Lemma 3.9.6. Let $f : X \to S$, $g : Y \to S$ be morphisms of schemes. Then we have $\text{size}(X \times _ S Y) \leq \max \{ \text{size}(X), \text{size}(Y)\} $.

**Proof.**
Let $S = \bigcup _{k \in K} S_ k$ be an affine open covering. Let $X = \bigcup _{i \in I} U_ i$, $Y = \bigcup _{j \in J} V_ j$ be affine open coverings with $I$, $J$ of cardinality $\leq \text{size}(X), \text{size}(Y)$. For each $i \in I$ there exists a finite set $K_ i$ of $k \in K$ such that $f(U_ i) \subset \bigcup _{k \in K_ i} S_ k$. For each $j \in J$ there exists a finite set $K_ j$ of $k \in K$ such that $g(V_ j) \subset \bigcup _{k \in K_ j} S_ k$. Hence $f(X), g(Y)$ are contained in $S' = \bigcup _{k \in K'} S_ k$ with $K' = \bigcup _{i \in I} K_ i \cup \bigcup _{j \in J} K_ j$. Note that the cardinality of $K'$ is at most $\max \{ \aleph _0, |I|, |J|\} $. Applying Lemma 3.9.5 we see that it suffices to prove that $\text{size}(f^{-1}(S_ k) \times _{S_ k} g^{-1}(S_ k)) \leq \max \{ \text{size}(X), \text{size}(Y))\} $ for $k \in K'$. In other words, we may assume that $S$ is affine.

Assume $S$ affine. Let $X = \bigcup _{i \in I} U_ i$, $Y = \bigcup _{j \in J} V_ j$ be affine open coverings with $I$, $J$ of cardinality $\leq \text{size}(X), \text{size}(Y)$. Again by Lemma 3.9.5 it suffices to prove the lemma for the products $U_ i \times _ S V_ j$. By Lemma 3.9.4 we see that it suffices to show that

We omit the proof of this inequality. $\square$

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