Sums of Arbitrary Sets

Let S be a a subset of the positive real numbers. Define

\sum S = \sup \{ \sum F : F \subseteq S, |F| < \infty \}

This is well-defined and non-negative, Since \sum \emptyset = 0.

lemma: \sum S < \infty implies S is countable.

Suppose \sum S < \infty. Then for any n there is a finite set F_n such that \sum S - n^{-1} < \sum F_n \le \sum S. Define S_n = \cup_{k \le n} F_k. Then \sum S - N^{-1} < \sum S_n \le \sum S for all n \ge N. Hence \lim_n \sum S_n = \sum S. Since \cup_n S_n = \cup_n F_n is at most countable, if S is not countable there is a \lambda \in S - \cup_n F_n. Let G_n = F_n \cup \{\lambda\}. Then \sum S + \lambda - n^{-1} < \sum G_n \le \sum S which is false when n > \lambda^{-1}. Hence no such \lambda exists, that is S = \cup_n F_n.

lemma: Show that when \sum S < \infty there is a bijection x : \mathbb{N} \rightarrow S such that \sum x = \sum S.

Let x be an enumeration of S. For any finite set F of S there is an n such that \sum F \le \sum_{i=1..n} x_i \le \sum_i x_i. Hence \sum S \le \sum_i x_i. But Since F_i = \{x_1,\ldots, x_i\} is a finite subset of S, \sum F_i = \sum_{j=1..i} x_j \le \sum S. Hence \sum_i x_i \le \sum S.

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