Product Space is T3 iff Factor Spaces are T3/Product Space is T3 implies Factor Spaces are 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 $\mathbb S$.

Let $T$ be a $T_3$ space.

Then:
 * for each $\alpha \in I$, $\struct{S_\alpha, \tau_\alpha}$ is a $T_3$ space.

Proof
Let $\alpha \in I$.

Let $x, y \in S_\alpha$.

By the Axiom of Choice, there exists $z \in S$.

Define $x', y' \in S$ by:
 * $x'_\beta = \begin{cases} z_\beta & \beta \ne \alpha \\ x & \beta = \alpha \end{cases}$

and
 * $y'_\beta = \begin{cases} z_\beta & \beta \ne \alpha \\ y & \beta = \alpha \end{cases}$

By definition of a $T_2$ (Hausdorff) space:
 * $\exists U, V \in \tau : x' \in U, y' \in V$ and $U \cap V = \O$

Recall that the topology on the product space $T$ is the Tychonoff topology.

By definition of the Tychonoff topology the natural basis is a basis for the Tychonoff topology.

Then:
 * $\exists U', V' \in \BB : x' \in U' \subseteq U, y' \in V' \subseteq V$

where $\BB$ is the natural basis for $\tau$.

From Subsets of Disjoint Sets are Disjoint:
 * $U' \cap V' = \O$

From Natural Basis of Tychonoff Topology:
 * $\displaystyle U' = \prod_{\beta \mathop \in I} U_\beta$ where:
 * for all $\beta \in I : U_\beta \in \tau_\beta$
 * for all but finitely many indices $\beta : U_\beta = X_\beta$

Similarly:
 * $\displaystyle V' = \prod_{\beta \mathop \in I} V_\beta$ where:
 * for all $\beta \in I : V_\beta \in \tau_\beta$
 * for all but finitely many indices $\beta : V_\beta = X_\beta$

From General Case of Cartesian Product of Intersections:
 * $\displaystyle \paren{\prod_{\beta \mathop \in I} U_\beta} \bigcap \paren{\prod_{\beta \mathop \in I} V_\beta} = \prod_{\beta \mathop \in I} \paren{U_\beta \cap V_\beta}$

Thus:
 * $\displaystyle \prod_{\beta \mathop \in I} \paren{U_\beta \cap V_\beta} = \O$

Now for all $\beta \in I$ such that $\beta \ne \alpha$:
 * $z_\beta = x'_\beta \in U_\beta$

and
 * $z_\beta = y'_\beta \in V_\beta$

So:
 * $\forall \beta \in I, \beta \ne \alpha : U_\beta \cap V_\beta \ne \O$

From Cartesian Product of Family is Empty iff Factor is Empty it follows that:
 * $U_\alpha \cap V_\alpha = \O$

Also we have:
 * $x = x'_\alpha \in U_\alpha$

and
 * $y = y'_\alpha \in V_\alpha$

Since $U_\alpha$ and $V_\alpha$ are open in $\struct {S_\alpha, \tau_\alpha}$ then $x$ and $y$ are separated by open sets.

Since $x$ and $y$ were arbitrary, then $\struct {S_\alpha, \tau_\alpha}$ is a $T_2$ (Hausdorff) space by definition.

Since $\alpha \in I$ was arbitrary, then for each $\alpha \in I$, $\struct {S_\alpha, \tau_\alpha}$ is a $T_2$ (Hausdorff) space.