Product Space is T3 1/2 iff Factor Spaces are T3 1/2/Product Space is T3 1/2 implies Factor Spaces are T3 1/2

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 $\mathbb S$.

Let $T$ be a $T_{3 \frac 1 2}$ space.

Then for each $\alpha \in I$, $\struct{S_\alpha, \tau_\alpha}$ is a $T_{3 \frac 1 2}$ space.

Proof
Suppose $T$ is a $T_{3 \frac 1 2}$ space.

Since $S_\alpha \ne \O$ we also have $S \ne \O$ by the axiom of choice.

Let $\alpha \in I$.

From Subspace of Product Space Homeomorphic to Factor Space, $\struct{S_\alpha, \tau_\alpha}$ is homeomorphic to a subspace $T_\alpha$ of $T$.

From $T_{3 \frac 1 2}$ property is hereditary then $T_\alpha$ is $T_{3 \frac 1 2}$.

From T3 1/2 Space is Preserved under Homeomorphism then $\struct{S_\alpha, \tau_\alpha}$ is $T_{3 \frac 1 2}$.

Since $\alpha \in I$ was arbitrary then the result follows.