In my recent question on Godel Completeness I mentioned that there was a related question I wanted to ask, but would keep separate. I have been recently studying "non-well ordered sets" and Chapter 7 of this paper is useful background. There are many mathematical questions about non-well ordered sets, but here I want to return to something I noticed in Logic.

There is a form of Godel Completeness known as "Generalised Completeness Theorem" which applies to uncountable languages. For this theorem to be proven it requires the Axiom of Choice. This is understandable, and is needed because the uncountable language needs to be ordered to construct the formulae ordering needed in proofs: $\phi_1, \phi_2, \phi_3, ...$

However I have only seen this point mentioned for uncountable languages - never as an assumption in connection with countable languages. EDIT Thus AC is invoked in proofs involving uncountable languages. But Sets (in ZF) can be "uncountable" for two distinct reasons: Let S be such a ZF set:

  1. S is well ordered, but has cardinality greater than $\aleph_0$
  2. S is not well ordered, and so again cannot be put into correspondence with $\omega$.

In the latter case giving a "size" to the set is harder. However a theorem called Hartog's theorem provides an associated ordinal - that is Hartogs(S) is an ordinal. This does not quite give us the notion of exact size like we had in ZFC, but it is a beginning. As a finite ordinal example

Hartogs$(n) = n+1$

This is what I shall mean in the Question by "low cardinality" non-well ordered sets - sets which would like to be countable but cannot be due to being non-well ordered. This idea is related to Conways "Counted - Countable" distinction mentioned in that paper (I think).

Such sets still have finite subsets (which are always well-ordered without any Choice) and examining these elements might lead one to infer that the main set was well ordered.

So why does Formal Language theory only consider "Well ordered languages?" END EDIT

My initial answer is that formal languages are often assumed built on a finite alphabet - hence all words and formulae can be lexicographically ordered without even Choice being required.

Beyond this case it all becomes more problematic. One can have an infinite alphabet. This has to be assumed well-ordered to start things off. In several of my Logic textbooks we are simply told: A is an alphabet with members - a1, a2, a3, .... So this implicitly (but not explicitly) assumes either (1) that Choice is invoked or (2) that A is an infinite well ordered set to start with. As a result all theorems (including the regular Godel Completeness) may now depend on Choice via this assumption? For those reading my related question this would seem to imply that e.g. Godel's Completeness theorem is never really independent of Choice (and not merely WKL)? In short Formal language texts are saying "Let A be a well ordered set ..."

The benefit of having a countable supply of variables could still be resolved in such a case: Let S be a "low cardinality" non-well ordered Set. Then let A = S U Nat. A is a non-well ordered alphabet, generating a non-well ordered language with (a class of) indexical terms.

The Amorphous Set construction in Answer to my previous question is an example of a non-well ordered set construction. Borrowing Conway's term one might want to distinguish between:

  1. Countable Languages (those with an enumerable alphabet, etc)
  2. Counted Languages (those which cannot be isomorphic to Nat)

I have never seen this subject discussed before, so I am looking for other comments.

EDIT: I have removed all references to Countable Choice in the first version of this question for reasons identified in comments below.

  • $\begingroup$ I am pretty sure the answer is: in most set theories the set of integers is well ordered (often they are defined to be the smallest inductive set). Countability is defined as bijection with the integers. This gives a natural well ordering. In particular, well-ordering of countable objects is independent of choice of any sort. (In other words, while the well-ordering theorem is equivalent to the axiom of choice, your assumption that countable well-ordering is equivalent to the axiom of countable choice is false.) $\endgroup$ – Willie Wong Feb 4 '14 at 8:40
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    $\begingroup$ @Willie: I haven't met a set theory where the set of integers wasn't internally well-ordered. :-) $\endgroup$ – Asaf Karagila Feb 4 '14 at 8:47
  • $\begingroup$ @Asaf: neither have I. But I am much more ignorant than you are in the matter, so I left open the possibility that someone came up with some weirdo theory in which it isn't. $\endgroup$ – Willie Wong Feb 4 '14 at 8:56
  • $\begingroup$ @Roy Simpson: as I point out in my comment below Asaf's answer, I think you may be confused about what the axiom of countable choice says, and what it can be used to prove. Like the axiom of choice, the axiom of countable choice makes all its choices simultaneously. The axiom of dependent choice makes choices one after another that can depend on previous choices, but the axiom of countable choice cannot do that. To invoke it you have to lay out the countable family of choices all at the same time. See en.wikipedia.org/wiki/Axiom_of_countable_choice $\endgroup$ – Carl Mummert Feb 4 '14 at 12:10
  • $\begingroup$ @WillieWong: As I remark to Asaf I am not assuming that $\omega$ cannot be well ordered, or that such well ordering requires any version of the Axiom of Choice. Ordinals are ordered by definition and construction. So the Cardinality Class of a Non-well ordered Set will not contain an ordinal. This makes them harder to count, but a Theorem called Hartog's theorem provides an associated ordinal, so it is still meaningful to talk of "low cardinality" such sets. Hence "non-well ordered languages" too. $\endgroup$ – Roy Simpson Feb 4 '14 at 13:15

