Number of equivalence classes based on a relation regarding a non-principal ultrafilter

Question

We have an equivalence relation on $\mathbb{N}^\mathbb{N}$ given by $$f\equiv g \iff \{n\in\mathbb{N}: f(n)=g(n)\}\in\mathbb{U},$$ where $\mathbb{U}$ is a non-principal ultrafilter on $\mathbb{N}$.

The question is to find the cardinality of the set of all equivalence classes. My guess would be $2^{\aleph_0}$, but, well, it is just a guess. I don't really know how to tackle this exercise, not even where to start...

What we know, I believe, is that all the sequences which differ on all the places but for one, are in separate equivalence classes (because the ultrafilter is non-principal). On the other hand, there might not be a $2$-element set in $\mathbb{U}$ as well, or $3$-element, $4$-element etc., which would give me that there are at most $$\sup_\limits{n\in\omega}\aleph_0^n = \aleph_0^{\aleph_0}=2^{\aleph _0}$$ equivalence classes (I count all the combinations of $n$-element sets). Hence my guess. Could be wrong though.

Do you have any ideas?

Answer 1

Your cardinal arithmetic is just wrong. $$\sup_{n\in\omega}\aleph_0^n=\sup_{n\in\omega}\aleph_0=\aleph_0\neq\aleph_0^{\aleph_0}.$$

For your question, first show there exists a set $\cal A\subseteq P(\Bbb N)$, such that:

1. If $A,B\in\cal A$ then either $A=B$ or $A\cap B$ is finite.
2. $|\mathcal A|=2^{\aleph_0}$.
3. If $A\in\cal A$, then $A$ is not in the ultrafilter.

Now use $\cal A$ to construct a family of pairwise-distinct equivalence classes.

(Note that the third property is the easiest to satisfy, the first two guarantee that at most one member of $\cal A$ can be in the ultrafilter, so we can remove it and get the third property.)

Answer 2

HINT: Let $\Bbb Q=\{q_n:n\in\Bbb N\}$. For each $x\in\Bbb R$ there is an injection $f_x:\Bbb N\to\Bbb N$ such that the sequence $\langle q_{f(n)}:n\in\Bbb N\rangle$ converges to $x$. Consider $\{f_x:x\in\Bbb R\}$.