Lemma 5.24.2. Let $\mathcal{I}$ be a cofiltered category. Let $i \mapsto X_ i$ be a diagram of spectral spaces such that for $a : j \to i$ in $\mathcal{I}$ the corresponding map $f_ a : X_ j \to X_ i$ is spectral.

1. Given nonempty subsets $Z_ i \subset X_ i$ closed in the constructible topology with $f_ a(Z_ j) \subset Z_ i$ for all $a : j \to i$ in $\mathcal{I}$, then $\mathop{\mathrm{lim}}\nolimits Z_ i$ is nonempty.

2. If each $X_ i$ is nonempty, then $X = \mathop{\mathrm{lim}}\nolimits X_ i$ is nonempty.

Proof. Denote $X'_ i$ the space $X_ i$ endowed with the constructible topology and $Z'_ i$ the corresponding subspace of $X'_ i$. Let $a : j \to i$ in $\mathcal{I}$ be a morphism. As $f_ a$ is spectral it defines a continuous map $f_ a : X'_ j \to X'_ i$. Thus $f_ a|_{Z_ j} : Z'_ j \to Z'_ i$ is a continuous map of quasi-compact Hausdorff spaces (by Lemmas 5.23.2 and 5.12.3). By Lemma 5.14.6 the space $\mathop{\mathrm{lim}}\nolimits Z'_ i$ is nonempty. Since $\mathop{\mathrm{lim}}\nolimits Z'_ i = \mathop{\mathrm{lim}}\nolimits Z_ i$ as sets we conclude. $\square$

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