Assuming that then a set is denumerable if it bijects with .
by j,k)\mapsto 2^j3^k" alt="fj,k)\mapsto 2^j3^k" /> is an injection.
Because any infinite subset of a denumerable set is denumerable, we are done.
Hi!
Problem:
Prove that the sets and are denumerable.
(If someone wants to give a precise definition of denumerable please do).
For a fixed the set is denumerable since is denumerable.
If we change to , the set is also denumerable.
So we have a denumerable amount of denumerable sets, which gives that is denumerable?
I tried proving in a similar way.
Help appreciated!
The codomain is an infinite subset of N- and so denumerable.
As Plato said before, "an infinite subset of a denumerable set is denumerable".
As for showing that Q is denumerable, now you can use: every positive number in Q, , can be written in the form where m and n are relatively prime integers in N. The mapping is an injection from to NxN showing the is denumerable. The set of negative rationals can be handled in the same way and the union of two denumerable sets is denumerable.