Show that the compactness theorem does not apply to infinite logic 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!

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}\}$.

