T2 Space is Preserved under Closed Bijection

Theorem
Let $T_A = \struct {S_A, \tau_A}$ and $T_B = \struct {S_B, \tau_B}$ 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:
 * $\forall x, y \in S_A, x \ne y: \exists U_A, V_A \in \tau_A: x \in U_A, y \in V_A: U_A \cap V_A = \O$

Suppose that $T_B$ is not Hausdorff.

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

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 $U_A = \phi^{-1} \sqbrk {U_B}, V_A = \phi^{-1} \sqbrk {V_B}, W_A = \phi^{-1} \sqbrk {W_B}$.

From Preimage of Intersection under Mapping, we have:
 * $\phi^{-1} \sqbrk {U_B \cap V_B} = \phi^{-1} \sqbrk {U_B} \cap \phi^{-1} \sqbrk {V_B}$

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:
 * $a \in U_B \implies \map {\phi^{-1} } a \in U_A$


 * $b \in V_B \implies \map {\phi^{-1} } b \in V_A$

Let $x = \map {\phi^{-1} } a, y = \map {\phi^{-1} } b$.

Thus we have that:


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

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

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