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Let $A_1,A_2,A_3,\ldots,A_n,\ldots$ be a collection of subsets in $[0,1]^{\Bbb N}$ and let $$A=A_1\times A_2\times A_3\times\ldots\subseteq[0,1]^{\Bbb N}\;.$$ Show that

$$\operatorname{Cl}A=\operatorname{Cl}A_1\times\operatorname{Cl}A_2\times\operatorname{Cl}A_3\times\ldots\times\operatorname{Cl}A_n\times\ldots$$

for any topology $t$ on $[0,1]^{\Bbb N}$ such that $t_T\subseteq t\subseteq t_B$, where $t_T$ is the usual product topology, and $t_B$ is the box topology.

Brian M. Scott
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jose maria
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    I find it difficult to understand your question. – Eric Auld Dec 17 '13 at 20:10
  • Jose, please check to make sure that I correctly interpreted everything. Also, do you really want the sets $A_n$ to be subsets of $[0,1]^{\Bbb N}$, or should they be subsets of $[0,1]$? – Brian M. Scott Dec 17 '13 at 20:34
  • @Yiorgios: While I think it very likely that the sets $A_n$ are supposed to be subsets of $[0,1]$, I’m not willing to change it without the OP’s consent. – Brian M. Scott Dec 17 '13 at 20:47
  • @Brian: The product of subset of $[0,1]^{\mathbb N}$ is not a subset of $[0,1]^{\mathbb N}$. – Yiorgos S. Smyrlis Dec 17 '13 at 21:15
  • @Yiorgios: Obviously. That does not justify making a substantive change to the OP’s question without his consent. Moreover, it is a subset of $([0,1]^{\Bbb N})^{\Bbb N}$, which can be identified naturally with $[0,1]^{\Bbb N}$, and while I very much doubt that this is what’s intended, it is possible. – Brian M. Scott Dec 17 '13 at 21:18

1 Answers1

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As the standard product topology (or weak topology) is weaker than the box topology, and hence the box topology has more closed sets, we have that $$ \mathrm{Cl}_{B}(A) \subset \mathrm{Cl}_{T}(A). $$ Note also that $\prod_{n\in\mathbb N}\mathrm{Cl}\, A_n$ is closed in both topologies, as an intersection of closed sets, and hence $$ \mathrm{Cl}_{B}(A) \subset \mathrm{Cl}_{T}(A) \subset \prod_{n\in\mathbb N}\mathrm{Cl}\, A_n. $$

Therefore, it suffices to show that $$ \prod_{n\in\mathbb N}\mathrm{Cl}\, A_n\subset\mathrm{Cl}_{B}(A). $$ This is no harder than proving that $$ \mathrm{Cl}\,A_1\times\mathrm{Cl}\,A_2\subset \mathrm{Cl}\,(A_1\times A_2). $$

Yiorgos S. Smyrlis
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