Show that every non-empty open subset of an irreducible topological space is dense.
I know a lemma that states that $U \subset$X is dense iff for all $A \in \tau$, $A \cap U \neq \emptyset$.
So then let U be an open set in $(X, \tau_{zar})$ that is irreducible. Then I want to show that for all $A \in \tau$, $A \cap U \neq \emptyset$. I don't know how to show this though, nor how the irreducibility fits in.
Best Answer
Let $U$ be a nonempty open subset of an irreducible topological space $X$. Denote by $\overline{U}$ the closure of $U$ in $X$. Then $(X - U, \overline{U})$ is a decomposition of $X$. Because $X$ is irreducible, one of these sets equals $X$. Since $U$ is nonempty,we have $$X \setminus U \neq X \implies \overline{U} = X.$$