Bourbaki

Exercise 11, § 8

General Topology, IV, p. 63

Bourbaki marks this one as harder than the rest.

Let EE be a separated topological space. Suppose that for every finite sequence ss whose terms are equal to 0 or to 1, there exists a nonempty subset A(s)A(s) of EE, satisfying the following conditions: 1° If ss is a sequence of nn terms (equal to 0 or 1), s′,s′′s', s'' the sequences of n+1n + 1 terms (equal to 0 or 1) whose first nn terms are equal to those of ss, one has A(s)=A(s′)∪A(s′′)A(s) = A(s') \cup A(s''); moreover, if s0s_0 is the empty sequence, A(s0)=EA(s_0) = E. 2° For every infinite sequence (un)n≥0(u_n)_{n \geq 0} whose terms are equal to 0 or to 1, if sns_n denotes the finite sequence (uk)0≤k≤n(u_k)_{0 \leq k \leq n}, the filter basis formed by the A(sn)A(s_n) converges towards a point of EE. Now prove that under these conditions there exists a continuous surjective mapping from the triadic set of Cantor KK into EE.

Solution

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Machine translation of the English, by gpt-5-mini. Not checked by a person. That model is a cut down version of the one the rest of this language was written by.