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Let $(X,\tau)$ be a topological space. We say ${\cal U}\subseteq \tau$ is an open cover if

  • $\bigcup {\cal U} = X$, and
  • $X\notin {\cal U}$.

${\cal U}$ is minimal if for all $U_0\in {\cal U}$ we have $\bigcup \big({\cal U}\setminus \{U_0\}\big) \neq X$. Clearly, every $T_1$-space on more than $1$ point possesses a minimal cover: pick $x\neq y\in X$ and let ${\cal U} = \big\{X\setminus\{x\}, X\setminus\{y\}\big\}$.

Question. Given any open cover of a Hausdorff space $(X,\tau)$ with $|X|>1$, does it have a refinement that is a minimal cover?

Note. I did not ask for subcovers in the question because of the following example: Let $X = \mathbb{R}$ with the Euclidean topology, then ${\cal U} = \big\{ \{x\in\mathbb{R}: x < n\} : n\in\mathbb{N}, n\geq 1 \big\}$ does not have a minimal subcover.

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    $\begingroup$ It is not hard to see that any countable cover has a minimal refinement. So the answer is yes, for $T_1$ Lindelof spaces. $\endgroup$ – Ramiro de la Vega Apr 26 '18 at 12:50
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No -- the space $\omega_1$ (the first uncountable ordinal with the order topology) is a counterexample.

Consider the open cover $\mathcal V$ of $\omega_1$ consisting of all bounded open sets. I claim that this cover has no minimal refinement. The proof is an application of Fodor's pressing down lemma.

Suppose $\mathcal U$ is a refinement of $\mathcal V$, which just means that $\mathcal U$ is an open cover of $\omega_1$ consisting of bounded sets. We'll show that $\mathcal U$ is not minimal. For each limit ordinal $\alpha$, choose some $U_\alpha \in \mathcal U$ with $\alpha \in U_\alpha$. Because $\alpha$ is a limit, there is some $\beta_\alpha < \alpha$ such that $$(\beta_\alpha,\alpha] \subseteq U_\alpha.$$ The function $\alpha \mapsto \beta_\alpha$ is regressive (i.e., it satisfies the hypothesis of Fodor's lemma), so there is some fixed $\beta$ and some uncountable set $S \subseteq \omega_1$ such that $\beta_\alpha = \beta$ for all $\alpha \in S$.

The interval $[0,\beta]$ is countable, so there is some $\alpha_0 \in S$ such that $$\bigcup \{[0,\beta] \cap U_\alpha : \alpha \in S\} = \bigcup \{[0,\beta] \cap U_\alpha : \alpha \in S, \alpha < \alpha_0\}.$$ (In other words, any ordinal $\leq\!\beta$ that appears in some $U_\alpha$, $\alpha \in S$, must appear by $\alpha_0$.) Because $U_{\alpha_0}$ is bounded and $S$ is uncountable, there is some $\alpha_1 \in S$ such that $\sup U_{\alpha_0} < \alpha_1$.

The set $U_{\alpha_0}$ can be broken into two pieces: $U_{\alpha_0} \cap [0,\beta]$ and $U_{\alpha_0} \cap (\beta,\omega_1)$. The first piece is contained in $\bigcup\{U_\alpha : \alpha \in S, \alpha < \alpha_0\}$, by our choice of $\alpha_0$. The second piece is contained in $(\beta,\alpha_1] \subseteq U_{\alpha_1}$, by our choice of $\alpha_1$. Both pieces are covered by $\mathcal U \setminus \{U_{\alpha_0}\}$. Thus $\mathcal U \setminus \{U_{\alpha_0}\}$ is still an open cover of $\omega_1$, and $\mathcal U$ is not minimal.

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I believe that $\omega_1$ with the order topology is a counter example. Call a subset of $\omega_1$ bounded if it is contained in some initial interval $[0,\alpha]$ and unbounded otherwise.

Lemma: Any cover of $\omega_1$ by bounded open subsets is not minimal.

I am sure that this is standard, but here is a proof. It contains way too many notations and is a bit clumsy, I am afraid, but I hope it is readable.

We can assume that such a cover is is of cardinality $\omega_1$. Let thus $\mathcal{V} = \{V_\alpha\,:\,\alpha\in\omega_1\}$ be a cover of $\omega_1$ by bounded open subsets. By a standard argument the sets below are closed and unbounded in $\omega_1$:

\begin{align*} C_1 &= \{\alpha\,:\, \cup_{\beta<\alpha} V_\beta\subset[0,\alpha)\}\\ C_2 &= \{\alpha\,:\, \cup_{\beta<\alpha} V_\beta\supset[0,\alpha)\} \end{align*} Hence their intersection $C = \{\alpha\,:\, \cup_{\beta<\alpha} V_\beta =[0,\alpha)\}$ is closed and unbounded as well. We may assume that each member of $C$ is a limit ordinal. When $\alpha\in \omega_1$, let $\alpha^*$ denote the smallest element of $C$ which is $>\alpha$. For each $\alpha\in C$ choose $V_{s(\alpha)}\ni\alpha$ with $s(\alpha)\in[\alpha,\alpha^*)$. Since $V_{s(\alpha)}$ is open and $\alpha$ is limit, $V_{s(\alpha)}$ contains some interval $[\gamma(\alpha),\alpha]$. By Fodor's Lemma, there is some $\gamma\in\omega_1$ such that the set $S$ of $\alpha\in C$ such that $\gamma(\alpha) = \gamma$ is stationary and in particular unbounded. For any $\alpha\in S$ with $\alpha>\gamma^*$, $V_{s(\alpha)}$ contains in particular the interval $[\gamma^*,\alpha]$. Fix now $\eta\in[\gamma^*,\gamma^{**})$. Choose $\alpha \ge \gamma^{**}$ in $S$. Then $$ V_\eta \subset V_{s(\alpha)} \cup \left(\cup_{\beta<\gamma^*}V_\beta\right). $$ Notice that $\gamma^*\le\eta<\alpha\le s(\alpha)$, hence the indices in the union at the righthand side do not contain $\eta$. It follows that $\cup (\mathcal{V} \backslash \{V_\eta\}) = \omega_1$.

Corollary: The cover of $\omega_1$ by the intervals $[0,\alpha]$, $\alpha\in\omega_1$, has no refinement that is a minimal cover.

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  • $\begingroup$ Ah, I see that Will Brian gave the same answer while I was typing. Sorry about the duplicate. $\endgroup$ – Mathieu Baillif Apr 25 '18 at 19:01
  • $\begingroup$ Thanks for this beautiful answer -- I would really like to accept both, I hope you don't mind me accepting Will's answer, he wrote it a bit earlier and it is also well comprehensible $\endgroup$ – Dominic van der Zypen Apr 26 '18 at 4:13
  • $\begingroup$ No worries, Will answered before and his answer is better redacted. $\endgroup$ – Mathieu Baillif Apr 26 '18 at 6:45

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