# Product of sigma compact spaces

Let $\{X_{i}: i \in I\}$ be a family of $\sigma$-compact but not compact topological spaces. How to show that $\prod_{i \in I} X_{i}$ is $\sigma$-compact if and only if $I$ is finite?

So here it goes, using the hints by Sam and jspecter.

The direction $\Leftarrow$ indeed follows by induction, it suffices to do the case $n=2$. Indeed if $X_{1},X_{2}$ are $\sigma$ compact and $X_{1}=\cup C_{n}$, $X_{2}=\cup D_{n}$ now $X_{1} \times X_{2}$ = $\cup C_{n} \times \cup D_{n}$. Now here's my question: how do we write this? in general the cartesian product does not distributes with respect the union, although I guess that doesn't matter because iwe would get a countable union of finite product of compact sets, so its $\sigma$-compact.

For the other direction, suppose $I$ is infinite and suppose $\prod X_{i} = \cup_{n \in \mathbb{N}} C_{n}$ where each $C_{n}$ is a compact subset of $\prod_{i \in I} X_{i}$. Denote the projection by $p_{n}$ then $p_{n}(C_{n})$ is a compact subset of $X_{i}$. Since $X_{i}$ is not compact then $X_{i} \setminus p_{n}(C_{n})$ is non-empty, so for each $i \in I$ pick $z_{i} \in X_{i} \setminus p_{n}(C_{n})$, then $(z_{i}) \in \prod_{i \in I} X_{i}$, but not in the union $\cup C_{n}$, OK?

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Hint: Prove $\Leftarrow$ by induction on the number of factors. You'll need the case of two factors, so start with this one. Remember that finite products of compact spaces are compact and finite products of countable index sets are countable. It's not that hard, I'm sure you can do it. ;) – Sam Aug 13 '11 at 20:10
To prove the other direction try a diagonalization argument. – jspecter Aug 13 '11 at 20:17
@user10: Regarding the first part, I would look at the family $\{C\times D: \exists n:C=C_n,\exists m:D=D_m\}$. – LostInMath Aug 13 '11 at 23:11
@user10: In the second part you want your projection to be $p_i$, not $p_n$; it’s $p_i[C_n]$ that’s a compact subset of $X_i$ for each $i\in I$ and $n\in\mathbb{N}$. Once you fix that, it’s okay. – Brian M. Scott Aug 13 '11 at 23:30
@Brian M. Scott: can you please post your reply as an answer and if it is not much to ask can you also ellaborate more on the first part? – user10 Aug 13 '11 at 23:32

Your second part is almost right, but you have to do a little more work even than I suggested in my comment. Since you’re assuming that $I$ is infinite, let $\{i_n:n \in \mathbb{N}\}$ be a set of distinct indices in $I$. Now for each $n \in \mathbb{N}$, $p_{i_n}[C_n]$ is a compact subset of $X_{i_n}$, so you can pick a point $z_{i_n} \in X_{i_n} \setminus p_{i_n}[C_n]$. For $i \in I \setminus \{i_n:n\in\mathbb{N}\}$ let $z_i$ be an arbitrary point of $X_i$. (It’s possible, of course, that $I \setminus \{i_n:n\in\mathbb{N}\} = \varnothing$, in which case there are no other $z_i$ to be picked.) Then $\langle z_i:i \in I \rangle \in \prod\limits_{i\in I}X_i \setminus \bigcup\limits_{n\in\mathbb{N}}C_n$, as desired. (The extra complication here is because the diagonalization requires assigning to each $n \in \mathbb{N}$ a single coordinate $i_n$ on which $z$ will avoid $C_n$, but $I$ might be an uncountable set, in which case there will necessarily be coordinates left over that weren’t needed for the diagonalization.)
For the first part you’ve already seen that it suffices to prove that the product of two $\sigma$-compact spaces is compact. I’ll use the notation that you’ve already established: $X_1 = \bigcup\limits_{n\in\mathbb{N}}C_n$, and $X_1 = \bigcup\limits_{n\in\mathbb{N}}D_n$, where the sets $C_n$ and $D_n$ are compact. You want to write $X_1 \times X_2$ as a countable union of compact sets.
For each pair $\langle m,n \rangle \in \mathbb{N}$ let $K(m,n) = C_m \times D_n$, and let $\mathscr{K} = \{K(m,n):m,n \in \mathbb{N}\}$; clearly $\mathscr{K}$ is a countable family of compact subsets of $X_1 \times X_2$, and I claim that $\cup\mathscr{K} = X_1 \times X_2$. (This, by the way, is exactly what LostInMath was suggesting.) To see this, let $\langle x_1,x_2 \rangle \in X_1 \times X_2$ be arbitrary; clearly $x_1 \in C_m$ and $x_2 \in D_n$ for some $m,n \in \mathbb{N}$. But then $\langle x_1,x_2 \rangle \in C_m \times D_n =$ $K(m,n)$, as claimed.