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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)

Michael Albanese
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FNH
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    You are absolutely correct. I think the point of the exercise is to compare the closure in the box topology with the closure in the product topology and the topology of uniform convergence - you get very different and surprising answers – user115940 Aug 19 '14 at 10:31
  • @user115940, Right , the exercise compares the closure on box and product topology. but not the uniform convergence as it's not provided so far in the text. – FNH Aug 19 '14 at 10:40
  • @user115940, Do you find my justification correct? – FNH Aug 19 '14 at 10:54
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    I think the solution is clean and correct – user115940 Aug 19 '14 at 10:58
  • is it true from your proof that we conclude that then $\forall x \in \mathbb{R}^{\infty} \implies x \in Cl({\mathbb{R}^{\infty}})$ – sophie-germain Oct 07 '14 at 05:58

1 Answers1

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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.

Michael Albanese
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