I want to prove: If $A_s$ is an $F_{\sigma}$-set in $X_s$ for every $s \in S$ and $|S| < \aleph_0$, then $\prod_{s\in S} A_s$ is an $F_{\sigma}$-set in the Cartesian product $\prod_{s\in S} X_s$. Note that the assumption about the cardinality of $S$ is essential.

Now I first wanted the proof it for two topological spaces $X, Y$. If I have two sets which are $F_{\sigma}$, i.e. $$ A_1 = \bigcup_{n=1}^{\infty} E_n \quad A_2 = \bigcup_{n=1}^{\infty} F_n $$ where $E_n$ is closed in $X$ and $F_n$ is closed in $Y$. Now I consider the set $A_1 \times A_2 \subseteq X \times Y$, my idea is to write $$ A_1 \times A_2 = \bigcup_{n=1}^{\infty} E_n \times A_2 = \bigcup_{n=1}^{\infty} (E_n \times A_2). $$ I want to show that $E_n \times A_2$ is closed in $X\times Y$, for that I have to show that the complement is open, the complement is $$ (E_n \times A_2)^C = (E_n^C \times A_2^C) \cup (E_n^C \times A_2) \cup (E_n \times A_2^C) $$ but from here I have no idea how to proceed, any hints?

EDIT: Think I got it. \begin{align*} A_1 \times A_2 & = \bigcup_{n=1}^{\infty} E_n \times A_2 \\ & = \bigcup_{n=1}^{\infty} (E_n \times A_2) \\ & = \bigcup_{n=1}^{\infty} (E_n \times \bigcup_{i=1}^{\infty} F_i) \\ & = \bigcup_{n=1}^{\infty} (\bigcup_{i=1}^{\infty} (E_n \times F_i)). \end{align*} And a finite union (and also countable unions) of countable unions are countable.

  • $\begingroup$ The edited version is fine. In the original version you couldn’t hope to show in general that $E_n\times A_2$ is closed in $X\times Y$, because there’s no reason to think that $A_2$ is closed in $Y$. $\endgroup$ – Brian M. Scott Jul 13 '13 at 19:59

The product of closed sets is closed, so $E_i \times F_j$ is closed in $X \times Y$ for all $i,j$. Then we have $$ A_1 \times A_2 = \bigcup_{i,j=1}^\infty E_i\times F_j $$ and hence $A_1 \times A_2$ is $F_\sigma$ in $X \times Y$. Note that this immediately generalizes to a countable product space. We need countable so that there are only countably many indices in the big union.


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