Lemma 86.10.3. Let $X$ be an affine formal algebraic space over a scheme $S$.

1. If $X$ is Noetherian, then $X$ is adic*.

2. If $X$ is adic*, then $X$ is adic.

3. If $X$ is adic, then $X$ is weakly adic.

4. If $X$ is weakly adic, then $X$ is classical.

5. If $X$ is weakly adic, then $X$ is countably indexed.

6. If $X$ is countably indexed, then $X$ is McQuillan.

Proof. Statements (1), (2), (3), and (4) follow by writing $X = \text{Spf}(A)$ and where $A$ is a weakly admissible (hence complete) linearly topologized ring and using the implications between the various types of such rings discussed in Section 86.7.

Proof of (5). By definition there exists a weakly adic topological ring $A$ such that $X = \mathop{\mathrm{colim}}\nolimits \mathop{\mathrm{Spec}}(A/I)$ where the colimit is over the ideals of definition of $A$. As $A$ is weakly adic, there exits in particular a countable fundamental system $I_\lambda$ of open ideals, see Definition 86.7.1. Then $X = \mathop{\mathrm{colim}}\nolimits \mathop{\mathrm{Spec}}(A/I_ n)$ by definition of $\text{Spf}(A)$. Thus $X$ is countably indexed.

Proof of (6). Write $X = \mathop{\mathrm{colim}}\nolimits X_ n$ for some system $X_1 \to X_2 \to X_3 \to \ldots$ of thickenings of affine schemes over $S$. Then

$A = \mathop{\mathrm{lim}}\nolimits \Gamma (X_ n, \mathcal{O}_{X_ n})$

surjects onto each $\Gamma (X_ n, \mathcal{O}_{X_ n})$ because the transition maps are surjections as the morphisms $X_ n \to X_{n + 1}$ are closed immersions. Hence $X$ is McQuillan. $\square$

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