Product Space is T1 iff Factor Spaces are T1

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 $I$.

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

Then $T$ is a $T_1$ (Fréchet) space each of $\struct {S_\alpha, \tau_\alpha}$ is a $T_1$ (Fréchet) space.

Necessary Condition
Suppose that for some $\beta$, $\struct {S_\beta, \tau_\beta}$ 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 = \family {x_\alpha}: x_\alpha = \begin{cases}

s_\alpha & : \alpha \ne \beta \\ a & : \alpha = \beta \end{cases}$


 * $z = \family {x_\alpha}: x_\alpha = \begin{cases}

s_\alpha & : \alpha \ne \beta \\ b & : \alpha = \beta \end{cases}$

That is, $y$ and $z$ match on all coordinates except that for $\beta$.

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

Then $z \in H$ as $\forall U_\beta \in \map {\pr_\beta} H: a \in U_\beta \implies b \in U_\beta$

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

Sufficient Condition
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 coordinate.

Suppose, $a_\alpha = p, b_\alpha = q$ for some coordinate $\alpha$.

Then for each $U_\alpha \in \tau_\alpha$:


 * $\ds a \in U_\alpha \times \prod_{\beta \ne \alpha} S_\beta \implies b \in U_\alpha \times \prod_{\beta \ne \alpha} S_\beta$

But:


 * $\ds a \in U_\alpha \times \prod_{\beta \ne \alpha} S_\beta$ $p \in U_\alpha$
 * $\ds b \in U_\alpha \times \prod_{\beta \ne \alpha} S_\beta$ $q \in U_\alpha$

from which we infer, for each $U_\alpha \in \tau_\alpha$:


 * $p \in U_\alpha \implies q \in U_\alpha$

It follows that $\struct {S_\alpha, \tau_\alpha}$ is not a $T_1$ (Fréchet) space.