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
- Mizar article TSP_1:def 6