# Compactness of $\operatorname{Proj}S$

It is well-known that for any ring $A$, $\operatorname{Spec}A$ is quasi-compact. Is it true in general that $\operatorname{Proj}S$ is quasi-compact, where $S$ is an $\mathbb{N}$-graded ring?

Since $D_+(h)\simeq\operatorname{Spec}S_{(h)}$ form an open covering of $\operatorname{Proj}S$, where $h$ is a homogeneous element of a positive degree, $\operatorname{Proj}S$ would be compact if $S_+$ is finitely generated. But what if $S_+$ is not finitely generated?

Let $k$ be a field. I claim that the infinite-dimensional projective space $\mathbb{P}^{\infty}_k := \mathrm{Proj}(k[x_0,x_1,\dotsc])$ is not quasi-compact. It is covered by the infinite-dimensional affine spaces $D_+(x_i) = \mathrm{Spec}(k[\dfrac{x_0}{x_i},\dfrac{x_1}{x_i},\dotsc]) \cong \mathbb{A}^{\infty}_k$. On $k$-valued points, this decomposition is $$\mathbb{P}^{\infty}_k(k) \cong (k^{\mathbb{N}} \setminus \{0\}) / k^* = \bigcup_{i=0}^{\infty} \{[x] : x_i \neq 0\} = \bigcup_{i=0}^{\infty} D_+(x_i)(k)$$ From this we see directly that no finite subcover suffices: If $n \in \mathbb{N}$, then $$[e_n] \in \mathbb{P}^{\infty}_k(k) \setminus \cup_{i<n} D_+(x_i)(k).$$