# Löwenheim number of $L_{\kappa, \lambda}$

Just to clarify, I don't mean the Löwenheim-Skolem number as it's found on wikipedia, but rather the name given, for instance, on one of the axioms of abstract elementary classes: a cardinal $$\kappa$$ s.t. one has $$X \subseteq B \prec A$$, $$\lvert B \rvert \leq \lvert X \rvert + \kappa$$.

I don't have much familiarity with infinitary logics so I'd like to ask for a resource on Löwenheim Numbers for $$L_{\kappa, \lambda}$$, considering some multi-sorted signature/vocabulary of a given arity and cardinality (assuming they exist, of course, and with possible changes on the above inequality). I tried searching on the internet but couldn't find anything that wasn't about second-order logic. I also tried thinking about a "standard" approach (Tarski-Vaught test) but I'm not very confident on the argument (and the cardinalities involved) I came up with, so I'd like to see a result like this in the literature.

Any help is appreciated!

EDIT:

Here's my attempt with cardinal arithmetic: Fixing a signature of arity $$< \lambda$$ and cardinality $$\gamma$$, we consider the case of the logic $$L_{\kappa, \lambda}$$.

First for terms, at level $$0$$ we have either a variable of some sort ($$\gamma$$ options, so we have enough variables, and we'll add new variables on the fly) or some element of $$\lvert X \rvert$$. At the first step, we have $$\gamma$$-many functions of a given arity $$< \lambda$$, so an upper bound is $$\gamma \cdot (\gamma + \lvert X \rvert)^{< \lambda}$$. Applying this again, we get $$\xi^\alpha$$, with $$\xi = \gamma + \lvert X \rvert$$, where $$\alpha = <\lambda$$ or $$\lambda$$ depending on whether $$\lambda$$ is regular or not.

Now, to the formulas, at level $$0$$ we have $$\gamma$$-many relations with their arities $$< \lambda$$, so we use same arithmetic and the bound stays at $$\xi^\alpha$$ in this case. For quantification, we have a choice of variables so we get $$2^{< \lambda} \cdot \xi^\alpha = \xi^\alpha$$ options. As for disjunction/conjunction, we have at most $$(\xi^{\alpha})^{< \kappa}$$ options. Since $$\kappa$$ is regular at successive steps that bound will stay the same. This completes the counting of formulas and witness formulas.

Now, we need $$\lambda$$ steps so we can use the Tarski-Vaught argument, but the bound stays the same at successor steps and limit steps, so we conclude that the upper bound for the size of the elementary substrucuture containing $$X$$ is

$$((\gamma + \lvert X \rvert)^{< \lambda})^{< \kappa} = (\gamma + \lvert X \rvert)^{< \kappa}$$ by the assumption $$\lambda \leq \kappa$$ or $$((\gamma + \lvert X \rvert)^{\lambda})^{< \kappa}$$ which can be simplified to $$(\gamma + \lvert X \rvert)^{\kappa}$$ or $$(\gamma + \lvert X \rvert)^{< \kappa}$$ whether $$\lambda = \kappa$$ or $$\lambda < \kappa$$, but the former is impossible because $$\kappa$$ is assumed regular.

Is everything correct? Is there a smaller bound?

• Maybe if you post your argument (at least the cardinal arithmetic that you came up with), someone can check it. Commented Oct 16, 2022 at 14:44
• @AlexKruckman Good suggestion, I've included my reasoning in an edit. Commented Oct 16, 2022 at 16:13