Why does the compactness theorem not apply to infinite subsets?

Question

I am trying to prove:

Let $T$ be a theory such that the closed-world assumption $\mathrm{CWA}(T)$ is consistent. The closed-world assumption $\mathrm{CWA}^P(T)$ relative to a predicate $P$ then also is consistent.

Edit: "consistent" means "has a model" or "is satisfiable".

I have established that $\mathrm{CWA}^P(T) \subseteq \mathrm{CWA}(T)$. To conclude I want to show that the consistency of $\mathrm{CWA}^P(T)$ follows because it is a subset of a consistent theory (compactness theorem)

However, the compactness theorem only applies to finite subsets (AFAIK), and I cannot guarantee that my closed-world assumption wrt. $P$ is finite.

1. Why does the compactness theorem only apply to finite subsets? Intuitively it should apply to all subsets?
2. Can I apply the compactness theorem here because I missed something?
3. Is there any other way to argue?

Answer 1

I don't know what the "closed-world assumption" is, but it sounds like you're misunderstanding the compactness theorem.

Any subset (finite or not) of a consistent theory is consistent. This is just a basic fact of logic; the proof depends on what you mean by "consistent" (has a model, or does not prove "$\perp$" in a fixed proof system - these two are the same post-completeness theorem, but not a priori), but under any interpretation, it isn't hard.

Compactness is a tool for showing that larger sets of sentences are consistent, based on the consistency of smaller ones. Namely, compactness says that if every finite subset of $\Gamma$ is consistent, then so is $\Gamma$.

For example, one standard application of compactness is to show that there is no set of sentences $\Gamma$ (in your favorite language) which is true in exactly the finite structures. Why? Well, consider the expanded language with new constant symbols $c_i$ ($i\in\mathbb{N}$), and let $$\Sigma=\Gamma\cup\{c_i\not=c_j: 1\le i<j\}.$$ Now consider the subsets $$\Sigma_n=\Gamma\cup\{c_i\not=c_j: 1\le i<j\le n\}.$$ Every $\Sigma_n$ has a model (namely, an expansion of any model of $\Gamma$ of size $n$), and any finite subset of $\Sigma$ is contained in some $\Sigma_n$; so by Compactness, $\Sigma$ itself has a model. But that model must be infinite, so $\Gamma$ has an infinite model (the reduct of any model of $\Sigma$), contradiction.

Now how we used Compactness here: we showed that a "large" set of sentences was consistent, based on the known consistency of its "small" subsets.