TITLE: Relationship between Continuum Hypothesis and Special Aleph Hypothesis under ZF
QUESTION [17 upvotes]: Special Aleph Hypothesis AH(0) is the claim $2^{\aleph_0}=\aleph_1$, i.e. there is a bijection from $2^{\aleph_0}$ to $\aleph_1$.
Continuum Hypothesis CH is the claim $\aleph_0 \leq \mathfrak{a}< 2^{\aleph_0} \Rightarrow \mathfrak{a}=\aleph_0$, i.e. if there is an injection from $\aleph_0$ to $\mathfrak{a}$, an injection from $\mathfrak{a}$ to $2^{\aleph_0}$ and no injection from $2^{\aleph_0}$ to $\mathfrak{a}$, then there is a bijection from $\aleph_0$ to $\mathfrak{a}$.
AH(0) and CH are known to be equivalent under ZFC. What if we don't assume the Axiom of Choice? Under ZF, is it known that AH(0) $\nRightarrow$ CH or CH $\nRightarrow$ AH(0)?

REPLY [5 votes]: It might also be worth mentioning also that it is known to be relatively consistent with $ZF+\neg AC$ that there are infinite Dedekind finite sets of reals, that is, a set $A\subset\mathbb{R}$ that is infinite, but which has no countable subset. Such a set is uncountable, with  cardinality strictly less than the continuum, but is incomparable in cardinality with $\aleph_0$. In other words, it is relatively consistent with $ZF+\neg AC$ that there is a cardinality $\frak{a}$ with $\frak{a}\lt 2^{\aleph_0}$ and $\frak{a}$ is not finite (and even $\frak{a}$ is uncountable), yet $\aleph_0\not\leq\frak{a}$. In other words, just knowing a set $A$ is uncountable, one cannot conclude in ZF (unless inconsistent) that $\aleph_0\leq |A|$. Many would regard the existence of such cardinalities as even worse than the kind of counterexamples for which your question is asking.