Product Space is T0 iff Factor Spaces are T0

Theorem
Let $T_A = \struct {S_A, \tau_A}$ and $T_B = \struct {S_B, \tau_B}$ be topological spaces.

Let $T = T_A \times T_B$ be the product space formed from $T_A$ and $T_B$.

Then $T$ is a $T_0$ (Kolmogorov) space $T_A$ and $T_B$ are themselves both $T_0$ (Kolmogorov) spaces.

Necessary Condition
Let $T$ be a $T_0$ (Kolmogorov) space.

From:
 * Subspace of Product Space is Homeomorphic to Factor Space
 * Separation Properties Preserved in Subspace
 * Separation Axioms Preserved under Homeomorphism

it follows that $T_A$ and $T_B$ are both $T_0$.

Sufficient Condition
Now suppose that $T$ is not $T_0$.

Then $\exists p, q \in S_A \times S_B$ such that:
 * $p \ne q$

and:
 * $\forall U \in \tau: \set {p, q} \subseteq U$ or $\set {p, q} \cap U = \O$

If $p \ne q$, then they are different in at least one coordinate.

, let $p_A \ne q_A$.

Let $V \in \tau_A$ such that $p_A \in V$.

Then from Projection from Product Topology is Continuous:
 * $\pr_A^{-1} \sqbrk V \in \tau$.

Since $p \in \pr_A^{-1} \sqbrk V$, we have that:
 * $q \in \pr_A^{-1} \sqbrk V$

From Projection is Surjection:
 * $\map {\pr_i} q = q_A \in V$

So:
 * $\forall V \in \tau_A: \set {p_A, q_A} \subseteq V$

or:
 * $\set {p_A, q_A} \cap V = \O$

Thus, by definition, $T_A$ is not $T_0$.