Bourbaki

Remark

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

General Topology, chapter I, § 1. Open sets, neighbourhoods, closed sets

The interior of a non-empty set can be empty; this is the case for a set consisting of a single point which is not open, for example on the rational line \* (or the real line) *.

Proposition 1 of no. 2 can be restated as follows:

A set is open if and only if it coincides with its interior.

The property (VII)(V_{II}) of no. 2 implies that every point which is an interior point of each of two subsets AA and BB is an interior point of A∩BA \cap B; consequently

(1)(1)
A˙∩B˙=A˙∩B˙. \dot{A} \cap \dot{B} = \dot{A} \cap \dot{B}.

Every point which is interior to the complement of a set AA is said to be an exterior point of AA, and the set of these points is called the exterior of AA in XX; a point x∈Xx \in X which is an exterior point of AA is therefore characterized by the property that xx has a neighbourhood which does not meet AA.

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

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Cites

Transcribed from the printed volume. Not translated.