Showing that the hyperintegers are uncountable In class, we constructed the hyperintegers as follows:
Let $N$ be a normal model of the natural number with domain $\mathbb{N}$ in the language $\{0, 1, +, \cdot, <, =\} $. Also let $F$ be a fixed nonprincipal ultrafilter on $\omega$. Then we have $N^*$ as the ultrapower $N^\omega /F$ with domain $\mathbb{N}^{\omega} / F$ . 
Now I need to show that $N^*$ is uncountable (this means that it's cardinality is $2^{\aleph_0}$ I guess). 
The domain of $N^*$ are the equivalence classes of functions $\{f_F : f \mbox{ a function from }\mathbb{N} \to \mathbb{N}\}$
Further on, we have $N^{\sharp}$ which is the equivalence classes of all the constant functions. 
This is my intuition: I know that $N$ and $N^{\sharp}$ are isomporphic (hence have the same cardinality, which is $\aleph_0$ since $N$ is a model for the natural numbers). Furthermore I know that $N^{\sharp}$ is elementary equivalent to $N^*$. But $N^*$ has an extra 'copy' of $N^\sharp$ on top of it, the hyperintegers. So then the cardinalty of $N^*$ should be $2^{\aleph_0}$ (does that even exist? And is that uncountable?). I hope someone can help me with a formal proof.

Here's a diagonalization argument. Let $\left(\mathbf{x}^{(m)} \right)_{m \in \mathbb{N}}$ denote a sequence of natural sequences $\left(x_{n}^{(m)} \right)_{n \in \mathbb{N}} \in \mathbb{N}^{\mathbb{N}}$. Define a sequence $\left(y_n \right)_{n \in \mathbb{N}}$ by
$$y_{n} = 1 + \max_{1 \leq m \leq n} \left(x_{n}^{(m)} \right) .$$
Let $\sim$ denote the ultrafilter equality, i.e. $(a_{k}) \sim (b_{k}) \iff \{ k \in \mathbb{N} : a_k = b_k \} \in F$.
I claim that $(y_n) \not \sim \left( x_{n}^{(m)} \right)$ for all $m$. To see this, note that $y_n > x_{n}^{(m)}$ for all $n \geq m$, so $\left\{ n : y_n = x_{n}^{(m)} \right\} \subseteq \{ 1, \ldots , m - 1 \}$, which means the set is finite. But a nonprincipal ultrafilter contains no finite sets, so $\left\{ n : y_n = x_{n}^{(m)} \right\} \not \in F$. Thus $(y_n) \not \sim \left(x_{n}^{(m)} \right)$ for all $m$. Thus $^{*} \mathbb{N} \setminus S$ is non-empty for all countable $S$.
Note that in fact we can say that $(y_n) > \left(x_{n}^{(m)} \right)$ as a nonstandard natural, so we can elaborate that every countable subset of $^{*} \mathbb{N}$ is bounded.
