Lemma 5.24.5. Let \mathcal{I} be a cofiltered index 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. Then the inverse limit X = \mathop{\mathrm{lim}}\nolimits X_ i is a spectral topological space and the projection maps p_ i : X \to X_ i are spectral.
Proof. The limit X = \mathop{\mathrm{lim}}\nolimits X_ i exists (Lemma 5.14.1) and is quasi-compact by Lemma 5.24.1.
Denote p_ i : X \to X_ i the projection. Because \mathcal{I} is cofiltered we can apply Lemma 5.14.2. Hence a basis for the topology on X is given by the opens p_ i^{-1}(U_ i) for U_ i \subset X_ i open. Since a basis for the topology of X_ i is given by the quasi-compact open, we conclude that a basis for the topology on X is given by p_ i^{-1}(U_ i) with U_ i \subset X_ i quasi-compact open. A formal argument shows that
as topological spaces. Since each f_ a is spectral the sets f_ a^{-1}(U_ i) are closed in the constructible topology of X_ j and hence p_ i^{-1}(U_ i) is quasi-compact by Lemma 5.24.1. Thus X has a basis for the topology consisting of quasi-compact opens.
Any quasi-compact open U of X is of the form U = p_ i^{-1}(U_ i) for some i and some quasi-compact open U_ i \subset X_ i (see Lemma 5.24.4). Given U_ i \subset X_ i and U_ j \subset X_ j quasi-compact open, then p_ i^{-1}(U_ i) \cap p_ j^{-1}(U_ j) = p_ k^{-1}(U_ k) for some k and quasi-compact open U_ k \subset X_ k. Namely, choose k and morphisms k \to i and k \to j and let U_ k be the intersection of the pullbacks of U_ i and U_ j to X_ k. Thus we see that the intersection of two quasi-compact opens of X is quasi-compact open.
Finally, let Z \subset X be irreducible and closed. Then p_ i(Z) \subset X_ i is irreducible and therefore Z_ i = \overline{p_ i(Z)} has a unique generic point \xi _ i (because X_ i is a spectral space). Then f_ a(\xi _ j) = \xi _ i for a : j \to i in \mathcal{I} because \overline{f_ a(Z_ j)} = Z_ i. Hence \xi = \mathop{\mathrm{lim}}\nolimits \xi _ i is a point of X. Claim: \xi \in Z. Namely, if not we can find a quasi-compact open containing \xi disjoint from Z. This would be of the form p_ i^{-1}(U_ i) for some i and quasi-compact open U_ i \subset X_ i. Then \xi _ i \in U_ i but p_ i(Z) \cap U_ i = \emptyset which contradicts \xi _ i \in \overline{p_ i(Z)}. So \xi \in Z and hence \overline{\{ \xi \} } \subset Z. Conversely, every z \in Z is in the closure of \xi . Namely, given a quasi-compact open neighbourhood U of z we write U = p_ i^{-1}(U_ i) for some i and quasi-compact open U_ i \subset X_ i. We see that p_ i(z) \in U_ i hence \xi _ i \in U_ i hence \xi \in U. Thus \xi is a generic point of Z. We omit the proof that \xi is the unique generic point of Z (hint: show that a second generic point has to be equal to \xi by showing that it has to map to \xi _ i in X_ i since by spectrality of X_ i the irreducible Z_ i has a unique generic point). This finishes the proof. \square
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Comment #3648 by Brian Conrad on
Comment #3744 by Johan on