S is uncountable := |$\mathbb{N}$| < |S|
S is noncountable := |S| $\not\leq |\mathbb{N}|$
(X,$T$) is a nice space := (X,$T$) is a compact Hausdorff space without isolated points

Does [ ZF / ZF + Countable Choice ] prove that every nice space is [ uncountable / noncountable ] ?
If not, is it known to prove that the statement implies some choice principle?
What if the spaces are additionally assumed to be metrizable?

Now, that's basically 12 questions, so I certainly don't anticipate answers for all of them.
If it matters, the one I'm most curious about is "Does ZF prove that every nice space is noncountable?".


One of the usual ways of proving in ZFC that every compact Hausdorff space $X$ without isolated points (perfect space) is uncountable is by proving that there is a copy of the Cantor space $2^\omega$ inside it, as follows. Pick two points and separate them with neighborhoods $U_0, U_1$ having disjoint closures. Inside each of these neighborhoods, pick two points and separating neighborhoods $U_{00},U_{01}\subset U_0$ and $U_{10},U_{11}\subset U_1$ having disjoint closures inside those neighborhoods, and so on proceeding inductively. Every infinite binary sequence $s\in 2^{\omega}$ determines a unique nesting sequence of these sets, which must be nonempty. And so we have continuum many points in $X$, so it is uncountable.

This proof, however, makes several uses of the axiom of choice. First, we have the choices involved with picking the points to be separated, and second, the choices involved with picking the separating neighborhoods. Although there are only countably many choices being made here, this is an instance of Dependent Choice, a stronger principle than mere countable choice, since the choices are being made in succession. Finally, third, a subtle point, we have the choices involved in picking for each binary sequence a single point from the intersection of the corresponding nested neighborhoods. After all, there could be many points in that intersection.

With some additional assumptions on $X$, however, we can get around these uses of choice, and thereby obtain answers to some of your questions. For example, if we only aim to prove that $X$ is noncountable, rather than uncountable, then we may assume towards contradiction that $X$ is countable, which provides for us a canonical way of picking points from the space. (In the case of the first use of choice, it would suffice if $X$ were separable, since we could just pick points from a fixed countable dense set.) If $X$ were a metric space, then we have a canonical way to pick neighborhoods of any given point. Also, by making these neighborhoods shrink to $0$ as the construction proceeds, we ensure that the intersection of the nested sets contains a single point.

Thus, this argument shows in ZF, without any choice, that every compact Hausdorff metric space having no isolated points is noncountable. More generally, it shows, again without any choice, that every separable compact Hausdorff metric space without isolated points has uncountable size at least continuum.

If we have Dependent Choice, then we can prove that every compact Hausdorff metric space is uncountable of size at least continuum, since DC allows us to overcome the first two uses of choice (picking the points and the neighborhoods), and by shrinking the neighborhoods we avoid the need for choice in the last step.

A clever person may be able to improve these arguments to cover additional cases.

Meanwhile, let me mention an interesting example on the other side of the question. This example illustrates that several of the usual equivalent formulations of compactness are no longer equivalent in the non-AC context. Namely, it is consistent with ZF that there is an infinite but Dedekind finite set $D$ of real numbers. That is, $D$ is infinite, but has no countably infinite subset. It follows that $D$ has at most finitely many isolated points, since otherwise we could enumerate the rational intervals and find these isolated points, thereby enumerating a countably infinite subset of $D$, which is impossible. Let us simply omit these finitely many isolated points and thereby assume without loss of generality that $D$ is an infinite Dedekind-finite set of reals having no isolated points. Since $D$ is Dedekind-finite, every sequence in $D$ has only finitely many values and hence has a convergent (constant) subsequence. Thus, $D$ is a sequentially compact set of reals. In other words, $D$ is a sequentially compact metrizable space with no isolated points. However, $D$ is not uncountable in the sense you mentioned, since we don't even have $|\mathbb{N}|\leq D$, as there is no countably infinite subset of $D$. Nevertheless, $D$ is noncountable.

  • $\begingroup$ Joel, I'm not following you when you say, "which provides for us a canonical way of picking points from the space". It sounds like you need to choose a bijection with $\mathbb{N}$ in order to proceed. $\endgroup$ – Todd Trimble Sep 12 '10 at 13:44
  • $\begingroup$ If $X$ is countable, you can fix an enumeration of $X$ for the rest of the proof, and use this to select points from $X$. $\endgroup$ – Joel David Hamkins Sep 12 '10 at 13:47
  • $\begingroup$ Yes. My head must be in topos theory too much, where we sometimes have difficulty making even just one choice. ;-) $\endgroup$ – Todd Trimble Sep 12 '10 at 14:00
  • 2
    $\begingroup$ Everything Joel said here is correct, but a warning for others about something that confused me once. We have to be careful about the term "perfect space". Sometimes people use the term for any space with no isolated points, like the Wikipedia article does. Without a compactness or completeness condition you can't prove in ZFC that every uncountable, second countable space $U$ contains a copy of Cantor space, or even has cardinality continuum, but you can prove $U$ has an uncountable closed subset with no isolated points. Proof sketch: throw out every countable basic open neighborhood of $U$. $\endgroup$ – Carl Mummert Sep 12 '10 at 23:44

I'd like to extend on Joel's answer, and point that ZF itself cannot prove that every "nice space" is uncountable.

It is consistent that the axiom of choice fails and there exists a Dedekind-finite set $X$ which can be topologized as follows:

  1. $X$ is Hausdorff compact;
  2. $X$ is strongly connected (every real valued function is constant);
  3. The topology is an order topology of a dense linear order with endpoints (if we remove the endpoints then this is a locally-compact space, and every closed interval is compact).

It follows that there are no isolated points, so this is a nice space. However the set itself is Dedekind-finite and so not uncountable (but it is still noncountable). Such topological space is called a Läuchli continuum.

For example if we consider Mostowski's ordered model, by adding two endpoints to the atoms the result is a Dedekind-finite $X$ with the above properties.

I elaborated on this in a recent math.SE answer which includes all the relevant references and more.

It should be pointed that in the great book of choice principles the existence of non-trivial Läuchli continua is the negation of Form 155; whereas countable choice is Form 8. I could not find much connection between the two forms in the site.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy