Bourbaki

Exercise 11, § 8

General Topology, IV

Bourbaki marks this one as harder than the rest.

Let XX be a Hausdorff topological space. Suppose that, for each finite sequence ss whose terms are equal to 0 or 1, there is a nonempty subset A(s)A(s) of XX which satisfies the following conditions: (i) If ss is a sequence of nn terms (each of which is 0 or 1) and s′,s′′s', s'' are the sequences of n+1n + 1 terms (each of which is 0 or 1) whose first nn terms are the same as those of ss, then A(s)=A(s′)∪A(s′′)A(s) = A(s') \cup A(s''); and if s0s_0 is the empty sequence, A(s0)=XA(s_0) = X. (ii) For each infinite sequence (un)n≥0(u_n)_{n \geq 0} whose terms are equal to 0 or 1, if sns_n denotes the finite sequence (uk)0≤k≤n(u_k)_{0 \leq k \leq n}, then the filter base formed by the A(sn)A(s_n) converges to a point of XX. Show that under these conditions there is a surjective continuous mapping of Cantor’s triadic set KK onto XX.

Solution

No solution has been written yet.

Transcribed from the printed volume. Not translated.