Exercise 13, § 3

In this exercise, we admit the continuum hypothesis, that is, we adjoin to the axioms of set theory the axiom « Card($\mathbf{R}$) is the smallest uncountable cardinal ». (1)

a) Every gauge on $\mathbf{R}$ is atomic (cf. Exer. 11).

b) Let X be a set and $m$ a gauge on X satisfying the following property: for every subset A of X such that $m(A) > 0$, there exists a subset B of A such that $0 < m(B) < m(A)$; then $m$ is zero (for every integer $n \geq 1$, there exists a finite partition $(X_{n,i})_{i \in I_n}$ of X such that $m(X_{n,i}) \leq 1/n$ for all $i \in I_n$; set $I = \prod_{n \geq 1} I_n$ and $X_\alpha = \bigcap_{n \geq 1} X_{n,\alpha_n}$ for every $\alpha = (\alpha_n)_{n \geq 1}$ in I; the family $(X_i)_{i \in I}$ is a partition of X, $m(X_i) = 0$ for every $i \in I$, and every gauge on I is atomic by a); conclude by means of Exer. 9 c)).

c) Let X be a set on which there does not exist any nontrivial Ulam ultrafilter; then every gauge on X is atomic (otherwise, by b), there would exist a diffuse gauge $m$ on X and a subset A of X such that $m(A) > 0$ and such that $m(B) = 0$ or $m(A - B) = 0$ for every subset B of A; the set $\mathscr{U}$ of subsets B of X such that $m(A) = m(A \cap B)$ would be a nontrivial Ulam ultrafilter on X).

d) Show that $2^c$ is an Ulam cardinal for every Ulam cardinal c (by c), it suffices to show that if C is a set with cardinal c, then every Ulam ultrafilter $\mathscr{U}$ on $X = \{0, 1\}^C$ is trivial; equip C with a well-ordering structure and define by transfinite induction a family $s = (s_\alpha)_{\alpha \in C}$ of elements of $\{0, 1\}$ such that set of $x = (x_\alpha)_{\alpha \in C}$ in X for which

(1) It comes to the same to say that $\mathbf{R}$ can be equipped with a well-ordering structure $\prec$ for which the set of $x \in \mathbf{R}$ such that $x \prec a$ is countable for every $a \in \mathbf{R}$. It has been shown (cf. P. COHEN, Proc. Nat. Acad. Sci. U.S.A. 50 (1963), 1143–1148 and 51 (1964), 105–110) that this axiom does not introduce any new contradiction into the theory of sets, and that it is moreover independent of the other axioms of set theory.

$x_\alpha = s_\alpha$ for all $\alpha \leq \beta$ belongs to $\mathscr{U}$ for every $\beta \in C$ (cf. Exer. 12); then $s$ belongs to every element of $\mathscr{U}$.(2)