Lemma 3.9.2. With notations $\text{size}$, $Bound$ and $\mathit{Sch}_\alpha$ as above. Let $S_0$ be a set of schemes. There exists a limit ordinal $\alpha$ with the following properties:

1. We have $S_0 \subset V_\alpha$; in other words, $S_0 \subset \mathop{\mathrm{Ob}}\nolimits (\mathit{Sch}_\alpha )$.

2. For any $S \in \mathop{\mathrm{Ob}}\nolimits (\mathit{Sch}_\alpha )$ and any scheme $T$ with $\text{size}(T) \leq Bound(\text{size}(S))$, there exists a scheme $S' \in \mathop{\mathrm{Ob}}\nolimits (\mathit{Sch}_\alpha )$ such that $T \cong S'$.

3. For any countable1 diagram category $\mathcal{I}$ and any functor $F : \mathcal{I} \to \mathit{Sch}_\alpha$, the limit $\mathop{\mathrm{lim}}\nolimits _\mathcal {I} F$ exists in $\mathit{Sch}_\alpha$ if and only if it exists in $\mathit{Sch}$ and moreover, in this case, the natural morphism between them is an isomorphism.

4. For any countable diagram category $\mathcal{I}$ and any functor $F : \mathcal{I} \to \mathit{Sch}_\alpha$, the colimit $\mathop{\mathrm{colim}}\nolimits _\mathcal {I} F$ exists in $\mathit{Sch}_\alpha$ if and only if it exists in $\mathit{Sch}$ and moreover, in this case, the natural morphism between them is an isomorphism.

Proof. We define, by transfinite induction, a function $f$ which associates to every ordinal an ordinal as follows. Let $f(0) = 0$. Given $f(\alpha )$, we define $f(\alpha + 1)$ to be the least ordinal $\beta$ such that the following hold:

1. We have $\alpha + 1 \leq \beta$ and $f(\alpha ) \leq \beta$.

2. For any $S \in \mathop{\mathrm{Ob}}\nolimits (\mathit{Sch}_{f(\alpha )})$ and any scheme $T$ with $\text{size}(T) \leq Bound(\text{size}(S))$, there exists a scheme $S' \in \mathop{\mathrm{Ob}}\nolimits (\mathit{Sch}_\beta )$ such that $T \cong S'$.

3. For any countable diagram category $\mathcal{I}$ and any functor $F : \mathcal{I} \to \mathit{Sch}_{f(\alpha )}$, if the limit $\mathop{\mathrm{lim}}\nolimits _\mathcal {I} F$ or the colimit $\mathop{\mathrm{colim}}\nolimits _\mathcal {I} F$ exists in $\mathit{Sch}$, then it is isomorphic to a scheme in $\mathit{Sch}_\beta$.

To see $\beta$ exists, we argue as follows. Since $\mathop{\mathrm{Ob}}\nolimits (\mathit{Sch}_{f(\alpha )})$ is a set, we see that $\kappa = \sup _{S \in \mathop{\mathrm{Ob}}\nolimits (\mathit{Sch}_{f(\alpha )})} Bound(\text{size}(S))$ exists and is a cardinal. Let $A$ be a set of schemes obtained starting with $\kappa$ as in Lemma 3.9.1. There is a set $CountCat$ of countable categories such that any countable category is isomorphic to an element of $CountCat$. Hence in (3) above we may assume that $\mathcal{I}$ is an element in $CountCat$. This means that the pairs $(\mathcal{I}, F)$ in (3) range over a set. Thus, there exists a set $B$ whose elements are schemes such that for every $(\mathcal{I}, F)$ as in (3), if the limit or colimit exists, then it is isomorphic to an element in $B$. Hence, if we pick any $\beta$ such that $A \cup B \subset V_\beta$ and $\beta > \max \{ \alpha + 1, f(\alpha )\}$, then (1)–(3) hold. Since every nonempty collection of ordinals has a least element, we see that $f(\alpha + 1)$ is well defined. Finally, if $\alpha$ is a limit ordinal, then we set $f(\alpha ) = \sup _{\alpha ' < \alpha } f(\alpha ')$.

