Union of Open Irreducible Non-Disjoint Subspaces is Irreducible

Theorem

Let $T = \struct {S, \tau}$ be an irreducible topological space.

Let $U$ and $V$ be open irreducible subspaces of $T$.

Let their intersection $U \cap V$ be non-empty.

Then their union $U \cup V$ is an irreducible subspace of $T$.

In view of the definition of the subspace topology $\tau_{U \cup V}$, it suffices to show that for all closed sets $A_1$ and $A_2$:

$U \cup V \subseteq A_1 \cup A_2 \implies \exists i_0 \in \set {1, 2} : U \cup V \subseteq A_{i_0}$

To this end, let $A_1$ and $A_2$ closed sets such that:

$U \cup V \subseteq A_1 \cup A_2$.

Then, in particular:

$U \subseteq A_1 \cup A_2$.

Since $U$ is irreducible, we have:

$(1):\quad \exists i_0 \in \set {1,2} : U \subseteq A_{i_0}$

In particular:

$\ds V$

$\subseteq$

$\ds \paren {V \setminus U} \cup U$

$\ds $

$\subseteq$

$\ds \paren {V \setminus U} \cup A_{i_0}$

by $(1)$

Thus:

$V = \paren {V \setminus U} \cup \paren {V \cap A_{i_0} }$

Here, both $V \setminus U$ and $V \cap A_{i_0}$ are closed in $V$ with respect to the subspace topology $\tau_V$.

Since $U \cap V$ is non-empty, we have:

$V \setminus U \subsetneq V$

That is, $V \setminus U$ is a proper subset of $V$.

Since $V$ is irreducible, $V \cap A_{i_0}$ is not a proper subset of $V$.

That is:

$V = V \cap A_{i_0}$

Hence:

$V \subseteq A_{i_0}$

Together with $(1)$, we conclude:

$U \cup V \subseteq A_{i_0}$

$\blacksquare$