[proofplan] We first recall why the analytic space $\mathbb{P}^n_{\mathbb{C}}$ is compact: it is the continuous image of the compact unit sphere in $\mathbb{C}^{n+1}$. We then prove directly that a projective zero locus is closed in the analytic topology by checking closedness on the standard affine charts, where homogeneous equations become ordinary polynomial equations. Finally, we use the open-cover definition of compactness to show that a closed subset of a [compact space](/page/Compact%20Space) is compact. [/proofplan]

[step:Realize complex projective space as a compact quotient of the unit sphere]Let $S^{2n+1}:=\{z=(z_0,\ldots,z_n)\in \mathbb{C}^{n+1}: |z_0|^2+\cdots+|z_n|^2=1\}$. The set $S^{2n+1}$ is compact in the Euclidean topology on $\mathbb{C}^{n+1}\cong \mathbb{R}^{2n+2}$ by the [Heine-Borel theorem](/theorems/309), since it is closed and bounded. Define $q:S^{2n+1}\to \mathbb{P}^n_{\mathbb{C}}, \qquad q(z_0,\ldots,z_n)=[z_0:\cdots:z_n]$. The map $q$ is continuous by the definition of the analytic topology on complex [projective space](/page/Projective%20Space): on each standard chart, let $U_i:=\{[z_0:\cdots:z_n]\in \mathbb{P}^n_{\mathbb{C}}: z_i\neq 0\}$. The chart map identifies $U_i$ with $\mathbb{C}^n$ by ratios $z_j/z_i$ for $j\neq i$, and these ratios are continuous on $q^{-1}(U_i)=\{z\in S^{2n+1}:z_i\neq 0\}$. The map $q$ is surjective. Indeed, if $[w_0:\cdots:w_n]\in \mathbb{P}^n_{\mathbb{C}}$, choose a representative $w=(w_0,\ldots,w_n)\in \mathbb{C}^{n+1}\setminus\{0\}$ and define $\lambda:=\left(|w_0|^2+\cdots+|w_n|^2\right)^{-1/2}\in \mathbb{R}_{>0}$. Then $\lambda w\in S^{2n+1}$ and $q(\lambda w)=[w_0:\cdots:w_n]$. Therefore $\mathbb{P}^n_{\mathbb{C}}=q(S^{2n+1})$ is compact as the continuous image of a compact space.[/step]

[guided]The goal of this step is to prove compactness of the ambient analytic space $\mathbb{P}^n_{\mathbb{C}}$ without using any algebraic geometry. We compare projective space to a familiar compact space: the unit sphere in $\mathbb{C}^{n+1}$. Define $S^{2n+1}:=\{z=(z_0,\ldots,z_n)\in \mathbb{C}^{n+1}: |z_0|^2+\cdots+|z_n|^2=1\}$. Under the identification $\mathbb{C}^{n+1}\cong \mathbb{R}^{2n+2}$, this is the ordinary Euclidean unit sphere. It is closed because it is the inverse image of $\{1\}$ under the [continuous function](/page/Continuous%20Function) $z\mapsto |z_0|^2+\cdots+|z_n|^2$, and it is bounded because every point on it has Euclidean norm $1$. Hence $S^{2n+1}$ is compact by the [Heine-Borel theorem](/theorems/315). Now define the map $q:S^{2n+1}\to \mathbb{P}^n_{\mathbb{C}}$, $q(z_0,\ldots,z_n)=[z_0:\cdots:z_n]$. This map sends a nonzero vector on the sphere to its complex line through the origin. Why is it continuous? The analytic topology on $\mathbb{P}^n_{\mathbb{C}}$ is described by the standard affine charts $U_i:=\{[z_0:\cdots:z_n]\in \mathbb{P}^n_{\mathbb{C}}: z_i\neq 0\}$. On $U_i$, the coordinates are the ratios $z_j/z_i$ for $j\neq i$. On the preimage $q^{-1}(U_i)=\{z\in S^{2n+1}:z_i\neq 0\}$, each ratio $z_j/z_i$ is a continuous complex-valued function because the denominator is nonzero throughout this [open set](/page/Open%20Set). Since continuity can be checked in these charts, $q$ is continuous. It remains to check that $q$ hits every point of projective space. Let $[w_0:\cdots:w_n]\in \mathbb{P}^n_{\mathbb{C}}$. Choose a representative $w=(w_0,\ldots,w_n)\in \mathbb{C}^{n+1}\setminus\{0\}$. Define $\lambda:=\left(|w_0|^2+\cdots+|w_n|^2\right)^{-1/2}\in \mathbb{R}_{>0}$. Then $\lambda w\in S^{2n+1}$, and multiplying a homogeneous coordinate vector by the nonzero scalar $\lambda$ does not change its projective point. Thus $q(\lambda w)=[w_0:\cdots:w_n]$. So $q$ is surjective. Since a continuous image of a compact [topological space](/page/Topological%20Space) is compact, $\mathbb{P}^n_{\mathbb{C}}=q(S^{2n+1})$ is compact.[/guided]

