# Cardinality of the union of all repeated Cartesian products of N with itself [duplicate]

This question already has an answer here:

Here is a silly question, but I am a silly person.

Consider the: Natural Numbers. Natural Numbers X Natural Numbers. Natural Numbers X Natural Numbers X Natural Numbers ...

Now take the union of all of these sets. In other words, this is the set of all sequences of natural numbers/tuples of any size, of natural numbers.

The cardinality is certainly at least that of the power set of the naturals, or the cardinality of the reals, because this is essentially like the power set except now, order matters/repetition is allowed.

But does having order matter, and allowing repetitions, increase cardinality of an infinite set? Will this set have a cardinality aleph-one, or will it be higher? I suspect it is the same, but maybe this does push the cardinality up in the same way that taking a power set would?

Thank you.

## marked as duplicate by Asaf Karagila♦, user127096, Claude Leibovici, Hanul Jeon, AvitusApr 5 '14 at 7:44

This set is countable. Let $\mathbb{N}^k$ denote the $k$ fold cartesian product of $\mathbb{N}$. By induction, $\mathbb{N}^k$ is countable. Hence
$$\bigcup_{k \in \omega} \mathbb{N}^k$$