# Show A is countable infinity

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 ...
• @Oria: I would say simply that it can be bigger than $k$, or that it can be made as large as you like by taking enough elements, but yes, that’s the idea. And yes, the union of countably many finite sets is countable. – Brian M. Scott Nov 6 '13 at 16:01
• But why is that true though? Like, consider the set $\{1,\frac{1}{2},\frac{1}{4},\frac{1}{8},...\}$ and so on, if you sum them all up, you won't get a value larger than 2. And still that is an infinite set. – Oria Gruber Nov 6 '13 at 16:10
• @Oria: But it’s not like the sets $A(n)$: they all have positive lower bounds, while the greatest lower bound of this set is $0$. This set has only finitely many elements greater than any given $\frac1n$. – Brian M. Scott Nov 6 '13 at 16:13