Recent content by snakesonawii

Exactly, hence why this was a seeming contradiction.

Yes, after thinking about it I came to the same conclusion. You can't put them in bijective correspondence because if you wanted to map the union onto X^\omega you could do so injectively by adding 0s but there's no way to make this mapping surjective.

I know that a countable union of countable sets is countable, and that a finite product of countable sets is countable, but even a countably infinite product of countable sets may not be countable. Let X be a countable set. Then X^{n} is countable for each n \in N. Now it should also be true...