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We say that a $T_2$-space $(X,\tau)$ has homeomorphic open sets if every non-empty open set $U\subseteq X$ endowed with the subspace topology is homeomorphic to $(X,\tau)$.

The rationals with the Euclidean topology are an example of a space with homemorphic open sets, as well as $\{0,1\}^\lambda$, where $\lambda$ is an infinite cardinal and $\{0,1\}$ carries the discrete topology.

Is there for every infinite cardinal $\kappa$ a space of cardinality $\kappa$ with homeomorphic open sets?


Edit. Apologies for falsely claiming that $\{0,1\}^\lambda$ is a space with homeomorphic open sets - Joel David Hamkins' comment below gives an argument refuting my claim.

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    $\begingroup$ If you remove a point from $2^\omega$, it is no longer compact, and so not homeomorphic to the whole space $2^\omega$. Doesn't this refute your claim about $\{0,1\}^\lambda$? $\endgroup$ – Joel David Hamkins May 15 '18 at 10:45
  • $\begingroup$ You are right - thanks Joel! I'll add this as a note in the question $\endgroup$ – Dominic van der Zypen May 15 '18 at 11:27
  • $\begingroup$ Do you have any uncountable examples of this phenomenon? $\endgroup$ – Joel David Hamkins May 15 '18 at 11:44
  • $\begingroup$ I thought I had :-) but now it looks like $\mathbb{Q}$ is the only example I have. It would also be interesting to see another countable example, not homeomorphic to $\mathbb{Q}$. $\endgroup$ – Dominic van der Zypen May 15 '18 at 11:47
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    $\begingroup$ I wonder whether $2^{<\kappa}$ for uncountable ordinals $\kappa$ would be an uncountable example analogous to $\mathbb{Q}$, but I worry that cofinality differences in the order could show up in the homeomorphism types of different open sets. $\endgroup$ – Joel David Hamkins May 15 '18 at 12:04
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The Baire space $\omega^\omega$ has homeomorphic open sets: it is an immediate consequence of the fact that it is the unique nonempty zero-dimensional Polish space, up to homeomorphism, every compact subset of whose has empty interior (theorem 7.7 in Kechris). In fact we actually have better:

Proposition. For every infinite set $X$ with the discrete topology, $X^\omega$ has homeomorphic open sets.

Proof. Let $U \subseteq X$ be nonempty open. I claim that $U$ is a union of exactly $\kappa$ disjoint cones, where $\kappa = |X|$. To see this, it is enough to show that $U$ is a union of $\leqslant \kappa$ disjoint cones, as a cone is itself a union of $\kappa$ disjoint cones. But if $T \subseteq X^{< \omega}$ is a nonempty tree such that $[T] = U^c$, then $U$ is the disjoint union of all cones $N_s$ where $s \notin T$ but $s_{\restriction |s| - 1} \in T$.

Now, writing $U$ as a union of $\kappa$ disjoint cones, we can get a homeomorphism between $U$ and $X$ by gluing together homeomorphism between these cones and cones of the form $N_{(x)}$, $x \in X$.

So the answer to Dominic's question is positive at least for cardinals of the form $\kappa^\omega$.

Edit: Will Brian remarked, in the comments, that the same argument actually enables to show that the subspace of $X^\omega$ whose elements are eventually constant also had homeomorphic open sets. So for every infinite cardinal, there exists a space with homeomorphic open sets.

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    $\begingroup$ Every nonempty compact subset of $\omega^\omega$ has nonempty interior? What about singletons? $\endgroup$ – Nik Weaver May 15 '18 at 12:59
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    $\begingroup$ @NikWeaver: I think it should say empty interior. $\endgroup$ – Will Brian May 15 '18 at 13:01
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    $\begingroup$ Yes @NikWeaver, Will Brian was right: it was "empty interior". I edit to correct the mistake. $\endgroup$ – N. de Rancourt May 15 '18 at 13:43
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    $\begingroup$ However @DominicvanderZypen, maybe you should not accept the answer yet, the question is still interesting as we have no example for a lot of cardinals. It would be nice if other people could look at this question. $\endgroup$ – N. de Rancourt May 15 '18 at 13:46
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    $\begingroup$ Doesn't your argument apply equally well to the subspace of $X^\omega$ consisting of eventually constant sequences? It's still true that the closed sets can be expressed as trees, and it's still true that every open set is a union of disjoint cones. But this subspace has cardinality $|X|$, so this version of the argument would work for any infinite cardinal. $\endgroup$ – Will Brian May 15 '18 at 14:14

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