0
$\begingroup$

Sometimes I see statements such as

Regarding what's required for a complete foundation, it's at least 99.9% correct to say set theory is enough.

I don't understand what that is supposed to mean, as this implies "we don't need more (of something)". More of what? Math obviously needs more axioms, e.g. the axioms of vector spaces.

The ZFC axioms postulate the existence of $\varnothing$ and state other things. On the other hand, the axioms of vector spaces postulate the existence of $0$ and state other things. Both sets of axioms are just sets of sentences in logic.

Aside from arguably being more widely used, what, if anything, makes the axioms of set theory special?

$\endgroup$
6
  • 1
    $\begingroup$ You're misunderstanding the role of the ZFC axioms. The idea is that the ZFC axioms describe the whole mathematical universe; vector spaces, for example, are objects in this universe. So e.g. we'd expect the ZFC axioms to prove, appropriately phrased, "Any two vector spaces over the same field with bases of the same cardinality are isomorphic." (Which they do.) $\endgroup$ – Noah Schweber Nov 9 '20 at 2:29
  • 3
    $\begingroup$ The “axioms of vector spaces” are not axioms in the same sense as the axioms of set theory; instead, they are really part of a “definition” of what we will call a vector space. They can be described entirely in terms of sets, functions of sets, and their properties. So we are really just saying “We will call $(V,k,+,\cdot)$ a vector space provided that $V$ and $k$ are sets, $+$, $\cdot$ are functions, and they satisfy the following properties.” So, in principle, you don’t “need” the axioms of vector spaces if you have those of set theory. $\endgroup$ – Arturo Magidin Nov 9 '20 at 2:29
  • 1
    $\begingroup$ By contrast, we never think of the vector space axioms as describing the whole mathematical universe. They're doing a totally different thing. $\endgroup$ – Noah Schweber Nov 9 '20 at 2:29
  • 1
    $\begingroup$ It's a good thing you asked a second question, so people could repeat what was said the first time... $\endgroup$ – David C. Ullrich Nov 9 '20 at 2:32
  • 3
    $\begingroup$ @MaxB: ZFC does not define set or membership; those are primitive notions. The empty set is a defined concept. The point is that you can define vector spaces and their properties in set theory (much like “Peano’s Axioms” are theorems of set theory, not axioms), but you can’t define sets and their axioms in an axiomatic linear algebra using what you are thinking of as “axioms of vector spaces”. As such, we don’t need “the axioms of vector spaces” if we have the axioms of set theory, which was, after all, your claim. $\endgroup$ – Arturo Magidin Nov 9 '20 at 3:08
5
$\begingroup$

You're misunderstanding the role of the $\mathsf{ZFC}$ axioms. Their goal is to basically be a "once-and-for-all" framework for doing mathematics; rather than describe any particular mathematical object necessarily, the idea is that they should describe the whole universe of mathematics. This is a universe which, among other things, contains vector spaces - so one of the things the $\mathsf{ZFC}$ axioms should be able to do is prove things about vector spaces. By contrast this just isn't something the vector space axioms are intended to do: we don't e.g. ask "Can the vector space axioms prove Fermat's last theorem?"

The $\mathsf{ZFC}$ axioms therefore need to be considered along with general implementation strategies - ideas for how a priori non-set-theoretic mathematical concepts can be implemented in set theory. These strategies - or rather strategy really, there's never much of a trick to it, it's just really tedious - mean that we can express all of mathematics in the language of sets alone in a very real sense.

Having fixed a way of expressing mathematical statements in the pure language of set theory, we now want to actually prove/disprove them. The success of $\mathsf{ZFC}$ is measured in large part by its ability to prove things about other mathematical objects, implemented according to the previously-mentioned strategies, which we can prove in "naive mathematics" already. This is what the quote is referring to: while there are indeed mathematical questions which the $\mathsf{ZFC}$ axioms do not resolve, they are few and far between (outside of mathematical logic anyways :P).

$\endgroup$
14
  • 3
    $\begingroup$ @MaxB No, you don't. For example, the sentence "Every vector space has a basis" can be formulated in terms of set theory alone, and proved just from the ZFC axioms. The point is that the conditions for being a vector space can be boiled down to set-theoretic properties alone, so you don't need to consider a separate set of axioms - you just put everything inside $\mathsf{ZFC}$. $\endgroup$ – Noah Schweber Nov 9 '20 at 3:03
  • 1
    $\begingroup$ This also addresses issues re: the limitations of first-order logic; see e.g. here. Essentially, trying to express or prove things about algebraic structures in a first-order way within those structures is often impossible; but we can do (almost) everything satisfactorily inside the first-order theory $\mathsf{ZFC}$. $\endgroup$ – Noah Schweber Nov 9 '20 at 3:13
  • 2
    $\begingroup$ @MaxB At this point I don't understand what you're asking. Let's start here: do you agree that there is a sentence, purely in the language of set theory, which intuitively corresponds to "Every vector space has a basis?" $\endgroup$ – Noah Schweber Nov 9 '20 at 3:28
  • 1
    $\begingroup$ Here's the point: there is a pretty straightforward (if really tedious) strategy for turning a natural-language mathematical statement, like "Every vector space has a basis," into a sentence purely in the language of set theory - with no additional symbols/notions - which intuitively has the same meaning. The vast majority of the translations, in this sense, of statements arising naturally in mathematics (outside logic anyways) are then provable or disprovable from the $\mathsf{ZFC}$ axioms alone. In this sense, $\mathsf{ZFC}$ suffices for the vast majority of mathematics. $\endgroup$ – Noah Schweber Nov 9 '20 at 3:34
  • 1
    $\begingroup$ @MaxB That's the cool bit: it turns out that you can! As my comments/answer allude to, there is a general strategy for doing this, and moreover the $\mathsf{ZFC}$ axioms prove almost everything we'd want to in these contexts. This is what the quote is about: in a very real sense, (almost) all of math can be done just inside $\mathsf{ZFC}$. $\endgroup$ – Noah Schweber Nov 9 '20 at 3:51

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.