Bourbaki

Exercise 2, § 1

Integration, III

Bourbaki marks this one as harder than the rest.

Let Y be a noncompact locally compact space that is countable at infinity and let (Yn)(Y_n) be an increasing sequence of compact subsets of Y forming a covering of Y and such that Yn⊂Y∘n+1Y_n \subset \overset{\circ}{Y}_{n+1} for all n, so that every compact subset of Y is contained in one of the YnY_n (GT, I, §9, No. 9, Cor. 1 of Prop. 15). Let βY\beta Y be the Stone–Čech compactification of Y (GT, IX, §1, Exer. 7), so that the space C(βY;R)\mathscr{C}(\beta Y; \mathbf{R}) may be identified with the space of bounded continuous real-valued functions on Y, and Y is a dense open set in βY\beta Y; for every function f∈C(βY;R)f \in \mathscr{C}(\beta Y; \mathbf{R}), Supp(f) is thus the closure in βY\beta Y of the set of y∈Yy \in Y such that f(y)≠0f(y) \neq 0. Recall that if Z1,Z2Z_1, Z_2 are two subsets of Y, closed in Y and disjoint, then their closures Z‾1,Z‾2\overline{Z}_1, \overline{Z}_2 in βY\beta Y are also disjoint (GT, IX, §4, Exer. 17).

Let ω\omega be a point of βY−Y\beta Y - Y and let X=βY−{ω}X = \beta Y - \{ \omega \}. Show that, on the space K(X;R)\mathscr{K}(X; \mathbf{R}), the direct limit topology T\mathscr{T} defined in No. 1 is identical to the topology of uniform convergence. (If p is a semi-norm that is continuous for T\mathscr{T}, show, by means of a contradiction argument, that there exists a number c>0c > 0 such that p(f)≤c∥f∥p(f) \leq c \|f\| for every function f∈K(X;R)f \in \mathscr{K}(X; \mathbf{R}). In the contrary case, there would exist a sequence (fn)(f_n) of functions in K(X;R)\mathscr{K}(X; \mathbf{R}) such that ∥fn∥≤1, p(fn)≥n, Supp(fn)∩Yn=∅\|f_n\| \leq 1 , \ p(f_n) \geq n , \ \mathrm{Supp}(f_n) \cap Y_n = \varnothing, and

Supp(fn)∩Supp(fk)=∅ \mathrm{Supp}(f_n) \cap \mathrm{Supp}(f_k) = \varnothing
for k<nk < n. Then let Zi (i=0,1)Z_i \ (i = 0, 1) be the union of the sets Y∩Supp(f2n+i)Y \cap \mathrm{Supp}(f_{2n+i}); show that one of the two sets Z0,Z1Z_0, Z_1 is relatively compact in X (GT, X, §4, Exer. 7), and from this deduce a contradiction.) From this, deduce that the space K(X;R)\mathscr{K}(X; \mathbf{R}) equipped with the topology T\mathscr{T} is not quasi-complete.

Solution

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Transcribed from the printed volume. Not translated.