I'm quite new to general topology, so I apologize if this question is trivial. The famed Heine-Borel theorem states that in $\mathbb{R}^n$, a set is compact if and only if it is closed and bounded. Clearly this result doesn't hold over all metric spaces, (non-complete ones are the usual go-to examples). There's another theorem that states that a metric space is compact if and only if it is complete and totally bounded. I have a couple questions about this.
1) As the second theorem requires completeness and total boundedness, could you show me an example of a metric space that is bounded, complete, but not compact?
2) Second, (a softer question), what kinds of spaces satisfy what I'll call the "Heine-Borel property" (i.e. subspaces of them are compact if and only if they are closed and bounded). Is $\mathbb{R}^n$ the only example of such a space? If not, what other things can we say about that class of spaces?
Best Answer
For $(1)$, consider the discrete metric (all distinct points are distance $1$ apart) on any infinite set. This is (trivially) complete and bounded, but not compact.
For $(2)$, the following are equivalent for a metric space $X$:
$X$ has the Heine-Borel property.
For every point $x\in X$, every closed ball of finite radius centered on $x$ is compact.
For some point $x\in X$, every closed ball of finite radius centered on $x$ is compact.
The implications $1\rightarrow 2\rightarrow 3$ are immediate; the interesting one is $3\rightarrow 1$. Suppose $a\in X$ has the property that every closed ball of finite radius centered on $a$ is compact, and let $S\subseteq X$ be closed and bounded; we want to show that $S$ is compact. Let $r=\sup\{d(a, v): v\in S\}$; then $S$ is a closed subset of the closed ball centered on $a$ of radius $r+1$. But any closed subset of a compact set is compact, and that closed ball is compact by assumption on $a$.
Note that this tells us that every space with the Heine-Borel property is $\sigma$-compact, that is, a union of countably many compact subsets. The converse is false: any countably infinite discrete metric space is trivially $\sigma$-compact, but doesn't have the Heine-Borel property (since it itself is closed-in-itself, bounded, but not compact).