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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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