Lemma 32.4.6. In Situation 32.4.5.

1. We have $S_{set} = \mathop{\mathrm{lim}}\nolimits _ i S_{i, set}$ where $S_{set}$ indicates the underlying set of the scheme $S$.

2. We have $S_{top} = \mathop{\mathrm{lim}}\nolimits _ i S_{i, top}$ where $S_{top}$ indicates the underlying topological space of the scheme $S$.

3. 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 $s$.

4. 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. $\square$

Comment #1254 by Michael on

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

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