Closure of Irreducible Subspace is Irreducible

Theorem
Let $X$ be a topological space.

Let $Y \subseteq X$ be an irreducible subspace.

Then its closure $\overline Y$ is also irreducible.

Proof
By the definition of an irreducible subset, $Y \subseteq X$ is irreducible any non-empty open set in $Y$ intersects.

Since the open sets in $\bar Y$ is the same as the open sets in $Y$, any two of them still trivially intersects in $\bar Y$, showing that $\bar Y$ is also irreducible.

More generally, we can also show that if $\bar Y$ is irreducible for a subset $Y \subseteq X$, then $Y$ is also irreducible in $X$. We prove it by contradiction.

$Y$ is not irreducible.

Then there exist two closed proper subsets $Y_1$, $Y_2$ such that $Y = Y_1 \cup Y_2$.

Then $\bar Y = \bar {Y_1}\cup \bar {Y_2}$, which contradicts with the assumption.

Also see

 * Closure of Connected Set is Connected