Product Space is T0 iff Factor Spaces are T0

Theorem
Let $\mathbb S = \left\{{\left({S_\alpha, \tau_\alpha}\right)}\right\}$ be a set of topological spaces for $\alpha$ in some indexing set $I$.

Let $\displaystyle T = \left({S, \tau}\right) = \prod \left({S_\alpha, \tau_\alpha}\right)$ be the product space of $\mathbb S$.

Then $T$ is a $T_0$ (Kolmogorov) space iff each of $\left({S_\alpha, \tau_\alpha}\right)$ is a $T_0$ (Kolmogorov) space.

Proof

 * Suppose $\exists \beta: \left({S_\beta, \tau_\beta}\right)$ is not a $T_0$ space.

Then $\exists a, b \in S_\beta$ such that $\forall U_\beta\in \tau_\beta$, either $a, b \in U_\beta$ or $a, b \notin U_\beta$.

Consider the elements $y, z \in S$ defined as:
 * $y = \left \langle {x_\alpha}\right \rangle: x_\alpha = \begin{cases}

s_\alpha & : \alpha \ne \beta \\ a & : \alpha = \beta \end{cases}$


 * $z = \left \langle {x_\alpha}\right \rangle: x_\alpha = \begin{cases}

s_\alpha & : \alpha \ne \beta \\ b & : \alpha = \beta \end{cases}$

That is, $y$ and $z$ match (arbitrarily) on all ordinates except that for $\beta$.

Let $H \subseteq S: y \in H$.

Then $z \in H$ as $\forall U_\beta \in \operatorname{pr}_\beta \left({H}\right): b \in U_\beta$

Similarly, let $K \subseteq S: z \in K$.

Then $y \in K$ as $\forall U_\beta \in \operatorname{pr}_\beta \left({H}\right): a \in U_\beta$

So $T$ is not a $T_0$ (Kolmogorov) space.


 * Suppose $T$ is not a $T_0$ (Kolmogorov) space.

Then $\exists a, b \in S, a \ne b$ such that for all $U \in \tau$, either $a, b \in U$ or $a, b \notin U$.

Then $a$ and $b$ are different in at least one ordinate.

Suppose, $a_m = p, b_m = q$ for some ordinate $m$.

Then either $a_m, b_m \in U_m$ or $a_m, b_m \notin U_m$.

It follows that $\left({S_m, \tau_m}\right)$ is not a $T_0$ (Kolmogorov) space.