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It is a remarkable property of uncountable compact metric spaces that each of them contains a homeomorphic copy of the Cantor set. In general, one cannot expect containment of Cantor cubes (in particular, in the case of scattered compact spaces) but instead different totally disconnected subspaces occur quite often. My question is the following

Let $K$ be a compact Hausdorff space. Does $K$ contain a closed, totally disconnected subspace having the same cardinality as $K$?

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EDIT: I tried to improve the presentation. I hope that it is a bit more readable now.

I think that the following construction gives a consistent counter-example. It uses the combinatorial principle $\diamondsuit$, which implies CH (the continuum hypothesis) and is independant from ZFC.

I'll work with ordinals, and since $\alpha<\omega_1$ (for instance) may be seen either as a point or as a an initial segment in $ \omega_1 $, I'll use the interval notation when I refer to the latter.

Start with a circle bundle over $\omega_1$, that is, a space $X$ with a continous map $\pi$ onto $\omega_1$, such that the fiber over each point $\pi^{-1}(\{\alpha\})$ is a circle. (Of course, $\omega_1$ is endowed with the order topology.) Say that a subset of $X$ is bounded if it is contained in $\pi^{-1}([0,\alpha])$ for some $\alpha<\omega_1$, and unbounded otherwise.

Nyikos showed in his article "The theory of non-metrizable manifolds" in the Handbook of set theoretic topology (Example 6.17) that using $\diamondsuit$, we can construct a circle bundle over $\omega_1$ with the following properties:

1) The underlying set of $X$ is $\omega_1\times\mathbb{S}^1$, but the topology is not the product topology.

2) $\pi^{-1}([0,\alpha])$ is homeomorphic to $[0,\alpha]\times\mathbb{S}^1$ with the usual product topology for any $\alpha<\omega_1$. In particular, $X$ is locally compact. We may fix an homeomorphism $\psi_\alpha:\pi^{-1}([0,\alpha])\to [0,\alpha]\times\mathbb{S}^1$ for each $\alpha<\omega_1$.

3) If $E\subset X$ is closed, then either $E$ is bounded, or $E$ contains $\pi^{-1}(C)$ for $C$ a closed and unbounded subset of $\omega_1$. In particular, in the latter case $E$ is not totally disconnected since it contains copies all the fibers (which are circles) above the member of $C$.

(Note: Actually, Nyikos builds a bundle over the long ray ${\mathbb{L}}_+$ which is even a surface, but we take only the restriction to $\omega_1$.)

Now, define a circle bundle $\pi':Y\to\omega_{\omega_1}$ as follows. The underlying set of $Y$ is $\omega_{\omega_1}\times\mathbb{S}^1$. $X$ will be included in $Y$ as the union of the fibers above the $\omega_\alpha$, for $\alpha<\omega_1$. The topology "between" $\omega_\alpha$ and $\omega_{\alpha+1}$ is the usual product topology. That is, if $\omega_\alpha < \gamma < \omega_{\alpha+1}$, given $x\in\mathbb{S}^1$, take $\beta,\beta'$ with $\omega_\alpha\le\beta<\gamma<\beta'< \omega_{\alpha+1}$ and an open $O\subset\mathbb{S}^1$ containing $x$, then $(\beta,\beta')\times O$ is a neighborhood of $\langle\gamma,x\rangle$.

We now define the neighborhoods of $\langle\omega_{\alpha},x\rangle$. In $X$, choose a neighborhood $U$ of $\langle \alpha,x\rangle$, and denote by $U^\beta$ the intersection $U\cap \pi^{-1}(\{\beta\})$. Set $V^{\omega_\beta}=U^\beta$, and if $\omega_\beta<\gamma<\omega_{\beta+1}$, set $V^{\gamma}=U^{\beta+1}$. That is: $V^\gamma$ is equal to the intersection of $U$ with the fiber over $\beta+1$. Then a neighborhood of $\langle\omega_{\alpha},x\rangle$ is given by the union of $\{\gamma\}\times V^{\gamma}$ for all $\gamma$ greater than some $\gamma'$.

Then $Y$ has the following properties:

2') For any $\alpha<\omega_{\omega_1}$, $(\pi')^{-1}([0,\alpha])$ is homeomorphic to $[0,\alpha]\times\mathbb{S}^1$ with the usual product topology (and thus $Y$ is locally compact). To see this, define the homeomorphism $\phi:(\pi')^{-1}([0,\alpha])\to [0,\alpha]\times\mathbb{S}^1$ by setting $\phi(\langle\omega_{\alpha},x\rangle)=\psi(\langle\alpha,x\rangle)$, and for $\omega_\alpha<\gamma<\omega_{\alpha+1}$, set $\phi(\langle\gamma,x\rangle)=\psi(\langle\alpha+1,x\rangle)$, where $\psi$ is defined in 2) above.

3') As 3) with $\omega_{\omega_1}$ rather than $\omega_1$.

Now, a bounded subset of $Y$ is contained in some $[0,\alpha]\times\mathbb{S}^1$, and has thus cardinality $\max\{|\alpha|,\omega_1\}<\omega_{\omega_1}$ (since $2^\omega=\omega_1$ by CH). Taking the one-point compactification yields a compact space such that a closed set of cardinality $\omega_{\omega_1}$ cannot be totally disconnected.

I hope that I did not miss something.

I don't know whether another construction can be done in ZFC, but this particular one needs at least something more than ZFC+CH, because Nyikos space $X$ would have to contain a copy of $\omega_1$ (which is impossible, and thus the space does not exist) in a model of ZFC+CH due to Eisworth and Nyikos.

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  • $\begingroup$ Mathieu, could you please say a word or two about why your space is locally compact Hausdorff (so that it has a one-point compactification) and why a closed set in the compactification containing the "infinite" point cannot be totally disconnected and of size $\omega_{\omega_1}$? $\endgroup$ – Ramiro de la Vega Apr 12 '13 at 14:03
  • $\begingroup$ Ramiro, I have tried to improve the presentation to address your questions, I hope it is clearer now. $\endgroup$ – Mathieu Baillif Apr 14 '13 at 19:37
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    $\begingroup$ Thanks for accepting my answer, but it might be somewhat prematurate since I gave only a partial answer. $\endgroup$ – Mathieu Baillif Apr 19 '13 at 11:37
  • $\begingroup$ The same kind of construction seems to be possible under the weaker axiom $\clubsuit_C$, see mathoverflow.net/questions/180670/…. $\endgroup$ – Mathieu Baillif Sep 16 '14 at 19:47

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