Open Set of Irreducible Space is Irreducible

Theorem

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

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

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

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

Aiming for a contradiction, suppose $U$ is not irreducible in $T$.

Then $U = V_1 \cup V_2$ for some closed sets $V_1$ and $V_2$ of $\struct {U, \tau_U}$.

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

Let $W_3 := W_2 \cup \paren {S \setminus U}$.

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.

$\blacksquare$