Show that the compactness theorem does not apply to infinite logic

Question

I am trying to understand why the compactness theorem does not apply in infinite logic and I wonder if anyone has a good example and explanation for this?

Edit: By infinite logic I mean logic that allows infinitely many conjunctions and disjunctions. More exactly:

• $M \models \bigvee \Gamma$ iff $M \models \varphi$ for some set of sentences $\varphi \in \Gamma$.
• $M \models \bigwedge \Gamma$ iff $M \models \varphi$ for some set of sentences $\varphi \in \Gamma$.

The compactness theorem:

The compactness theorem states that a set of first-order sentences has a model if and only if every finite subset of it has a model.

Thanks in advance!

Answer 1

This is called "infinitary logic." For every pair of infinite cardinals $\kappa\ge\lambda$ there is a logic $\mathcal{L}_{\kappa,\lambda}$ gotten by closing first-order logic under conjunctions and disjunctions of size $<\kappa$ and universal and existential quantification over tuples of length $<\lambda$. The most common infinitary logics are of the form $\mathcal{L}_{\kappa,\omega}$ - so only finitary quantification is allowed, although we permit "big" Boolean combinations.

The logic $\mathcal{L}_{\omega,\omega}$ is just first-order logic itself. The first infinitary logic is $\mathcal{L}_{\omega_1,\omega}$, where we expand first-order logic by allowing countably infinite conjunctions and disjunctions. Here we already see a failure of compactness: consider the sentence $$(*)\quad\bigvee_{n\in\mathbb{N}}[\forall x_1,...,x_n(\bigvee_{1\le i<j\le n}x_i=x_j)].$$ This is true in a structure iff that structure is finite. But this yields a counterexample to compactness (think about the proof that every first-order theory with arbitrarily large finite models has an infinite model):

Consider the theory $\{(*)\}\cup\{\mbox{"There are at least $n$ elements"}: n\in\mathbb{N}\}$.

Answer 2

Suppose that an infinitely long disjunction (1) exists.

\begin{equation}\tag{1}a_1\lor a_2\lor a_3\lor\dots\end{equation}

Then consider an infinite collection of formulae (2), consisting of the negation of each instance of the disjunction.

\begin{equation}\tag{2}\{\lnot a_1,\lnot a_2,\lnot a_3,\dots\}\end{equation}

A theory $T$ consisting of (1) and all the formulae in (2) will necessarily be unsatisfiable, as each negated propositional atom $a_i$, cannot be satisfied unless the atom is false.

However, every finite subset of formulae in $T$ will be satisfiable, (e.g. $\lnot a_1$ and $a_1\lor a_2\lor a_3\lor\dots$). If the compactness theorem applied, then the set comprising all formulae in the theory will be satisfiable.

This is a contradiction, so it is not the case that the compactness theorem applies in what you have called infinite logic.