Closure of Irreducible Subspace is Irreducible/Proof 1

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

By definition, $Y$ is an irreducible subset of $S$ in $T$ if and only if the subspace $\struct {Y, \tau_Y}$ is an irreducible topological space.

That is, such that two arbitrary non-empty open sets of $\struct {Y, \tau_Y}$ are not disjoint.

The open sets of $T$ in $Y^-$ are the same as the open sets of $\struct {Y, \tau_Y}$.

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Hence by definition of irreducible space, any two open sets in $Y^-$ are not disjoint in $Y^-$.

That is $Y^-$ is also irreducible.

More generally, we can also show that if $Y^-$ is irreducible for a subset $Y \subseteq S$, then $Y$ is also irreducible in $T$.

Aiming for a contradiction, suppose $Y$ is not irreducible.

Then there exist two proper subsets $Y_1$, $Y_2$ of $Y$ which are closed in $Y$ such that $Y = Y_1 \cup Y_2$.

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Then:

$Y^- = {Y_1}^- \cup {Y_2}^-$

which contradicts the assumption.

$\blacksquare$