K is a basic elementary class if and only if both K and the complement of K are closed under ultraproducts and elementary equivalence.

logicmodel-theory

I am reading this logic book and was trying to proof the second theorm :

Theorem 13.3: Let K be an arbitrary class of models. Then:

  1. K is an elementary class if and only if K is closed under ultraproducts and elementary
    equivalence.
  2. K is a basic elementary class if and only if both K and the complement of K are closed
    under ultraproducts and elementary equivalence.

At the end of the proof of (1) it states : (2) This follows easily from (1) and the compactness theorem. We leave the details for an
exercise.

This is a sketch of a proof of (2) but I didn't use compactness so I would like some feedback:

Assume K is a basic elementary class, then thehre is a sentense $\varphi$ of $\mathscr L$ such that K is the class of all models of $\varphi$.

  1. Closure under ultraproudcts:
    Let {A$_i$} be a family of models in K, we will show that $\prod_DA_i$ is also in K. A$_i$ satisfies $\varphi$ (as K is the class of models satisfying $\varphi$) every element in the ultra product also satisfies $\varphi$ which implies that $\prod_DA_i$ is also in K hence K is closed under ultraproudcts.
    Let {B$_i$} be a family of models in K$^c$ since K is basic elemantry class exist a sentence $\varphi$ such that K is precisely the class of models satisfying $\varphi$, since B$_i$ is in K$^c$ for each i B$_i$ doesn't satisfey $\varphi$, now consider $\not$$\varphi$ since B$_i$ doesn't satisfey $\varphi$ it satisfies $\not$$\varphi$ which implies that $\prod_DB_i$ doesn't satisfey $\varphi$ hence $\prod_DB_i$ satisfies $\not$$\varphi$ hence K$^c$ is closed under ultraproducts.

  2. Closure under elementary equivalence: Let A be a model in K$^c$ and let B be a model such that B$\equiv$A.B$\equiv$A hence B and A satisfies the same sentences, as A doesn't satisfey $\varphi$ also B doesn't satifey $\varphi$ hence B is in K$^c$ by definition hence K$^c$ is closed under elementary equivnalnce. (For K its the same argument…)

For the other $\to$ : if both K and K$^c$ are closed under ultraproudcts and elementary equivalnce, we can constract a sentence $\varphi$ such that K is the class of models of $\varphi$ and K$^c$ is the class of models of $\not$$\varphi$ thus K is a basic elemntary class and the proof is complete.

Best Answer

The implication "if both $K$ and the complement of $K$ are closed under ultraproducts and elementary equivalence, then $K$ is a basic elementary class" is the nontrivial one. You write:

If both $K$ and $K^c$ are closed under ultraproducts and elementary equivalence, we can construct a sentence $\varphi$ such that $K$ is the class of models of $\varphi$ and $K^c$ is the class of models of $\lnot \varphi$. Thus $K$ is a basic elementary class and the proof is complete.

It should be obvious to you that this is no proof at all! "We can construct a sentence $\varphi$..." How?

... well, this is probably where you need to use compactness...

Hint: Suppose $K$ and $K^c$ are both closed under ultraproducts and compactness. By (1), $K$ and $K^c$ are both elementary classes. Suppose $T_1$ is a theory axiomatizing $K$ and $T_2$ is a theory axiomatizing $K^c$. What can you say about the theory $T_1\cup T_2$? Remember, your goal is to replace $T_1$ by a single sentence axiomatizing $K$.

Your proof of the other direction looks fine (though the writing and spelling and grammar are pretty sloppy). But I want to point out that this direction really does follow from (1) with no extra work. Assume $K$ is a basic elementary class. Then $K^c$ is also basic elementary (since if $\varphi$ is a sentence axiomatizing $K$, then $\lnot \varphi$ axiomatizes $K^c$). In particular, both $K$ and $K^c$ are elementary classes. So by (1), both $K$ and $K^c$ are closed under ultraproducts and elementary equivalence.

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