# Closure of Irreducible Subspace is Irreducible

## 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 1

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 $\struct {Y, \tau_Y}$.

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 $Y = Y_1 \cup Y_2$.

Then:

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

which contradicts the assumption.

$\blacksquare$

## Proof 2

Observe that for each subset $V \subseteq Y^-$ which is closed in $T$:

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

Indeed:

 $\ds Y \setminus V$ $=$ $\ds \O$ $\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 $(1)$:

$Y \setminus V_1 \ne \O$

and:

$Y \setminus V_2 \ne \O$

In particular:

$V_1 \cap Y \subsetneqq Y$

and:

$V_2 \cap Y \subsetneqq Y$

Because:

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

we have also:

$V_1 \cap Y \ne \O$

and:

$V_2 \cap Y \ne \O$

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.

$\blacksquare$