I would like to show that if I have a metric space $(M,d)$ and that $M$ is totally bounded, that its closure $cl(M)$ is also bounded.

My general strategy is to show that for every $\epsilon > 0$, there exist $x_1,...,x_n \in M$ such that $M = \cup B_{\epsilon/2}(x_i)$. And then I use the fact that this works for $\epsilon$ also since it works for $\epsilon /2$. However, should I be having two cases? I am listing the cases as follows:

1st case: $cl(M) = M$

2nd case: $cl(M) \neq M$.

Thank you!


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.


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 $x$.


Note firstly that an open ball is contained in a closed ball of the same radius, and a closed ball is contained in an open ball of any larger radius.

  1. By definition, if $S$ is totally bounded it is covered by a finite number of open balls radius $\epsilon/2$ centered on points in $S$.
  2. Closed balls of the same radius $\epsilon/2$ on the same centres therefore also cover $S$.
  3. Their finite union, $C$ is a closed set containing $S$.
  4. By definition, $cl(S)$, the closure of $S$ is the intersection of all closed sets containing $S$, and is a subset of every closed set that contains $S$ (in particular, $C$).
  5. So, $C$ contains $cl(S)$.
  6. Open balls of radius $\epsilon$ on the same centres cover the corresponding closed balls, therefore cover $C$ , therefore cover $cl(S)$.

So, $cl(S)$ is totally bounded.

(adapted from https://math.stackexchange.com/q/693968)


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