Several forcing axioms imply $2^{\aleph_0 }= \aleph_2$. What about $2^{\aleph_1}$?

Question

On the one hand, it seems intuitive that $2^{\aleph_1 }> 2^{\aleph_0}$, because $\aleph_1 > \aleph_0$. However, I also know that, like many things involving the continnum function, that's actually independent of ZFC. So, since these forcing axioms seem to resolve what $2^{\aleph_0}$ is, do they also resolve $2^{\aleph_1}$? If so, is it also $\aleph_2$, or is it something bigger?

Answer 1

Martin's axiom, $\mathsf{MA}$, already implies that $2^\kappa=2^{\aleph_0}$ for every $\kappa<2^{\aleph_0}$. This means that any forcing axiom contradicting $\mathsf{CH}$ directly implies $2^{\aleph_1}=2^{\aleph_0}$.

Now as you observe, "strong" forcing axioms - $\mathsf{PFA}$, $\mathsf{MM}$, etc. - not only contradict $\mathsf{CH}$ but also imply $2^{\aleph_0}=\aleph_2$. Consequently, we get the general rule of thumb that strong forcing axioms imply $2^{\aleph_0}=2^{\aleph_1}=\aleph_2$.