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May 16, 2018 at 22:40 comment added Henno Brandsma And a zero-dimensional space $X$ is called strongly homogeneous when every non-empty clopen subset of $X$ is homeomorphic to $X$.
May 16, 2018 at 22:39 comment added Henno Brandsma 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).
May 16, 2018 at 22:26 comment added Henno Brandsma 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.
May 15, 2018 at 17:31 vote accept Dominic van der Zypen
May 15, 2018 at 13:40 vote accept Dominic van der Zypen
May 15, 2018 at 14:33
May 15, 2018 at 12:40 answer added N. de Rancourt timeline score: 20
May 15, 2018 at 12:04 comment added Joel David Hamkins 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.
May 15, 2018 at 11:47 comment added Dominic van der Zypen 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 $\mathbb{Q}$.
May 15, 2018 at 11:44 comment added Joel David Hamkins Do you have any uncountable examples of this phenomenon?
May 15, 2018 at 11:32 history edited Dominic van der Zypen CC BY-SA 4.0
my example of $2^\lambda$ is false
May 15, 2018 at 11:27 comment added Dominic van der Zypen You are right - thanks Joel! I'll add this as a note in the question
May 15, 2018 at 10:45 comment added Joel David Hamkins 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 $\{0,1\}^\lambda$?
May 15, 2018 at 8:14 history asked Dominic van der Zypen CC BY-SA 4.0