Irreducible Subspace is Contained in Irreducible Component

Theorem
Let $T = \struct {S_1, \tau}$ be a non-empty topological space.

Let $S_2$ be an irreducible subset of $S_1$.

Then there exists an irreducible component $S_3$ of $T$ such that $S_2 \subseteq S_3$.

Outline of Proof
We apply Zorn's Lemma to the set of irreducible subspaces, ordered by the subset relation.

Proof
By definition, an irreducible component of $X$ is an irreducible subspace that is maximal among the irreducible subspaces, ordered by the subset relation.

Consider the ordered set $\struct {\AA, \subseteq}$, where:
 * $\AA := \leftset{S : S}$ irreducible subset of $S_1$ such that $\rightset{S_2 \subseteq S}$

We need to show that $\AA$ has a maximal element.

As $S_2 \in \AA$, we have $\AA \ne \O$.

Thus, in view of Zorn's Lemma, it suffices to show:
 * each non-empty chain in $\AA$ has an upper bound in $\AA$.

Let $C$ be a non-empty chain in $\AA$.

Let:
 * $M := \bigcup C$

We now show that $M$ is an upper bound of $C$.

By construction of $M$, it suffices to show:
 * $M \in \AA$

As $C \subseteq \AA$, we have:
 * $S_2 \subseteq M$

So, if remains to show that $M$ is an irreducible subset of $S_1$.

Let $U,V$ are non-empty open sets of $\struct {M, \tau_M}$.

In view of, we need to show:
 * $U \cap V \ne \O$.

By definition of $\tau_M$, there are $U_1, V_1 \in \tau$ such that:
 * $U = U_1 \cap M$

and:
 * $V = V_1 \cap M$

Since:
 * $\bigcup _{N \in C} \paren {U_1 \cap N} = U_1 \cap M \ne \O$

there exists $N_U \in C$ such that:
 * $U_1 \cap N_U \ne \O$

Similarly, there exists $N_V \in C$ such that:
 * $V_1 \cap N_V \ne \O$

As $C$ is a chain, we have either:
 * $N_U \subseteq N_V$ or $N_V \subseteq N_U$

, suppose:
 * $N_U \subseteq N_V$

Let:
 * $U' := U_1 \cap N_V$

and:
 * $V' := V_1 \cap N_V$

Then $U', V'$ are non-empty open sets of $N_V$.

Since $N_V$ is irreducible, we have:
 * $U' \cap V' \ne \O$

This implies:
 * $U \cap V \ne \O$

Also see

 * Point is Contained in Irreducible Component