Completeness theorem for Predicate logic with uncountable sentences?

Question

I know the proof of completeness theorem for Predicate derivateion. And it's used that sentences of PLE(Predicate logic with identity) is countable for proving that for any consistent set A in PLE, there is a maximally consistent set of PLE that contains A. (In this post, consistent means consistent in PDE.)

But let's think that there are uncountable senteces of PLE. (I think it can be done by just using uncountably many symbols. We can regard two symbols different if they have the same shape but different sizes, and the the range of size is over real numbers. Also, (not surely) we may think there is a way such that for any cardinal number K, we can make K number of symbols.) And I thought, just for the of the completeness of the PDE with uncountable sentences, there are some troubles.

(1) If there are uncountably many sentences of PLE, for any consistent set of sentences of PLE, how can we construct or (prove the existence of) a maximally consistent set that contains the consistent set? (Or is it possible?)

(2) Let's suppose (1) is possible. (i.e. we can prove the existence of a maximally consistent set for any consistent set.) Then, is maximally consistent set quantificationally consistent in PLE?

(I think (1) may be depend on axiom of choice and (2) would be true in ZF. Also, I think soundness theorem of PLE with uncountable sentences and compactness theorem will hold.)

Answer 1

Your second paragraph seems confused. Two symbols are different if... well... they are different. This has nothing to do with their shape (though two symbols of different shape are certainly different), and I'm not even sure what the size of a symbol would mean. (Edit: I was assuming you meant something different by 'shape' too, so my previous sentence doesn't applies to that, not what you actually meant. See the comments.)

For instance, you can have a language with the symbols $\le $ and $E$. These are two different binary relation symbols (cause we say they're different just like in another mathematical context we might posit that $x$ and $y$ are two different variables). Perhaps we have chosen the notation suggestively and $\le$ will be interpreted as a partial ordering and $E$ will be interpreted as an equivalence relation.

Perhaps, provided our axioms don't preclude this, we will even interpret them as the same relation... that doesn't change the fact that they are two distinct symbols, just like $x$ and $y$ are different variables but they may represent the same number.

We might have another language where we have a set of binary relation symbols $\{E_r:r\in \mathbb R\},$ one for each real number. For $r\ne r',$ $E_r$ and $E_{r'}$ are different symbols. Again this is just because we say so... it is how we are defining the language. We can have a set of symbols of any cardinality we want just by indexing with a set of that cardinality.

But as for your question, if we assume the axiom of choice, the proof can go largely the same as the one I presume you've seen. You can well-order your set of sentences and then use transfinite induction rather than regular induction on the natural numbers to construct the maximal consistent set. This can also be replaced by an equivalent Zorn's lemma argument where you argue that the consistent extensions of the theory are a partially ordered set closed under chain unions (since any inconsistency needs to come from a finite number of statements). Note that choice here can be replaced with the assumption that the particular language under consideration is well-orderable.

If we don't have choice and the language cannot be assumed well-orderable (like in the case of the reals above), then this proof won't go through. It turns out that the statement that the completeness theorem holds for an arbitrary language is equivalent to the ultrafilter lemma, which is a weaker version of choice, so some use of choice is unavoidable.

And in regard to your second question, yes, the only place choice is used in the proof is in the construction of the maximal consistent extension. Once we have such an extension of the Henkin completion of the original theory, we can construct a term model. Neither the Henkin completion step nor the term model construction step are complicated significantly by an uncountable language.