Proposition 1

Tag 0B3Z, permanent. Cite it as [Bourbaki, Tag 0B3Z].

General Topology, chapter I, § 6. Filters

A necessary and sufficient condition that there should exist a filter on $X$ containing a set $\mathfrak{G}$ of subsets of $X$ is that no finite subset of $\mathfrak{G}$ has an empty intersection.

The filter $\mathfrak{G}''$ defined above is said to be generated by $\mathfrak{G}$ , and $\mathfrak{G}$ is said to be a subbase of $\mathfrak{G}''$ .

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

Cited by

Theorem 1, § 10
Definition 4, § 6
Example, § 6
Corollary 2, § 6
Proposition 7, § 6
Proposition 8, § 7
Definition 1, § 9

Cites

Nothing.