Every locally finite collection is also point finite

Question

We are given the following two definitions for a collection $\mathscr{U} = \{U_\alpha\}_{\alpha \in A}$ of subsets of a topological space $X$.

$\mathscr{U}$ is locally finite if every point $x\in X$ has neighborhood that meets finitely many $U_\alpha$

and

$\mathscr{U}$ is point finite if every point $x\in X$ lies in only finitely many $U_\alpha$

My textbook claims that it's apparent that every locally finite collection is also point finite without proof.

However, let's consider the collection of open intervals $\{(n, n+1) : n\in \mathbb{Z}\}$ in $\mathbb{R}$ under the usual topology.

Is this set not locally finite? For any any $r \in \mathbb{R}$, we consider the the neighborhood $U_r = (r-\frac{1}{4}, r+\frac{1}{4})$. Then clearly $U_r$ can only meet at most $2$ of those open intervals. However, this collection is not point finite since $n \in \mathbb{Z}$ is not contained in any of those open intervals.

Where am I going wrong here?

Answer 1

Your collection is point finite. Each $n\in\mathbb Z$ belongs to zero of those intervals, and therefore the set of intervals to which $n$ belongs is emtpy. And $\emptyset$ is a finite set.

Answer 2

The omitted prooflet:

Let $\mathcal{U}$ be a locally finite collection.

Let $x \in X$. Then by local finiteness we have $O(x)$,an open neighbourhood of $x$, such that $$\{U \in \mathcal{U}\mid U \cap O(x) \neq \emptyset \}\text{ is finite. }$$

But as $$\{U \in \mathcal{U}: x \in U\} \subseteq \{U \in \mathcal{U}\mid U \cap O(x)\neq \emptyset \}$$ because if $x \in U$ for sure $O(x) \cap U \neq \emptyset$, the former set is also finite and so $\mathcal{U}$ is point-finite. Your example shows that the inclusion can be strict: for $n \in \Bbb N$ the left hand set is finite and the right hand one has 2 elements. Both finite sets. No contraction..

You example family is both locally finite and point finite.