# Interesting consequences of $2^{\aleph_0} = \aleph_2$?

So reading some other questions here, I noticed that the continuum hypothesis $$2^{\aleph_0} = \aleph_1$$ has some strange consequences for analysis. The negation of the Axiom of Symmetry is one, as well as this Fubini's theorem counterexample.

Now, I also know that $$2^{\aleph_0}$$ can be just about any aleph number, but are there any interesting or counterintuitive consequences of it specifically being $$\aleph_2$$? What about the more generalized version of this hypothesis, $$2^{\aleph_\alpha} = \aleph_{\alpha+2}$$? Is that even consistent?

• I have heard that the continuum hypothesis is equivalent to the uniqueness of the ultrapower $\mathbb{R}^\omega/U$ up to isomorphism. Oct 3, 2018 at 9:01

I don't know any interesting consequences of $$2^{\aleph_0}=\aleph_2$$ or related statements alone. However, there are reasonably natural principles which imply $$2^{\aleph_0}=\aleph_2$$ and also imply many interesting statements, my favorite being the Proper Forcing Axiom (PFA) and its variants; see this paper by Moore.
Meanwhile, the principle "For all $$\alpha$$, $$2^{\aleph_\alpha}=\aleph_{\alpha+2}$$" is consistent (relative to large cardinals, necessarily), but that's far from obvious; see this paper of Magidor and Woodin, as well as this Mathoverflow question.
That said, again I don't know any interesting consequences from this alone. In general, my understanding is that simple arithmetic facts like these don't tend to have lots of interesting consequences on their own. (Obviously when I say "simple" I'm referring to the form of the statement; as the results mentioned above indicate, "$$\forall\alpha(2^{\aleph_\alpha}=\aleph_{\alpha+2})$$" is extremely complicated once we really dive intothings.)
• Wasn't there some paper on set theory in an AMS journal a few years ago arguing that $2^{\aleph_0} = \aleph_2$ was the "right" resolution of the CH problem philosophically in combo with a PFA variant? By a well-known guy. Woodin? Oct 3, 2018 at 21:58
• @HennoBrandsma Woodin argued for $2^{\aleph_0}=\aleph_2$ previously - in this survey (maybe what you recall?) - but that now he favors CH being true, partly due to confidence in "Ultimate $L$" and partly due to Sargsyan's refutation of aspects of his argument against CH (see the end of, and the comments below, this MO answer, and really that whole question is interesting). Oct 3, 2018 at 22:18