$\kappa$ is compact $\implies$ $\kappa$ is regular

Question

Definitions

The Stanford Encylopedia of Philosophy makes the following definition in their article on Infinitary Logic https://plato.stanford.edu/entries/logic-infinitary/

Let κ be an infinite cardinal. A language L is said to be κ-compact (resp. weakly κ-compact) if whenever Δ is a set of L-sentences (resp. a set of L-sentences of cardinality ≤ κ) and each subset of Δ of cardinality < κ has a model, so does Δ. Notice that the usual compactness theorem for L is precisely the assertion that L is ω-compact.

Let us call a cardinal κ compact if the language L(κ,κ) is κ-compact

Additionally, I define the cofinality of an ordinal to be the least ordertype of cofinal subsets.

Take the Axiom of Choice for granted.

Question

How can we show that a compact cardinal $\kappa$ must be regular?

My initial thought was to assume that, if not, then there is an increasing cofinal $f: \alpha \to \kappa$ where $\alpha= \operatorname{cf}(\kappa) < \kappa$. Then I was thinking of constructing a "creative" set of sentences $\{\phi_i\}_{i \in \alpha}$ and considering a corresponding set $\{\phi_{f(i)}\}_{i \in \alpha}$ and somehow taking advantage of satisfiability. I'm doubtful that this idea would work, however.

Answer 1

Suppose $\kappa$ is singular, so we have $\kappa=\bigcup_{i<\alpha}A_i$ for some $\alpha<\kappa$ and some sets $A_i$ whose cardinalities are $<\kappa$. Let $p_\xi$ be a propositional variable for each ordinal $\xi<\kappa$, and consider the set $\Sigma$ of the following formulas: $\neg p_\xi$ for each $\xi<\kappa$ and the one additional formula $\bigvee_{i<\alpha}\big(\bigvee_{\xi\in A_i}p_\xi\big)$. Note that each of the disjunctions here has $<\kappa$ disjuncts, so all $\kappa$ of the sentences in $\Sigma$ are in $L_{\kappa,\kappa}$ (in fact in $L_{\kappa,1}$ since no quantifiers are used). $\Sigma$ is not satisfiable because the one additional formula requires at least one $p_\xi$ to be true. But any subset of $\Sigma$ of size $<\kappa$ (indeed, any proper subset of $\Sigma$) is stisfiable.

Answer 2

Suppose $\kappa$ is singular, and let $(\alpha_i)_{i\in I}$ be a cofinal sequence of ordinals less than $\kappa$, with $|I|<\kappa$.

We will work in the language of strict order, $\{<\}$. Observe the following:

1. There is a sentence $\varphi$ of $L_{\kappa,\kappa}$ (even of $L_{\omega_1,\omega_1}$) axiomatizing the well-orders. Just take the conjunction of the first-order linear order axioms with a sentence asserting that there is no descending $\omega$-sequence.
2. For every ordinal $\alpha<\kappa$, there is a sentence $\theta_\alpha$ of $L_{\kappa,\kappa}$ such that for a well-order $M$, $M\models \theta_\alpha$ if and only if the order type of $M$ is $\geq \alpha$: $$\exists (x_\beta)_{\beta<\alpha} \bigwedge_{\beta<\beta'} x_\beta < x_{\beta'}$$

Now look at the set $\Delta$ consisting of $\varphi$, $\theta_\alpha$ for every $\alpha<\kappa$, and the sentence $$\bigvee_{i\in I} \lnot\theta_{\alpha_i}.$$ Every subset of $\Delta$ of size $<\kappa$ has a model, but $\Delta$ has no model, so $L_{\kappa,\kappa}$ is not $\kappa$-compact. Since $|\Delta|=\kappa$, $L_{\kappa,\kappa}$ is not even weakly $\kappa$-compact.