-1
$\begingroup$

Interesting reply to this question at Quora from Alex Sadovsky, Ph.D. Mathematics & Biomechanics, University of California, Irvine

Nothing. All of modern mathematics is described in the language of naive set theory. Every area of mathematics, whether category theory, representation theory, functional analysis, or algebraic geometry, is just a specialization. It does not introduce anything that could not be described by set theory.

Category theory specializes in Morphisms.

Comments from readers here?

$\endgroup$
6
  • $\begingroup$ I mean, categories can be (and are usually) described in terms of set theory, so clearly anything that can be encoded with categories can also be encoded with sets. $\endgroup$ – Captain Lama Apr 13 '20 at 15:37
  • 1
    $\begingroup$ “I recall that, at the 1963 meeting devoted to Logic, Methodology and Philosophy of Science in Jerusalem, Bill Lawvere proposed basing mathematics on categories rather than sets. Alfred Tarski, who was in the audience, objected: what is a category if not a set of objects and a set of arrows? Lawvere replied: set theory deals with the binary relation of membership, category theory with the ternary relation of composition. Apparently, Tarski was satisfied with the answer.” $\endgroup$ – fosco Apr 13 '20 at 16:00
  • 2
    $\begingroup$ Can you link to the Quora question? $\endgroup$ – Noah Schweber Apr 13 '20 at 16:25
  • $\begingroup$ @NoahSchweber Link added. $\endgroup$ – Dan Christensen Apr 13 '20 at 17:17
  • $\begingroup$ Please don't post "fwd from Quora" stuff here. If you have a question, ask the question. This is not a site to "get comments" or "encourage discussion". It's a Q&A site. $\endgroup$ – Asaf Karagila Apr 14 '20 at 13:23
6
$\begingroup$

The answer is right, and we can make this precise. As this answer says as well, there is the Elementary Theory of the Category of Sets (ETCS), which is in the language of category theory, and there is ZFC, the well-known axiomatisation of set theory. ZFC is really stronger than ETCS, the answer I linked goes in more detail, but point is that we can add axioms (a variant of replacement) to ETCS to make it in fact equiconsistent with ZFC.

Edit. From a comment from Noah Schweber: a survey of possibly-relevant weaker fragments of ZFC can be found here. I thought this might be relevant to point out here as well.

Edit 2. Upon reading the Quora answer again I should say that I agree with the technical part of the answer, namely that set theory and category theory can encode the same things. That was the point of my answer here. I do not agree that everything in mathematics is a "specialisation" of (naive) set theory. For example, I highly doubt many mathematicians view the natural numbers as the von Neumann ordinals.

$\endgroup$
24
  • $\begingroup$ If it would be necessary to add axioms to the ETCS in order to make it equivalent to the ZFC, does that not imply that the ETCS is more general and set theory is a special case with more restrictions? Which would mean that the answer is wrong, i.e. the ETCS can not be captured in the framework of set theory? $\endgroup$ – Felix B. Apr 13 '20 at 16:07
  • 3
    $\begingroup$ @FelixB. There's a difference between framework and specific theory. ZF lacks the axiom of choice; does that mean ZF can't be captured in the framework of ZFC? The point is that ETCS fits, up to appropriate equivalence, into a "context of theories" which ZFC-style set theory captures exactly. (A survey of possibly-relevant weaker fragments of ZFC can be found here.) $\endgroup$ – Noah Schweber Apr 13 '20 at 16:16
  • 3
    $\begingroup$ Although that said, I take huge issue with the "almost equiconsistent" bit - dropping replacement results in a gigantic loss of consistency strength. The right thing to compare ETCS to if I recall correctly is Z (or ZC). To see the drop in consistency strength, keep in mind that ZFC proves that $V_\lambda\models ZC$ for every infinite limit cardinal $\lambda$. This doesn't affect the broad point of the answer of course, since Z itself is a ZFC-style set theory, but I think it's worth noting. $\endgroup$ – Noah Schweber Apr 13 '20 at 16:18
  • 1
    $\begingroup$ @spaceisdarkgreen: You can construct a structure that is isomorphic to $(V_\omega,\in)$ in ZC. You just can't get a structure where the relation is actually $\in$ itself. So from the perspective of "ordinary" mathematics where you only care about studying structures up to isomorphism, you can do pretty much everything you might care about as long as you don't need any sets whose cardinality is too big (basically, anything that involves strictly increasing an infinite cardinality infinitely many times). $\endgroup$ – Eric Wofsey Apr 13 '20 at 19:15
  • 1
    $\begingroup$ @spaceisdarkgreen: Well, for instance, you can define a relation $R$ on $\omega$ such that $(\omega,R)\cong (V_\omega,\in)$, by encoding the combinatorics of hereditarily finite sets with numbers in any of the usual ways. $\endgroup$ – Eric Wofsey Apr 13 '20 at 19:21

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