Product Space is T3.5 iff Factor Spaces are T3.5

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 iff 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 $S_\alpha \neq \varnothing$ we also have $S\neq\varnothing$.

From Subspace of Product Space Homeomorphic to Factor Space, every $\left({S_\alpha,\tau_\alpha}\right)$ is homeorphic 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 $x\in S$ and $F$ be a closed subset of $S$ such that $x\not\in F$.

We can then find a neighborhood

$\operatorname{pr}_{\alpha_{1}}^{-1}(U_1)\cap...\cap\operatorname{pr}_{\alpha_{n}}^{-1}(U_n)$

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

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

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

$f_k : S_{\alpha_{k}} \to [0,1]$ such that $f_k(x_{\alpha_{k}}) = 1$ and $f_{k}(S_{\alpha_{k}} \setminus U_{k}) = 0$.

We define $g: S \to [0,1]$ by setting

$g(y)= \operatorname{min}\{f_{k}(y_{\alpha_{k}}) : k = 1,...,n\}$.

Then $g$ is continuous and we have $g(x)=1$ and $g(S\setminus F)=0$.

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

$\blacksquare$

Sources

• Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (1970)... (previous)... (next): $\text{I}: \ \S 2$: Functions, Products, and Subspaces