The following definitions and Theorems come from M. Magidor's paper "On the Role of Supercompact and Extendible Cardinals in Logic" (Israel J. Math., Vol. 10, 1971):


"Definition: Logic is called $\kappa$ compact iff for every set of formulae A in this logic, if every subset of A of cardinality $\lt$$\kappa$ has a model, then A has a model.


The $L^{n}\_{\kappa}$ logic is like the n-th order logic, except that we allow conjunction and disjunctions of less than $\kappa$ formulae. The usual second order logic is of course $L^{2}\_{\omega}$.


Theorem 4. $\kappa$ is extendible iff $L^{2}\_{\kappa}$ is $\kappa$ compact. $\kappa$ is the first extendible iff it is the first $\alpha$ such that second order logic is $\alpha$ compact." He also proves this theorem as well: "if $\kappa$ is extendible it is supercompact". However, nowhere in his paper does he say whether he is using full semantics or Henkin models when he refers to $\kappa$ compactness. I therefore will go out on a limb here (so to speak) and presume (rightly or wrongly) that he is using full semantics rather than Henkin models (after all, he does speak of "the usual second order logic" and that, I suppose, might be a subtle indication that he means to use full semantics).


Under this assumption, it is well known that $L^{2}\_{\omega}$ is not compact. Under this assumption, it seems one can prove the noncompactness of $L^{2}\_{\omega}$ using Theorem 4 and "if $\kappa$ is extendible then it is supercompact". I argue in the following manner: by Theorem 4, if $L^{2}\_{\omega}$ is $\omega$ compact (i.e. compact) then $\omega$ is extendible. If $\omega$ is extendible it is supercompact. However, in his answer to arsmath's mathoverflow question, "$\aleph\_0$ (arsmath writes it "Aleph 0") as a large cardinal", Amit Kumar Gupta shows that $\omega$ is not supercompact, hence by this, the chain of contrapositives of the implications I use and modus ponens, $L^{2}\_{\omega}$ is not $\omega$ compact (i.e. not compact).


This, however, raises the following question: even if $L^{2}\_{\omega}$ is not compact, by Theorem 4, if there exists some extendible $\kappa$ ($\gt$$\omega$) then $L^{2}\_{\kappa}$ is $\kappa$ compact (i.e. second order logic satisfies some compactness theorem). Those who do not believe in the existence of large cardinals should ask themselves whether second order logic should be $\kappa$ compact under full semantics. Another question that arises is this: since $L^{2}\_{\omega}$ is compact under Henkin models, does Theorem 4 hold over Henkin models? 



A question regarding extendible cardinals and a result of M. Magidor
For your second question, the answer is "no." Since, with Henkin semantics, second-order logic is just re-syntacted first-order logic, of course theorem 4 fails for Henkin semantics: $L\_\kappa^2$ with Henkin semantics is compact iff $L\_{\kappa,\omega}$ is compact - that is, iff $\kappa$ has the *tree property* (which is implied by, but not equivalent to, $\kappa$ being weakly compact - weakly compact = tree property *plus inaccessible*). 


Meanwhile, I'm not sure what your first question is asking. It is quite reasonable to have a sequence of stronger logics $\mathcal{L}\_\alpha$ ($\alpha\in ON$) such that compactness fails early ($\mathcal{L}\_\omega$ is not $\omega$-compact) but holds occasionally later on ($\mathcal{L}\_\kappa$ is $\kappa$-compact for some $\kappa$). For instance, take $\mathcal{L}\_\alpha=L\_{\aleph\_\alpha,\omega}$. Then $\mathcal{L}\_\alpha$ is compact iff $\aleph\_\alpha$ has the tree property - in particular, it is enough for $\alpha$ to be weakly compact. So I don't see what challenge this raises for "those who do not [or do?] believe in the existence of large cardinals."


