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

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Theorem

Let $\mathbb S = \left\{{\left({S_\alpha, \tau_\alpha}\right)}\right\}$ be a set of topological spaces for $\alpha$ in some indexing set $I$ with $S_\alpha \neq \varnothing$ for every $\alpha \in I$.


Let $\displaystyle T = \left({S, \tau}\right) = \prod \left({S_\alpha, \tau_\alpha}\right)$ be the product space of $\mathbb S$.


Then $T$ is a $T_{3 \frac 1 2}$ space if and only if each of $\left({S_\alpha, \tau_\alpha}\right)$ is a $T_{3 \frac 1 2}$ space.


Proof

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

Since $S_\alpha \ne \varnothing$ we also have $S \ne \varnothing$.

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

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

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


Suppose every $\left({S_\alpha, \tau_\alpha}\right)$ 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:

$\operatorname{pr}_{\alpha_1}^{-1} \left({U_1}\right) \cap \dotsb \cap \operatorname{pr}_{\alpha_n}^{-1} \left({U_n}\right)$

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 $\left({S_\alpha, \tau_\alpha}\right)$ is a $T_{3 \frac 1 2}$ space, there exists a continuous mapping:

$f_k: S_{\alpha_k} \to \left[{0 \,.\,.\, 1}\right]$

such that:

$f_k \left({x_{\alpha_k} }\right) = 1$

and:

$f_k \left({S_{\alpha_k} \setminus U_k}\right) = 0$

We define $g: S \to \left[{0 \,.\,.\, 1}\right]$ by setting:

$g \left({y}\right) = \min \left\{ {f_k \left({y_{\alpha_k} }\right): k = 1, \dotsc, n}\right\}$

Then $g$ is continuous and we have:

$g \left({x}\right) = 1$

and:

$g \left({S \setminus F}\right) = 0$

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

$\blacksquare$


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