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.

In fact, there is an injection from the set of reals to the hypernaturals: for given real $r$ consider
$$f(r) = \langle \lfloor rn\rfloor : n=1, 2,\cdots\rangle.$$
You can see that if $r\neq s$ then the sequence $f(r)$ and $f(s)$ are eventually componentwise different. Therefore the map $r \mapsto [f(r)]$ is injective.
