TITLE: Is there a "tree-like" proof of compactness theorem in the case of uncountably many variables?
QUESTION [15 upvotes]: I like proofs using trees and König's lemma, since they are very visual.
One of the applications of König's lemma you can show to students is proving compactness theorem for propositional calculus, which says that a set of formula is satisfiable if and only if every finite subset is satisfiable. 
In this proof you have to order the statements in your theory into a sequence and order the variables accordingly. 
 Then you construct a tree in which each vertex corresponds to truth values of a variables from the statements $T_1,\dots,T_n$ and then infinite branch gives you interpretation of all variables which is consistent with the whole theory.
In this way you can prove the compactness theorem under the assumption that the set of variables is countable.

Is there a proof in the similar spirit which works for arbitrary sets of variables too? (Or some technique which can be used to obtain uncountable version if we already have countable version?)

I am interested mainly in the proofs avoiding the use of completeness theorem.

REPLY [7 votes]: Here is a way to prove compactness for propositional logic in terms of trees.  As Andreas mentions in the comments, the tree property for $\omega$ is not enough.  Instead, we can use a two-cardinal variation of the tree property.
If $\kappa$ is a regular cardinal and $\lambda \ge \kappa$, a $(\kappa,\lambda)$ tree $T$ is a set of $2$-valued functions whose domains are elements of $\mathcal{P}_\kappa(\lambda)$ (that is, subsets of $\lambda$ of size $<\kappa$) such that 

the restriction of every function in $T$ is in $T$, and
every element of $\mathcal{P}_\kappa(\lambda)$ is the domain of some function in $T$.

A cofinal branch of $T$ is a function $f:\lambda \to 2$ such that $f \restriction s \in T$ for every $s \in \mathcal{P}_\kappa(\lambda)$.  There is a principle TP$(\kappa,\lambda)$, isolated by Weiss, that says that every thin $(\kappa,\lambda)$ tree has a branch.  We don't need to use the definition of "thin", because it is automatic if $\kappa = \omega$ or more generally if $\kappa$ is a strong limit cardinal. 
If $\kappa$ is a strongly inaccessible cardinal, then TP$(\kappa,\lambda)$ holds if and only if $\kappa$ is $\lambda$-compact, so it is a large cardinal property.  However, TP$(\omega,\lambda)$ can be proved to hold in ZFC: Take an ultrafilter $U$ on $\mathcal{P}_\omega(\lambda)$ that is fine, meaning that it contains the set $\{s \in \mathcal{P}_\omega(\lambda) : \alpha \in s\}$ for every $\alpha < \lambda$.  Because we are not requiring any amount of completeness, such an ultrafilter exists by Zorn's lemma.  Given an $(\omega,\lambda)$-tree $T$, choose for each set $s \in \mathcal{P}_\kappa(\lambda)$ some function $f_s \in T$ with domain $s$.  Then we can define a branch $f$ of $T$ by $f(\alpha) = 1$ if and only if $f_s(\alpha) = 1$ for $U$-almost every $s \in \mathcal{P}_\omega(\lambda)$.
Now we can use this tree property to find a truth assignment for a set $S$ of propositional formulas.  By enlarging $S$ we may assume that it is closed under subformulas and in particular that every propositional variable appearing in a formula in $S$ is itself in $S$. Define an $(\omega, S)$-tree $T$ consisting of 
all consistent truth assignments defined on finite subsets $s$ of $S$. (Strictly speaking we ought to fix a bijection between $S$ and $|S|$ here.)  By consistent I mean, for example, that if the formulas $\varphi$, $\psi$, and $\varphi \wedge \psi$ are all in $s$, then $\varphi \wedge \psi$ is assigned the value 1 if and only if both $\varphi$ and $\psi$ are.
Assuming TP$(\omega,|S|)$, this tree has a branch, and such a branch is a truth assignment that satisfies all the formulas in $S$.