Let $(X,\tau)$ be a topological space, and let ${\cal U}$ be an open cover. We say that ${\cal U}$ is thick if for all $x\in X$ we have $$|\{V\in {\cal U}: x\in V\}| = |X|.$$ Is there a Hausdorff space $(X,\tau)$ with $|X|>1$ and an open cover ${\cal U}$ of $X$ such that every refinement of ${\cal U}$ is thick?
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asked May 25, 2018 at 13:47
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2 Answers
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This is impossible for any $T_1$ space $X$. For suppose $X$ is $T_1$ and $\mathcal U$ is an open cover of $X$, and fix $x \in X$. If $U$ is some member of $\mathcal U$ containing $x$, then $\{U\} \cup \{V \setminus \{x\} : V \in \mathcal U,\, V \neq U\}$ is a refinement of $\mathcal U$, but it is not thick because only one set in this refinement contains $x$.
answered May 25, 2018 at 17:44
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1$\begingroup$ Of course! For the question it suffices to handle only one point, not all points, as in my answer. I wonder, however, about generalizations of the theorem in my answer to uncountable spaces. $\endgroup$ Commented May 25, 2018 at 17:50
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2$\begingroup$ @JoelDavidHamkins: Good question! A similar question was already asked by Dominic here on MathOverflow (my assumption is that thinking about it further partially motivated this question) -- mathoverflow.net/questions/298639/…. My answer to that question shows that the countability assumption in your theorem is indispensable. $\endgroup$ Commented May 25, 2018 at 17:52
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This is a partial answer. I would like to note merely that this is not possible in a countable space.
Theorem. In any countable Hausdorff space $X$ and any open cover $U$, there is a refinement of $U$ to an open cover $U'$ such that every point $x\in X$ is in only finitely many elements of $U'$.
Proof. Enumerate the space as $X=\{x_n\mid n\in\omega\}$. At each stage $n$, consider $x_n$, and let $U_n$ be a set in $U$ with $x_n\in U_n$. Let $U_n'$ be a refinement of $U_n$ that excludes the points $x_k$ for $k<n$. This is possible by the separation axiom. Let $U'$ be the set of all $U_n'$. This refines the original cover, and $x_n$ is an element of at most $n+1$ many elements of $U'$, since it is excluded after stage $n$. $\Box$
answered May 25, 2018 at 16:04
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$\begingroup$ Thanks Joel for this partial answer! I was thinking about how to carry on the induction to larger cardinals, but it seems like one would get a "predecessor problem" (in your argument, you have only finitely many $k<n$). $\endgroup$ Commented May 25, 2018 at 16:30
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$\begingroup$ I guess this argument uses only $T_1$ and not fully Hausdorff. $\endgroup$ Commented May 25, 2018 at 17:09
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2$\begingroup$ Spaces satisfying the conclusion of your theorem are called metacompact. The ordinal $\omega_1$ is not metacompact since countably compact metacompact spaces are compact. $\endgroup$ Commented May 25, 2018 at 19:53
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2$\begingroup$ By the way, $\omega_1$ is not meta-Lindelof either (I believe for ordinals, meta-Lindelof and metacompact coincide). So the theorem cannot be pushed up even changing "finitely many" by "countably many". $\endgroup$ Commented May 25, 2018 at 20:36