The Stacks project

Proof. Let $Y = \bigcup _{j \in J} V_ j$ be an affine open covering of $Y$ with $|J| \leq \text{size}(Y)$. By Lemma 3.9.5 it suffices to bound the size of the inverse image of $V_ j$ in $X$. Hence we reduce to the case that $Y$ is affine, say $Y = \mathop{\mathrm{Spec}}(B)$. For any affine open $\mathop{\mathrm{Spec}}(A) \subset X$ we have $|A| \leq \max (|B|, \aleph _0) = \text{size}(Y)$, see remark above and Lemma 3.9.4. Thus it suffices to show that $X$ has at most $\text{size}(Y)$ affine opens. This is clear if $X$ is quasi-compact, whence case (1) holds. In case (2) the number of isomorphism classes of $B$-algebras $A$ that can occur is bounded by $\text{size}(B)$, because each $A$ is of finite type over $B$, hence isomorphic to an algebra $B[x_1, \ldots , x_ n]/(f_1, \ldots , f_ m)$ for some $n, m$, and $f_ j \in B[x_1, \ldots , x_ n]$. However, as $X \to Y$ is a monomorphism, there is a unique morphism $\mathop{\mathrm{Spec}}(A) \to X$ over $Y = \mathop{\mathrm{Spec}}(B)$ if there is one, hence the number of affine opens of $X$ is bounded by the number of these isomorphism classes.

To prove the final statement of the lemma consider the ring $B = \prod _{n \in \mathbf{N}} \mathbf{F}_2$ and set $Y = \mathop{\mathrm{Spec}}(B)$. For every ultrafilter $\mathcal{U}$ on $\mathbf{N}$ we obtain a maximal ideal $\mathfrak m_\mathcal {U}$ with residue field $\mathbf{F}_2$; the map $B \to \mathbf{F}_2$ sends the element $(x_ n)$ to $\mathop{\mathrm{lim}}\nolimits _\mathcal {U} x_ n$. Details omitted. The morphism of schemes $X = \coprod _\mathcal {U} \mathop{\mathrm{Spec}}(\mathbf{F}_2) \to Y$ is a monomorphism as all the points are distinct. However the cardinality of the set of affine open subschemes of $X$ is equal to the cardinality of the set of ultrafilters on $\mathbf{N}$ which is $2^{2^{\aleph _0}}$. We conclude as $|B| = 2^{\aleph _0} < 2^{2^{\aleph _0}}$. $\square$