Irreducible Component is Closed

Theorem

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

Let $Y$ be an irreducible component of $T$.

Then $Y$ is closed in $T$.

Proof

By Closure of Irreducible Subspace is Irreducible, the closure $Y^-$ of $Y$ is irreducible.

By Set is Subset of its Topological Closure, $Y \subseteq Y^-$.

Because $Y$ is an irreducible component, we must have $Y = Y^-$.

By Set is Closed iff Equals Topological Closure, $Y$ is closed in $T$.

$\blacksquare$