Closure of Irreducible Subspace is Irreducible

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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 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

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:

$Y^- \subseteq A_{i_0}$

$\blacksquare$


Proof 3

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

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

Indeed:

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


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.

$\blacksquare$


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