# Munkres Exercise 39.6. Showing that a collection is countably locally finite but not locally finite.

This is Munkres Exercise 39.6.

Consider $$\mathbb{R}^\omega$$ in the uniform topology. Given $$n$$, let $$\mathscr{B}_n$$ be the collection of all subsets of $$\mathbb{R}^\omega$$ of the form $$\Pi A_i$$, where $$A_i = \mathbb{R}$$ for $$i \le n$$ and $$A_i$$ equals either $$\{0\}$$ or $$\{1\}$$ otherwise. Show that the collection $$\mathscr{B}=\cup \mathscr{B}_n$$ is countable locally finite, but neither countable nor locally finite.

The collection is clearly not countable. I have trouble with showing that it is countably locally finite but not locally finite.

The idea that I have is, we may focus on points that have, for all but finitely many coordinates, values between $$0$$ and $$2$$. This is because if they have infinitely many coordinates with values $$\ge 2$$ than a radius $$1$$ ball will not intersect any of the elements from any collection.

Thus, take a point $$x$$ that takes values $$(0,2)$$ from some $$n$$. Then, if all $$x_i\ge 1$$ for $$i \ge n$$, then take the radius $$1$$ ball and it intersects only the set $$\mathbb{R}^n \times 1 \times 1 \cdots$$ from $$\mathscr{B}_n$$. Hence we may consider $$x$$ that takes values between $$0$$ and $$1$$ for some coordinate $$\ge n$$. In this case take the radius $$1/2$$ ball. Then for the coordinates in $$(0,1/2)$$ it will intersect only the point $$0$$ and for the coordinates in $$(1/2,1)$$ it will intersect only $$1$$, while if it is $$1/2$$ then it will not intersect any. Hence in any case, it will intersect only finitely many elements $$A_i$$.

To show that it is not locally finite, we simply note that if $$x$$ is a point with a neighborhood that intersects finite($$>0$$) number of elements from $$\mathscr{B}_n$$, then it intersects with some elements from $$\mathscr{B}_m$$, $$m\ge n$$. Hence we have intersection with countably infinite number of elements.

Is this idea sound? I would appreciate any help.

Your idea for proving the countably local finiteness of $$\mathscr R$$ is correct. To prove the non-local-finiteness of $$\mathscr R$$, observe every neighborhood of $$(0,0,...)$$ will intersect $$\Bbb R^n \times \{0\}\times \{0\}\times...$$ for each $$n\in\Bbb N$$.