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(Take ZFC as background.)

The following two statements both follow from GCH:

ICF. Injective continuum function.

The continuum function (i.e. $\kappa \mapsto 2^\kappa)$ is injective.

NJA. No jumping axiom.

For all infinite cardinals $\kappa$ and $\nu$, we have: $$\kappa < 2^\nu \rightarrow 2^\kappa \leq 2^\nu.$$

To see that NJA follows from GCH, assume $\kappa < 2^\nu$. Hence $\kappa < \nu^+$. So $\kappa \leq \nu$. Thus $2^\kappa \leq 2^\nu$.

In fact, the converse holds too; we can actually prove GCH from the above two axioms. To see this, notice that GCH can be interpreted as saying that from $\kappa < 2^\nu$, we can derive $\kappa \leq \nu$. So assume $\kappa < 2^\nu$. Then from NJA, we deduce $2^\kappa \leq 2^\nu$. So from ICF, it follows that $\kappa \leq \nu$, as required.

Another interesting axiom that seems to be related to NJA is:

BA. Beth axiom.

For all infinite cardinals $\kappa$, there exists an ordinal $\alpha$ such that $2^\kappa = \beth_\alpha$.

I haven't been able to puzzle out whether or not NJA and BA imply each other, so:

Question 0. Is there a relationship between NJA and BA?

Okay. My motivation for factoring GCH as ICF+NJA is that I'm interested in axioms for set theory that determine the structure of the cardinal numbers (like GCH), but which don't trivialize the cardinal characteristics of the continuum (unlike GCH).

One approach to find such axioms is to look for statements that prove the truth of NJA, but which don't prove ICF. An obvious first axiom to consider is:

DCF. Degenerate continuum function.

For all infinite cardinal numbers $\kappa$, the following hold.

  • If $\kappa$ is regular, then $2^\kappa = 2^{\kappa^+}$.

  • $2^\kappa$ is regular.

This clearly refutes ICF. For example, under DCF, we can prove that $$2^{\aleph_0} = 2^{\aleph_1} = 2^{\aleph_2} = \cdots$$

and even that $$2^{\aleph_0} = 2^{\aleph_\omega}.$$

But, I haven't been able to puzzle out whether DCF implies either and/or both of NJA or BA.

Question 1. Does DCF imply either and/or both of NJA or BA?

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    $\begingroup$ What about NINJA (Non-injective No Jump Axiom)? Or NRITHA (No Running In The Hallways Axiom)? Or NARNIA (Not Again! Restore Natural Injectivity Axiom)? :-) $\endgroup$ – Asaf Karagila Apr 15 '16 at 8:33
  • $\begingroup$ @AsafKaragila, but if NRITHA was true in Plato's "true" model of set theory, I wouldn't have run head-first into that brick wall all those years ago. And if NINJA was true, this girl I kind of like wouldn't have seen me cough and sneezing like an idiot while trying to ninja away from her. It seems most likely, therefore, that NARNIA is real :) $\endgroup$ – goblin Apr 15 '16 at 8:37
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First of all, NJA implies BA. To see this, take a $\kappa$ and consider whether it is a beth number or not. If $\kappa=\beth_\alpha$ then of course $2^\kappa=\beth_{\alpha+1}$. Otherwise we can fit $\kappa$ between two beth numbers, $\beth_\alpha<\kappa<\beth_{\alpha+1}$, whence NJA implies that $2^\kappa=\beth_{\alpha+1}$.

On the other hand, the implication is not reversible: simply look at a model where $2^{\omega}=\omega_2, 2^{\omega_1}=2^{\omega_2}=\omega_3$ and GCH holds above. Here BA holds, since $2^{\omega_1}=\beth_2$, but NJA clearly fails at $\omega_1<2^{\omega}$.

For your second question, DCF does not imply BA (and therefore also not NJA). To get this, start with a model where $2^{\kappa}=\kappa^{+\omega+1}$ for every regular $\kappa$ (we can get this by Easton's theorem). This model is easily seen to satisfy DCF. Now force over this model to get $2^\omega=\aleph_{\omega\cdot 2+1}$ and $2^{\aleph_{\omega+1}}=\aleph_{\omega\cdot 2+2}$. In the end we stil have DCF (we only changed exponentiation on two blocks of $\omega$ many cardinals) as well as $\beth_1=\aleph_{\omega\cdot 2+1}$ and $\beth_2=\aleph_{\omega\cdot 3+1}$. But if we look at $\kappa=\aleph_{\omega+1}$ the cardinal $2^\kappa=\aleph_{\omega\cdot 2+2}$ is not a beth number, so BA fails.

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  • $\begingroup$ Miha, it's clear that NJA proves $2^{<\kappa} = \kappa$. Do you know whether or not the converse holds? $\endgroup$ – goblin Apr 25 '16 at 13:11
  • $\begingroup$ @goblin It's not true that NJA proves $2^{<\kappa}=\kappa$: a model with $2^{\omega}=2^{\omega_1}=\omega_2$ and GCH above satisfies NJA but $2^{<\omega_1}=\omega_2$. Rather, NJA proves $2^{<2^{\kappa}}=2^{\kappa}$, and it's easy to see that the converse also holds. $\endgroup$ – Miha Habič Apr 26 '16 at 14:58
  • $\begingroup$ Ahh yes, sorry. You're completely right, of course. $\endgroup$ – goblin Apr 27 '16 at 11:01

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