One more question about set theory:
$A\subseteq R$ is an infinite set of positive numbers.
Assume there is a value $k \in Z$ such that for any $B \subseteq A$:
$\sum_{i=0}^\infty b(i) \le k$ where $b \in B$
Show that A is of countable cardinality
Hint: Look at the sets $A(n) = \{a\in A | a>\frac{1}{n}\}$
What I tried doing: I understand that when n goes to infinity, A(n) gets closer and closer to A. And if I could show that for all values of n, A(n) is countable, then I could show that the unification:
$A(1) \cup A(2) \cup ... \cup A(n) \cup A(n+1) \cup...=A$ is countable.
But why can you say that if there exists such a value k, such that the summary of all values of A(n) is less than k, then A(n) is countable?
Prove that each $A(n)$ must actually be finite. HINT: If $A(n)$ is infinite, and all of its elements are bigger than $\frac1n$, then sums of those elements can be ...