Absoluteness of Countability Let M be a countable transitive model for ZFC, P is a partial order in M. Notions like "partial orders" and "dense" are absolute. Consider the following set
$S$={$D\in M: D$ is dense in $P$} = {$D: D$ is dense in $P$}$^M$, the superscript notion denotes relativization. The remark is the set is usually not countable in $M$. (Note: since $M$ is countable, from $V$, the class of all sets, anything lies in $M$ should be countable with respect to $V$). I know from Skolem's Paradox, countability is not absolute. However, if considering the following function:
f: $S$ $\rightarrow$ $\omega$ and f is in $V$ and f is injective. Since $f\in P(S\times \omega)$, and $S, \omega \in M$, therefore, by the fact that $M$ is a transitive model of ZFC $f \in M$. The only conclusion I can draw at this point is when relativized to $M, f$ is not one-to-one. However, I feel confused and could not see how this is the case. I am asking for somewhat better example illustrating the non-absoluteness of notion of countability. 
By the way, the question originates from Chapter VII Section 2 on Kunen's Set Theory (1980).
 A: You asked for an example to show that countability is not
absolute to transitive models of set theory.
If there is any transitive model of set theory, then by the
Lowenheim-Skolem theorem this model has a countable
elementary substructure, whose Mostowski collapse provides
a countable transitive model $M$ of set theory (enough of
set theory to prove that the reals are uncountable, say).
The model $M$ is countable externally, but inside this
model there are uncountable objects, such as
$\mathbb{R}^M$. So this is a set of real numbers, which in
the real world we can see is countable, because it is a
subset of the countable set $M$, but inside $M$ it is
thought to be uncountable.
Meanwhile, it is perhaps important to mention that the non-absoluteness of countability is much worse than this. The method of forcing shows that for any model of set theory $M$ and any set $x\in M$, there is a larger model $N$, which can be constructed by the method of forcing, such that $x$ is countable in $N$. In slogan form, "any set can be made countable by forcing." The forcing notion is the partial order consisting of finite partial functions from $\omega$ to $x$, ordered by inclusion. A simple density argument shows that any generic filter for this partial order provides a function from $\omega$ onto $x$, thereby witnessing that $x$ has become countable in the forcing extension. 
One of the particular steps in your argument where I would
object is where you say "therefore, by the fact that $M$ is
a transitive model of ZFC, $f\in M$". It simply does not
follow from $M$ being transitive that it contains all such
functions $f$. You seem to suggest that since $f$ is a
subset of $S\times \omega$ that it should be in $M$, but
transitivity means that $M$ contains all
elements-of-elements, not necessarily all subsets of
elements. In particular, the power set of $S\times\omega$
inside $M$ is much smaller than the power set of
$S\times\omega$ in $V$. Indeed, the contradiction of your
argument proves this, since $f$ itself is not necessarily
in $M$. Thus, not only is countability not absolute, but also the power set operation is not absolute. 
