Jónsson cardinals

Jónsson cardinals are some really strange creatures. They are large cardinals and consistency-wise they lie above 0^\sharp, but they don’t have to be strongly inaccessible, or even regular. It turns out that K nevertheless computes the successors of all Jónsson cardinals correctly! But before delving into that story, let’s try to understand how these cardinals behave.

A good place to start would probably be to define these odd beasts. As with several of the large cardinals in this domain, several equivalent definitions arise.

Definition. A Jónsson cardinal \kappa is a cardinal satisfying one, equivalently all, or the following:

  1. Any algebra (that is, a structure \left<A,f_n\right>_{n<\omega}) of size \kappa has a proper subalgebra of size \kappa.
  2. For any first-order language \mathcal L and any \mathcal L-structure \mathcal M of size \kappa there is a proper elementary substructure \bar{\mathcal M}\prec\mathcal M of size \kappa.
  3. For any predicate A, there is an elementary substructure \left<X,\in,\bar A\right>\prec\left<V_\kappa,\in,A\right> of size \kappa and such that X\cap\kappa\neq\kappa.
  4. \kappa\to[\kappa]^{<\omega}_\kappa.

Here the fourth definition means that given any colouring c:[\kappa]^{<\omega}\to\kappa we can find a subset H\subset\kappa such that c"[H]^n\neq\kappa for each n<\omega — that is, c will admit at least one colour when colouring subsets of size n of H. Jónsson cardinals lie in the “Ramsey-area” of the large cardinal hierarchy:

Consistency strength of Jónsson cardinals, beautifully illustrated in MS Paint

A quick fun fact we can deduce is that Jónsson cardinals have to be uncountable.

Fun fact. \omega is not Jónsson.

Proof. We’re going to use definition 1 here. Let f:\omega\to\omega be defined as f(n):=n-1. Then \left<\omega,f\right> is an algebra with no proper infinite subalgebras. QED

We can push this lower bound for the least Jónsson a bit higher:

Proposition (Chang, Rowbottom, Erdös-Hajnal). If \kappa isn’t Jónsson then neither is \kappa^+.

Proof. Now we’re using definition 4. Let c_\alpha be colourings witnessing that \alpha\not\to[\kappa]_\alpha^{<\omega}. Define then a colouring c:[\kappa^+]^{<\omega}\to\kappa^+ as c(s):=c_\alpha(s-\{\alpha\}) if \alpha\leq\sup s and c(s):=0 otherwise. We have to show that c uses all colours on any H\in[\kappa^+]^{\kappa^+}, so let \xi<\kappa^+ be a colour. Pick \alpha>\xi such that |H\cap\alpha|=\kappa. Then by definition of c_\alpha, there is some s\in[H\cap\alpha]^{<\omega} such that c_\alpha(s)=\xi, as c_\alpha uses all colours. But then c(s\hat{\ }\{\alpha\})=c_\alpha(s)=\xiQED

This then means that \aleph_\omega is now the lowest bound for the least Jónsson. Whether or not it actually is Jónsson, is open. We do have the following results however:


  1. (Shelah) \aleph_{\omega+1} is not Jónsson.
  2. (Rowbottom-Devlin) The least Jónsson has either countable cofinality or is weakly inaccessible.

Instead of proving the theorem, let’s try to establish the above position of Jónsson cardinals in the large cardinal hierarchy. Ramsey cardinals are always Jónsson, using definition 4 above, so we’re left to prove that Jónsson cardinals consistency-wise implies that every real has a sharp. But this is again a direct implication: Let \mathcal M\prec\left<L_\kappa[x],\in,x\right> be a proper elementary substructure with \kappa Jónsson, x a real and |\mathcal M|=\kappa. Then by elementarity we get that \mathcal M\cong\left<L_\kappa[\bar x],\in,\bar x\right> for some real \bar x. But since \text{trcl}(x)\subseteq\mathcal M, \bar x=x. Then we have a non-trivial elementary embedding j:L_\kappa[x]\to L_\kappa[x], which is equivalent to x^\sharp by a famous result by Kunen.

In the next post we’ll take a look at how these cardinals are being treated inside the core model K.