- #1

joeboo

- 35

- 0

Let [itex]X_\alpha[/itex] be an indexed collection of spaces, and [itex]A_\alpha \subset X_\alpha[/itex] be a collection of subsets.

Under the product topology, show that, as a subset of [itex]X = \prod_\alpha{X_\alpha}[/itex]

[tex]\overline{\prod_\alpha{A_\alpha}} = \prod_\alpha{\overline{A_\alpha}}[/tex]

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

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

[tex]\overline{\prod_\alpha{A_\alpha}} \subset \prod_\alpha{\overline{A_\alpha}}[/tex]

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

The following is my attempt at a proof:

Let [itex]x = (x_\alpha) \in \prod_\alpha{\overline{A_\alpha}}[/itex], and [itex]U \subset X[/itex] be a neighborhood of [itex]x[/itex]. We can assume [itex]U[/itex] is a basis element for the box topology on [itex]X[/itex]

Then [itex]\exists \hspace{3} U_\alpha \subset X_\alpha[/itex] such that [itex]U = \prod_\alpha{U_\alpha}[/itex], where [itex]U_\alpha[/itex] are open.

Then we have:

[tex]x = (x_\alpha) \in U \longrightarrow x_\alpha \in U_\alpha[/tex]

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

[tex]U_\alpha \cap A_\alpha \neq \varnothing[/tex] for all [itex]\alpha[/itex]

so that:

[tex]\prod_\alpha{U_\alpha} \cap \prod_\alpha{A_\alpha} \neq \varnothing[/tex]

That, and [itex]x \in U[/itex] gives:

[tex]x \in \overline{\prod_\alpha{A_\alpha}}[/tex]

and therefore:

[tex]\prod_\alpha{\overline{A_\alpha}} \subset \overline{\prod_\alpha{A_\alpha}}[/tex]

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.