T0 Space is Preserved under Closed Bijection

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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 $T_0$ (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:

$\exists x, y \in S_B: \forall V \in \tau_B: \left\{{x, y}\right\} \subseteq V \lor \left\{{x, y}\right\} \cap V = \varnothing$


From Image of Subset under Relation is Subset of Image: Corollary 3 it follows that:

$\forall V \in \tau_B: \phi^{-1} \left[{\left\{{x, y}\right\}}\right] \subseteq \phi^{-1} \left[{V}\right] \lor \phi^{-1} \left[{\left\{{x, y}\right\}}\right] \cap \phi^{-1} \left[{V}\right] = \varnothing$

that is:

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

But by definition of continuous mapping, $U = \phi^{-1} \left[{V}\right]$ is open in $T_A$ if and only if $V$ is open in $T_B$.

Thus:

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

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.

$\blacksquare$


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