Lemma 87.18.3. Let $S$ be a scheme. Let $X$ be a formal algebraic space over $S$. Then $X$ is an affine formal algebraic space if and only if its reduction $X_{red}$ (Lemma 87.12.1) is affine.

**Proof.**
By Lemmas 87.16.1 and 87.17.1 and Definitions 87.16.3 and 87.17.2 we see that $X$ is quasi-compact and quasi-separated. By Yasuda's lemma (Lemma 87.18.1) we can write $X = \mathop{\mathrm{colim}}\nolimits X_\lambda $ as a filtered colimit of thickenings of algebraic spaces. However, each $X_\lambda $ is affine by Limits of Spaces, Lemma 70.15.3 because $(X_\lambda )_{red} = X_{red}$. Hence $X$ is an affine formal algebraic space by definition.
$\square$

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