Characterization of T0 Space by Closures of Singletons

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Theorem

Let $T = \left({S, \tau}\right)$ be a topological space.

Then

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

where $\left\{{y}\right\}^-$ denotes the closure of $\left\{{y}\right\}$.


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 that

$x \in \left\{{y}\right\}^- \land y \in \left\{{x}\right\}^-$

Then:

$\left\{{x}\right\} \subseteq \left\{{y}\right\}^- \land \left\{{y}\right\} \subseteq \left\{{x}\right\}^-$

By Topological Closure of Subset is Subset of Topological Closure:

$\left\{{x}\right\}^- \subseteq \left({\left\{{y}\right\}^-}\right)^- \land \left\{{y}\right\}^- \subseteq \left({\left\{{x}\right\}^-}\right)^-$

By Closure of Topological Closure equals Closure:

$\left\{{x}\right\}^- \subseteq \left\{{y}\right\}^- \land \left\{{y}\right\}^- \subseteq \left\{{x}\right\}^-$

Then by definition of set equality:

$\left\{{x}\right\}^- = \left\{{y}\right\}^-$

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

$\left\{{x}\right\}^- \ne \left\{{y}\right\}^-$

This contradicts the equality.

Thus the result by Proof by Contradiction

$\Box$


Necessary Condition

Assume that:

$(1): \quad \forall x, y \in S: x \ne y \implies x \notin \left\{{y}\right\}^- \lor y \notin \left\{{x}\right\}^-$

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

$\forall x, y \in S: x \ne y \implies \left\{{y}\right\}^- \ne \left\{{x}\right\}^-$

Let $x, y \in S$ such that

$x \ne y$

Aiming for a contradiction suppose that

$(2): \quad \left\{{y}\right\}^- = \left\{{x}\right\}^-$

By definition of singleton:

$x \in \left\{{x}\right\} \land y \in \left\{{y}\right\}$

By Set is Subset of its Topological Closure

$\left\{{x}\right\} \subseteq \left\{{x}\right\}^- \land \left\{{y}\right\} \subseteq \left\{{y}\right\}^-$

Then by definition of subset and $(2)$:

$x \in \left\{{y}\right\}^- \land y \in \left\{{x}\right\}^-$

This contradicts the assumption $(1)$.

Thus the result by Proof by Contradiction.

$\blacksquare$


Sources