Characterization of T0 Space by Closures of Singletons

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Theorem

Let $T = \struct {S, \tau}$ be a topological space.

Then

$T$ is a $T_0$ space if and only if:
$\forall x, y \in S: x \ne y \implies x \notin \set y^- \lor y \notin \set x^-$

where $\set y^-$ denotes the closure of $\set y$.


Proof

Sufficient Condition

Let $T$ be a $T_0$ space.

Let $x, y \in S$ such that

$x \ne y$


Aiming for a contradiction, suppose

$x \in \set y^- \land y \in \set x^-$

Then:

\(\ds x\) \(\in\) \(\ds \set y^-\)
\(\, \ds \land \, \) \(\ds \set y\) \(\subseteq\) \(\ds \set x^-\)
\(\ds \leadsto \ \ \) \(\ds \set x\) \(\subseteq\) \(\ds \set y^-\) Definition of Singleton
\(\, \ds \land \, \) \(\ds \set y\) \(\subseteq\) \(\ds \set x^-\)
\(\ds \leadsto \ \ \) \(\ds \set x^-\) \(\subseteq\) \(\ds \paren {\set y^-}^-\) Topological Closure of Subset is Subset of Topological Closure
\(\, \ds \land \, \) \(\ds \set y^-\) \(\subseteq\) \(\ds \paren {\set x^-}^-\)
\(\ds \leadsto \ \ \) \(\ds \set x^-\) \(\subseteq\) \(\ds \set y^-\) Closure of Topological Closure equals Closure
\(\, \ds \land \, \) \(\ds \set y^-\) \(\subseteq\) \(\ds \set x^-\)
\(\ds \leadsto \ \ \) \(\ds \set x^-\) \(=\) \(\ds \set y^-\) Definition 2 of Set Equality


By Characterization of $T_0$ Space by Distinct Closures of Singletons:

$\set x^- \ne \set y^-$

This contradicts the equality.

Thus the result by Proof by Contradiction

$\Box$


Necessary Condition

Assume that:

$(1): \quad \forall x, y \in S: x \ne y \implies x \notin \set y^- \lor y \notin \set x^-$

By Characterization of $T_0$ Space by Distinct Closures of Singletons it suffices to prove

$\forall x, y \in S: x \ne y \implies \set x^- \ne \set y^-$

Let $x, y \in S$ such that $x \ne y$.

Aiming for a contradiction, suppose

$(2): \quad \set x^- = \set y^-$

Then:

\(\ds x\) \(\in\) \(\ds \set x\) Definition of Singleton
\(\, \ds \land \, \) \(\ds y\) \(\in\) \(\ds \set y\)
\(\ds \set x\) \(\subseteq\) \(\ds \set x^-\) Set is Subset of its Topological Closure
\(\, \ds \land \, \) \(\ds \set y\) \(\subseteq\) \(\ds \set y^-\)
\(\ds \leadsto \ \ \) \(\ds x\) \(\in\) \(\ds \set x^-\) Definition of Subset
\(\, \ds \land \, \) \(\ds y\) \(\in\) \(\ds \set y^-\)
\(\ds \leadsto \ \ \) \(\ds x\) \(\in\) \(\ds \set y^-\) by hypothesis: $(2)$
\(\, \ds \land \, \) \(\ds y\) \(\in\) \(\ds \set x^-\)

This contradicts the assumption $(1)$.

Thus the result by Proof by Contradiction.

$\blacksquare$


Sources

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