Closure of Irreducible Subspace is Irreducible/Proof 2
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Theorem
Let $T = \struct {S, \tau}$ be a topological space.
Let $Y \subseteq S$ be a subset of $S$ which is irreducible in $T$.
Then its closure $Y^-$ in $T$ is also irreducible in $T$.
Proof
In view of Definition of Irreducible Subset, it suffices to show that for all closed sets $A_1$ and $A_2$ in $T$:
- $Y^- \subseteq A_1 \cup A_2 \implies \exists i_0 \in \set {1, 2} : Y^- \subseteq A_{i_0}$
To this end, let $A_1$ and $A_2$ be closed sets in $T$ such that:
- $Y^- \subseteq A_1 \cup A_2$
Then, in particular:
- $Y \subseteq A_1 \cup A_2$
Since $Y \subseteq S$ is an irreducible subset:
- $\exists i_0 \in \set {1, 2} : Y \subseteq A_{i_0}$
By Closure of Subset of Closed Set of Topological Space is Subset:
- $Y^- \subseteq A_{i_0}$
$\blacksquare$