Let \(X\) be an affine variety over \(\CC\). Show that every nonempty Zariski-open subset \(U\subseteq X\) is dense in \(X\).
Solution
1. Any two nonempty Zariski-open subsets of \(X\) intersect.
An affine variety is irreducible by definition. Let \[ U,V\subseteq X \] be nonempty open subsets. If \[ U\cap V=\emptyset, \] then \[ X=(X\setminus U)\cup(X\setminus V). \] Both complements are proper closed subsets because \(U\) and \(V\) are nonempty. This expresses \(X\) as the union of two proper closed subsets, contradicting irreducibility. Hence \[ U\cap V\neq\emptyset. \]
2. Every nonempty open subset \(U\subseteq X\) satisfies\[ \boxed{\overline U=X.} \]
1. Therefore \(\overline U=X\), so \(U\) is dense.
3. Q.E.D.
Step 2 is exactly the assertion of the problem.