Topological Closure of Singleton is Irreducible
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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 1
Follows from:
$\blacksquare$
Proof 2
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 $\set x^- \subseteq X_2$
- $X_1 \subseteq \set x$ and $X_2 \subseteq \set x$
Thus by definition of set equality:
- this contradicts $\set x^- \ne X_1$ and $\set x^- \ne X_2$
$\blacksquare$
Sources
- Mizar article YELLOW_8:17