# Löwenheim number ℓ(L) ≤ Hanf number h(L)?

For an arbitrary set of L-sentences T $$\subseteq$$ L[$$\tau$$] (where $$\tau$$ is the vocabulary we are working in), T 'pins down the cardinal' $$\kappa$$ iff T has a model of cardinality k but does not have models of arbitrarily high cardinality. The hanf number h(L) of a logic is defined as the supremum of all cardinals that are pinned down Model-Theoretic Logics, Barwise (see page 64)

The Löwenheim number of a logic L is defined as:

$$\ell$$(L) = sup{ $$\kappa_φ$$ : φ is a sentence in L } where $$\kappa_φ$$ is the smallest cardinality of a model of an L-sentence φ.

I.e., e smallest cardinal κ such that if an arbitrary sentence of L has any model, the sentence has a model of cardinality no larger than κ. (https://en.wikipedia.org/wiki/L%C3%B6wenheim_number)

An extensions of the Löwenheim number, the Löwenheim-Skolem number LS(L) can also be defined using an arbitrary set of L-sentences.

It would seem to me the relation goes like this $$\ell$$(L) $$\le$$ LS(L) $$\le$$ h(L) for any logics L since how can you have the cardinal at which models of arbitrary high cardinality are guaranteed below the cardinal where an arbitrary L-sentence is guaranteed to have a model? But in Model-Theoretic Logics, Barwise page 67 where it states:

On the other hand h(L) may be smaller than the Löwenheim number $$\ell$$(L) as defined below, which may itself be smaller than 2^$$\aleph_0$$.

This doesn't seem to make sense as the Löwenheim number should be at most the hanf number (for most logics outside of first order logic the hanf number is in fact almost always larger), is this some kind of error in the book? Or have I misinterpreted some things?

Mini-question: In my private research into Löwenheim-Skolem properties I may have also found another possible error in a powerpoint by John T. Baldwin Calculating Hanf Numbers where on slide 19 it states:

The hanf number for first order theories with vocabulary of size $$\kappa$$ is $$\kappa$$

This result should be false since by compactness we can prove that "any theory with an infinite model has a model for all infinite cardinalities", and hence arbitrary models, so the Hanf number should be $$\aleph_0$$. Also by vocabulary I think he means 'language' since you can have a finite signature/vocabulary but not a finite language Signature (logic).