Lemma 32.4.6 (01YY)—The Stacks project

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Table of contents
Part 2: Schemes
Chapter 32: Limits of Schemes
Section 32.4: Descending properties
Lemma 32.4.6 (cite)

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Lemma 32.4.6. In Situation 32.4.5.

We have $S_{set} = \mathop{\mathrm{lim}}\nolimits _ i S_{i, set}$ where $S_{set}$ indicates the underlying set of the scheme $S$ .
We have $S_{top} = \mathop{\mathrm{lim}}\nolimits _ i S_{i, top}$ where $S_{top}$ indicates the underlying topological space of the scheme $S$ .
If $s, s' \in S$ and $s'$ is not a specialization of $s$ then for some $i \in I$ the image $s'_ i \in S_ i$ of $s'$ is not a specialization of the image $s_ i \in S_ i$ of $s$ .
Add more easy facts on topology of $S$ here. (Requirement: whatever is added should be easy in the affine case.)

Proof. Part (1) is a special case of Lemma 32.4.1.

Part (2) is a special case of Lemma 32.4.2.

Part (3) is a special case of Lemma 32.4.4. $\square$

Comments (2)

Comment #1254 by Michael on January 09, 2015 at 11:35

there is a typo in the proof of (2): deisred should be desired

Comment #1265 by Johan on January 22, 2015 at 19:41

OK, thanks! Fixed here.

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