Proposition 2
Tag 0AZ7, permanent. Cite it as [Bourbaki, Tag 0AZ7].
General Topology, chapter I, § 1. Open sets, neighbourhoods, closed sets
If to each element of a set there corresponds a set of subsets of such that the properties (VI), (V_{II}), (V_{III}) and (V_{IV}) are satisfied, then there is a unique topological structure on such that, for each , is the set of neighbourhoods of in this topology.*
By Proposition 1, if there is a topology on satisfying these conditions, the set of open sets for this topology is necessarily the set of subsets of such that for each we have ; hence the uniqueness of this topology if it exists.
The set certainly satisfies axioms (O_I) and (O_{II}): for (O_I), this follows immediately from (V_I), and for (O_{II}), from (V_{II}). It
!Diagram showing sets V, U, W, and points x, y, z
Figure 1.
remains to show that, in the topology defined by , is the set of neighbourhoods of for each . It follows from (V_I) that every neighbourhood of belongs to . Conversely, let be a set belonging to , and let be the set of points such that ; if we can show that and , then the proof will be complete. We have since ; also , for every point belongs to by reason of (V_III) and the hypothesis . It remains to show that , i.e. that for each ; now (Fig. 1) if then by (V_IV) there is a set such that for each we have ; since means that , it follows that , and therefore, by (V_I), that .
Q.E.D.
Proposition 2 shows that a topology on can be defined by means of the sets of neighbourhoods of points of , subject only to the axioms (V_I), (V_II), (V_III) and (V_IV).
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.