[step:Show the projective zero locus is closed in the analytic topology] Since $X=V_+(S)$, a point $[z_0:\cdots:z_n]\in \mathbb{P}^n_{\mathbb{C}}$ belongs to $X$ exactly when $f(z_0,\ldots,z_n)=0$ for every $f\in S$. For each $i\in\{0,\ldots,n\}$, let $\varphi_i:U_i\to \mathbb{C}^n$ be the standard affine chart $\varphi_i([z_0:\cdots:z_n])=\left(z_0/z_i,\ldots,z_{i-1}/z_i,z_{i+1}/z_i,\ldots,z_n/z_i\right)$. For a [homogeneous polynomial](/page/Homogeneous%20Polynomial) $f\in S$ of degree $d_f\in \mathbb{N}\cup\{0\}$, define its dehomogenization on the $i$-th chart as the polynomial map $f_i:\mathbb{C}^n\to \mathbb{C}$ obtained by substituting $1$ for the $i$-th homogeneous coordinate. More explicitly, if $y=(y_1,\ldots,y_n)\in \mathbb{C}^n$ denotes the affine chart coordinates, then $f_i(y)$ is $f$ evaluated at the corresponding homogeneous coordinate vector with $i$-th coordinate equal to $1$. For $[z]\in U_i$, homogeneity gives $f(z)=z_i^{d_f} f_i(\varphi_i([z]))$. Since $z_i\neq 0$ on $U_i$, this implies $U_i\cap X=\bigcap_{f\in S}\varphi_i^{-1}(f_i^{-1}(\{0\}))$. Each $f_i$ is continuous in the Euclidean topology on $\mathbb{C}^n$, and $\{0\}\subset \mathbb{C}$ is closed. Hence $f_i^{-1}(\{0\})$ is closed in $\mathbb{C}^n$, so $U_i\cap X$ is closed in $U_i$ for every $i$. Because the standard charts $U_0,\ldots,U_n$ cover $\mathbb{P}^n_{\mathbb{C}}$, and closedness in a space covered by open sets can be checked locally on that [open cover](/page/Open%20Cover), $X$ is closed in the analytic topology on $\mathbb{P}^n_{\mathbb{C}}$. [/step]

[step:Transfer compactness from the ambient projective space to the closed subspace] Let $\mathcal{U}$ be an open cover of $X$ in the [subspace topology](/page/Subspace%20Topology). For every $U\in\mathcal{U}$, choose an open set $\widetilde{U}\subset \mathbb{P}^n_{\mathbb{C}}$ such that $U=\widetilde{U}\cap X$. Since $X$ is closed in $\mathbb{P}^n_{\mathbb{C}}$, its complement $\mathbb{P}^n_{\mathbb{C}}\setminus X$ is open. Therefore $\{\mathbb{P}^n_{\mathbb{C}}\setminus X\}\cup\{\widetilde{U}:U\in\mathcal{U}\}$ is an open cover of $\mathbb{P}^n_{\mathbb{C}}$. The space $\mathbb{P}^n_{\mathbb{C}}$ is compact by the first step, so there exist $U_1,\ldots,U_m\in\mathcal{U}$ such that $\mathbb{P}^n_{\mathbb{C}}=(\mathbb{P}^n_{\mathbb{C}}\setminus X)\cup \widetilde{U}_1\cup\cdots\cup \widetilde{U}_m$. Intersecting both sides with $X$ gives $X=U_1\cup\cdots\cup U_m$. Thus every open cover of $X$ has a [finite subcover](/page/Finite%20Subcover). Hence $X$ is compact in the subspace topology inherited from the analytic topology on $\mathbb{P}^n_{\mathbb{C}}$. [/step]