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 $\struct {Y, \tau_Y}$.
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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.
$\blacksquare$
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.
$\blacksquare$