It is well-known that the continuum hypothesis (CH) cannot be proven under the standard axioms (i.e. independence from ZFC).

However, to (non-expert, beginning student of the field) me, it seems like we could add some "natural" axioms until we see that the CH is true or false.

Are there explicit axiom sets under which CH is true or false? Particularly, are there "convincing" sets, i.e. ones where the axioms seem natural and inarguable that allow proof of it, which might reasonably convince someone to believe in its truth or falsity? (Of course, it would also be cool to see more exotic ones too)

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    $\begingroup$ Add CH as an axiom? $\endgroup$ – Hagen von Eitzen Dec 5 '16 at 19:33
  • $\begingroup$ $V=L$ (The Gödel model) comes to mind. But natural I would not call it. I recall reading a paper on this exact question by Gödel, but I don't recall the title. $\endgroup$ – Henno Brandsma Dec 5 '16 at 19:35
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    $\begingroup$ Of interest: logic.harvard.edu/efi.php $\endgroup$ – Henno Brandsma Dec 5 '16 at 19:38
  • $\begingroup$ @HagenvonEitzen Fair enough, but hard to say if that's natural or not (or this question might not exist) :) $\endgroup$ – user3658307 Dec 5 '16 at 19:43
  • $\begingroup$ In addition to what @Henno linked, let me tout my own horn, and link this blog post. $\endgroup$ – Asaf Karagila Dec 5 '16 at 21:55

The Axiom of Determinacy (AD) implies CH in the form "There is no set intermediate in cardinality between $\mathbb{N}$ and $\mathbb{R},\!"$ which I believe is the original form of CH. (Sometimes you'll see CH written as $2^{\aleph_0}=\aleph_1,$ which is equivalent in ZFC, but not in ZF alone.)

Of course, to accept AD, you need to be willing to forego the Axiom of Choice (in fact, under AD, the set of real numbers is not well-orderable, so $2^{\aleph_0}$ does not equal $\aleph_1).$

It's interesting to note that, although AD has some strange consequences, this one is natural; the proof that AD implies CH follows the idea of a proof that Cantor originally hoped would work in general: Every uncountable set of reals contains a perfect subset. (This is sufficient since every perfect set of reals has the cardinality of the continuum.)

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    $\begingroup$ AD is a good argument. $\endgroup$ – Asaf Karagila Dec 5 '16 at 22:09

This is an ongoing issue in the world of Platonists and set theory.

Some people take "canonical inner models" (which imply $\sf GCH$) as a reasonable additional axiom, Woodin for example is a proponent of the $V=\text{Ultimate }L$ axiom he formulated. This axiom allows very large cardinals to exist in the universe of sets, but at the same time this axiom still implies the continuum hypothesis and much more.

Some people, however, would argue that forcing axioms like $\sf PFA$ are more natural. These have natural consequences on the structure of sets of reals (e.g. every reasonably definable set of reals is Lebesgue measurable), and at the same time they also imply that $2^{\aleph_0}=\aleph_2$. Woodin, some years ago, was a strong proponent of these sort of axioms which had some sort of canonicity to them.

The real problem lies somewhere else, though. Once you've shown that some statement is neither provable nor disprovable, taking it becomes either an issue of faith, or choosing not to be a Platonist and not subscribing to a fixed notion of mathematical "truth". So non-Platonists generally don't have a problem that $\sf CH$ is undecidable. Or the Axiom of Choice, because why accept it and not its negation? But Platonists will generally have a hard time deciding whether or not some statement is true or not.

If you look at the historical references, Choice was a controversial axiom, and mathematicians like Lebesgue and Borel rejected Choice, as its consequences mismatched their views (e.g. all sets of reals should be measurable). Nowadays, however, we accept Choice because it entered the mathematical canon. It is perfectly reasonable to hope that in a few decades something like $\sf CH$ or its negation will enter the mathematical canon, but maybe it won't. We can't tell the future.


There is "diamond",often denoted by $\diamond,$ which is: There exists $\{A_x:x\in \omega_1\}$ with $A_x\subset x$ for all $x\in \omega_1,$ such that for any $S\subset \omega_1,$ the set $\{x\in \omega_1:S\cap x=A_x\}$ is stationary in $\omega_1.$

$\diamond$ is consistent with ZFC because $V=L\implies \diamond\;,$ but $\diamond$ does not imply $V=L.$

$\diamond \implies CH\;$: For $S\subset \omega$ and $\omega\leq x<\omega_1$ we have $S\cap x =S,$ and the set $\{x\in \omega_1:S\cap x=A_x\}$ is stationary, hence unbounded, in $\omega_1.$ So there exists $x\geq \omega$ with $A_x=S\cap x=S.$ Let $f(S)$ be the least (or any) $x \geq \omega$ such that $A_x=S\cap x=S.$ We have $f(S)\ne f(T)$ for distinct $S,T \subset \omega,$ because $A_{f(S)}=S\ne T=A_{f(T)}.$ So $f: P(\omega)\to \omega_1$ is injective.

  • $\begingroup$ \lozenge or \diamondsuit produce $\lozenge$ which is a better symbol. $\endgroup$ – Asaf Karagila Dec 27 '16 at 14:01
  • $\begingroup$ @AsafKaragila. Thank you. $\endgroup$ – DanielWainfleet Dec 27 '16 at 21:30
  • $\begingroup$ \Diamond works too: $\Diamond$ $\endgroup$ – Noah Schweber Dec 28 '16 at 1:10
  • $\begingroup$ @NoahSchweber. Thank you. $\endgroup$ – DanielWainfleet Dec 28 '16 at 1:12

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