Show that the $\mathbb{R^N} $ with the box topology is a regular space.

Question

What was I thinking

There is a neighborhood W=$\prod_i (-r_i,r_i)$ that does not intersect A . Consider a linear transformation $h_i:x_i \to x_i/r_i$ (linear transformation is a homeomorphism in both the product and box topology on $\mathbb{R^N}$). Also a product of continuous functions is continuous so, that the product of homeomorphisms is a homeomorphism. If y=h(x) $\in$ V , V is open in the box topology, then there is a basis neighborhood y $\in$ V′=$\prod_i V′_i \subseteq V$ and h(z) $\in$ V′ iff for all i : $h_i(z_i) \in V′_i$ iff for all i : $z_i \in h^{−1}_i(V′_i)$ . Therefore, for all z $\in \prod_i h^{−1}_i(V′_i)$ which is open in the box topology if all $h_i$ are continuous, h(z)$ \in V′ \subseteq V$, it is also true for the linear transformation, and if was possible separate 0 from h(A) by a continuous function f , then $f \circ h$ would separate 0 and A . We Know that h(A) is closed and does not have any points in W′=$\prod_i(−1,1)$ for if it had some than A would have a point in W . We also know that $\mathbb{R^N}_{uniform}$ is completely separable, and there is a continuous function f that separates 0 and X−B(0,1) $ \subseteq $ W′. So, f is continous in the finer box topology as well, and separates 0 from h(A). This way we show that 0 can be separated by a continuous function from a closed set A that does not contain it.

Any comment woud be appreciated.

Thanks in advance.

The box topology coincides with the usual product topology for finite products. Do you know the theorem for the usual topology on $\mathbb{R}^n$? — vujazzman, Nov 16 '20 at 15:00
I think your ideas are essentially correct but they are written in a way hard to read. You haven't defined $A$, $V$, $V_i$ and $h$ for instance. Sometimes it can help to remind the reader what you are arguing: "we must show..." etc. — vujazzman, Nov 16 '20 at 15:08

score 1 · Accepted Answer · answered Feb 11 '21 at 20:20

If $\langle X_i:i\in I\rangle$ is any $I$-indexed family of regular spaces, then the box product $Y=\prod_{i\in I}X_i$ is regular. To see this, let $y=\langle x_i:i\in I\rangle \in Y$, with $C\subseteq Y\setminus\{y\}$ a closed set. Then it is easy to check that, for each $i\in I$, there is a closed $C_i\subseteq X_i$ with $C\subseteq \prod_{i\in I}C_i\subseteq Y\setminus\{y\}$. For each index $i$, we know that $x_i\not\in C_i$; so, using regularity of $X_i$, let $U_i,V_i$ be disjint open sets in $X_i$ such that $x_i\in U_i$ and $C_i\subseteq V_i$. Then the sets $U=\prod_{i\in I}U_i$ and $V=\prod_{i\in I}V_i$ are open in the box topology, they're disjoint, $y\in U$, and $C\subseteq V$. Thus $Y$ is regular.

Show that the $\mathbb{R^N} $ with the box topology is a regular space.

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