Product Space is T1 iff Factor Spaces are T1

Theorem

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

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 $T_1$ (Fréchet) space iff each of $\left({S_\alpha, \tau_\alpha}\right)$ is a $T_1$ (Fréchet) space.

Proof

• Suppose $\exists \beta: \left({S_\beta, \tau_\beta}\right)$ is not a $T_1$ space.

Then $\exists a, b \in S_\beta$ such that $\forall U_\beta \in \tau_\beta$, $a \in U_\beta \implies b \in U_\beta$.

Consider the elements $y, z \in S$ defined as:

$y = \left \langle {x_\alpha}\right \rangle: x_\alpha = \begin{cases} s_\alpha & : \alpha \ne \beta \\ a & : \alpha = \beta \end{cases}$

$z = \left \langle {x_\alpha}\right \rangle: x_\alpha = \begin{cases} s_\alpha & : \alpha \ne \beta \\ b & : \alpha = \beta \end{cases}$

That is, $y$ and $z$ match (arbitrarily) on all ordinates except that for $\beta$.

Let $H \subseteq S: y \in H$.

Then $z \in H$ as $\forall U_\beta \in \operatorname{pr}_\beta \left({H}\right): a \in U_\beta \implies b \in U_\beta$

So $T$ is not a $T_1$ (Fréchet) space.

$\Box$

• Suppose $T$ is not a $T_1$ (Fréchet) space.

Then $\exists a, b \in S, a \ne b$ such that for all $U \in \tau$, $a \in U \implies b \in U$.

Then $a$ and $b$ are different in at least one ordinate.

Suppose, $a_m = p, b_m = q$ for some ordinate $m$.

Then either $a_m \in U_m \implies b_m \in U_m$.

It follows that $\left({S_m, \tau_m}\right)$ is not a $T_1$ (Fréchet) 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