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

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.

• You don't have all the sequences, only the finite ones. The sequence of all the odd integers is in none of your products. You have a countable union of countable sets, which therefore is countable. – Nate Eldredge Apr 5 '14 at 1:37
• If you want the "set of all sequences", you want something more like the set of all functions from N to itself. That one has cardinality continuum. – Nate Eldredge Apr 5 '14 at 1:39
• Ok. So the construction of the set is just not at all what I wanted it to be. Thank you. – Jonathan Hebert Apr 5 '14 at 1:47

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
$$\bigcup_{k \in \omega} \mathbb{N}^k$$