Irreducibility can be characterized as follows: a subset $V\subseteq X$ is irreducible if and only if every non-empty open set in $V$ is dense (in $V$).
Suppose that $V$ is irreducible, and let $U\subseteq \overline{V}$ be a non-empty open set in the closure $\overline{V}$ of $V$. Then $U\cap V$ open in $V$, and non-empty because $V$ is dense in $\overline{V}$. The irreducibility of $V$ implies $\overline{U\cap V}\cap V=V$. Hence, $V\subseteq \overline{U\cap V}\subseteq \overline{U}$, which implies $\overline{V}=\overline{U}$, in other words $U$ is in fact dense in $\overline{V}$.
Does the converse implication hold? If $\overline{V}$ is irreducible, can we conclude that $V$ is irreducible as well (and maybe more generally, we could work with a dense subset of an irreducible space). If $U\subseteq V$ non-empty open, then $U=V\cap W$ where $W$ is non-empty and open in $\overline{V}$. We can then again deduce that $W$ is dense in $\overline{V}$. I need to show that $U$ is dense in $V$, in other words $V\subseteq \overline{U}$. Does this relation hold?
Thanks
Best Answer
Yes, if $\overline V$ is irreducible then so is $V$. For if $A_1$ and $A_2$ are disjoint, non-empty, open subsets of $V$, write $A_i=V\cap B_i$, with $B_i$ open in $\overline V$. Since the latter is irreducible, $B_1\cap B_2$ is a non-empty open subset of $\overline V$ hence the density of $V$ there implies that $$ \emptyset \neq B_1\cap B_2\cap V = A_1\cap A_2, $$ a contradiction.