1
$\begingroup$

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?

$\endgroup$
2
$\begingroup$

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 ...

$\endgroup$
  • $\begingroup$ It can be infinity. And this is a contradiction with what we were given about the number k in the question $\endgroup$ – Oria Gruber Nov 6 '13 at 16:00
  • $\begingroup$ Aaahh I see...if A(n) is infinite, then such a value k does not exist. so all A(n)s are finite. and the infinite unification of finite sets is an infinite set of countable infinity? $\endgroup$ – Oria Gruber Nov 6 '13 at 16:01
  • $\begingroup$ @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. $\endgroup$ – Brian M. Scott Nov 6 '13 at 16:01
  • $\begingroup$ 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. $\endgroup$ – Oria Gruber Nov 6 '13 at 16:10
  • $\begingroup$ @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 $\frac1n$. $\endgroup$ – Brian M. Scott Nov 6 '13 at 16:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.