Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Is there such a thing as starting with a weak kind of set theory and making an argument to show that it should be expanded into a stronger set theory such as ZFC? Something similar to going from the rationals to the reals? With number systems, we have evidence which forces us into a stronger system (sqrt of 2). Is there such a thing with set theory?

share|cite|improve this question
    
A very vague question, possibly too unclear, but possibly answerable anyway. – 6005 4 hours ago
    
I can think of at least 5 completely different answers to this question. 1. Intuistical motivations like 'the collection scheme obviously is true'. 2. Technical requirements like 'without the powerset axiom, replacement is just weird.' 3. Historical motivations. 'We've taken xyz for granted all through the 19th century, so xyz should be true in our axiomitization.' 4. Convenience. It's often desirable to work with a strong (and natural) theory that you often don't have to think about while dealing with it. 5. Compatibilty. ZFC seems strong and simple enough for most mathemeticians. – Stefan 4 hours ago
    
Okay... 4. and 5. are not 'completely different' and - in this very short presentation - almost seem the same. However, they have different intended foci. In item 4. I was thinking of the working set theorist, who doesn't want to constantly worry about the minor details of technical constructions (at least this is a common complaint about fine structure). Item 5. was intended to say that the average mathematican usually doesn't have to worry about his background theory at all. He uses constructions that seem justified and just runs with them. The modelling in ZFC doesn't concern him. – Stefan 4 hours ago

For category theory, sometimes you could need to handle the set of all sets and this set does not exist in $ZF$. But using $NBG$, you can speak of the class of all sets which would be similar to the set of all sets but untouched by the same problems that arise when you speak about the same concept in $ZF$. So $NBG$ possess a larger ontology than $ZF$ and indeed, $ZF$ can be constructed as a subsystem of $NBG$. I'm just not sure if this is the kind of completing you're looking for, but if this is, take a look at the first chapter of Goldblatt's: The categorial analysis of logic.

I guess that these theories can be completed by generating a new theory in which certain sets that were not accepted in the previous theories exist for the new ones, just as the set I mentioned before. Take $ZF$ and remove the axiom of the power set, it will be a less complete theory, that is: You won't be able to construct certain sets that could be built with $ZF$.

This reminds me of an exercise in the beginning or Jech/Hrbáček's: Introduction to Set Theory They give the axioms of existence, extensionality and comprehension. A little later in the book, they ask you to prove that with these axioms alone, you can only construct $\emptyset$.

Start reading what I suggested, soon Asaf will emerge from the darkness and settle this down.

share|cite|improve this answer

While your analogy with the case of rationals $\rightarrow$ reals is (in my opinion) somewhat strained, I think this is a very good question. Perhaps it would be better to recast it as:

What are criteria which might make us want to add axioms to our set theory?

There are three broad cases I can think of:

  • Accomodating informal set-theoretic arguments. Basically, the idea is that any "reasonable" argument in naive set theory should have a counterpart in the formal set theory. This was a driving force behind adding the axiom of replacement (due to Fraenkel); Zermelo set theory without replacement (ZC) is vastly weaker than ZFC. But this process appears to have ended with ZFC.

  • Solving mathematical problems. We might argue, for instance, that the Continuum Hypothesis should be resolvable in set theory. This would mean we need to add axioms, since our current set of axioms are insufficient. Similarly, a desire to develop a robust theory of projective sets leads to large cardinals. This is a major driving force in modern set-theoretic research, although whether it will ever result in a change of the underlying axioms is highly dubious (in my opinion).

  • Filling in holes. This is the closest analogy with the reals case. The idea is: if some set-theoretic object can be "well-approximated" by objects which our current axioms can prove exist, then we should be able to prove that that object exists itself. While large cardinals themselves can be viewed as a goal of this philosophy, I think it's maybe not the most exact instance available to us. The forcing axioms provide a robust example of this philosophy: basically, they say that the set-theoretic universe contains "highly generic" objects. Martin's Maximum can be viewed as a natural stopping point of this line of inquiry. Interestingly, forcing axioms and large cardinals are deeply intertwined.


I have to leave my computer unfortunately, but I'll add to this answer later tonight.

share|cite|improve this answer
    
@Hurkyl yes, ZC denotes ZFC without replacement but with choice. note that unlike ZF vs ZFC, there is a huge difference in consistency strength between ZC and ZFC (look at $V_{\omega+\omega} $). – Noah Schweber 1 hour ago
    
I've made the fix then – Hurkyl 59 mins ago

Going from the rationals to the reals makes the first-order theory of the ordered algebraic structures weaker: the first-order theory of the rationals is not decidable but the first-order theory of the reals is. You are confusing the logical strength of a theory with its apparent ontological requirements: in the example you give, the theory of $\Bbb{Z}[\sqrt{2}]$ can easily be interpreted in the theory of $\Bbb{Z}$. It is wrong to say that the theory of $\Bbb{Z}[\sqrt{2}]$ is stronger than the theory of $\Bbb{Z}$.

share|cite|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.