Characterization of T0 Space by Distinct 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
$\forall x, y \in S: x \ne y \implies \left\{{x}\right\}^- \ne \left\{{y}\right\}^-$

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

$x \ne y$

By definition of $T_0$ space:

$\left({\exists U \in \tau: x \in U \land y \notin U}\right) \lor \left({\exists U \in \tau: x \notin U \land y \in U}\right)$

WLOG: Suppose

$\exists U \in \tau: x \in U \land y \notin U$

By definition of singleton:

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

By Set is Subset of its Topological Closure:

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

Then by definition of subset:

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

Hence by definition of intersection:

$\left\{{x}\right\}^- \cap U \ne \varnothing$

By definition of singleton:

$\forall z: z \in \left\{{y}\right\} \implies z = y$

Then by definition of intersection:

$\left\{{y}\right\} \cap U = \varnothing$

Hence by Disjoint Open Sets remain Disjoint with one Closure

$\left\{{y}\right\}^- \cap U = \varnothing$

Thus:

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

$\Box$


Necessary Condition

Assume that:

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

By Characterization of $T_0$ Space by Closed Sets it suffices to prove that

$\left({\exists F \subseteq S: F}\right.$ is closed $\left. {\land x \in F \land y \notin F}\right)$

or

$\left({\exists F \subseteq S: F}\right.$ is closed $\left. {\land x \notin F \land y \in F}\right)$

Then assume:

$(2): \quad \forall F \subseteq S: F$ is closed $\land x \in F \implies y \in F$

Define $F := \left\{{y}\right\}^-$

By Topological Closure is Closed:

$F$ is closed.

We will prove now that

$x \notin F$

Aiming for a contradiction suppose that

$x \in F$

Then:

$\left\{{x}\right\} \subseteq F$

By Topological Closure of Subset is Subset of Topological Closure:

$\left\{{x}\right\}^- \subseteq F^-$

By Set is Closed iff Equals Topological Closure:

$(3): \quad \left\{{x}\right\}^- \subseteq F$

By definition of singleton:

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

By Set is Subset of its Topological Closure:

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

Then by definition of subset:

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

Because $\left\{{x}\right\}^-$ is closed therefore by $(2)$:

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

Then:

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

By Topological Closure of Subset is Subset of Topological Closure:

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

Then by Closure of Topological Closure equals Closure:

$F \subseteq \left\{{x}\right\}^-$

Hence by $(3)$ and definition of set equality:

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

This contradicts the assumption $(1)$.

Thus $x \notin F$.

By definition of singleton:

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

By Set is Subset of its Topological Closure:

$\left\{{y}\right\} \subseteq \left\{{y}\right\}^- = F$

Thus by definition of subset:

$y \in F$

Hence

$\exists F \subseteq S: F$ is closed $\land x \notin F \land y \in F$

$\blacksquare$


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