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!

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.
