Bourbaki

Exercise 4, § 8

General Topology, IV

Bourbaki marks this one as harder than the rest.

Let (un)(u_n) be a summable sequence of numbers > 0 in R\mathbf{R} which satisfies the following conditions : un+1≤unu_{n+1} \leq u_n and un≤∑k=1∞un+ku_n \leq \sum_{k=1}^{\infty} u_{n+k} for all n≥0n \geq 0. Show that for every number aa such that 0<a≤s=∑n=0∞un0 < a \leq s = \sum_{n=0}^{\infty} u_n, there exists a subset I of N\mathbf{N} such that a=∑n∈Iuna = \sum_{n \in I} u_n. These conditions are satisfied in particular if un+1≤un≤2un+1u_{n+1} \leq u_n \leq 2u_{n+1} for all nn. Consider the case of dyadic expansions.

Solution

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Transcribed from the printed volume. Not translated.