Pick $\beta _0$ such that $S_0 \subset V_{\beta _0}$. By construction $f(\beta ) \geq \beta$ and we see that also $S_0 \subset V_{f(\beta _0)}$. Moreover, as $f$ is nondecreasing, we see $S_0 \subset V_{f(\beta )}$ is true for any $\beta \geq \beta _0$. Next, choose any ordinal $\beta _1 > \beta _0$ with cofinality $\text{cf}(\beta _1) > \omega = \aleph _0$. This is possible since the cofinality of ordinals gets arbitrarily large, see Proposition 3.7.2. We claim that $\alpha = f(\beta _1)$ is a solution to the problem posed in the lemma.

The first property of the lemma holds by our choice of $\beta _1 > \beta _0$ above.

Since $\beta _1$ is a limit ordinal (as its cofinality is infinite), we get $f(\beta _1) = \sup _{\beta < \beta _1} f(\beta )$. Hence $\{ f(\beta ) \mid \beta < \beta _1\} \subset f(\beta _1)$ is a cofinal subset. Hence we see that

$V_\alpha = V_{f(\beta _1)} = \bigcup \nolimits _{\beta < \beta _1} V_{f(\beta )}.$

Now, let $S \in \mathop{\mathrm{Ob}}\nolimits (\mathit{Sch}_\alpha )$. We define $\beta (S)$ to be the least ordinal $\beta$ such that $S \in \mathop{\mathrm{Ob}}\nolimits (\mathit{Sch}_{f(\beta )})$. By the above we see that always $\beta (S) < \beta _1$. Since $\mathop{\mathrm{Ob}}\nolimits (\mathit{Sch}_{f(\beta + 1)}) \subset \mathop{\mathrm{Ob}}\nolimits (\mathit{Sch}_\alpha )$, we see by construction of $f$ above that the second property of the lemma is satisfied.

Suppose that $\{ S_1, S_2, \ldots \} \subset \mathop{\mathrm{Ob}}\nolimits (\mathit{Sch}_\alpha )$ is a countable collection. Consider the function $\omega \to \beta _1$, $n \mapsto \beta (S_ n)$. Since the cofinality of $\beta _1$ is $> \omega$, the image of this function cannot be a cofinal subset. Hence there exists a $\beta < \beta _1$ such that $\{ S_1, S_2, \ldots \} \subset \mathop{\mathrm{Ob}}\nolimits (\mathit{Sch}_{f(\beta )})$. It follows that any functor $F : \mathcal{I} \to \mathit{Sch}_\alpha$ factors through one of the subcategories $\mathit{Sch}_{f(\beta )}$. Thus, if there exists a scheme $X$ that is the colimit or limit of the diagram $F$, then, by construction of $f$, we see $X$ is isomorphic to an object of $\mathit{Sch}_{f(\beta + 1)}$ which is a subcategory of $\mathit{Sch}_\alpha$. This proves the last two assertions of the lemma. $\square$

 Both the set of objects and the morphism sets are countable. In fact you can prove the lemma with $\aleph _0$ replaced by any cardinal whatsoever in (3) and (4).

Comment #144 by on

In 3, put the footnote either after "countable" or after "category".

Comment #608 by Wei Xu on

The proof does not show that if the limit or colimit of $F:\mathcal{I}\to Sch_{\alpha}$ does not exist in $Sch$, then so does in $Sch_{\alpha}$.

In the proof part, to define $f(\alpha)$, I guess it should be added one more property (4): For any countable diagram category $\mathcal{I}$ and any functor $F: \mathcal{I}\to Sch_{f(\alpha)}$, if the limit of $F$ or the colimit of $F$ does not exist in $Sch$, then the limit of $F$ or the colimit of $F$ does not exist in $Sch_{\beta}$. (Basically, successively adding counterexamples.)

And then show that the $\alpha=f(\beta_1)$ is a solution to the problem posed in the lemma.

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