Product Space is T3 1/2 iff 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 topological spaces for $\alpha$ in some indexing set $I$ with $S_\alpha \neq \O$ for every $\alpha \in I$.

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

Then $T$ is a $T_{3 \frac 1 2}$ space each of $\struct{S_\alpha, \tau_\alpha}$ 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 $S \ne \O$.

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

By $T_{3 \frac 1 2}$ property is hereditary we then find that $\struct{S_\alpha, \tau_\alpha}$ is $T_{3 \frac 1 2}$.

This obviously holds for every $\alpha \in I$.

Suppose every $\struct{S_\alpha, \tau_\alpha}$ is a $T_{3 \frac 1 2}$ space.

Let $x \in S$.

Let $F$ be a closed subset of $S$ such that $x \notin F$.

We can then find a neighborhood:


 * $\map {\pr_{\alpha_1}^{-1}} {U_1} \cap \dotsb \cap \map {\pr_{\alpha_n}^{-1}} {U_n}$

of $x$ which is disjoint from $F$.

Here every $U_k$ is open in $S_{\alpha_k}$ for all $1 \le k \le n$.

Since every $\struct{S_\alpha, \tau_\alpha}$ is a $T_{3 \frac 1 2}$ space, there exists a continuous mapping:


 * $f_k: S_{\alpha_k} \to \closedint {0} {1}$

such that:
 * $\map {f_k} {x_{\alpha_k} } = 1$

and:
 * $\map {f_k} {S_{\alpha_k} \setminus U_k} = 0$

We define $g: S \to \closedint {0} {1}$ by setting:


 * $\map g y = \min \set{\map {f_k} {y_{\alpha_k} }: k = 1, \dotsc, n}$

Then $g$ is continuous and we have:
 * $\map g x = 1$

and:
 * $\map g {S \setminus F} = 0$

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