Product Space is T0 iff Factor Spaces are T0/General Result

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Theorem

Let $\SS = \family {\struct{S_\alpha, \tau_\alpha}}_{\alpha \mathop \in I}$ be an indexed family of topological spaces where $I$ is an arbitrary index set.


Let $\displaystyle T = \struct{S, \tau}= \prod \struct{S_\alpha, \tau_\alpha}$ be the product space of $\SS$.


Then $T$ is a $T_0$ (Kolmogorov) space if and only if each of $\struct{S_\alpha, \tau_\alpha}$ is a $T_0$ (Kolmogorov) space.


Proof

Sufficient Condition

Suppose $\exists \beta: \struct{S_\beta, \tau_\beta}$ is not a $T_0$ space.

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

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 (arbitrarily) on all ordinates except that for $\beta$.

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

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

Similarly, let $K \subseteq S: z \in K$.

Then $y \in K$ as $\forall U_\beta \in \map {\pr_\beta} H : a \in U_\beta$

So $T$ is not a $T_0$ (Kolmogorov) space.

$\Box$


Necessary Condition

Suppose $T$ is not a $T_0$ (Kolmogorov) space.

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

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

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

Then either $a_m, b_m \in U_m$ or $a_m, b_m \notin U_m$.

It follows that $\struct{S_m, \tau_m}$ is not a $T_0$ (Kolmogorov) space.

$\blacksquare$


Sources