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Let $\{A_i\}_{i\in S}$ be an open cover for a regular topological space $X$. The family $\{A_i\}_{i\in S}$ is called locally finite if for every point $x\in X$ there exists a neighborhood $U$ of $x$ such that the set $\{s\in S : U \cap A_s\neq \emptyset\}$ is finite.

Let the family $\{A_i\}_{i\in S}$ have the property that every point of $X$ is contained in only finitely many of $A_i'$s. Can we deduce that the family $\{A_i\}_{i\in S}$ is locally finite?

bof
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Deroty
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1 Answers1

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In other words, you're asking if a point-finite open cover is locally finite. For a simple counterexample, take $X=\mathbb R,$ and the open cover $\mathcal A=\{\mathbb R,(\frac12,1),(\frac13,\frac12),(\frac14,\frac13),\dots\}.$

bof
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    Beause a nbhd of the point $0$.......+1 – DanielWainfleet Dec 31 '16 at 09:47
  • I gave you a +1 . That means I gave an up-vote. Doesn't it? – DanielWainfleet Dec 31 '16 at 11:11
  • Thank you for your answer bof. Is the inverse true, that is , if a locally finite space is point - finite? – Deroty Dec 31 '16 at 11:31
  • Of course a locally finite cover is point-finite. Consider the contrapositive: if the cover is not point-finite, some point $x$ belongs to infinitely many members of the cover, and then every neighborhood of $x$ meets infinitely many members of the cover, so it's not locally finite. – bof Dec 31 '16 at 11:36