# Open subset in Irreducible Topological Space is dense.

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.

• You mean "... iff for all $A\in \tau$ with $A\ne\emptyset$, $A\cap U\ne \emptyset$" (and in fact this is hardly a lemma, but a possible definition of dense. – Hagen von Eitzen Mar 30 '14 at 10:22

## 2 Answers

Just show that every two open nonempty subsets intersect.

If not, take complements to show the space is reducible.

• "every two open nonempty subsets intersect" is a possible definition of irreducible topological space. – Hagen von Eitzen Mar 30 '14 at 10:21
• @HagenvonEitzen Well he didn't state his definition, so I assumed it's "cannot be union of 2 proper closed subsets". Why would he ask this question if he had your definition? – user2345215 Mar 30 '14 at 10:22
• Acknowledged. Then again, every "good" definition should start with a theorem "The following properties are equivalent" (e.g. not union opf proper closed subsets; any two nonempty open sets intersect, any nonempty open subset is dense) to motivate a subsequent definition "We call something ... if it has one (and hence all) of the above properties". Thus in a "good" exposition, the exercise should be void to begin with ... – Hagen von Eitzen Mar 30 '14 at 10:26

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 $$X \setminus U \neq X \implies \overline{U} = X.$$

• Should it be $X\setminus U\neq X$? – math112358 Jun 9 at 23:10