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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
    Commented May 15, 2018 at 10:45
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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
    Commented May 15, 2018 at 12:04
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    $\begingroup$ You might be interested in the classic Toronto space problem: is there an uncountable Hausdorff space $X$ such that any $Y \subseteq X$ with $|Y| = |X|$ is homeomorphic to $X$? It's still open, even for size $\aleph_1$, AFAIK. Note that your problem is different as it asks for open sets etc. $\endgroup$
    – Henno Brandsma
    Commented May 16, 2018 at 22:26
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    $\begingroup$ There are related notions of $B$-homogenous: a space is called that if there is a base $\mathcal{B}$ for the open sets such for every $B_1, B_2 \in \mathcal{B}$ there is a homeomorphism $h$ of $X$ such that $h[B_1] = B_2$. Open problem: is every topological vector space $B$-homogenous (yes for locally convex). $\endgroup$
    – Henno Brandsma
    Commented May 16, 2018 at 22:39
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    $\begingroup$ And a zero-dimensional space $X$ is called strongly homogeneous when every non-empty clopen subset of $X$ is homeomorphic to $X$. $\endgroup$
    – Henno Brandsma
    Commented May 16, 2018 at 22:40

1 Answer 1

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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
    Commented May 15, 2018 at 12:59
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    $\begingroup$ @NikWeaver: I think it should say empty interior. $\endgroup$
    – Will Brian
    Commented May 15, 2018 at 13:01
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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
    Commented May 15, 2018 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
    Commented May 15, 2018 at 14:14
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    $\begingroup$ Of course @WillBrian you are right! That's nice, maybe you should post it as an answer? Or if you prefer, I can add it on my answer. $\endgroup$
    – N. de Rancourt
    Commented May 16, 2018 at 9:57

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