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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.

user138864
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2 Answers2

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

Léreau
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user149792
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Just show that every two open nonempty subsets intersect.

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

user2345215
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  • "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
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    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