Bourbaki

Exercise 12, § 9

General Topology, I, p. 106

Bourbaki marks this one as harder than the rest.

Let X_0 be an uncountable well-ordered set, having a greatest element, a its smallest element, b the smallest of the elements x ∈ X_0 such that the interval [a, x[ is uncountable. We denote by X the interval [a, b[ endowed with the topology T_−(X).

a) The space X is locally compact and non-compact (I, p. 105, Exercise 2) and every sequence in X admits a value of adherence (note that every countable subset of X is bounded above).

b) Let f be a mapping from X into itself such that, for every sufficiently large x, one has f(x) < x; now prove that there exists c ∈ X such that, for every x ∈ X, there exists y ≥ x such that f(y) ≤ c. (Reason by contradiction, by forming a sequence (z_n) of points of X such that z_{n+1} is the smallest of the elements z′ ∈ X such that for every x ≥ z′, one has f(x) ≥ z_n.)

c) Deduce from b) that X is not paracompact.

¹ This example (unpublished) was communicated to us by D. Zelinsky.

Solution

No solution has been written yet.

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.