Finite Product of Sigma-Compact Spaces is Sigma-Compact

Theorem
Let $n \in \Z_{>0}$ be a (strictly) positive integer.

Let $\set {\struct {S_i, \tau_i}: 1 \le i \le n}$ be a finite set of topological spaces.

Let $\ds \struct {S, \tau} = \prod_{i \mathop = 1}^n \struct {S_i, \tau_i}$ be the product space of $\set {\struct {S_i, \tau_i}: 1 \le i \le n}$.

Let each of $\struct {S_i, \tau_i}$ be $\sigma$-compact.

Then $\struct {S, \tau}$ is also $\sigma$-compact.

Proof
Let $\struct {S_1, \tau_1}$ and $\struct {S_2, \tau_2}$ be $\sigma$-compact.

Then $\ds S_1 = \bigcup_{n \mathop = 1}^\infty C_n$ and $\ds S_2 = \bigcup_{n \mathop = 1}^\infty D_n$ where all the $C_n$ and $D_n$ are compact sets.

From Tychonoff's Theorem, the set $\KK = \set {C_n \times D_m: n, m \in \N}$ is a countable set of compact sets of $S_1 \times S_2$ with the product topology.

Consider $\tuple {a, b} \in S_1 \times S_2$.

Then $a \in S_1$ and $b \in S_2$.

Using the $\sigma$-compactness of $S_1$ and $S_2$, $\exists n, m$ such that $a \in C_n$ and $b \in D_m$.

So $\tuple {a, b} \in C_n \times D_m \in \KK$.

Thus, $\bigcup \KK = S_1 \times S_2$ and then $S_1 \times S_2$ is $\sigma$-compact.

Using induction on the number of factors of the product we get the result.