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?
Best Answer
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.