Finite number of cardinal in a model of $ZFC + \neg CH$ ? Hello everybody, little question for logicians:
Considering $ZFC + \neg CH$, is it possible to construct a model $V \vDash ZFC + \neg CH$
such that there exists
a finite number of cardinal between $\aleph_0$ and $2^{\aleph_0}$ ?
Same question with exactly one cardinal between $\aleph_0$ and $2^{\aleph_0}$ ?
Thanks.
 A: Yes!  Almost anything is possible.  You can force over a model of ZFC + CH to create a new model where $2^{\aleph_0}$ is $\aleph_2$, for example, so that there is one cardinal between $\aleph_0$ and the continuum.  The idea is to create new binary sequences, new real numbers, with a partial order (a notion of forcing) and allow a generic filter to make it coherent and take you to the new universe where the continuum is a new size. You could have 3 or 4 or any finite number of cardinals between $\aleph_0$ and $2^{\aleph_0}$ by adding new subsets of natural numbers.  
See the discussion below about how the answer to the question "What size can the continuum be?" is due to Cohen, Solovay, and Easton.  Also, see in the comments how the continuum could reach as far up as $\aleph_{2^{\aleph_0}}$, so there are continuum many cardinals between omega and $2^{\aleph_0}$. Hamkins' paper on the Multiverse shows that the ability to force to create models which have a variety of sizes of the continuum settles the continuum hypothesis. You can read all about how to add new reals to create a new model in Thomas Jech's Set Theory or Kenneth Kunen's book on the same subject.  
