon$. Therefore, $\operatorname{cl}(M)\subseteq\bigcup_i B_\varepsilon(x_i)$.
Therefore, $\operatorname{cl}(M)$ is totally bounded.
Solution 2:
For the proof above, there is no need to do two cases.
Let $\epsilon > 0$. Since $M$ is totally bounded, there exists a finite collection of open balls $\left\{B_i\right\}_{i=1}^n$ of radius $\frac{\epsilon}{2}$ covering $M$. Each ball $B_i$ is centred at a point $x_i$.
Let $x\notin M$ be a limit point of $M$. Since $x$ is a limit point, the punctured ball of radius $\frac{\epsilon}{2}$ around $x$ contains a point $y \in M$. Furthermore, $y\in B_k$ for some $k$.
We can create a new cover $\left\{C_i\right\}_{i=1}^n$ where $C_i$ is the ball centred at $x_i$ with radius $\epsilon$. Note that $C_k$ contains $x$.