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 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 $x, y \in S : x \ne y$.

Then $x_\alpha \ne y_\alpha$ for some $\alpha \in I$.

Since $\struct {S_\alpha, \tau_\alpha}$ is Hausdorff then:
 * $\exists U, V \in \tau_\alpha: x_\alpha \in U, y_\alpha \in V : U \cap V = \O$

Let $\pr_\alpha: S \to S_\alpha$ be the projection of $S$ to $S_\alpha$.

Then:

By definition of the projection $\pr_\alpha$:
 * $\map {\pr_\alpha} x = x_\alpha \in U$

By definition of the preimage under $\pr_\alpha$:
 * $x \in \map {\pr_\alpha^\gets} U$

Similarly:
 * $y \in \map {\pr_\alpha^\gets} V$

By definition of the Tychonoff topology $\tau$:
 * $\map {\pr_\alpha^\gets} U, \map {\pr_\alpha^\gets} V \in \tau$

Since $x, y \in S$ were arbitrary, it follows that $T$ is a $T_2$ (Hausdorff) space.