Irreducible Component is Closed

Theorem
Let $X$ be a topological space.

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

Then $Y$ is closed.

Proof
By Closure of Irreducible Subspace is Irreducible, the closure $\overline Y$ of $Y$ is irreducible.

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

Because $Y$ is an irreducible component, we must have $Y = \overline Y$.

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