Open Set of Irreducible Space is Irreducible

Theorem
Let $T = \left({S, \tau}\right)$ be an irreducible topological space.

Let $U$ be a non-empty open set of $T$.

Then $U$ is irreducible in its induced subspace topology.

Proof
Let $T = \left({S, \tau}\right)$ be an irreducible topological space.

Let $U$ be a non-empty open set of $T$.

$U$ is not irreducible in $T$.

Then $U = V_1 \cup V_2$ for some closed sets $V_1$ and $V_2$ of $\left({U, \tau_U}\right)$.

By definition of subspace topology:
 * $V_1 = U \cap W_1$

and:
 * $V_2 = U \cap W_2$

for some closed sets $W_1$ and $W_2$ of $T$.

Because $W_1 = S \implies V_1 = U \cap W_1 = U$, it follows that:
 * $W_1 \ne S$

This is a contradiction, because $V_1$ is a proper subset of $U$.

Now $U \ne \varnothing$ implies that $S \setminus U$ is a proper subset of $S$ which is a closed set of $T$.

Let $W_3 := W_2 \cup \left({S \setminus U}\right)$.

Then $W_3$ is a closed set of $T$.

Also, $W_3 \ne S$, because otherwise $V_2 = U \cap W_2 = U \cap W_3 = U$.

Thus $S = W_1 \cup W_3$ shows that $T$ is not irreducible.

The result follows by Proof by Contradiction.

Also see

 * Space is Irreducible iff Open Sets are Connected
 * Closed Set of Ultraconnected Space is Ultraconnected