Show that every Zariski-closed subset of \[ \PP^n_\CC \] is compact in the usual Hausdorff topology.
Solution
Let \[ S^{2n+1} = \left\{ z=(z_0,\ldots,z_n)\in\CC^{n+1}: \sum_{i=0}^n\abs{z_i}^2=1 \right\}. \]
1. The map\[ q:S^{2n+1}\longrightarrow\PP^n_\CC, \qquad z\longmapsto[z_0:\ldots:z_n] \]is continuous and surjective.
The usual topology on complex projective space is the quotient topology on \[ (\CC^{n+1}\sm\{0\})/\CC^\times. \] Restricting the quotient map to the unit sphere gives \(q\), so \(q\) is continuous.
Every projective point \[ [z_0:\ldots:z_n] \] has a nonzero representative \(z\in\CC^{n+1}\). Dividing by its Euclidean norm produces a representative on \(S^{2n+1}\). Hence \(q\) is surjective.
2. The complex projective space \(\PP^n_\CC\) is compact.
The sphere \(S^{2n+1}\) is compact by the Heine–Borel theorem. By step 1, \[ \PP^n_\CC=q(S^{2n+1}). \] A continuous image of a compact space is compact, so \(\PP^n_\CC\) is compact.
3. Let\[ X\subseteq\PP^n_\CC \]be Zariski closed. Then\[ q^{-1}(X)\subseteq S^{2n+1} \]is closed.
Choose homogeneous polynomials \[ \{F_\alpha\}_{\alpha\in A} \subseteq \CC[x_0,\ldots,x_n] \] whose common projective zero locus is \(X\): \[ X=V(F_\alpha:\alpha\in A). \] Then \[ q^{-1}(X) = \left\{ z\in S^{2n+1}: F_\alpha(z)=0 \text{ for every }\alpha \right\}. \] Each restriction \[ F_\alpha|_{S^{2n+1}}:S^{2n+1}\longrightarrow\CC \] is continuous, so \[ (F_\alpha|_{S^{2n+1}})^{-1}(\{0\}) \] is closed. Therefore \[ q^{-1}(X) = \bigcap_{\alpha\in A} (F_\alpha|_{S^{2n+1}})^{-1}(\{0\}) \] is closed as an arbitrary intersection of closed subsets.
4. Every Zariski-closed subset\[ X\subseteq\PP^n_\CC \]is compact.
By step 3, \(q^{-1}(X)\) is closed in the compact sphere \(S^{2n+1}\), hence is compact. By construction, \[ q(q^{-1}(X))=X. \] Since \(q\) is continuous, \(X\) is a continuous image of a compact space. Therefore \[ \boxed{X\text{ is compact}.} \]
5. Q.E.D.
Step 2 proves that \(\PP^n_\CC\) is compact, and steps <1>3–<1>4 prove the assertion for every Zariski-closed subset.