# Closures and Infinite Products

( This is from an exercise in Munkres' Topology )

Let $X_\alpha$ be an indexed collection of spaces, and $A_\alpha \subset X_\alpha$ be a collection of subsets.
Under the product topology, show that, as a subset of $X = \prod_\alpha{X_\alpha}$

$$\overline{\prod_\alpha{A_\alpha}} = \prod_\alpha{\overline{A_\alpha}}$$

This part I have no problem with. However, he then asks if this holds under the box topology on $X$

Clearly ( closure being the intersection of all containing closed sets ):

$$\overline{\prod_\alpha{A_\alpha}} \subset \prod_\alpha{\overline{A_\alpha}}$$

The reverse inclusion, on the other hand, has me in a twist.

The following is my attempt at a proof:

Let $x = (x_\alpha) \in \prod_\alpha{\overline{A_\alpha}}$, and $U \subset X$ be a neighborhood of $x$. We can assume $U$ is a basis element for the box topology on $X$
Then $\exists \hspace{3} U_\alpha \subset X_\alpha$ such that $U = \prod_\alpha{U_\alpha}$, where $U_\alpha$ are open.

Then we have:
$$x = (x_\alpha) \in U \longrightarrow x_\alpha \in U_\alpha$$

Because the $U_\alpha$ are neighborhoods of the $x_\alpha$, and $x_\alpha \in \overline{A_\alpha}$ for all $\alpha$, we have:

$$U_\alpha \cap A_\alpha \neq \varnothing$$ for all $\alpha$

so that:

$$\prod_\alpha{U_\alpha} \cap \prod_\alpha{A_\alpha} \neq \varnothing$$

That, and $x \in U$ gives:

$$x \in \overline{\prod_\alpha{A_\alpha}}$$

and therefore:

$$\prod_\alpha{\overline{A_\alpha}} \subset \overline{\prod_\alpha{A_\alpha}}$$

I can't see the flaw in this proof, yet I can't somehow shake the feeling that this inclusion shouldn't hold. I know the box topology can give some funky results ( like the closure of the set of sequences with finitely many non-zero entries, or the product of continuous functions not necessarily being continuous ), so I'm a bit weary of it. That, and the way Munkres states the question gives me the suspicion it's a trick question.
So, is there a flaw with my proof? Or am I just being paranoid ( which I often tend to be )

Thanks in advance for any comments, and also for putting up with my anal-latex exactness.