Let $(X, d)$ be a metric space.
Question: Is the closed ball of radius $r > 0$ centered at some $x \in X$, i.e., $\overline{B_r(x)} \subseteq X$ totally bounded?
By definition, a set $V \subseteq X$ is totally bounded if for any $\epsilon > 0$, there is a finite covering of $V$ with balls of radius $\epsilon$.
- When $X = \mathbb{R}^n$, the answer is yes: Because, $\overline{B_r(x)} \subseteq X$ is closed and bounded, hence compact (by Heine-Borel). We know that in metric spaces, compactness is equivalent to completeness and total boundedness.
- For infinite-dimensional normed spaces, we can use Riesz Lemma to show that the answer is false. Example: https://math.stackexchange.com/a/1774263
- For a general metric space $X$, I am visualizing that we can cover the closed ball as follows. I placed the closed ball in a surrounding square so that the visualization is easier.
where small $\epsilon$-balls are interlocking, to ensure that there the "gaps":
Clearly, for any $\epsilon > 0$, such a construction will require only a finite number of $\epsilon$ balls. However, there are several questions:
- Ignoring those $\epsilon$-balls that lie outside the closed ball, we need to ensure that remaining $\epsilon$ balls $B_\epsilon(y)$ have their center points inside $\overline{B_r(x)}$. If not, we need to "replace" $B_\epsilon(y)$ by finitely many $\epsilon$ balls with centers in $\overline{B_r(x)}$ that cover $B_\epsilon(y)$. Can we always do that?
- Positive answer: Is there an explicit formulation for the balls?
- Negative answer: If $\overline{B_r(x)}$ is not totally bounded, what part of this construction is faulty? (I am guessing (1) might be the culprit) Moreover, what are some examples where the closed ball is totally bounded, and examples where the closed ball is not totally bounded?
