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).
Best Answer
Yes this does appear to be a typo, the notion you had of the relation $\ell$(L) ≤ LS(L) ≤ h(L) is correct. Page 67 in Model Theoretic Logics references result Väänänen [1982a], which is the following result
From Abstract logic and set theory. II. Large cardinals
This result states that the hanf number of 2nd order logic (the context is that L$^*$ = L$^2$) is less than the Löwenheim number of some logic with härtig's quantifier. A very different from the one in Model Theoretic Logics. It also doesn't make logical sense for the Hanf number to be less than the Löwenheim number or the Löwenheim-Skolem number as pointed out (if you are going with the above definition, there have been other definitions where for an arbitrary L-sentence $\phi$ instead of an arbitrary set of L-sentences, for which this is incomparable with Löwenheim-Skolem number, there is also a notion of Hanf number for properties).
As for the 'mini question', the result is true. The Hanf number for any first order theory T with signature of cardinality $\kappa$ is $\kappa$ (assuming the signature is of infinite size). This is because we can show via the Löwenheim-Skolem theorem that if T has an infinite model then it has a model of 'other' infnite cardinalities (this 'other' forming a proper class), although they may be 'gaps' in the cardinalities for uncountable signatures where models of certain cardinalities (less than the size of the language) are not guaranteed to exist.
The Hanf number is $\aleph$$_0$ only when considering a first order theory with a countable signature.