Separation Properties Preserved under Topological Product

Theorem
Let $\mathbb S = \left \langle {\left({S_i, \tau_i}\right)}\right \rangle_{i \in I}$ be a family of topological spaces where $I$ is an arbitrary index set.

Let $\displaystyle T = \left({S, \tau}\right) = \prod \left({S_i, \tau_i}\right)$ be the product space of $\mathbb S$.

Then $T$ has one of the following properties iff each of $\left({S_i, \tau_i}\right)$ has the same property:


 * $T_0$ (Kolmogorov) Property


 * $T_1$ (Fréchet) Property


 * $T_2$ (Hausdorff) Property


 * $T_{2 \frac 1 2}$ (Completely Hausdorff) Property


 * $T_3$ Property


 * $T_{3 \frac 1 2}$ Property

If $T = \left({S, \tau}\right)$ has one of the following properties then each of $\left({S_i, \tau_i}\right)$ has the same property:


 * $T_4$ Property


 * $T_5$ Property

but the converse does not necessarily hold.

Corollary
$T = \left({S, \tau}\right)$ has one of the following properties iff each of $\left({S_i, \tau_i}\right)$ has the same property:


 * Regular Property


 * Tychonoff (Completely Regular) Property

If $T = \left({S, \tau}\right)$ has one of the following properties then each of $\left({S_i, \tau_i}\right)$ has the same property:


 * Normal Property


 * Completely Normal Property

but the converse does not necessarily hold.

Proof

 * Product Space is $T_0$ iff Component Spaces are $T_0$


 * Product Space is $T_1$ iff Component Spaces are $T_1$


 * Product Space is $T_2$ iff Component Spaces are $T_2$


 * Product Space is $T_{2 \frac 1 2}$ iff Component Spaces are $T_{2 \frac 1 2}$


 * Product Space is $T_3$ iff Component Spaces are $T_3$


 * Product Space is $T_{3 \frac 1 2}$ iff Component Spaces are $T_{3 \frac 1 2}$


 * Component Spaces are $T_4$ if Product Space is $T_4$


 * Component Spaces are $T_5$ if Product Space is $T_5$

Proof of Corollary

 * A regular space is a topological space which is both a $T_0$ (Kolmogorov) space and a $T_3$ space.

Hence from:
 * Product Space is $T_0$ iff Component Spaces are $T_0$

and
 * Product Space is $T_3$ iff Component Spaces are $T_3$

it follows that $T$ is a regular space iff each of $\left({S_i, \tau_i}\right)$ is a regular space.


 * A Tychonoff (completely regular) space is a topological space which is both a $T_0$ (Kolmogorov) space and a $T_{3 \frac 1 2}$ space.

Hence from:
 * Product Space is $T_0$ iff Component Spaces are $T_0$

and
 * Product Space is $T_{3 \frac 1 2}$ iff Component Spaces are $T_{3 \frac 1 2}$

it follows that $T$ is a Tychonoff space iff each of $\left({S_i, \tau_i}\right)$ is a Tychonoff space.


 * A normal space is a topological space which is both a $T_1$ (Fréchet) space and a $T_4$ space.

Hence from:
 * Product Space is $T_1$ iff Component Spaces are $T_1$

and
 * Component Spaces are $T_4$ if Product Space is $T_4$

it follows that if $T$ is a normal space then each of $\left({S_i, \tau_i}\right)$ is a normal space.


 * A completely normal space is a topological space which is both a $T_1$ (Fréchet) space and a $T_5$ space.

Hence from:
 * Product Space is $T_1$ iff Component Spaces are $T_1$

and
 * Component Spaces are $T_5$ if Product Space is $T_5$

it follows that if $T$ is a completely normal space then each of $\left({S_i, \tau_i}\right)$ is a completely normal space.