Closure of Irreducible Subspace is Irreducible/Proof 2

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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 subset $V \subseteq Y^-$ which is closed in $T$:

$(1): \quad Y^- \setminus V \ne \O \implies Y \setminus V \ne \O$


\(\ds Y \setminus V\) \(=\) \(\ds \O\)
\(\ds \leadsto \ \ \) \(\ds Y\) \(\subseteq\) \(\ds V\)
\(\ds \leadsto \ \ \) \(\ds Y^-\) \(\subseteq\) \(\ds V\) Closure of Subset of Closed Set of Topological Space is Subset
\(\ds \leadsto \ \ \) \(\ds Y^- \setminus V\) \(=\) \(\ds \O\)

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

That is, there exist proper subsets $V_1, V_2$ of $Y^-$ which are closed subsets in $T$ such that:

$Y^- = V_1 \cup V_2$

Then, from $(1)$:

$Y \setminus V_1 \ne \O$


$Y \setminus V_2 \ne \O$

In particular:

$V_1 \cap Y \subsetneqq Y$


$V_2 \cap Y \subsetneqq Y$


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

we have also:

$V_1 \cap Y \ne \O$


$V_2 \cap Y \ne \O$

Therefore $V_1 \cap Y$ and $V_2 \cap Y$ are proper closed subsets of $Y$, that form a cover of $Y$.

This contradicts the fact that $Y$ is irreducible.