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It is an amazing and well-known fact that the Continuum Hypothesis is logically independent of Zermelo-Frankel set theory with the Axiom of Choice (ZFC), assuming it is consistent. In a similar vein, the Axiom of Choice itself is logically independent of Zermelo-Frankel set theory (ZF), assuming it is consistent.

I'm not well-versed in axiomatic set theory. In my mind, the set theory you choose (ZF, ZFC or some other one) defines for you the rules of the game we call mathematics. To say that a statement is independent of the given set theory is to say that the statement defines a new rule which is not a combination of the previous rules (so it is genuinely a new rule), but also forbidding what new rule allows is not a combination of the old rules (so you can't be in the situation that the new set of rules contradicts themselves unless they contradicted themselves to begin with). This may be a very coarse (and possibly inaccurate) view of set theory, but I hope it is not too much of an oversimplification for the question I have in mind.

I was trying to think of a simple mathematical analogy for the notion of a logically independent statement and I came up with the following:

Let $x \in (-1, 1]$. The statement "$x \in (-1, 0]$" is logically independent because neither it, nor its negation ("$x \in (0, 1]$") can be logically deduced from the given information (i.e. the analogue of a set theory).

Is this even close to being representative of the notion of logically independent? If not, why not? If it is, what details does it overlook?

Links to the relevant terms (it looked very strange if I linked to them all in the paragraph):

Asaf Karagila
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Michael Albanese
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  • My answer here is somewhat related. – Asaf Karagila Feb 05 '13 at 14:50
  • As a largely unrelated comment, I don't think choosing a set theory defines the rules of the game of mathematics. Most mathematics has the same rules regardless of what sets are behind it, and if after all ZF turned out to be inconsistent, the majority of modern mathematics would survive in an essentially unchanged form. – Ben Millwood Feb 05 '13 at 15:10
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    Why did you come up with such a contrived example of "independence" when you have likely already studied a perfectly good one? Note that the basic axioms of Group Theory are perfectly good first-order statements, and you can also formulate the Axiom of Commutativity (AC) in this language. We know that AC is independent of GT, and that by adding AC or $\neg$AC to GT we get the theory of Abelian and non-Abelian groups, respectively. There are some metamathemtical issues that make independence in set-theory seem more daunting, but this has little to do with independence qua independence. – user642796 Feb 05 '13 at 15:18
  • @BenMillwood: I didn't intend for this sentence to have such an impact, I was just trying to fit it into the analogy of a game. What I mean is that the set theory you are working with effects what you are and are not allowed to do, and hence the results you can obtain. For example, it is well known that you cannot construct a non-measurable set in ZF but you can in ZFC. – Michael Albanese Feb 06 '13 at 02:13
  • @ArthurFischer: This is a really nice example and it does make the entire notion of independence clearer to me. I suppose I was hoping for something simple enough to explain to a non-mathematician. – Michael Albanese Feb 06 '13 at 02:17
  • @Michael: In the spirit of the part added to my answer, I think that the best example to a math ignorant person would be $\frac12$. It is obvious that you can't find an integer which will give us $1$ when we add it to itself, but we can find a rational number that has this property. – Asaf Karagila Feb 06 '13 at 03:10
  • Brouwer and Cohen both explicitly pointed out that independence is never absolute - but always relative to an axiom system. Not just set theory but also logic itself, eg with or without Excluded Middle. Without, you can't conclude existence of an object by contradicting the assumption that it doesn't exist. – alancalvitti Feb 06 '13 at 03:30

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It's not a bad analogy. But it can only take you so far. Similar analogies are comparing forcing to the construction of field extensions (in particular the algebraic closure). Both the analogies have some of the essence of the idea we want to state:

Sometimes you have insufficient data to decide whether a statement is true or false.

But ultimately these examples are very different because there is a very canonical way we see the real numbers and the interval $(-1,1]$, but there is absolutely no canonical way we see models of set theory (except, of course, those who feel it is inconsistent, then they see it canonically as having no models).

If you are trying to understand, or explain the idea of independence to someone who is not familiar with set theory, or even mathematics, enough to grasp "axioms" and "models" and so on -- then the analogy could serve you well if you use it properly (i.e. not stretching it too far, due to the point I made above).

Another analogy which I like very much which fits very well for independence is asking a question of the form "How many numbers satisfy the equation $x^3=2$?" an immediate answer would be one, or three, but it would also be zero. We didn't specify what "number" means here, and if we only think about it as a formula in the language of ring theory, or field theory if you will, then there are different models -- which are far from bizarre -- in which the answers are different.

