Topological Closure of Singleton is Irreducible/Proof 2

Theorem

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

Let $x$ be a point of $T$.

Then:

$\set x^-$ is irreducible

where $\set x^-$ denotes the topological closure of $\set x$.

Proof

Aiming for a contradiction, suppose that

$\set x^-$ is not irreducible.

By Set is Subset of its Topological Closure:

$\set x \subseteq \set x^-$

By definitions of singleton and Subset:

$x \in \set x^-$

By definition of irreducible:

$\exists X_1, X_2 \subseteq S: X_1, X_2$ are closed

and:

\(\ds \set x^-\)

\(=\)

\(\ds X_1 \cup X_2\)

\(\, \ds \text {and} \, \)

\(\ds \set x^-\)

\(\ne\)

\(\ds X_1\)

\(\, \ds \text {and} \, \)

\(\ds \set x^-\)

\(\ne\)

\(\ds X_2\)

By definition of union:

$x \in X_1$ or $x \in X_2$

By definitions of singleton and subset:

$\set x \subseteq X_1$ or $\set x \subseteq X_2$

By definition of closure :

$\set x^- \subseteq X_1$ or $\set x^- \subseteq X_2$

By Set is Subset of Union:

$X_1 \subseteq \set x$ and $X_2 \subseteq \set x$

Thus by definition of set equality:

this contradicts $\set x^- \ne X_1$ and $\set x^- \ne X_2$

$\blacksquare$