I was messing around with the intuition behind the size of weakly compact cardinals (in their usual characterization). I found an interesting, seemingly weaker LCA which still implies weak inaccessibility.

I started by making an intuitively powerful property that can be stated as an $\mathcal{L}_{\kappa,\kappa}$ scheme. I ended up with three properties:

  • $\text{LZ}_\lambda$ for $\lambda<\kappa$ is the $\mathcal{L}_{\kappa,\kappa}$-sentence "$\exists_{\alpha<\lambda}x_\alpha(\bigwedge_{\alpha<\lambda}\bigwedge_{\beta\neq\alpha}x_\alpha\neq x_\beta)$" which is actually true in a structure $\mathcal{M}$ iff it's universe has size at least $\lambda$ (it's not hard to see why).
  • $\text{LA}_\lambda$ for $\lambda<\kappa$ is the $\mathcal{L}_{\kappa,\kappa}$-sentence "$\exists_{\alpha<\lambda}x_\alpha\forall X(\bigvee_{\alpha<\lambda}x_\alpha=X)$" which is true in a structure $\mathcal{M}$ iff it's universe has size at most $\lambda$.

So with $\mathcal{L}_{\kappa,\kappa}$ we can actually bound the size of our structure from above and below to cardinals $<$$\kappa$. It's a strong failure of the generalized Lowenheim-Skolem below $\kappa$.

Applying it to Compactness

If $\kappa$ is weakly compact, and $T$ is an $\mathcal{L}_{\kappa,\kappa}$-theory with at most $\kappa$-many symbols, and $T$ is satisfiable, then $T+S$ for a scheme $S=\{A_0,A_1...A_{n<\omega}...\}$ is satisfiable if and only if $T+A_0$ is satisfiable and $T+A_0+A_1$ is satisfiable (etc.)

We can use this to our power. Let $\text{LS}_\kappa$ be the $\mathcal{L}_{\kappa,\kappa}$-scheme which has $\text{LZ}_\lambda$ for all $\lambda<\kappa$. Clearly, $\mathcal{M}\models\text{LS}_\kappa$ iff $|M|\geq\kappa$.

However, because $\kappa$ is weakly compact, $T+\text{LS}_\kappa$ is satisfiable (there is a model of $T$ of size at least $\kappa$) if and only if $T+\text{LZ}_\lambda$ is satisfiable for every $\lambda<\kappa$. However, that is only true if and only if for any $\lambda<\kappa$, there is a model of $T$ larger than $\lambda$. In other words, if there are arbitrarily large models of $T$ below $\kappa$, there is a model $\mathcal{M}\models T$ with $|M|\geq\kappa$.

Therefore, if $\kappa$ is weakly compact and $T$ is an $\mathcal{L}_{\kappa,\kappa}$-theory of at most $\kappa$-many symbols with arbitrarily large models of $T$ below $\kappa$, there is a model $\mathcal{M}\models T$ with $|M|\geq\kappa$.

A similar argument shows that if $\kappa$ is strongly compact and $T$ is ANY $\mathcal{L}_{\kappa,\kappa}$-theory with arbitrarily large models of $T$ below $\kappa$, there is a model $\mathcal{M}\models T$ with $|M|\geq\kappa$.

Furthermore, if $\kappa$ is extendible and $T$ is any $\mathcal{L}^n_{\kappa,\kappa}$-theory for finite $n$ with arbitrarily large models of $T$ below $\kappa$, there is a model $\mathcal{M}\models T$ with $|M|\geq\kappa$.

Turing Into a Unique Property

This property makes $\kappa$ intuitively quite large. If you look at the "shadow" of $\kappa$ left by $\aleph_0$ (because remember, this also applies to $\aleph_0$) then you find that any first-order finitary theory $T$ with arbitrarily large finite models must also have an infinite model.

Intuitively, this is saying that "first-order finitary logic cannot talk about infinite models very well, but it sure can talk about finite models." The generalization to this new property is that "$\mathcal{L}_{\kappa,\kappa}$ cannot talk about $\geq$$\kappa$-sized models very well, but it sure can talk about $<$$\kappa$-sized models."

Hey, but wait! We haven't named this property yet! Let's do that.


Let $T$ be a theory. $T$ is $\kappa$-unboundedly satisfiable iff for any $\lambda<\kappa$, there is a model of $T$ of size at least $\lambda$.

Let $\kappa$ be weakly Skolem iff $\kappa$ is uncountable and regular and every $\kappa$-unboundedly satisfiable $\mathcal{L}_{\kappa,\kappa}$-theory $T$ of at most $\kappa$-many symbols has a model of size at least $\kappa$.

Let $\kappa$ then be Skolem iff every $\kappa$-unboundedly satisfiable $\mathcal{L}_{\kappa,\kappa}$-theory $T$ has a model of size at least $\kappa$ (no restrictions on the symbols of $T$).

