Bourbaki

Exercise 14, § 6

General Topology, IX, p. 122

Bourbaki marks this one as harder than the rest.

(a) Let XX be a separated space, SS a Souslinian subspace of XX. Now prove that the set SS is approachable in XX (IX, p. 69). (Let PP be a Polish space, gg a continuous mapping of PP onto SS, (Cn,pn,φn)(C_n, p_n, \varphi_n) a sieve of PP and ff the corresponding continuous mapping of L(C)L(C) onto PP (with the notations of IX, p. 65); put h=g∘fh = g \circ f, and for every c∈Cnc \in C_n, denote by qn(c)q_n(c) the subspace of L(C)L(C) formed by the sequences (ck)(c_k) such that cn=cc_n = c. Let Fn(c)=h(qn(c))F_n(c) = h(q_n(c)), Xn(c)=g(φn(c))⊃Fn(c)X_n(c) = g(\varphi_n(c)) \supset F_n(c), for every c∈Cnc \in C_n; denote by Zn(c)Z_n(c) the union of D(Fn(c))D(F_n(c)) (IX, p. 111, exerc. 3) and of a meagre set in XX, such that Fn(c)⊂Zn(c)⊂Xn(c)‾F_n(c) \subset Z_n(c) \subset \overline{X_n(c)}, so that Zn(c)Z_n(c) is approachable in XX. Now prove that

Zn(c)∩⋃pn(c′)=cZn+1(c′)=Yn(c) Z_n(c) \cap \bigcup_{p_n(c') = c} Z_{n+1}(c') = Y_n(c)
is meagre in XX, by observing that Fn(c)=⋃pn(c′)=cFn+1(c′)F_n(c) = \bigcup_{p_n(c') = c} F_{n+1}(c'); prove on the other hand that the set Zn(c)−Fn(c)Z_n(c) - F_n(c) is contained in the union of the sets Ym(d)Y_m(d), where mm runs through the set of integers ≥n\geq n and for each mm, dd runs through the set of elements of CmC_m such that pnm(d)=cp_{nm}(d) = c. Conclude that Fn(c)F_n(c) is approachable). (b) Give an example of a continuous mapping ff of I={0,1}I = \{0, 1\} into itself and of an approachable set B⊂IB \subset I such that f(B)f(B) is not approachable. (Use exerc. 16 b) of IV, p. 64, and show that one can take for BB a subset of the Cantor set KK such that f(B)=Zf(B) = Z has the property described in exerc. 17 d) of IX, p. 114).

Solution

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Machine translation of the English, by gpt-5-6-mini. Not checked by a person. That model is a cut down version of the one the rest of this language was written by.