Closure of Irreducible Subspace is Irreducible/Proof 3

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 $T$.

Observe that for each closed set $V$ in $T$:

$(1): \quad V \subsetneqq Y^- \implies V \cap Y \subsetneqq Y$

Indeed:

$\ds V \cap Y$

$=$

$\ds Y$

$\ds \leadsto \ \ $

$\ds Y \setminus V$

$=$

$\ds \O$

$\ds \leadsto \ \ $

$\ds Y$

$\subseteq$

$\ds V$

$\ds \leadsto \ \ $

$\ds Y^-$

$\subseteq$

$\ds V$

$\ds \leadsto \ \ $

$\ds V$

$\subsetneqq$

$\ds Y^-$

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

That is, there exist $V_1, V_2 \subsetneqq Y^-$, closed in $\struct {Y^-, \tau_{Y^-} }$, such that:

$Y^- = V_1 \cup V_2$

$V_1$ and $V_2$ are also closed in $T$, since $Y^-$ is closed in $T$,

Then, from $(1)$:

$V_1 \cap Y \subsetneqq Y$

and:

$V_2 \cap Y \subsetneqq Y$

Therefore $V_1 \cap Y$ and $V_2 \cap Y$ are proper subsets of $Y$ such that:

$Y = \paren {V_1 \cap Y} \cup \paren {V_2 \cap Y}$

Furtheremore, $V_1 \cap Y$ and $V_2 \cap Y$ are closed in $\struct {Y, \tau_Y}$.

This contradicts the fact that $Y$ is irreducible.

$\blacksquare$