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Note: I know next to nothing about mathematical logic, so please don't feel that you have to explain any answer rigorously or technically in order for me to accept it.

Question: Is metric (Cauchy) completeness "outside the realm" of first order logic?

In particular, is the Dedekind-Cantor axiom necessarily a statement of second order logic?

Background: This is a follow-up to my previous question and my attempt to answer that question.

Tarski proved at least two interesting theorems which seem to suggest that first order logic is "unable to distinguish" between (some) metrically complete spaces and metrically incomplete spaces. For example, he showed that any real closed field is elementarily equivalent to the real numbers (which are metrically complete), including the real algebraic numbers (which are not metrically complete). He also proved the Lefschetz principle, which (I think) states something like all algebraically closed fields of characteristic zero are elementarily equivalent.

As I said before, I know next to nothing about mathematical logic, to the extent that I don't really know or understand the difference between first and second order logic, except that it maybe has something to do with what sets quantifiers are allowed to range over.

In particular, I do not understand the reason why the distinction between the two should be particularly interesting. However, if one type is unable to make a distinction, between Cauchy incomplete and Cauchy complete metric spaces, that is important to analysis, then I would begin to understand somewhat on a naive level the importance of the distinction. Hence this question.

Note: I hope this question doesn't seem too stupid because I "should" know that the answer is "yes" because of the elementary equivalence of all real closed fields. Consider the answers to this (very) related question. One answer to that question would imply that the answer to my question is "yes", because it says "a structure with the first order properties of the real numbers may not satisfy the completeness axiom, which is not first order". However, another answer to that question (actually two other answers) to that question point out that ZFC can be formulated in first order logic by something called the Lowenheim-Skolem theorem, and since analysis (including metric completeness) can be formulated in terms of ZFC, which is first order, the answer to my question should be "no". Obviously there is a subtle point that I am completely oblivious to here, hence my confusion and uncertainty. Any patience you may have for my naivete would be greatly appreciated.

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Chill2Macht
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Yes, it is. This follows from the (lower) Löwenheim–Skolem property of first order logic.

ZFC is a first-order theory and it can express many concepts, including but not limited to completeness of the reals, and it is sufficient to prove that reals are complete. So if $M$ is a model of ZFC, then $M$ believes that ${\bf R}^M$, the set it thinks is the set of real numbers, is complete, because ZFC proves that the reals are, in fact, complete.

However, it does not really mean that ${\bf R}^M$ is actually complete. For example, if $M$ is a countable model, it cannot be complete: indeed, ${\bf R}^M$ will also be countable (it is, after all, a subset of $M$), and a dense linear ordering. Such a space is never complete, but $M$ will not know it: the reason is that it does not know about all the Cauchy sequences (it only knows about countably many).

You may want to read about Skolem's paradox, this is a very similar phenomenon.

It's not so easy in general (you need to properly formalize a bunch of things), but I think it's true that there is no first order structure with a "definable metric" (or even "definable Hausdorff topology") whose models are all complete and not discrete.

tomasz
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    +1. Re: your last sentence, yes that's true: use Lowenheim-Skolem and the fact that every complete non-discrete metric space is uncountable. (This assumes a countable language, which is a natural assumption; but even that is unnecessary - for an arbitrary language, we can use the Omitting Types Theorem to build a model with a Cauchy sequence which doesn't converge.) – Noah Schweber Mar 10 '17 at 17:43
  • @NoahSchweber Doesn't the omitting types theorem also depend on countability of the language? Or is there some more powerful version I'm not familiar with? –  Mar 12 '17 at 07:20
  • Why doesn't $M$ know all Cauchy sequences in $M$? Does power set axiom not produce all possible sequences in $M$? Or is it because the $\mathbb{N}$ according to standard construction in ZFC, is a non-standard $\mathbb{N}$? – Violapterin Oct 03 '20 at 16:36
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    @Violapterin: If $M$ is a transitive model, then the natural numbers according to $M$ are the true natural numbers. However, the power set axiom does only what you say: it produces all sequences in $M$. However, there are sequence of elements of $M$ which are not themselves in $M$. – tomasz Oct 04 '20 at 23:18
  • I think I got it after reading Skolem's Paradox, and it really blows my mind. – Violapterin Oct 05 '20 at 08:27
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    @Violapterin: Don't worry. Once you get your head around the basic ideas, you get used to it real fast. – tomasz Oct 05 '20 at 08:33