If $2^{\aleph_0}=\aleph_2$, what is known about $2^{\aleph_1}$?

Question

This means: if $V\models{\sf ZFC}$ in such a way that $2^{\aleph_0}=\aleph_2$, can $2^{\aleph_1}$ be anything it ought to be?

By the way, can $2^{\aleph_0}=\aleph_2$ be considered the 'simplest' case of $\lnot{\sf CH}$ in some sense?

Answer 1

The behaviour of the continuum function on regular cardinals can be anything "not obviously contradictory". More precisely:

Theorem (Easton): Let $\text{Reg}$ be the class of regular cardinals. Supose $F:\text{Reg}\to \text{OR}$ is a cardinal-valued function such that

• $\kappa<\lambda \implies F(\kappa)\le F(\lambda)$
• $\text{cf}(F(\kappa))>\kappa$ for every $\kappa\in \text{Reg}$.

Then there is a model of ZFC such that $2^\kappa=F(\kappa)$ for every $\kappa\in \text{Reg}$.

The first restriction on $F$ is obvious, while the second one follows from König's lemma.

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If $\neg CH$, then $2^{\aleph_0}>\aleph_1$ (because $2^{\aleph_0}\ge \aleph_1$ just follows from Cantor's Theorem), so $\aleph_2$ is the smallest possible value of the continuum under $\neg CH$. There are some axioms superseding ZFC which imply $2^{\aleph_0}=\aleph_2$. The Proper Forcing Axiom is an example of this.