Is the collection of compact sets in a metric space equal to the set of closed and bounded sets, as long as the metric can go to infinity?

Question

In Munkres' Topology, Theorem 27.3 says:

A subspace $A \subset \mathbb R^n$ is compact if and only if it is closed and is bounded in the euclidean metric $d$ or the square metric $\rho$.

The proof makes sense to me. But right after theorem 27.3, he says:

Students often remember this theorem as stating that the collection of compact sets in a metric space equals the collection of closed and bounded sets. This statement is clearly ridiculous as it stands, because the question as to which sets are bounded depends for its answer on the metric, whereas which sets are compact depends only on the topology of the space.

A set $X$ being compact means that for any open covering of $X$, it has a finite subcollection that covers $X$ also. So I see why that depends only on the topology.

I was trying to think of a counterexample to the ridiculous statement above, and I think one could be the set $X = \mathbb R$, for both the discrete metric, and the Euclidean metric: $X$ isn't compact, $X$ is closed because it's the full space, but in the discrete metric case it is bounded, while in the Euclidean case it's not. So would that be an example where the same non-compact set is closed and bounded for one metric but not the other?

Assuming that's right, it's still a bit unsatisfying to me because the discrete metric feels somewhat "weird". So I'm trying to figure out what kind of metrics (other than the two in theorem 27.3) would obey the "sets are compact iff closed and bounded" rule.

Is there a clear cut answer? Is there some common property that make them obey the rule?

Answer 1

In your example the Euclidean metric and the discrete metric do not induce the same topology, thus it is not really relevant for your question.

Define a metric $d$ on $\mathbb R$ by $$d(x,y) = \min(\lvert x - y \rvert, 1 ) .$$ This metric is bounded, thus $\mathbb R$ is closed and bounded with respect to this metric.

It is easy to see that the Euclidean metric and $d$ induce the same topology (the standard topology) because they have the same open balls $B_r(x)$ for $r \le 1$. But $\mathbb R$ is not compact in the standard topology which provides a nice example for Munkres's remark.

By the way, the above construction works for any metric space $(X, d)$. Defining $d'(x,y) = \min(d(x, y), 1 )$ yields a bounded metric on $X$ which induces the same topology as $d$.

Munkres's Theorem 27.3 is true for all finite-dimensional normed vector spaces (any norm!). For infinite-dimensional normed vector spaces it is wrong. One can show that the closed unit ball is not compact (but clearly closed and bounded).

Answer 2

Paul Frost has already answered the first part of your question, and given the important example of finite dimensional normed vector spaces.

Regarding your final question, when a metric space has the property you describe, that closed bounded sets are compact, it is said to be "proper".

To determine when a metric space is proper, we start with freakish's comment on the original question, which gives the criterion for an arbitrary metric space to be compact - it must be complete and totally bounded.

From this, you can quickly deduce that a metric space is proper if and only if the metric space is complete, and balls are totally bounded.

Regarding conditions that guarantee a space is proper, we have already discussed finite dimensional normed spaces. A more general condition, of key importance in analysis on metric spaces, is the doubling property, which says that every ball of radius $R$ can be covered by at most $C$ balls or radius $\frac{R}{2}$, for some constant $C$. It is not hard to check that this doubling condition implies balls are totally bounded, so that every complete doubling metric space is proper.

In some sense, this isn't too far a step beyond the setting of finite dimensional normed spaces. The reason I say this is that if $X$ is doubling, a theorem of Assouad (strengthened by Naor and Neiman at the link) implies that for each $s\in(\frac{1}{2},1)$, $(X,d^s)$ embeds bi-Lipschitzly into some finite dimensional Euclidean space.

However, it should be noted that a metric space can be proper, or even be compact itself, without being doubling or embedding into any finite dimensional space, even topologically. For example, $X=[0,1]^{\mathbb N}$ is metrizable, and compact by Tychonoff's theorem, yet infinite dimensional, hence no matter what metric is given to it, it certainly cannot be doubling or embed into a finite dimensional Euclidean space.

Answer 3

Consider $X=\mathbb Q$, and the closed and bounded set $[0,\sqrt 2]\cap\mathbb Q=[0,\sqrt 2)\cap\mathbb Q$. This set is not compact as the open cover $\{(-1,\frac{n}{n+1}\sqrt 2)\cap\mathbb Q:n<\omega\}$ has no finite subcover.