Closure of Irreducible Subspace is Irreducible

Theorem
Let $X$ be a topological space.

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

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

Proof
By the definition of an irreducible subset, $Y\subset X$ is irreducible if and only if any nonempty open subset in $Y$ intersects.

Since the open subsets in $\bar{Y}$ is the same as the open subsets 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 $Y\subset X$, then $Y$ is also irreducible in $X$. We prove it by contradiction.

Assume $Y$ 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