Product Space is T3 iff Factor Spaces are T3/Factor Spaces are T3 implies Product Space is T3

Theorem
Let $\mathbb S = \family{\struct{S_\alpha, \tau_\alpha}}_{\alpha \mathop \in I}$ be an indexed family of non-empty topological spaces for $\alpha$ in some indexing set $I$.

Let $\displaystyle T = \struct{S, \tau} = \displaystyle \prod_{\alpha \mathop \in I} \struct{S_\alpha, \tau_\alpha}$ be the product space of $\mathbb S$.

For each $\alpha \in I$, let $\struct{S_\alpha, \tau_\alpha}$ be a $T_3$ space.

Then $T$ is a $T_3$ space.

Proof
Let $U$ be open in $T$ and $x \in U$.

From Natural Basis of Tychonoff Topology, there exists $\displaystyle U' = \prod_{\alpha \mathop \in I} U'_\alpha$ such that:
 * for all $\alpha \in I : U_\alpha \in \tau_\alpha$
 * $J = \set {\alpha \in I: U_\alpha \ne X_\alpha}$ is finite
 * $x \in U'$
 * $U' \subseteq U$

By definition of a $T_3$ space then for all $\alpha \in J$:
 * $\exists V_\alpha \in \tau_\alpha : x_\alpha \in V_\alpha$ and $V^-_\alpha \subseteq U'_\alpha$

For $\alpha \notin J$ let:
 * $V_\alpha = X_\alpha$

and let:
 * $\displaystyle V = \prod_{\alpha \mathop \in I} V_\alpha$

By definition of the Cartesian product V:
 * $x \in V$.

From Natural Basis of Tychonoff Topology:
 * $V$ open in $T$.

Now consider the Cartesian product of closed sets:
 * $\displaystyle \prod_{\alpha \mathop \in I} V^-_\alpha$

From Product of Closed Sets is Closed, $\displaystyle \prod_{\alpha \mathop \in I} V^-_\alpha$ is closed.

From Underlying Set of Topological Space is Closed, for $\alpha \notin J$:
 * $V^-_\alpha = X^-_\alpha = X_\alpha$

From Cartesian Product of Family of Subsets, it follows that:
 * $\displaystyle V = \prod_{\alpha \mathop \in I} V_\alpha \subseteq \prod_{\alpha \mathop \in I} V^-_\alpha \subseteq \prod_{\alpha \mathop \in I} U'_\alpha \subseteq U$

It follows that $T$ is a $T_3$ space by definition.