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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.

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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 $n\in\Bbb N$.

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