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This question is motivated by this one, where no simple solution within ZFC seems to exist. Let me ask a weaker question then.

Suppose that $K$ is a compact, Hausdorff, non-metrizable space. Does it contain a closed, non-metrizable, totally disconnected subspace?

Note that the cardinality of the (consistent) counterexample in Mathieu Baillif's answer is ludicrously large ($=\omega_{\omega_1}$). Hence my question.

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  • $\begingroup$ Did you intend to require Hausdorff as before? (There seem to be some easy counterexamples if not.) $\endgroup$ Sep 11, 2014 at 21:16
  • $\begingroup$ Thank you for pointing this out. Yes, I follow Engelking's convention that compact spaces are Hausdorff. Corrected. $\endgroup$
    – spooky
    Sep 11, 2014 at 21:20
  • $\begingroup$ Actually, the space $X$ (a circle bundle over $\omega_1$ due to Nyikos) I gave in my answer you cite is a consistent counter-example: it has the property that bounded subsets are metrizable, and closed unbounded subsets are not totally disconnected. The one point compactification of this space then does the job. $\endgroup$ Sep 12, 2014 at 8:18

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EDIT (24th feb. 2015): I added a paragraph at the end to answer a question of Tomek Kania.

As I wrote in a comment above, the Nyikos space $X$ I used in my answer to the question cited above gives a consistent counter-example (when taking the one-point compactification). This space is built with the help of the axiom $\diamondsuit$. I just noticed that a construction can be made using only the axiom $\clubsuit_C$ ("club for clubs") which is much weaker. In particular, $\clubsuit_C$ is compatible with MA + $\neg$CH. I am 100% sure that this construction has been made before, by Nyikos probably, but I did not find it in print, so here is a sketch of how it works.

Let $\Lambda$ be set of limit ordinals in $\omega_1$. The axiom $\clubsuit_C$ says:

$\clubsuit_C$: There exists $\langle S_\lambda:\lambda\in\Lambda\rangle$ such that $S_\lambda$ is an increasing $\omega$-sequence converging to $\lambda$, with the property that if $C\subset\omega_1$ is closed and unbounded (abbreviated by "club"), there is some $\lambda\in\Lambda$ with $S_\lambda\subset C$.

The idea is the following. The underlying set of the space $Z$ is $\omega_1\times[0,1]$ (in the end we take the one-point compactification and basically add $\{\omega_1\}\times\{0\}$). Write $S_\lambda=\langle s^\lambda_1, s^\lambda_2,\dots\rangle$ in increasing order. We define the topology by induction such that $Z_\lambda=\lambda\times[0,1]$ is homeomorphic to a closed subspace of $\lambda\times[0,1]$ with the product topology. Thus, $Z_{\lambda+1}$ is compact metrizable. Moreover, the homeomorphism preserves the fibers. The way the topology on $Z_{\lambda+1}$ (for limit $\lambda$) is defined is summarized in this picture.

Topology in $Z_{\lambda+1}$

We arrange such that for any $\omega$-sequence $x_n$ in $[0,1]$, the sequence $\langle s_n^\lambda,x_n\rangle$ has all of $\{\lambda\}\times[0,1]$ in its closure.

If $\lambda$ is limit, the topology of $Z_\lambda$ is just the one given by the increasing union, and $Z_{\lambda+2}$ is the disjoint union of $Z_{\lambda+1}$ and $\{\lambda+1\}\times[0,1]$.

Taking $Z=\cup_{\lambda<\omega_1}Z_\lambda$, we obtain a countably compact first countable locally compact and locally metrizable space. If $E\subset Z$ is closed non-metrizable, then it is not contained in any $Z_\lambda$ and one sees easily that it must intersect the fibers in a closed and unbounded subset $C$ of $\omega_1$. But then, by $\clubsuit_C$, $E$ must contain a sequence $\langle s_n^\lambda,x_n\rangle$ and thus all of $\{\lambda\}\times[0,1]$ for some $\lambda$. Thus $E$ is not totally disconnected. The one point compactification of $Z$ has the same property.

