An example presented in my course notes is that the set of all infinite sequences of integers is uncountable.
To prove this, my professor elected to assume that this set were countable, and provided a contradiction similar to Cantor's diagonalization argument, defining a sequence which could not be equal to some sequence in our denumerable set.
Out of curiosity however I think I came up with another proof employing cardinal arithmetic, and wanted to know if it looks okay. My primary concern is whether it is correct and reasonable to simply state that this set is expressible by $\mathbb{N}^{Z}$.
Proof:
This set of infinite sequences is expressible via $\mathbb{N}^{\mathbb{Z}}$. Then
$$
\left|\mathbb{N}^{\mathbb{Z}}\right| = \left|\mathbb{N}\right|^{|\mathbb{Z}|} = \aleph_{0}^{\aleph_{0}} \leq (2^{\aleph_{0}})^{\aleph_{0}} = 2^{(\aleph_{0}\cdot \aleph_{0})} = 2^{\aleph_{0}} \leq \aleph_{0}^{\aleph_{0}}
$$
shows that $\left|\mathbb{N}^{\mathbb{Z}}\right| = c$ and thus it is uncountable. $\square$
Best Answer
Yes this is fine. Although just to prove that $\Bbb{N^Z}$ is uncountable you only need to argue that $\aleph_0^{\aleph_0}\geq 2^{\aleph_0}$, which is trivial, and then that $2^{\aleph_0}$ is already uncountable.