another countable question,

If A and B are countable, then A x B is countable.

If A and B infinitely countable, then:

A = {a_1, a_2, a_3, ...}
B = {b_1, b_2, b_3, ...}

A x B =

(a_1, b_1) (a_1, b_2) (a_1, b_3) ....

(a_2, b_1) (a_2, b_2) (a_2, b_3) ...

(a_3, b_1) (a_3, b_2) (a_3, b_3) ...

...

I think there are several ways to come up with the mapping (I think the book
does a similar proof where it maps a_ij-->2^i * 3^j which is a subset of Z
and by a previous theorem is countable).

But I think there is a diagonal argument that I want to understand too. I
know how to count them diagonally, but can a formula for the mapping be
obtained by that, or is it more of an "algorithm" proof that just tells you
how to do it?



.