Separation Properties Preserved under Topological Product

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

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

Then $T$ has one of the following properties 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.

Proof

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


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


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


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


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


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


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


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