Product Space is T3 1/2 iff Factor Spaces are T3 1/2/Factor Spaces are T3 1/2 implies Product Space is 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 $\mathbb S$.

For each $\alpha \in I$, let $\struct{S_\alpha, \tau_\alpha}$ be a $T_{3 \frac 1 2}$ space.

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

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

Let $x \in S$.

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

By definition of a closed subset, $S \setminus F \in \tau$

By definition of the Tychonoff topology there exists an open set $B$ of the natural basis containing $x$ which is disjoint from $F$.

By definition of the natural basis, $B$ is of the form:
 * $\map {\pr_{\alpha_1}^{\gets}} {U_1} \cap \dotsb \cap \map {\pr_{\alpha_n}^{\gets}} {U_n}$

where:
 * $\pr_{\alpha_k}$ is the $\alpha_k$-th projection from $S$
 * $U_k$ is open in $S_{\alpha_k}$ for all $1 \le k \le n$

By definition of a $T_{3 \frac 1 2}$ space, for each $1 \le k \le n$ there exists a continuous mapping:
 * $f_k: S_{\alpha_k} \to \closedint {0} {1}$

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

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

Let $g_k = f_k \circ \pr_{\alpha_k}$ be the composite mapping of $f_k$ with $\pr_{\alpha_k}$ for each $1 \le k \le n$.

From Composite of Continuous Mappings is Continuous each $g_k: S \to \closedint {0} {1} $ is a continuous mapping.

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


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

From Min Rule for Continuous Functions:
 * $g$ is continuous.

Now:

Let $y \in F$.

By definition of disjoint:
 * $\exists j \in \set{1, \dotsc, n} : y \notin \map {\pr_{\alpha_j}^{\gets}} {U_j}$

By definition of the inverse image mapping of $\pr_{\alpha_j}$:
 * $y_{\alpha_j} \notin U_j$

Thus:
 * $\map {f_j} {y_{\alpha_j}} = 0$

So:

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