Can the real numbers be forced to have arbitrary cardinality? Let $\mathbb R$ be the real numbers in a given model of set theory.   
Given an arbitrary cardinal number $\kappa$, does forcing produce a larger model in which the cardinality of $\mathbb R$ is equal to $\kappa$?  In particular, is there always (or ever) a model in which $\mathbb R$ is countable?
 A: Let us assume ZFC is our set theory.
Given a model of set theory, $M$, we always have that $M$ thinks $\mathbb R^M$ (what $M$ thinks is the real numbers) is of cardinality $2^{\aleph_0}$. It is a theorem of ZF that $\mathbb R\sim{\cal P}(\mathbb N)$, so from the internal point of view $M\models|\mathbb R| = 2^{\aleph_0}$. This, however, does not say much about what $\alpha$ we have $|\mathbb R|=\aleph_\alpha$.
Using forcing we can always ensure there is some $M\subseteq M'$ such that $M'\models|\mathbb R|=\aleph_1$, we may have added real numbers in the process, but we can ensure that in $M'$ this is the situation.
Now given $\kappa$ which is uncountable we only have one requirement from $\kappa$ that $\operatorname{cf}(\kappa)>\aleph_0$, we can now add $\kappa$ many real numbers via forcing to have $M\subseteq M'\subseteq M''$ such that $M''\models |\mathbb R|=\kappa$.
This covers all the cardinals except those with countable cofinality. Suppose now that we started with $M$ such that $M\models CH$, we may collapse $\aleph_1$ to be countable, namely find $M\subseteq M'$ such that $M'\models|\omega_1^M|=\aleph_0$. In the new model we have added real numbers, however externally we have that $M$ is a model in which there are only countably many real numbers.
I cannot give you a concrete answer about singular cardinals of countable cofinality, however I can make a remark about a symmetric extension (i.e. a submodel of a forcing extension):
Start with a model of CH, $M$, and collapse all the $\aleph_n$'s to be countable. We effectively added $\aleph_{\omega+1}$ many real numbers. So we have $M'$ such that $\aleph_{\omega+1}^M=\aleph_1^{M'}$.
Now there is a submodel, $M\subseteq N\subseteq M'$, in which the axiom of choice does not hold, where the continuum is a countable union of countable sets(!!). From $M'$ view (an external point of view) $N$ has $\aleph_\omega$ many real numbers. I believe that this argument can be done with every singular cardinal of countable cofinality, however I did not check that.
