# $\neg \textsf{AC}+ \neg\textsf{CH}$

1. Is there some interesting\surprising results that have only been proven by assuming $\neg \textsf{AC}$ and $\neg\textsf{CH}$ ?
2. Is there some interesting\surprising results implying both $\neg \textsf{AC}$ and $\neg\textsf{CH}$ ?

Edit :

My definition of $\textsf{CH}$ in $\textsf{ZF}$ is $\aleph_0<\mathfrak m\rightarrow 2^{\aleph_0}\leq\mathfrak m$

• If $AC$ fails at the level of the reals, doesn't that already imply the failure of $CH$? I'm asking, because I'm not entirely certain what $CH$ is supposed to mean in absence of a well-order of the reals. – Stefan Mesken Dec 9 '15 at 19:05
• There's not much you can prove from $\neg AC$ without more specific knowledge of how AC is failing. For instance, you might have an inaccessible cardinal $\kappa$ such that AC holds in $V_\kappa$ but fails somewhere above $\kappa$; then pretty much everything you might want to say about "ordinary mathematics" behaves as though AC were true. – Eric Wofsey Dec 9 '15 at 19:11
• @Stefan How about "There exists a set $X$ and injective maps $\Bbb \omega\to X\to2^omega$, but no no injective maps $2^\omega\to X$ or $X\to \omega$." – Hagen von Eitzen Dec 9 '15 at 19:11
• math.stackexchange.com/questions/472957/… It looks like it would be consistent. I don't know if there would be "interesting results", but I would guess probably not. I am under the impression that assuming AC leads to more interesting/useful results. Rejecting AC would result in a lot of sets being non-contructible. – Trevor Norton Dec 9 '15 at 19:11
• What is CH for you without choice? – Asaf Karagila Dec 9 '15 at 19:24

If you consider the continuum hypothesis to be $\aleph_0<\mathfrak m\rightarrow 2^{\aleph_0}\leq\mathfrak m$, then one possible result worth mentioning is:

There exists an infinite Dedekind-finite set of reals which is Borel.

If $A$ is such set, then $A\cup\Bbb N$ is a witness for the failure of $\sf CH$ and $\sf AC$. What is perhaps much more surprising is the fact that such set can be Borel. This is true in Cohen's first model.

You can actually replace the infinite Dedekind-finite set by all sort of sets which are "morally smaller than the continuum" (e.g. a set which is the countable union of countable sets, but not countable in itself).

Here is a surprising result, which is a bit in line of what you ask, although not entirely.

Assuming $\sf ZF+DC$, then we do not know of any model in which there is a non-measurable set and $\sf CH$ holds. In particular there is a model in which every set has the Baire Property, but non-measurable sets exist.

This is not entirely what you ask for, because it might be consistent that there are non-measurable sets while every set has size continuum and that $\aleph_1\neq2^{\aleph_0}$; however Shelah's model where every set of reals has the Baire Property (and this contradicts $\sf AC$), but there are non-measurable sets is one where $\aleph_1<2^{\aleph_0}$. The reason that we involve $\sf DC$ in all that is to have a reasonable way of defining a $\sigma$-additive and atomless measure on the reals to begin with.