Product of Hausdorff Factor Spaces is Hausdorff

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Theorem

Let $T_\alpha = \left({S_\alpha, \tau_\alpha}\right)$ and $T_\beta = \left({S_\beta, \tau_\beta}\right)$ be topological spaces.

Let $T = T_\alpha \times T_\beta$ be the product space of $T_\alpha$ and $T_\beta$

Let $T_\alpha$ and $T_\beta$ both be $T_2$ (Hausdorff) spaces.


Then $T$ is also a $T_2$ (Hausdorff) space.


General Result

Let $\mathbb S = \left\{{\left({S_\alpha, \tau_\alpha}\right)}\right\}$ 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 $\mathbb S$.

Let each of $\left({S_\alpha, \tau_\alpha}\right)$ for $\alpha \in I$ be $T_2$ (Hausdorff) spaces.


Then $T$ is a $T_2$ (Hausdorff) space.


Proof

Let $T_\alpha$ and $T_\beta$ be Hausdorff spaces.

Let $\left({a, b}\right)$ and $\left({c, d}\right)$ be two distinct points of the product space $T$.

Let $a = c$.

Then as $\left({a, b}\right) \ne \left({c, d}\right)$ it follows that $b \ne d$.

As $T_\beta$ is Hausdorff, there exists two disjoint open sets $U, V \subseteq T_\beta$ such that $b \in U, d \in V$.

Then $T_\alpha \times U$ and $T_\alpha \times V$ are two open disjoint sets in the product space $T$ containing the original points.

Let $a \ne c$. As $T_\alpha$ is Hausdorff, there exists two disjoint open sets $U, V \subseteq T_\alpha$ such that $a \in U, c \in V$.

Then $U \times T_\beta$ and $V \times T_\beta$ are two open disjoint sets in the product space $T$ containing the original points.



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