Product Space is T0 iff Factor Spaces are T0

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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 if and only if $T_A$ and $T_B$ are themselves both $T_0$ (Kolmogorov) spaces.


General Result

Let $\SS = \family {\struct{S_\alpha, \tau_\alpha}}_{\alpha \mathop \in I}$ be an indexed family of topological spaces where $I$ is an arbitrary index set.


Let $\displaystyle T = \struct{S, \tau}= \prod \struct{S_\alpha, \tau_\alpha}$ be the product space of $\SS$.


Then $T$ is a $T_0$ (Kolmogorov) space if and only if each of $\struct{S_\alpha, \tau_\alpha}$ is a $T_0$ (Kolmogorov) space.


Proof

Necessary Condition

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

From:

Subspace of Product Space Homeomorphic to Factor Space
Separation Properties Preserved in Subspace
Separation Axioms Preserved under Homeomorphism

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

$\Box$


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.

Without loss of generality, 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 $T_0$.

$\blacksquare$