Find the closure of $\mathbb{R}^{\infty}$ in $\mathbb{R}^{\omega}$ under the box topology.
Note: $\mathbb{R^{\infty}}$ is the set of all sequences $(t_1,t_2,\dots)$ such that $t_i\neq0$ for only finitely many values of $i$, and $\mathbb{R}^{\omega}=\mathbb{R} \times \mathbb{R}\times \mathbb{R} \times \dots$
In Munkres' Topology, this is part of an exercise. My answer is that the set $\mathbb{R}^{\infty}$ is closed in the box topology hence, its closure is itself.
Here is my justification of my answer:
Suppose $x=(x_1,x_2,\dots)$ is not in $\mathbb{R}^{\infty}$.
So for some $i_1,i_2,i_3,\dots$ we get, $x_{i_j} \neq 0$ for $j=1,2,3,\dots$It's clear that for every $x_{i_j}$, there is a neighborhood $U_{i_j}$ of $x_{i_j}$ in $\mathbb{R}$ such that $0 \not\in U_j$.
Now, let $U=\prod_{i=1}^{\infty} A_i$ where $A_{i_j}=U_{i_j}$ and $A_i$ is any open set of $x_i$ for $i \neq i_j$ for $j=1,2,3,\dots$
Now, $U$ is a neighborhood of $x$ but $U\cap \mathbb{R}^{\infty}=\emptyset$ so $x$ is not a limit point of $\mathbb{R}^{\infty}$ hence is not in the closure of $\mathbb{R}^{\infty}$.
Why is $U\cap \mathbb{R}^{\infty}=\emptyset$? Because for every element $y=(y_1,y_2,\dots)$ of $U$, there are an infinite number of $y_i$'s such that $y_i\not=0$ (by definition) so $y\not\in \mathbb{R^{\infty}}$.
My question is, is this answer true? If not, where does my argument go wrong? In this case, could you give me any hints to figure out the right answer myself? (hints not solution)
Best Answer
As mentioned in the comments, your argument is sound and the conclusion is correct.
Note however that you don't need to mention limit points. For any $x \in \mathbb{R}^{\omega}\setminus\mathbb{R}^{\infty}$, you have found an open neighbourhood $U$ of $x$ with $U\cap\mathbb{R}^{\infty} = \emptyset$, so $\mathbb{R}^{\omega}\setminus\mathbb{R}^{\infty}$ is open and therefore $\mathbb{R}^{\infty}$ is closed.