# Product Space is T1 iff Factor Spaces are T1

## Theorem

Let $\mathbb S = \left\{{\left({S_\alpha, \tau_\alpha}\right)}\right\}_\alpha$ 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 if and only if each of $\left({S_\alpha, \tau_\alpha}\right)$ is a $T_1$ (Fréchet) space.

## Proof

### Necessary Condition

Suppose that for some $\beta$, $\left({S_\beta, \tau_\beta}\right)$ 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 on all coordinates except that for $\beta$.

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

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

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

$\Box$

### 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 $U_\alpha \in \tau_\alpha$:

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

But:

$\displaystyle a \in U_\alpha \times \prod_{\beta \ne \alpha} S_\beta$ if and only if $p \in U_\alpha$
$\displaystyle b \in U_\alpha \times \prod_{\beta \ne \alpha} S_\beta$ if and only if $q \in U_\alpha$

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

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

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

$\blacksquare$