In the rational numbers there are no numbers with this property; in the real numbers there is just one; and in the complex numbers there is a set of three numbers with this property. We learn from this two things, the first is that the axioms of the field cannot prove the existence or the uniqueness (if it does exist) of such number, and that context is important. But when passing to set theory, or just any theory $T$, the context is lost because context is in the semantics and this is now a game of syntax (which by the completeness theorem we can turn into a game of semantics as well), and with only syntax to support us we learn that the axioms of field theory are too weak for this sort of proof. And this is the subtleties which I find discerning the examples of independence which we meet but often fail to recognize, and the independence which is "in your face" in set theory - which is also filled with technicalities and fine points about internal and external objects, and so on.

I like this proof because everyone knows (or should know) about irrational numbers, about the complex numbers and so on. It's easier to explain to the layman compared to group theory (which is easy to explain to a mathematician, but not to the common man).

But as I learned during the previous year (I was writing my masters) -- there are no real shortcuts in understanding all the delicate points, even in the big picture.

Asaf Karagila
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  • I think your highlighted sentence is what I should have in mind when considering logical independence. Do you know of a reference which outlines the analogy between forcing and field extensions? – Michael Albanese Feb 06 '13 at 02:22
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    @Michael: I gave some outlines on this very site. I'll hook you up with links in a few minutes... – Asaf Karagila Feb 06 '13 at 02:23
  • @Michael: Apparently I never wrote this analogy here. I did give other analogies (which I love so much more than the one given in your post and in the first link I provided). For example here, which I later improved here. I think the forcing-field extension analogy might be a bit more technical, it might be better to avoid it for now... – Asaf Karagila Feb 06 '13 at 02:37
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    @Michael: I updated my answer with the outline of the said analogy (not about forcing, though. That's for a whole other topic, I believe). – Asaf Karagila Feb 06 '13 at 02:48
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A comment on a passing remark in the question (though interesting enough to deserve comment, I think). The OP writes

In my mind, the set theory you choose (ZF, ZFC or some other one) defines for you the rules of the game we call mathematics.

Two observations.

First, most working mathematicians who haven't taught a set theory course recently won't even be able to tell you what the axioms of ZFC are, and seem pretty cheerful in their ignorance. So there is a real question in what sense mathematicians doing arithmetic combinatorics (for example) are playing a game defined by a set theory that seemingly plays no overt part in their work.

Most mathematicians smile indulgently when set theorists claim their theory has a special 'foundational' role: in practice, set theory is largely treated as just one mathematical enquiry among many. So there's certainly more that needs to be said, more cautiously, about the role of set theory than that, baldly, that it "defines the rules of the game".

Second, if you have set theorists who disagree (say one who accepts the axiom of determinacy and the other who doesn't) then, still, surely they are both still doing mathematics. It would be very odd for one to say "for me, ZF + AD defines the rules of the game I call mathematics, so you [who accept the incompatible ZFC] aren't doing mathematics". I think the ZFC-ist would be rightly a bit cheesed off at that definition of mathematics!

Peter Smith
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    One observation: When the real numbers are a countable union of countable sets the rules change for the "working" mathematician. You can't talk about countable union of countable sets for no reason; and you can't talk about measure theory as easily as before; and so on. When the real numbers have a dense Dedekind-finite set you can't say that they are a Lindelof space anymore, and you can't say that continuity by $\varepsilon$-$\delta$ is equivalent to sequential continuity; and so on. The rules do change, the rules of inference don't. But we can take care of that just as well. :-) – Asaf Karagila Feb 05 '13 at 16:52
  • @AsafKaragila Oh yes. Things are indeed interestingly complex! I was just gently complaining about a certain over-hastiness in the OP's passing remark (as you hear that sort of thing often). – Peter Smith Feb 05 '13 at 17:05
  • Just wanted to make sure everyone is on the same page. :-) – Asaf Karagila Feb 05 '13 at 17:10
  • Out of curiosity, what axiomatic system makes the real numbers a countable union of countable sets ? – M.G Feb 05 '13 at 18:30
  • @M.G: It is consistent with ZF that the real numbers are a countable union of countable sets, but you have to require this as an explicit addition to your axioms. I should also point out that ZF itself proves that the real numbers are uncountable, it just can't prove that countable unions of countable sets are countable. – Asaf Karagila Feb 06 '13 at 02:21
  • I didn't mean for that sentence to cause so much trouble. I don't think one set theory is the right one and the rest are useless, but I am aware that, depending on your axioms, there are things that you can or cannot do (i.e. constructing a non-measurable set). I don't think people who use ZF as opposed to ZFC or any other set theory are doing a different version of mathematics. What I meant was that everyone is doing mathematics, but the set theory that you are working with (either explicitly or implicitly) can affect what you can deduce as Asaf has pointed out. – Michael Albanese Feb 06 '13 at 02:33