An open cover that is not locally finite

Question

I could not understand why example $13.4$ is not locally finite.

Can you give me an explanation please.

Answer 1

An open cover $\mathscr{U}$ of a space $X$ is point-finite if $\{U\in\mathscr{U}:x\in U\}$ is finite for each $x\in X$: each point of $X$ belongs to only finitely many members of $\mathscr{U}$. Every locally-finite cover of a space is automatically point-finite as well. And this open cover of $\Bbb R$ is not even point-finite. In fact, it fails in the worst way to be point-finite: each $x\in\Bbb R$ belongs to infinitely many members of the cover.

To see this, let $x\in\Bbb R$. For each $n\in\Bbb Z^+$ there is a rational number $q_n$ such that $\left|x-q_n\right|<\frac1n$, and we may further assume that $q_n\notin\{q_k:k<n\}$, since there are infinitely many rational numbers within $\frac1n$ of $x$. But then $x\in U_{q_n,n}$ for each $n\in\Bbb Z^+$, and the sets $U_{q_n,n}$ are all distinct, so $x$ belongs to infinitely many members of the cover.

(This is a real distinction, by the way: there are point-finite covers that are not locally finite.)

Answer 2

This open cover is very far from being locally finite. In fact, no real number has a neighbourhood which intersects only finitely many $U_{r,n}$.

For example, if $U$ is any (open) neighbourhood of $0$, then clearly $0 \in U \cap U_{0,n}$ (and so $U \cap U_{0,n} \neq \emptyset$) for all $n \in \mathbb{Z}^+$ (and there are infinitely many of these).