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: \begin{enumerate} \item We have $\alpha + 1 \leq \beta$ and $f(\alpha) \leq \beta$. \item For any $S \in \Ob(\Sch_{f(\alpha)})$ and any scheme $T$ with $\text{size}(T) \leq Bound(\text{size}(S))$, there exists a scheme $S' \in \Ob(\Sch_\beta)$ such that $T \cong S'$. \item For any countable diagram category $\mathcal{I}$ and any functor $F : \mathcal{I} \to \Sch_{f(\alpha)}$, if the limit $\lim_\mathcal{I} F$ or the colimit $\colim_\mathcal{I} F$ exists in $\Sch$, then it is isomorphic to a scheme in $\Sch_\beta$. \end{enumerate} To see $\beta$ exists, we argue as follows. Since $\Ob(\Sch_{f(\alpha)})$ is a set, we see that $\kappa = \sup_{S \in \Ob(\Sch_{f(\alpha)})} Bound(\text{size}(S))$ exists and is a cardinal. Let $A$ be a set of schemes obtained starting with $\kappa$ as in Lemma \ref{lemma-bounded-size}. 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 $f(\alpha) = \sup_{\alpha' < \alpha} f(\alpha')$. \medskip\noindent 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 \ref{proposition-exist-ordinals-large-cofinality}. We claim that $\alpha = f(\beta_1)$ is a solution to the problem posed in the lemma. \medskip\noindent The first property of the lemma holds by our choice of $\beta_1 > \beta_0$ above. \medskip\noindent 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 $S \in \Ob(\Sch_\alpha)$. We define $\beta(S)$ to be the least ordinal $\beta$ such that $S \in \Ob(\Sch_{f(\beta)})$. By the above we see that always $\beta(S) < \beta_1$. Since $\Ob(\Sch_{f(\beta + 1)}) \subset \Ob(\Sch_\alpha)$, we see by construction of $f$ above that the second property of the lemma is satisfied. \medskip\noindent Suppose that $\{S_1, S_2, \ldots\} \subset \Ob(\Sch_\alpha)$ 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 $\{S_1, S_2, \ldots\} \subset \Ob(\Sch_{f(\beta)})$. It follows that any functor $F : \mathcal{I} \to \Sch_\alpha$ factors through one of the subcategories $\Sch_{f(\beta)}$. 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 $\Sch_{f(\beta + 1)}$ which is a subcategory of $\Sch_\alpha$. This proves the last two assertions of the lemma.