First of all, note that in first-order logic you usually have $\omega$ free variables at your disposal, so the language is never truly finite. This is because you want to be able to write open formulas and open terms, and not just closed terms and sentences involving them.

Now, note that the fact that we assume that the alphabet is countable (which is what the assumption that $A=\{a_1,a_2,a_3,\ldots\}$ means) does not require the axiom of countable choice. It's just an assumption that the alphabet can be indexed using the natural numbers.

Now, given an infinite well-ordered set $A$ we can definably enumerate all the finite sequences of elements of $A$ (let's denote this set as $Seq(A)$ for now), and we can prove that $A$ and $Seq(A)$ have the same cardinality. So $Seq(A)$ can be well-ordered as well.

Moreover, we can identify the well-formed formulas, the well-formed terms, and so on, within $Seq(A)$. So we can prove that if $\cal L$ is a language (including the logical symbols and all) which can be indexed using an ordinal, then the set of well-formed formulas can be indexed by an ordinal. And all this without any assumption of choice. Just that the set of symbols can be well-ordered.

As I mentioned in my answer to your previous question, over $\sf ZF$ the completeness theorem is equivalent to the ultrafilter lemma. There are models of $\sf ZF$ where these hold, but the axiom of countable choice fails. Cohen's first model, for example. Also, as Willie points out correctly in the comments, assuming that the axiom of countable choice is equivalent to the fact that $\omega$ is well-ordered is false. The integers, in $\sf ZF$, are defined as an ordinal, so they are always well-ordered. And countable sets are, by definition, sets which can be indexed using (possibly all the) finite ordinals.

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    $\begingroup$ There is another phenomenon here with countable choice that is hard to state properly. Because the powerset of any nonfinite set is uncountable, if we apply the usual proof via AC to well-order a nonfinite set, we will not be able to do that via countable choice. (I would want to say "countable choice cannot well order a countably infinite set", but that's nonsense.) The axiom of countable choice allows for making countably many simultaneous choices, but the proof that a well ordering exists requires simultaneous choices for the entire powerset of the set being ordered (except 0). $\endgroup$ – Carl Mummert Feb 4 '14 at 12:06
  • $\begingroup$ @Asaf Karagila: I will examine the Countable choice topic further, but I am not "assuming that the ACC is equivalent to the fact that $\omega$ is well ordered".I understand $\omega$ (like all ordinals) to be well ordered by construction. Infinite Sets can exist which cannot be well ordered (like the amorphous), therefore cannot be put into correspondence with $\omega$. In ZFC we can safely call such sets "uncountable": not in ZF. Low cardinality such sets could provide a set of symbols and a Language. Its finite subsets would be well-ordered of course, and the error is in extrapolation. $\endgroup$ – Roy Simpson Feb 4 '14 at 13:06
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    $\begingroup$ In ZF as well, an infinite set is countable if it is in bijection with $\omega$, and uncountable otherwise. The main difference when the axiom of choice fails is that cardinalities are no longer linearly ordered. The phrase in the question " implicitly (but not explicitly) assumes that Countable Choice is invoked" does seem to suggest that countable choice is somehow related to the ability to well order the alphabet... @Roy Simpson $\endgroup$ – Carl Mummert Feb 4 '14 at 13:14
  • $\begingroup$ @CarlMummert: I said "safely" there because in ZF "uncountable" now has two meanings, it has one in the more common ZFC. $\endgroup$ – Roy Simpson Feb 4 '14 at 13:20
  • $\begingroup$ @CarlMummert: OK I will change the offending wording about Countable Choice in that part of the Question. Please re-read that part of the question when I am done. $\endgroup$ – Roy Simpson Feb 4 '14 at 13:24

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