Bourbaki

Exercise 4, § 4

General Topology, I, 126

Bourbaki marks this one as harder than the rest.

Let KK be the discrete space consisting of the two numbers 0,10, 1; let AA be an infinite set, and let XX be the product space KAK^A. Let

V=∏α∈AVα V = \prod_{\alpha \in A} V_\alpha
be an elementary set in XX; if hh is the number of indices α\alpha such that Vα≠KV_\alpha \neq K, put μ(V)=2−h\mu(V) = 2^{-h}.

a) Show that if U1,…,UnU_1, \ldots, U_n are pairwise disjoint elementary sets, then

∑k=1nμ(Uk)≤1. \sum_{k=1}^n \mu(U_k) \leq 1.
[Write each UkU_k in the form Wk×KBW_k \times K^B, where BB (independent of kk) is the complement of a finite subset of AA, and WkW_k is a finite set; count up the number of points in WkW_k].

b) Deduce from a) that XX satisfies the condition (DIV)(D_{IV}) of Exercise 7 of § 1. [If there were an uncountable family (Uλ)(U_\lambda) of pairwise disjoint elementary sets, there would exist an integer p>0p > 0 such that μ(Uλ)≥1/p\mu(U_\lambda) \geq 1/p for an infinite number of indices λ\lambda].

c) Show that if AA is uncountable, then XX does not satisfy the condition (DIII)(D_{III}) of Exercise 7 of § 1 (cf. Exercise 3).

Solution

No solution has been written yet.

Transcribed from the printed volume. Not translated.