Martin's Axiom implies that $2^{\aleph\_0}$ is a regular cardinal. But can $2^{\aleph\_0}$ be a singular cardinal? 


By Konig's Lemma, it can never be $\aleph\_{\omega}$ since cf($2^{\aleph\_0}$)>$\aleph\_0$ but under what conditions can it be $\aleph\_{\omega\_1}$? It is even possible it can be $\aleph\_{\omega\_1}$?



Can the continuum be a singular cardinal?
Yes, but it must have uncountable cofinality. So if it is to be singular, the smallest possibility is $\aleph\_{\omega\_1}$. 


The basic fact is that if $\kappa$ is any cardinal such that $\kappa^\omega=\kappa$, then there is a forcing extension $V[G]$ in which $2^\omega=\kappa$. The forcing to achieve this is $\text{Add}(\omega,\kappa)$, which consists of finite partial functions from $\kappa\times\omega$ to $2$.


In particular, if you start with GCH in the ground model, and add $\aleph\_{\omega\_1}$ many Cohen reals, then you will have $2^\omega=\aleph\_{\omega\_1}$ in the forcing extension. The proof uses the following facts: (1) adding any number of Cohen reals is c.c.c. and therefore preserves all cardinals. (2) The forcing clearly adds at least that many reals. (3) A nice name argument shows that every real in the extension has a nice name in the ground model, and there are only $\aleph\_{\omega\_1}$ many such names. So the continuum of the extension is exactly $\aleph\_{\omega\_1}$. The same ideas work for any $\kappa$ for which $\kappa^\omega=\kappa$. 


