0
$\begingroup$

First to mention, I am just an undergraduate student with only very basic knowledge about logic and set theory. This is my general idea of how we define sets:

We first have first-order logic, which is like an "language"; it allows us to write down axioms to build up ZFC, then we have the notion of "set". Primitive notions are required in these steps. Similarly, we can define classes, which may not be sets.

Now, I want to build a new math object, which is not a set, but is not entirely independent from our ZFC set theory. (For example, given a set $A$, I want to define a math object $B$ such that $A\cup B=\varnothing.$ In this case, my new theory may need to extend the meaning of $\cup$ and $=$.) What are the steps that I have to go through, in order to build a new complete theory, and which probably doesn't contradict our original ZFC theory? What qualify my new theory to be a valid one?

P.s. Sorry if my question is unclear, I will like to clarify it if you spot any problem.

$\endgroup$
  • 1
    $\begingroup$ So if $B$ is not a set, how can you do a union (or any set operation) with $A$? $\endgroup$ – iiivooo Mar 28 '16 at 15:13
  • $\begingroup$ Presumably this is axiomatic. $\endgroup$ – Eric Stucky Mar 28 '16 at 15:14
  • $\begingroup$ @ iiivooo: So my new theory will need to extend the notion of union. $\endgroup$ – A. Chu Mar 28 '16 at 15:16
6
$\begingroup$

You are trying to have your cake and eat it too.

If you want the context of $\sf ZFC$, then every object is a set. Period. That's it, moving on.

You might want to read about $\sf ZFA$ which is $\sf ZF$ relaxed to allow non-set objects called atoms (or urelements). But in this context $\cup$ is not defined for non-sets, so $A\cup B$ when $B$ is not a set is not a well-defined expression (note that $\cup$ is not even in the language of set theory, it's a short hand for a formula which the axioms of $\sf ZFC$ prove has a unique solution to any two given sets).

Continuing with this, you can just define a new language in first-order (or whatever order) logic which includes two (or more) sorts of objects, sets and other types. Then you write the axioms for each object, with whatever relations and function symbols you've decided to add to your language (don't forget $\in$). You can also add interactions between the different sorts (like the union of a set and a non-set). In the case of $\sf ZFA$ we postulate that sets behave as they would in $\sf ZFC$, and you can do that also.

Then you're about done. Godspeed!

$\endgroup$
  • $\begingroup$ Generally the answer is simple. You want to improve upon set theory, you first need to study more set theory to understand why it needs improvement (and why your idea is a good one). After you've done that, often it is the case you realize set theory is great and more is less in that context. But in any case, you can't escape knowing more about logic and set theory if you want to do something like that. $\endgroup$ – Asaf Karagila Mar 28 '16 at 15:20
1
$\begingroup$

Your question seems to rest on the assumption that we need to "construct" these objects somehow. But your question also suggests that you know something about FOL and ZFC. In this context, a 'set' is nothing more than an object satisfying certain axioms, so it's not clear why you should expect your new objects ('anti-sets', perhaps?) to be anything more than that.

In short: there's nothing to define: you just have to figure out what you want your objects to do, write these as axioms, and then see if this ends up screwing everything up.

If it does screw everything up, there's a decent chance you'll notice fairly quickly.

But if it's not "obviously wrong" in this sense, and the theory turns out to be interesting, there's probably not much to say about consistency. Gödel informs us that self-consistent theories prove themselves as such, if they are sufficiently strong (and an extension of ZFC is certainly 'sufficiently strong' :P). So you would either have to invent yet another extension to prove consistency, or come to terms with the uncertainty.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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