Let $\kappa$ then be Extendibly Skolem iff every $\kappa$-unboundedly satisfiable $\mathcal{L}^n_{\kappa,\kappa}$-theory $T$ has a model of size at least $\kappa$.

Note - every weakly compact cardinal is weakly Skolem, every strongly compact cardinal is Skolem, and every extendible cardinal is extendibly Skolem. Every extendibly Skolem cardinal is Skolem, and every Skolem cardinal is weakly Skolem.

Proof of Weak Inaccessibility

To show that every weakly Skolem cardinal is inaccessible, I just have to show that it is a limit cardinal. This is easy enough.

Let $\kappa=\lambda^+$ be weakly Skolem. Then, $\text{LA}_\lambda$ is an $\mathcal{L}_{\kappa,\kappa}$-sentence. Clearly, $\lambda\models\text{LA}_\lambda$, and for any $\mu<\kappa$, $\mu\leq\lambda$. Therefore, there are arbitrarily large models of $\text{LA}_\lambda$ below $\kappa$.

This would imply there is a model $\mathcal{M}$ of $\text{LA}_\lambda$ of size at least $\kappa$, which is impossible by definition of $\text{LA}_\lambda$.

The Questions

  1. Are these cardinals strongly inaccessible? If not, are they equivalent to weak inaccessibility?
  2. Is weak Skolem-ness equivalent to weak compactness? Similarly with Skolem-ness and extendible Skolem-ness
  3. What properties do these cardinals have?

EDIT: By User Yair Hayut, weak Skolem-ness need not imply strong inaccessibility. However, any strongly inaccessible weakly Skolem cardinal is weakly compact.

  • 2
    $\begingroup$ The definition of Skolem cardinal is unclear to me. If we can take $T$ to be as large as you want, why can't we just take $T$ to be the (first order) theory with $\mu$ constants $\langle c_\alpha \mid \alpha < \mu\rangle$, $\mu > \kappa$, and the formulas $c_\alpha \neq c_\beta$ for all $\alpha < \beta < \mu$. Then, $T$ clear has models of sizes as large as you wish, but never less than $\mu$, so in particular it has no model of size $\kappa$. $\endgroup$
    – Yair Hayut
    Sep 5, 2018 at 7:45
  • 1
    $\begingroup$ Oh, good spotting! That was a typo. It doesnt need to have a model of size $\kappa $, just a model of size at least $\kappa $. $\endgroup$ Sep 5, 2018 at 14:39
  • 1
    $\begingroup$ Do you want also in the weakly Skolem cardinals case to require only the existence of a model of size at least $\kappa$, or do you want to require the size to be exactly $\kappa$ in this case? $\endgroup$
    – Yair Hayut
    Sep 5, 2018 at 14:46
  • 1
    $\begingroup$ Nope, also at least $\kappa $. If you want to you can do work on "exactly $\kappa $". $\endgroup$ Sep 5, 2018 at 14:50

1 Answer 1


Theorem: If $\kappa$ is weakly Skolem then the tree property holds at $\kappa$.

Proof: let $\mathcal T$ be a $\kappa$-tree. Let us define two sequences of constants $\langle d_\alpha \mid \alpha < \kappa\rangle$ and $\langle d_x \mid x \in \mathcal T\rangle$. Let us consider the theory $T$ with the following statements:

  • $d_\alpha \in \{0,1\}$,
  • For every $\alpha < \beta$, $d_\alpha \leq d_\beta$.
  • If $d_\alpha = 1$ and $x \in T$ of level $\alpha$, then $x = 0$.
  • $d_\alpha = 0$ if and only if there is $x\in \mathcal T$ of level $\alpha$ such that $c_x \neq 0$ (this is the only non first order statement).
  • If $c_x, c_y \neq 0$ then $x \leq_{\mathcal T} y$ or $y \leq_{\mathcal T} x$ and $c_x \neq c_y$.

It is routine to verify that if $M$ is a model of size $\kappa$ then $d_\alpha^M = 0$ for all $\alpha$. In this case the set $\{x\in \mathcal T \mid c_x \neq 0\}$ is a cofinal branch. Also, there is a model with more than $\lambda$ many elements for $T$ for every $\lambda < \kappa$ (which is obtained by taking $d_\alpha = 1$ for all $\alpha \geq \lambda$, picking an element of level $\lambda$, $t\in \mathcal T$, and taking $c_x = 0$ for all $x \in \mathcal{T}$ except $x \leq_{\mathcal T} t$, for which we take them to be all different. QED

In particular any weakly Skolem cardinal is weakly compact in $L$ and any strongly inaccessible weakly Skolem cardinal is weakly compact.

