T2 Space is Preserved under Closed Bijection

Theorem

Let $T_A = \left({S_A, \tau_A}\right), T_B = \left({S_B, \tau_B}\right)$ be topological spaces.

Let $\phi: T_A \to T_B$ be a closed bijection.

If $T_A$ is a $T_2$ (Hausdorff) space, then so is $T_B$.

Proof

Let $T_A$ be a $T_2$ (Hausdorff) space.

Then:

$\forall x, y \in S_A, x \ne y: \exists U_A, V_A \in \tau_A: x \in U_A, y \in V_A: U_A \cap V_A = \varnothing$

Suppose that $T_B$ is not Hausdorff.

Then:

$\exists a, b \in S_B: a \ne b: \forall U_B, V_B \in \tau_B: a \in U_B, b \in V_B \implies U_B \cap V_B \ne \varnothing$

That is, there exists at least one pair of points $a$ and $b$ for which all the open sets containing $a$ and $b$ are not disjoint.

As $\tau_B$ is a topology, it follows that $W_B = U_B \cap V_B$ is also an open set.

Let $U_A = \phi^{-1} \left({U_B}\right), V_A = \phi^{-1} \left({V_B}\right), W_A = \phi^{-1} \left({W_B}\right)$.

From Preimage of Intersection under Mapping, we have:

$\phi^{-1} \left({U_B \cap V_B}\right) = \phi^{-1} \left({U_B}\right) \cap \phi^{-1} \left({V_B}\right)$

that is:

$U_A \cap V_A = W_A$

By Bijection is Open iff Closed, $\phi$ is an open bijection.

By Bijection is Open iff Inverse is Continuous, it follows that $\phi^{-1}$ is continuous.

As $\phi$ is continuous, all of $U_A, V_A, W_A$ are open in $T_A$.

From the bijective nature of $\phi$, we have that:

$a \in U_B \implies \phi^{-1} \left({a}\right) \in U_A$
$b \in V_B \implies \phi^{-1} \left({b}\right) \in V_A$

Let $x = \phi^{-1} \left({a}\right), y = \phi^{-1} \left({b}\right)$.

Thus we have that:

$\exists x, y \in S_A: x \ne y: \forall U_A, V_A \in \tau_A: x \in U_A, y \in V_A \implies U_A \cap V_A \ne \varnothing$

contradicting the fact that $T_A$ is a $T_2$ (Hausdorff) space.

Hence $T_B$ must after all be a $T_2$ (Hausdorff) space.

$\blacksquare$