joeboo
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( 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.
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.