I don't know whether this type of things can be done in ZFC alone. The PFA (and weaker axioms as well) impedes such a construction with a first countable countably compact $Z$ because it imposes the presence of a copy of $\omega_1$ in it. And there is a complete forest of results about compact spaces under PFA which might imply a positive answer, but I am too ignorant to know.

Notice that using $Z$ instead of $X$ and any cardinal $\kappa> 2^\omega$ of cofinality $\omega_1$ instead of $\omega_{\omega_1}$ in my answer here we obtain a counter-example for that question as well under $\clubsuit_C$.


Tomek Kania asked whether there is a hereditarily separable such example. After some browsing of the litterature, I found one which exists under $\diamondsuit$. It is described on pages 652--655 of Nyikos's paper "The theory of nonmetrizable manifolds" in the Handbook of set-theoretic topology. The idea (due to Rudin-Zenor) is to use $CH+\clubsuit$ (which is equivalent to $\diamondsuit$) to obtain a $2$-dimensional nonmetrizable hereditarily separable manifold $M=\cup_{\alpha<\omega_1}M_\alpha$ such that $M_\alpha$ is a metrizable dense submanifold of $M$. ($M$ is in fact perfectly normal and countably compact.) Moreover, $M_{\alpha+1}$ is obtained by "insterting" a half open interval $[0,1)$ inside $M_\alpha$ in so that the following holds (see p.654):

Claim: If $F$ is a closed non-Lindelöf subset of $M$, then $F$ contains a copy of $[0,1)$, and is therefore not totally disconnected.

To finish, notice that if $F$ is non-metrizable, then it is non-Lindelöf. Then, take the one point compactification of $M$ to obtain the desired example.

I know that the original Rudin-Zenor construction of a perfectly normal nonmetrizable (hereditarily separable) manifold was done with $CH$ alone, I'd have to check whether the above claim also holds in their construction.

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  • $\begingroup$ Mathieu, is the space you've constructed separabe? I believe not, yet I don't see how to prove it. $\endgroup$ Feb 21, 2015 at 21:57
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    $\begingroup$ You are right, it is not separable. Notice that the $Z_\lambda$ are clopen. Since a countable set is contained in some $Z_\lambda$ (if it does not contain the point at infinity), its closure is contained in $Z_{\lambda}$ and is thus not dense, since it does not intersect $Z_{\lambda+2}-{Z_{\lambda}}$. $\endgroup$ Feb 22, 2015 at 12:57
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    $\begingroup$ I have no idea, I'm afraid. (By the way, what is a simple example of totally disconnected, non-metrizable, hereditarily separable compact space ?) $\endgroup$ Feb 22, 2015 at 13:16
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    $\begingroup$ The double arrow space (aka the split interval) is the canonical example. $\endgroup$ Feb 22, 2015 at 13:31
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    $\begingroup$ @Tomek Kania. Having thought a little bit about it, I believe that a variant of the construction of the Kunen line or the hereditarily separable (perfectly normal) non-metrizable manifold of Rudin-Zenor might yield such a counter-example. I'd have to spend more time on it, though. $\endgroup$ Feb 24, 2015 at 13:07
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The following is Question 374 in Open Problems in Topology II, and it is attributed to S. Todorcevic:

Is it consistent that every nonmetrizable perfect compactum contains a closed subspace with uncountably many clopen sets?

By the context I think "perfect compactum" means "perfecly normal compact space". Note that a closed, non-metrizable, totally disconnected subspace of a compact space must have uncountably many clopen sets. So it seems that either the consistency of a positive answer to the OP´s question is an open problem (at least in 2007), or there is a real counterexample that we are missing (note that the OP´s question does not require the space to be $T_6$ and requires the subspace to be zero-dimensional, so it should be easier to find a counterexample).

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