Closure of Irreducible Subspace is Irreducible

Theorem
Let $X$ be a topological space.

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

Then its closure $Y^-$ is also irreducible.

Proof
By the definition of an irreducible subspace, $Y \subseteq X$ is irreducible any two non-empty open sets in $Y$ are not disjoint.

The open sets in $Y^-$ are the same as the open sets in $Y$.

Hence by definition of irreducible space, any two open sets in $Y^-$ are not disjoint in $Y^-$.

That is $Y^-$ is also irreducible.

More generally, we can also show that if $Y^-$ is irreducible for a subset $Y \subseteq X$, then $Y$ is also irreducible in $X$.

$Y$ is not irreducible.

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

Then:
 * $Y^- = {Y_1}^- \cup {Y_2}^-$

which contradicts the assumption.

Also see

 * Closure of Connected Set is Connected