What's the interior of $R^{\infty}$ in the $R^{\omega}$ with the box topology and the production topology?

Question

What's the interior of $R^{\infty}$ in the $R^{\omega}$ with the box topology and the production topology? With $R^{\infty}$ being the set of all eventually zero sequences. About the question in Prob. 5, Sec. 20 in the book Topology by James R. Munkres, 2nd edition.

Answer 1

Let $(x_n)$ be a point from $\Bbb R^\infty$, taking $\Bbb R^\omega$ in the box topology.

Suppose that there is a basic box open set $U=\prod_n U_n$ such that $(x_n) \in U \subseteq \Bbb R^\infty$, with all $U_n$ open in $\Bbb R$ and of course $x_n \in U_n$. This means that for each $n$ there is an $r_n>0$ such that $(x_n -r, x_n+r) \subseteq U_n$.

As $(x_n)$ is eventually constant, there is an index $N$ such that $x_n=0$ for all $n \ge N$. Then the sequence $(y_n)_n$ with $y_n = x_n$ for $n < N$ and $y_n = \frac{r}{2}$ for $n \ge N$, is not eventually constant but is in $U$ by definition, contradicting $U \subseteq \Bbb R^\infty$ (now refuted by $(y_n)$).

So no point of $\Bbb R^\infty$ is interior point in the box topology so its interior is empty. And if $(x_n)_n$ would have been an interior point in the product topology, it would also have been one in the box topology (as this topology is larger) and so the interior in the product topology is also empty.

BTW, $\overline{\Bbb R^\infty} = \Bbb R^\omega$ in the product topology, while $\overline{\Bbb R^\infty} = \Bbb R^\infty$ in the box topology.