Lemma 42.11.2. Let $(S, \delta )$ be as in Situation 42.7.1. Let $X$, $Y$ be locally of finite type over $S$. Let $f : X \to Y$ be a morphism. Assume $f$ is quasi-compact, and $\{ Z_ i\} _{i \in I}$ is a locally finite collection of closed subsets of $X$. Then $\{ \overline{f(Z_ i)}\} _{i \in I}$ is a locally finite collection of closed subsets of $Y$.

Proof. Let $V \subset Y$ be a quasi-compact open subset. Since $f$ is quasi-compact the open $f^{-1}(V)$ is quasi-compact. Hence the set $\{ i \in I \mid Z_ i \cap f^{-1}(V) \not= \emptyset \}$ is finite by a simple topological argument which we omit. Since this is the same as the set

$\{ i \in I \mid f(Z_ i) \cap V \not= \emptyset \} = \{ i \in I \mid \overline{f(Z_ i)} \cap V \not= \emptyset \}$

the lemma is proved. $\square$

I think that one should note that (1) the quasi-compact open subsets form a basis for the topology, and (2) therefore a collection of subspaces $Z_i$ of $X$ is locally finite iff for all quasi-compact open subspaces $U \subset X$, the number of $Z_i$ intersecting $U$ is finite; one uses (1) for the reverse implication.

Comment #126 by on

OK, I tried to improve the exposition. The confusion I think arose because we are not using the most general definition of locally finite collection of subspaces, i.e., we have incorporated the thing with the quasi-compact opens into Definition 42.8.1. We really should define locally finite collections of subspaces in complete generality for subsets of topological spaces and then have some lemmas about this notion...

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