# Does ACH+NJA$(\mathbb{R})$ decide the Suslin problem?

Lets start with a couple of axioms in the language of ZFC.

ACH. Anticontinuum Hypothesis.

The cardinal $|\mathbb{R}|$ is an aleph fixed points: that is,

$$|\mathbb{R}| = \aleph_{|\mathbb{R}|}$$

(Compare with CH, which says $|\mathbb{R}| = \aleph_1$.)

NJA$(\mathbb{R})$. No jumping axiom for the real line.

For all cardinals $\kappa,$ we have:

$$\kappa < |\mathbb{R}| \rightarrow 2^\kappa \leq |\mathbb{R}|$$

Taken together, these axioms give us a huge wilderness of cardinals $\kappa$ such that $2^\kappa = |\mathbb{R}|$. Here's a tiny initial portion of them:

$$\aleph_0, \aleph_1,\aleph_2, \ldots \aleph_{\omega},\aleph_{\omega+1},\ldots, \aleph_{\aleph_1}, \ldots\aleph_{\aleph_\omega},\ldots$$

Okay. It is known that the Suslin problem is independent of both CH and $\neg$CH. I'm hoping that the combination of ACH (which is significantly stronger than $\neg$CH) and the assumption that a crapload of cardinals now satisfy $2^\kappa = |\mathbb{R}|$ will be a little more opinionated on this issue.

Question. Can we somehow use these cardinals to either construct a Suslin line, or else prove that no Suslin line exists?

In other words, does ACH+NJA$(\mathbb{R})$ decide the Suslin problem?

• What about the ACHE axiom stating an anticontinuum hypothesis everywhere, and its various extensions like STOMACHACHE, HEADACHE, and HEARTACHE? – Asaf Karagila Apr 22 '16 at 12:55
• @AsafKaragila, nice. I'd add the BUTTACHE axiom, except that it's a real pain in the arse ;) – goblin Apr 22 '16 at 15:04

No, these axioms do not suffice: They are consistent with $\mathsf{MA}$ (in the presence of inaccessibles, say), which implies that there are no Suslin lines. But adding a Cohen real adds Suslin lines (this is a result of Shelah) and does not affect cardinal arithmetic.
• Let me check if I got this right: Your forcing is just the usual finite product forcing to make $\operatorname{MA}$ true and $2^{\aleph_0} = \kappa$, where $\kappa$ is the inaccessible in your ground model? – Stefan Mesken Apr 22 '16 at 12:06
• Yes, that gives us the first part. For the second, just add one Cohen real to this model of $\mathsf {MA}$. – Andrés E. Caicedo Apr 22 '16 at 12:11
• @user254665 Yes, we can. Start in a model with MA and let $\kappa$ be a $\beth$-fixed point of uncountable cofinality. If you now add $\kappa$ many random reals, the resulting model satisfies ACH and NJA(R) and also has no Suslin trees by a result of Laver (R. Laver, Random reals and Souslin trees. Proc. Am. Math. Soc 100, no. 3 (1987)). – Miha Habič Apr 27 '16 at 15:18