What is the cardinality of $\left|C_s\right|$? Let $$C_s = \left\{ f\in \mathbb{N}/S \to \mathbb{N} : \forall M\in \mathbb{N} / S. f(M)\in M \right\}$$
Where $S$ is an equivalence class.
I need to prove $$\left| C_s\right| > \aleph_0 \implies \left| C_s\right| = \mathfrak{c}$$
And I am not allowed to assume there's no $\aleph_0 < a < \mathfrak{c}$ (Continuum hypothesis).
Start of proof:
$\left|\mathbb{N}/S\right|$ can't be finite because then, $\left| C_s\right|$ was also finite/countable.
Therefore, $\left|\mathbb{N}/S\right|$ must be infinite.
Since $\left|\mathbb{N}/S\right|$ is infinite but $\left|\mathbb{N}/S\right| \le \left|\mathbb{N}\right|$ it implies $\left|\mathbb{N}/S\right| = \aleph_0$
So far so good.
Now, the author claims:
"A finite number of equivalence classes with an infinite cardinality isn't possible since, if so then $\left| C_s\right|=\aleph_0$. And therefore, we have a set with $\aleph_0$ equivalence classes with the cardinality of $\aleph_0$. Hence, $\left| C_s\right| \ge \mathfrak{c}$"
I don't understand this claim completely and I'd be glad if someone could explain it me.
Thanks.
My guess:
Is the claim true because the cardinality of all finite subset of $\mathbb{N}$ is $\aleph_0$?
 A: I think the author's claim is wrong:
Let $M_n = \{2n,2n+1\}$ for $n\in\mathbb{N}$ and let
$\cal P$ be the partition $\{M_n \mid n \in \mathbb{N} \}$ of
$\mathbb{N}$.  (And note that we assume $0\in\mathbb{N}$ as is usual
in set theory.)  So, ${\cal
  P}=\{\{0,1\},\{2,3\},\{4,5\},\dots\}$.  This is
the set of equivalence classes of some equivalence relation $S$ on
$\mathbb N$ and all equivalence classes (all elements of $\cal P$) are
finite.
Now, for $g\in \{0,1\}^\mathbb{N}$ define $f_g:{\cal P}\rightarrow\mathbb{N}$ as $f_g(M_{n}) = \min M_{n}$ if $g(n)=0$ and 
$f_g(M_{n}) = \max M_{n}$ if $g(n)=1$.  Obviously, $f_g \in C_S$ for all $g\in \{0,1\}^\mathbb{N}$ and $f_g\neq f_h$ for $g\neq h$ which implies
$|C_s|\geq 2^{\aleph_0} = \mathfrak{c}$.
A: Here's an idea how to prove the original claim:
If $C_S$ is uncountable, then it is not possible that only finitely
many equivalence classes of $S$ have more than one element.  Because
otherwise let $M_0,M_1,M_2,\dots$ be an enumeration of $\mathbb{N}/S$.
By assumption there'd be a $k\in\mathbb{N}$ with $|M_n|=1$ for all
$n\geq k$.  This would imply $|C_S|\leq |\prod_{n=0}^{k-1} M_n| \leq
\aleph_0^k = \aleph_0$ (as two distinct elements of $C_s$ can only
differ on at most the first $k$ sets $M_0,\dots,M_{k-1}$).
In other words, you know you can find infinitely many equivalence
classes with at least two elements and you can now proceed as in my
previous answer.  (Use the second element instead of the maximum as
the maximum won't always exist.)  This shows $|C_s|\geq\mathfrak{c}$.
$|C_s|\leq\mathfrak{c}$ is obvious because of
$|C_s|\leq\aleph_0^{\aleph_0}=2^{\aleph_0}=\mathfrak{c}$.
