This is a follow-up on an older question.
Let $\Box_{i\in I} X_i$ denote the box product of the spaces $X_i$. Is there a Hausdorff space $(X,\tau)$ with $|X|>1$ such that $\Box_{n\in\omega}X$ connected?
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asked Apr 11, 2018 at 7:16
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3$\begingroup$ While this does not completely answer your question, theorem 1.3(iii) of Scott W. William's chapter "Box Products" (chapter 4) in Kunen & Vaughan's Handbook of Set Theoretical Topology (1984) states that the box product of an infinite family of non-discrete Hausdorff completely regular topological spaces is never connected. $\endgroup$– Gro-TsenCommented Apr 11, 2018 at 7:44
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$\begingroup$ Interesting - thanks for this hint! $\endgroup$– Dominic van der ZypenCommented Apr 11, 2018 at 9:24
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$\begingroup$ @Gro-Tsen, Of course if $X$ has an isolated point then $\Box_{n\in\omega}X$ is not connected. It also seems to me that William only uses regularity of $X$ in the theorem that you mention. So one has to look only for connected Hausdorff non-regular spaces. $\endgroup$– Ramiro de la VegaCommented Apr 11, 2018 at 23:39
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1 Answer
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If $X$ is the Irrational slope topology then the closures of any two non-empty open sets must intersect. It easily follows that $\Box_{n\in\omega}X$ is connected. Note that $X$ is Hausdorff but not regular. It seems (see the comments to the OP) that there are no $T_3$ examples.
answered Apr 11, 2018 at 23:52
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$\begingroup$ Great, Ramiro - I didn't know about that topology, thanks for making me aware of it! $\endgroup$ Commented Apr 12, 2018 at 6:45