Irreducible Space is Connected

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

Then $T$ is connected.

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

Then:
 * $\forall U_1, U_2 \in \tau: U_1, U_2 \ne \varnothing \implies U_1 \cap U_2 \ne \varnothing$

So trivially there are no two open sets that can form a separation of $T$.

The result follows from definition of connected.

Also see

 * Irreducible Component is Contained in Connected Component
 * Space is Irreducible iff Open Subsets are Connected