T0 Space is Preserved under Closed Bijection

Theorem
Let $T_A = \left({X_A, \vartheta_A}\right), T_B = \left({X_B, \vartheta_B}\right)$ 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:


 * $\forall x, y \in X_A$, either:
 * $\exists U \in \vartheta_A: x \in U, y \notin U$
 * or:
 * $\exists U \in \vartheta_A: y \in U, x \notin U$

Suppose that:
 * $\exists x, y \in X_B: \forall V \in \vartheta_B: x, y \in V \lor x, y \notin V$

That is:


 * $\exists x, y \in X_B: \forall V \in \vartheta_B: \left\{{x, y}\right\} \subseteq V \lor \left\{{x, y}\right\} \cap V = \varnothing$

From Subset of Image it follows that:
 * $\forall V \in \vartheta_B: \phi^{-1} \left({\left\{{x, y}\right\}}\right) \subseteq \phi^{-1} \left({V}\right) \lor \phi^{-1} \left({\left\{{x, y}\right\}}\right) \cap \phi^{-1} \left({V}\right) = \varnothing$

that is:


 * $\forall V \in \vartheta_B: \phi^{-1} \left({x}\right), \phi^{-1} \left({y}\right) \in \phi^{-1} \left({V}\right) \lor \phi^{-1} \left({x}\right), \phi^{-1} \left({y}\right) \notin \phi^{-1} \left({V}\right)$

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

Thus:
 * $\forall U \in \vartheta_A: \phi^{-1} \left({x}\right), \phi^{-1} \left({y}\right) \in U \lor \phi^{-1} \left({x}\right), \phi^{-1} \left({y}\right) \notin U$

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.