0

Consider a set X = $\bigcup_{n=1}^\infty$$S^n$, where $S^n$ = S x S x S x S ... (n times)

Question: Is X countable?

My attempt:

For two countable sets S and T, I found it easy to prove that their Cartesian product is countable. I defined a function f: S x T -> $\Bbb N$, f(m,n) = $2^m$$3^n$ and proved it was an injection, then defined a function g: $\Bbb N$ -> S x T, g(n) = (0,n) and showed it was also an injection.

Therefore by Cantor-Bernstein, there exists a bijection. So | S x T | = |$\Bbb N$| and S x T is countable.

Is there some way to extend this idea to the infinite union of Cartesian products of this countable set S ? My first thought was to define some function f($s_{1}$, ..., ) and in some way show that the injection that existed for S x T will always exist for any amount of Cartesian products. I also know that the for some countable sets A and B, C = B-A --> AUB = AUC and its easy to show that AUB is countable (considering both cases: C is finite or C is infinite and countable)

Completely stuck on marrying the two ideas though! Any help appreciated

Asaf Karagila
  • 393,674
  • countable union of countable sets is countable. Just consider a matrix, with i-th row containing elements of the i'th set... – Peter Franek Apr 18 '18 at 06:31
  • "Just". The proof uses the axiom of choice. –  Apr 18 '18 at 06:34
  • Are you hinting at Cantor's diagonal argument? or is the matrix something else entirely –  Apr 18 '18 at 06:42

0 Answers0