# Compact subset of an open set

Let $X$ be complete separable metric space and $A\subset X$ is open. Does it mean that there is a compact subset of $A$? My solution is the following: since $A$ is open there is $B(x,r)\subset A$, then $\overline{B(x,r/2)}\subset A$ where the closed ball is compact.

• How about taking just a point (if $A$ is nonempty)? Also: Why is a closed ball compact (that's not true in general)? E.g. a closed ball in an infinite-dimensional Banach space is never compact. – t.b. Jun 16 '11 at 15:52
• I consider the question interesting, although the presented argument is not good. "Is there any complete separable metric space with no nontrivial (non-finite) compact subsets?" This is just the statement with $A=X$ and removing the easy case of finite compact sets. – Beni Bogosel Jun 16 '11 at 16:03
• Theo's comment also implies that your space isn't locally compact in general, so you might not even have compact sets with nonempty interior. – Miha Habič Jun 16 '11 at 16:06
• To elaborate on my and Miha's comments: A metric space is said to be proper if closed balls are compact. A proper metric space is locally compact but the converse is not true: the real line with the metric $d(x,y) = \min{\{|x-y|,1\}}$ is locally compact (it carries the usual topology) but not proper because each closed ball of radius $\geq 1$ is the entire real line. Your strategy works as is, if the space is proper and if it is locally compact, just take $r$ small enough to ensure that the ball is compact. – t.b. Jun 16 '11 at 16:36
• @Beni: if $X$ is discrete (and countable, as we need a separable space) then $X$ has only the finite subsets as compact subsets. Otherwise we always have non-trivial convergent sequences (converging to a non-isolated point) which are always compact. So for the original question: if there is a non-isolated point in $A$, then there is an infinite compact subset of $A$, otherwise $A$ is an at most countable set of isolated points, hence without any non-finite compact subsets. – Henno Brandsma Jun 16 '11 at 17:44

Some observations:

1. $X$ is finite, then it is trivially compact and every subset is compact.
2. $X$ is discrete, then it has to be countable, and a subset is compact if and only if it is finite, and then we are in trouble.
3. $X$ is non-discrete countable, then it is homeomorphic to some countable ordinal with the order topology, then every open set contains some interval which contains an isolated point which is compact.
We cannot really hope for a lot more, since if the ordinal has infinitely many limit ordinals below it we can find a discrete infinite subset which is open, which reduces the case to the previous one (e.g. $\omega^\omega$ with the set $\{\omega^n+4\mid n<\omega\}$, then the set is a discrete and open subspace)
4. $X$ is uncountable, then it has a perfect subspace and a countable scattered part. If $A$ intersects with the scattered part of $X$ then we reduce to one of the previous cases.

This means that we only really need to deal with the case where $X$ is perfect, which is indeed the hard case.

• Sorry to ask, but what question exactly are you answering? – t.b. Jun 17 '11 at 7:29
• @Theo: Last night there was a long answer covering the case of a perfect Polish space, I complemented the other cases, by morrow that was gone. I figured I should probably just leave my observation so the next person trying to approach this question might find it useful. Do you believe I should delete the answer as well, as it is rather incomplete? – Asaf Karagila Jun 17 '11 at 7:53
• You can't, as it is accepted now :) I saw the other answer with very interesting facts, such as "every countable metric space is discrete" and other major inaccuracies... Since it is still unclear to me what the OP actually wanted to ask I think your answer is just fine as it is. Anyway, thanks for the explanation. – t.b. Jun 17 '11 at 8:00

In the remaining case, in which the space has no isolated points, it's entirely possible that all compact sets have empty interior.

Let $(\mathbb P, \mathcal T)$ be the space of irrational numbers with the usual topology. Suppose that $K$ is a subset of $\mathbb P$ with non-empty interior; then there are rational numbers $p$ and $q$ such that $p < q$ and $(p,q) \cap \mathbb P \subseteq K$, where $(p,q)$ is the usual open interval in $\mathbb R$. Fix $r \in (p,q) \cap \mathbb Q$, choose $n \in \omega$ so that $2^{-n} < \min\{r-p,q-r\}$, and for $k>n$ let $F_k = [r-2^{-k},r+2^{-k}] \cap \mathbb P$, a clopen subset of $\mathbb P$. Clearly $\bigcap \limits_{k>n} [r-2^{-k},r+2^{-k}] = \{r\}$, so $\bigcap \limits_{k>n} F_k = \emptyset$. Thus, $K$ contains a nested family of non-empty closed sets with empty intersection and so cannot be compact. That is, every compact subset of $\mathbb P$ has empty interior.

On the other hand, $\mathbb P$ is a $G_\delta$ in the space of reals with the usual metric, which is complete, so $(\mathbb P, \mathcal T)$ is completely metrizable by some metric $d$, and $(\mathbb P, d)$ is then a complete, separable metric space without isolated points in which all compact sets have empty interior. (In fact $(\mathbb P, \mathcal T)$ can be characterized as the unique topologically complete, zero-dimensional, nowhere locally compact, separable metric space; this is a classic result of Alexandroff and Urysohn.)

On the other hand, every non-empty open set $U$ in a topologically complete, separable metric space without isolated points does contain a Cantor set. To see this, construct a tree of open sets $V_s \subseteq U$ indexed by finite sequences of zeroes and ones in such a way that for each index sequence $s$, $\text{cl } V_{s0} \cap \text{cl } V_{s1} = \emptyset$, $\text{cl } V_{s0} \cup \text{cl } V_{s1} \subseteq V_s$, and $\text{diam}(V_s) < 2^{-|s|}$, where the diameter is taken with respect to a complete metric on the space. For each infinite sequence $\sigma$ of zeroes and ones let $p_\sigma$ be the unique point in $\bigcap \{V_s:s \text{ is an initial segment of } \sigma\}$; completeness of the metric ensures the existence of each $p_\sigma$. Let $K = \{p_\sigma:\sigma \in 2^\omega\}$; the construction ensures that the map $h:2^\omega \to K:\sigma \mapsto p_\sigma$ is a homeomorphism and hence that $K$ is a Cantor set.

Thus, if $U$ is a non-empty open set in a complete, separable metric space, either $U$ contains a compact set with non-empty interior, or $U$ contains a Cantor set. That seems to be about the best that we can do in general by way of finding 'large' subsets of $U$.