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From the book "Introduction to Smooth Manifolds", by John M. Lee, page 9:

Definition. Let $M$ be a topological space. A collection $\mathcal{X}$ of subsets of $M$ is said to be locally finite if each point of $M$ has a neighborhood that intersects at most finitely many of the sets in $\mathcal{X}$. Given a cover $\mathcal{U}$ of $M$; another cover $\mathcal{V}$ is called a refinement of $\mathcal{U}$ if for each $V \in \mathcal{V}$ there exists some $U \in \mathcal{U}$ such that $V \subseteq U$. We say that $M$ is paracompact if every open cover of $M$ admits an open, locally finite refinement.


Claim. Let $\mathcal{X}$ be a locally finite collection of subsets of $M$, and let $$x \in \bigcup_{ \mathcal{U} \in \mathcal{X} } \mathcal{U} \subseteq M$$ Then $x$ belongs to at most finitely many $\mathcal{U} \in \mathcal{X}$.

Proof. Suppose on contrary that there exists an infinite set $I$ of indexes;
such that $x \in \mathcal{U}_i$ for every $i \in I$. But notice that every neighborhood of $x$ intersects with each $\mathcal{U}_i$;
which contradicts the assumption that $\mathcal{X}$ is locally finite.


I myself suspect in the truth of the above claim.
But if this is true, then we can give an equivalent definition of "locally finite" as follows:

[Definition.] A collection $\mathcal{X}$ of subsets of $M$ is said to be locally finite if each point of $M$ is contained in at most finitely many of the sets in $\mathcal{X}$.

The above fake-definition looks like it is much simpler to understand; at least to me.

  • If my proof is true, then why didn't Lee use this other definition?

  • If my proof is false, where is the bug in my proof? Can you give me a counter-example, i.e. a collection $\mathcal{X}$ of subsets of $M$, such that: each point of $M$ has a neighbourhood that intersects at most finitely many of the sets in $\mathcal{X}$, but there exists a point $x$ which is contained in infinitely many sets in $\mathcal{X}$?


In short: Can you give an example of a topological space $M$ and a collection $\mathcal{X}$ of subsets of $M$, such that each point of $M$ has a neighbour that intersects at most finitely many of the sets in $\mathcal{X}$, but there exists a point $x$ which is contained in infinitely many sets in $\mathcal{X}$?

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    $\begingroup$ Your definition is that of a "point-finite" family. That is a weaker condition than being locally finite [Lee proves that local finiteness implies point-finiteness; or is that your proof?]. $\endgroup$ Aug 22, 2017 at 13:23
  • $\begingroup$ Surely any neighbourhood $U$ of such an $x$ intersects infinitely many of the sets in $\mathcal{X}$ since it contains $x$ which lies in infinitely many of the sets in $\mathcal{X}$. Such a pair $(M\mathcal{X})$ cannot exist. $\endgroup$
    – Tyrone
    Aug 22, 2017 at 13:26
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    $\begingroup$ You're claiming you have an "equivalent" definition. In order to prove that A and B are equivalent, it's not enough to prove "if A then B", you also have to prove "if B then A", which you have not done. $\endgroup$
    – bof
    Aug 22, 2017 at 14:29
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    $\begingroup$ No, quite the other way round. Every locally finite family is point-finite, but not all point-finite families are locally finite. For example $$\mathcal{X} = \{(-1,1)\} \cup \Biggl\{ \biggl(\frac{1}{n+1},\frac{1}{n-1}\biggr) : n \in \mathbb{N}\setminus \{0,1\}\Biggr\}$$ is a point-finite family that isn't locally finite. (The space is $\mathbb{R}$ with the standard topology.) $\endgroup$ Aug 22, 2017 at 14:31
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    $\begingroup$ Related discussion on meta: Should the overuse of colored text be discouraged? (Since your question was mentioned there as an example, I thought it might be a good idea to let you know about this discussion.) $\endgroup$ Sep 2, 2017 at 8:14

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If any $x\in M$ belonged to every member of $Y,$ where $Y$ is an infinite subset of $X,$ then every nbhd $U$ of $X$ would have non-empty intersection with every member of $Y$ (because $x\in U\cap Z$ for all $Z\in Y$), implying that $X$ is not locally-finite.

A family $F$ of subsets of $M$ is called point-finite iff every $x\in M$ belongs to only finitely many members of $F.$ So a locally-finite family is point-finite. Point-finite is not equivalent to locally-finite. For example if $M=\mathbb R$ then $F=\{\{x\}:x\in M\}$ is point-finite but not locally-finite.

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  • $\begingroup$ Subject reference : Meta-compact. $\endgroup$ Aug 23, 2017 at 4:07

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