2
$\begingroup$

While reading up on ordinals I came across the following lines on Wikipedia:

Essentially, an ordinal is intended to be defined as an isomorphism class of well-ordered sets: that is, as an equivalence class for the equivalence relation of "being order-isomorphic". There is a technical difficulty involved, however, in the fact that the equivalence class is too large to be a set in the usual Zermelo–Fraenkel (ZF) formalization of set theory. But this is not a serious difficulty.

And more:

The original definition of ordinal numbers, found for example in the Principia Mathematica, defines the order type of a well-ordering as the set of all well-orderings similar (order-isomorphic) to that well-ordering: in other words, an ordinal number is genuinely an equivalence class of well-ordered sets. This definition must be abandoned in ZF and related systems of axiomatic set theory because these equivalence classes are too large to form a set.

These lines had no citation so I was unable to follow them back to a source, nor could I find any information regarding sets 'too large' for ZF. So what did the author mean by this?

I have not formally studied set theory so have minimal knowledge of ZF(C) and ordinals, so I apologise if this question makes no sense/has been answered already, I couldn't find any information on it.

$\endgroup$
  • 2
    $\begingroup$ I am certain that this appears here before. But I can't be bothered to look for that right now, and it might as well be quicker to write an elaborate answer. The answer, in short, is that some collections are definable in ZF, but provably do not form a set. These are called proper classes, and the informal analogy is that these are collections which are "too big" to be sets (like how $\Bbb N$ is too big to be a finite set). The class of ordinals is an example. But also the "naive" definition of an ordinal as an equivalence class over all sets. $\endgroup$ – Asaf Karagila Mar 28 '18 at 17:10
2
$\begingroup$

None of the axioms of ZFC allow you to create sets with arbitrary elements. This is a deliberate restriction to avoid Russell's paradox and related issues. The closest four we have are:

  1. The axiom schema of separation: Allows taking arbitrary subsets of any set.
  2. The axiom of the power set: Allows taking the power set (set of all subsets) of any set.
  3. The axiom schema of replacement: Allows taking the image of any set under any definable function (i.e. "replace each element of $S$ with [insert expression]").
  4. The axiom of infinity: $\mathbb{N}$ is a set and can be used as a "starting point" for the above operations (as well as the rest of ZFC).

With these and the rest of the ZFC axioms, we can construct any "typical" set, without being able to construct stranger objects like "the set of all sets." The fact that we cannot construct these sets would be insufficient to prove they are not sets, so the axiom of regularity explicitly forbids them.

When we claim that something is "too big" to be a set, we typically mean that it cannot be constructed by any sequence of applications of the axioms of ZFC, and it may also violate the axiom of regularity by containing an infinite sequence of elements-within-elements.

$\endgroup$
1
$\begingroup$

The collection of all sets (which I will call $V$) is the prototypical example of a collection of sets that is a proper class — that is, a collection that is not a set.

In some sense $V$ can be viewed as being larger than every set, since every other set is a subcollection of $V$; thus, the 'problem' with $V$ is often interpreted as being too large.

One can even adopt the axiom of limitation of size; that the only way for a collection of sets to be a proper class is if that collection can be put in bijective correspondence with $V$.

Since the possibility of bijective correspondence is how we define the notion of size of a set; we can thus interpret proper classes as being the same size as $V$; that is, too large to be a set.

$\endgroup$

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.