Cofinality and the Continuum

Question

I know we know what the cofinality of the Continuum is not, i.e. it is not $\aleph_0$. Is there a "favorable" method out there for indicating (I don't want to say "showing") either of the following:

(cf)c = X

(cf)X = c

... where X is some otherwise given cardinal (not the same in both equalities listed; I'm not saying the cofinality of the continuum has to be equivalent to the continuum, or vice versa)? I.e. have sound and valid arguments been given for at least one value of said X, or at least a "plausibility range" for values of X?

Answer 1

In a precise sense, $\mathfrak{c}$ can have whatever cofinality we want it to, other than $\omega$:

Suppose $M$ is a countable model of ZFC and $\alpha$ is (in the sense of $M$) an uncountable regular cardinal. Then there is a forcing extension $M[G]$ with the same regular cardinals in which $cf(\mathfrak{c})=\alpha$.

(The countability hypothesis is only to guarantee that forcing extensions actually exist; if you're familiar with the approach to forcing via Boolean valued models, we can appropriately drop the countability hypothesis.)

This is a consequence of the flexibility of the exact value of the continuum: by adding $\kappa$-many Cohen reals for $\kappa$ sufficiently large (that is, greater than $\mathfrak{c}$), we ensure $cf(\mathfrak{c})^{M[G]}=cf(\kappa)^{M[G]}=\alpha$ (the second inequality being a consequence of the very nice properties of Cohen forcing).

So, for example, there's no hope of proving something like $cf(\mathfrak{c})=\omega_1$ or similar. Indeed, very strange phenomena are possible - for instance, $\mathfrak{c}$ could be singular!