Using a similar argument to the one in these slides of Joel Hamkins, one can start with a weakly compact cardinal $\kappa$ and add $\kappa^+$ Cohen reals. In this model, one can show that $\kappa$ remains weakly Skolem (even though it is no longer strongly inaccessible). Using a different argument, from the same slides, one can show that a weakly Skolem cardinal is never of the form $2^\lambda$ for some cardinal $\lambda$ (take the theory that enumerates $2^\lambda$ by constants, and say that there is something outside the enumeration).

If we require the models in the definition of weakly Skolem cardinal to have size exactly $\kappa$, then $\kappa$ has to be strongly inaccessible. Indeed, if $\kappa < 2^\lambda$ then there is a single $\mathcal{L}_{\kappa,\kappa}$ sentence such that all its models have size at least $2^\lambda > \kappa$:

Work in the language of set theory, $$\forall \langle x_\alpha \mid \alpha < \lambda\rangle, x_\alpha \in \{0,1\},\, \exists f\colon \lambda \to 2, \bigwedge_{\alpha < \lambda} f(\alpha) = x_\alpha.$$

Theorem: $\kappa$ is Skolem if and only if it is measurable.

Proof: Let us assume that $\kappa$ is Skolem. Let us consider a language with constants $\langle c_X \mid X \subseteq \kappa\rangle$ and a unary predicate $U$. Let $T$ be the theory saying that $\{A \subseteq \kappa \mid U(c_A)\}$ is a $\kappa$-complete ultrafilter (this is an $\mathcal{L}_{\kappa,\kappa}$-formula) together with the formulas $LZ_\lambda \rightarrow \neg U(c_\lambda)$ for all $\lambda < \kappa$. We do not require the constants $c_X$ to be evaluated as distinct elements.

Let us verify that $T$ has a model of size $\geq \lambda$ for every $\lambda < \kappa$ infinite. Indeed, take a model with exactly $\lambda$ elements, and let us define $U(c_X) \iff \lambda^+ \in X$. This is possible by letting the constants $c_Y$ to obtain exactly two values.

Since $\kappa$ is Skolem, $T$ has a model of size $\geq\kappa$. In this model $U = \{A \subset \kappa \mid U(c_A)\}$ is a non-principle $\kappa$-complete ultrafilter, and thus $\kappa$ is measurable.

Let us assume now that $\kappa$ is measurable. Let $T$ be an $\mathcal{L}_{\kappa,\kappa}$ theory. Let us assume that for every $\lambda < \kappa$ there is a model $M_\lambda$ of size $\geq \lambda$. Let $U$ be a $\kappa$-complete normal ultrafilter on $\kappa$ and let $j \colon V \to Ult(V, U)$ be the ultrapower embedding.

Let us consider $M = j(\langle M_\alpha \mid \alpha < \kappa\rangle)(\kappa)$. $M$ is a model of $j(T)$ in the model $Ult(V,U)$. Let $\mathcal{L}$ be the language of the theory $T$. For every $\varphi \in T$, $j(\varphi)$ is an $\mathcal{L}_{\kappa,\kappa}$ sentence in $j `` \mathcal{L}$. By the closure of $Ult(V,U)$ under $\kappa$-sequences, since $Ult(V,U) \models$"$M \models j(\varphi)$", we conclude that $V \models$"$M \models j(\varphi)$", and by restricting $M$ to the language $j `` \mathcal{L}$ and using the natural identification between $j ``\mathcal{L}$ and $\mathcal{L}$ we get that $V \models $"$ M \models \varphi$''. This is true for every $\varphi\in T$ and thus $M$ is (equivalent to) a model of $T$. $M$ has size at least $\kappa$ in $Ult(V,U)$ and thus also in $V$. QED

  • 1
    $\begingroup$ It might be interesting to see if weakly Skolem cardinal is always weakly Mahlo and has stationary reflection. $\endgroup$
    – Yair Hayut
    Sep 5, 2018 at 17:53
  • $\begingroup$ So weak Skolem-ness is equiconsistent with weak compactness. Huh! That's really intriguing. Can a similar argument show that Skolem-ness + strong inaccessibility -> strong compactness? I doubt so, because the standard proof of "strong compactness = ultrafilter characterization" is quite different than this proof. $\endgroup$ Sep 5, 2018 at 18:04
  • $\begingroup$ Nevermind, a quite simple argument shows that all weakly compact cardinals are Skolem. $\endgroup$ Sep 6, 2018 at 4:28
  • $\begingroup$ How do you show that the Skolem cardinals are the weakly compact cardinals? $\endgroup$
    – Yair Hayut
    Sep 6, 2018 at 5:51
  • 2
    $\begingroup$ @YairHayut It seems to me there's a simpler argument that measurable implies Skolem. Just take the ultrapower $\prod_{\lambda < \kappa} M_\lambda / U$. This models $T$ by infinitary Los' theorem, and it has cardinality $\geq \kappa$ since it satsifes $LZ_\lambda$ for each $\lambda<\kappa$. $\endgroup$
    – Tim Campion
    Oct 4, 2018 at 19:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.