Let \(X \neq \emptyset\) be a topological space. Prove that \(X\) is irreducible if and only if every nonempty open subset of \(X\) is connected.
Solution
1. If \(X\) is irreducible, then any two nonempty open subsets of \(X\)intersect.
Suppose \(U,V\subseteq X\) are nonempty open subsets with \[ U\cap V=\emptyset. \] Then \[ X=(X\sm U)\cup(X\sm V), \] where both closed subsets are proper. This contradicts irreducibility.
2. If \(X\) is irreducible, every nonempty open subset of \(X\) isconnected.
1. Hence \(U\) isconnected.
3. Conversely, suppose every nonempty open subset of \(X\) is connected.Then \(X\) is irreducible.
Suppose instead that \[ X=F\cup G \] for proper closed subsets \(F,G\subsetneq X\). Then \[ A=X\sm F, \qquad B=X\sm G \] are nonempty open subsets of \(X\). Moreover, \[ A\cap B = X\sm(F\cup G) = \emptyset. \] Their union is \[ A\cup B = X\sm(F\cap G), \] which is a nonempty open subset of \(X\). Since \(A\) and \(B\) are disjoint nonempty open subsets whose union is \(A\cup B\), this open subset is disconnected, contradicting the hypothesis.
Therefore no such decomposition \(X=F\cup G\) exists, and \(X\) is irreducible.
4. Hence\[ \boxed{ X\text{ is irreducible} \iff \text{every nonempty open subset of }X\text{ is connected}. } \]
Step 2 proves the forward implication and step 3 proves the reverse implication.
5. Q.E.D.
Step 4 is exactly the asserted equivalence.