2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.