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Let $\mathcal{U}$ be a locally finite open cover of a connected topological space $X$ such that every $U \in \mathcal{U}$ is precompact and nonempty. We have that every $U$ in $\mathcal{U}$ intersect only finitely many other element of $\mathcal{U}$ (because locally finiteness + precompact sets).

Question: Does $\mathcal{U}$ is a countable open cover?

I think yes, here below an attempt to prove it.

Fix $U_0 \in \mathcal{U}$ and consider $\mathcal{V} \subseteq \mathcal{U}$ defined as the smallest collection of open sets such that:

  • Contains $U_0$;
  • If $V \in \mathcal{V}$ then $\mathcal{V}$ include the finite set $\{U \in \mathcal{U} \, \colon V \cap U \neq \emptyset\}$,

this definition yields a countable subcover of $\mathcal{U}$ (because $X$ is connected).

Are the two covers the same cover? Take $U \in \mathcal{U}$ and $x \in U$, we have that exists $V \in \mathcal{V}$ such that $x \in V$ so $U \cap V \neq \emptyset$ which implies that $U \in \mathcal{V}$. So yes, the two covers are the same cover.

donovan
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  • You want "Let $\mathcal{U}$ be a locally finite open cover". – Paul Frost Apr 09 '22 at 12:49
  • @PaulFrost yes! Corrected, thank you – donovan Apr 09 '22 at 13:31
  • I have the feeling that this is not true. Did you think about what happens on "the long line"? https://en.m.wikipedia.org/wiki/Long_line_(topology) – PhoemueX Apr 09 '22 at 14:25
  • The two covers are the same. Take the union of the members of $\mathcal{V}$ (this is clopen) and use connectedness to conclude using your final paragraph. – Tyrone Apr 09 '22 at 19:50
  • Even if both covers are the same, I am not convinced of the reasoning for $\mathcal{V}$ to be countable. – PhoemueX Apr 10 '22 at 05:41
  • @PhoemueX I just posted an answer to tell the reason why I think $\mathcal{V}$ is countable – donovan Apr 16 '22 at 23:55

1 Answers1

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Take $\mathcal{V}_0 = \{U_0\}$ and for every $n \ge 1$ consider

$$\mathcal{V}_{n+1} =\{U \in \mathcal{U} \mid \exists V \in \cup_{\,i=0}^{\,n} \mathcal{V}_i \;\;\text{s.t.} \; U \cap V \neq \emptyset \} - \cup_{\,i=0}^{\,n} \mathcal{V}_i.$$

We see that $\mathcal{V}_i$ is finite for all $i \ge 0$. If $\mathcal{V}_0, \dots, \mathcal{V}_n$ are finite sets then $\cup_{\,i=0}^{\,n} \mathcal{V}_i$ is also finite. Using the fact that every $V \in \cup_{\,i=0}^{\,n} \mathcal{V}_i \subseteq \mathcal{U}$ intersect only finitely many other element of $\mathcal{U}$ then also $\mathcal{V}_{n+1}$ is finite and follows by induction that $\mathcal{V}_i$ is finite for all $i \ge 0$.

We have that $\bigcup_{\,i=0}^{\,\infty} \mathcal{V}_i$ is a countable cover of $X$, it is a cover because $X$ is connected.

If we prove that $\mathcal{V} = \bigcup_{\,i=0}^{\,\infty} \mathcal{V}_i$ then we are done (see definition of $\mathcal{V}$ in the question). The non trivial inclusion is "$\subseteq$". Take $V \in \mathcal{V}$ with $V \neq U_0$ and consider $x \in V$ and $y \in U_0$. The space $X$ is connected so exists a finite sequence $V_0, \dots, V_k \in \mathcal{V}$ such that $V_i \cap V_{i+1} \neq \emptyset$ for all $i \in \{0,\dots,k-1\}$ and $x \in V_0$, $y \in V_k$, this proves that $V$ is an element of $\bigcup_{\,i=0}^{\,\infty} \mathcal{V}_i$.

donovan
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  • I don't follow the "because $X$ is connected" argument. For this you would need that $\bigcup_{i=0}^\infty V_i$ is both open (this is OK) and closed. I don't see why it should be closed. – PhoemueX Apr 19 '22 at 08:52
  • @PhoemueX I learned from this answer the following fact: a topological space $X$ is connected if and only if for every open cover $\mathcal{U}$ of $X$ exists a finite chain of open sets $U_1, \dots, U_n \in \mathcal{U}$ such that $x \in U_1$ and $y \in U_n$ for every $x,y \in X$ and $U_i \cap U_{i+1} \neq \emptyset$ for every $i \in {1,\dots,n-1}$. – donovan Apr 19 '22 at 17:44
  • Looks good. Thanks for pointing out the other answer. This was interesting (although I should have known it)! – PhoemueX Apr 19 '22 at 18:17
  • @PhoemueX We apply this equivalence in the following way. Take $x \in X$ and $y \in U_0$. The space $X$ is connected so exists a finite chain $U_1,\dots,U_n \in \mathcal{U}$ such that $x \in U_1$ and $y \in U_n$. We have obtained a chain $U_0, U_1, \dots, U_n \in \mathcal{U}$ s.t. $x \in U_0$, $y \in U_n$ and $U_i \cap U_{i+1} \neq \emptyset$ for every $i \in {0,\dots,n-1}$. We have also that $U_1 \in \mathcal{V}0 \cup \mathcal{V}_1 \subseteq \mathcal{V}$ because $U_0 \cap U_1 \neq \emptyset$, similarly $x \in U_n \in \bigcup{i=0}^{n}\mathcal{V}_i$ so $\mathcal{V}$ is a open cover of $X$. – donovan Apr 19 '22 at 18:18