T0 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 $T_0$ (Kolmogorov) space, then so is $T_B$.

Proof
Let $T_A$ be a $T_0$ (Kolmogorov) space.

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.

By definition:


 * $\forall x, y \in S_A$, either:
 * $\exists U \in \tau_A: x \in U, y \notin U$
 * or:
 * $\exists U \in \tau_A: y \in U, x \notin U$

Suppose that:
 * $\exists x, y \in S_B: \forall V \in \tau_B: x, y \in V \lor x, y \notin V$

That is:


 * $\exists x, y \in S_B: \forall V \in \tau_B: \set {x, y} \subseteq V \lor \set {x, y} \cap V = \O$

From it follows that:
 * $\forall V \in \tau_B: \phi^{-1} \sqbrk {\set {x, y} } \subseteq \phi^{-1} \sqbrk V \lor \phi^{-1} \sqbrk {\set {x, y} } \cap \phi^{-1} \sqbrk V = \O$

that is:


 * $\forall V \in \tau_B: \map {\phi^{-1} } x, \map {\phi^{-1} } y \in \phi^{-1} \sqbrk V \lor \map {\phi^{-1} } x, \map {\phi^{-1} } y \notin \phi^{-1} \sqbrk V$

But by definition of continuous mapping, $U = \phi^{-1} \sqbrk V$ is open in $T_A$ $V$ is open in $T_B$.

Thus:
 * $\forall U \in \tau_A: \map {\phi^{-1} } x, \map {\phi^{-1} } y \in U \lor \map {\phi^{-1} } x, \map {\phi^{-1} } y \notin U$

This contradicts the condition that $T_A$ is a $T_0$ (Kolmogorov) space.

It follows that $T_B$ must also be a $T_0$ (Kolmogorov) space.