I know that $\mathbb{R}^\omega$ is not locally compact in the product topology. Since the box topology is finer than the product topology, does that mean that $\mathbb{R}^\omega$ is not locally compact in the box topology?
Alternatively, trying to prove that it is locally compact. We know that $\mathbb{R}^\omega$ is Hausdorff in any topology, locally compactness is equivalent to finding for each $x$, a neighborhood $U$ of $x$ such that there is another neighborhood $V$ of $x$ such that $\bar{V}\subset U$.
Well let $x=(x_1,x_2,x_3,\dots)$. Let $U$ be some neighborhood of $x$. Can't I always find a neighborhood $V$ inside $U$ (i.e. in some way in between $x$ and $U$), such that the closure of $V$ is contained in $U$?
What about for the uniform topology?
Thanks in advance!
Best Answer
If $X=\square_{i=0}^\infty\Bbb R$ were locally compact, then it would be locally compact Hausdorff, so each point would be contained in an open box $U$ such that $\overline U$ is compact. $\overline U$ is then a compact Hausdorff space with the box topology, and this topology is strictly finer (on $\overline U$) than the product topology. However, $\overline U$ is also compact Hausdorff in the product topology. The point is now that a compact Hausdorff space is minimally Hausdorff, it means there cannot be a coarser topology under which the space remains Hausdorff.
As for the uniform topology: Assume that $X=\left(\prod\limits_{i=0}^\infty\Bbb R,\tau_{un}\right)$ with the uniform topology were locally compact. Then $x=(x_n)_n\in\prod\limits_{i=0}^\infty\Bbb R$ would have a basic neighborhood $U$ of the form $\prod\limits_{i=0}^\infty(x_i-\varepsilon,x_i+ε)$ whose closure $\prod\limits_{i=0}^\infty[x_i-ε,x_i+ε]$ is compact. However such a set isn't compact in the uniform topology. To show this, assume wLoG that the set in question is $\prod\limits_{i=0}^\infty[0,1]=:Y$. Define $y^n=(y^n_k)_k$, where $y^n_k=\delta_{nk}$. Then the set $\{y^n\mid n\in\Bbb N\}\subset Y$ is an infinite set without a limit point. It follows that $\prod\limits_{i=0}^\infty[0,1]$ isn't compact, not even countably compact. Hence $X$ fails to be locally compact.