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Davood
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Ambiguity in the definition of " $\color{Red}{\text{locally finiteness}}$ of a collection of subsets of a topological space".

In the book " $\color{Blue}{\text{Introduction to Smooth Manifolds}}$ " , by $\color{Blue}{\text{John M. Lee}}$ ; page 9:


Let $M$ be a topological space. $\color{Purple}{\text{A collection}}$ $\mathcal{X}$ $\color{Purple}{\text{of subsets of}}$ $M$ $\color{Purple}{\text{is said to be}}$ $\color{Red}{\text{locally finite}}$ $\color{Purple}{\text{if each point of}}$ $M$ $\color{Purple}{\text{has a neighborhood that intersects at most finitely many of the sets in}}$ $\mathcal{X}$. 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.


$\color{Teal}{\text{Claim}}$ :
Let $\mathcal{X}$ be a $\color{Red}{\text{locally finite}}$ collection of subsets of $M$ , and let $x \in \cup_{ \mathcal{U} \in \mathcal{X} } \mathcal{U} \subseteq M$ ;
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 $\color{Red}{\text{locally finite}}$.


I myself suspect in the truth of the above $\color{Teal}{\text{Claim}}$.
$\color{Red}{\text{But}}$ if this is true; then we can give an equivalent definition of $\color{Red}{\text{locally finite}}$ as follows:

$\color{Green}{\text{Definition}}$ :

$\color{Purple}{\text{A collection}}$ $\mathcal{X}$ $\color{Purple}{\text{of subsets of}}$ $M$ $\color{Purple}{\text{is said to be}}$ $\color{Red}{\text{locally finite}}$ $\color{Purple}{\text{if each point of}}$ $M$ $\color{Teal}{\text{contained in}}$ $\color{Purple}{\text{at most finitely many of the sets in}}$ $\mathcal{X}$.


The above $\color{Red}{\text{fake}}$-$\color{Teal}{\text{definition}}$ look likes much more simpler to understnading; at least for me.

  • If my proof is $\color{Green}{\text{true}}$; then why $\color{Blue}{\text{John M. Lee}}$ did not use this $\color{Teal}{\text{definition}}$?

  • If my proof is $\color{Red}{\text{false}}$; where is the bug in my proof.
    In this case give me a counter-example;
    i.e. A collection $\mathcal{X}$ of subsets of $M$ , such that:
    $\color{Purple}{\text{each point of}}$ $M$ $\color{Purple}{\text{has a neighborhood}}$
    $\color{Purple}{\text{that intersects at most finitely many of the sets in}}$ $\mathcal{X}$ ;
    $\color{Red}{\text{but}}$ $\color{Green}{\text{there exist a point}}$ $x$ $\color{Green}{\text{which is contained in infinitely many sets in}}$ $\mathcal{X}$.




In short
give me a topological space $M$ and a collection $\mathcal{X}$ of subsets of $M$ , such that:
$\color{Purple}{\text{each point of}}$ $M$ $\color{Purple}{\text{has a neighborhood}}$
$\color{Purple}{\text{that intersects at most finitely many of the sets in}}$ $\mathcal{X}$ ;
$\color{Red}{\text{but}}$ $\color{Green}{\text{there exist a point}}$ $x$ $\color{Green}{\text{which is contained in infinitely many sets in}}$ $\mathcal{X}$.

Davood
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