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Fix a set theory, say ZFC. Now suppose you have a statement $P(x)$ written in the language of that theory, which has precisely one free variable $x$. We can conceive of the class of all $x$ in the domain of discourse such that the statement $P(x)$ holds. It is my understanding that this class may or may not itself correspond to an object in the domain of discourse. My questions are as follows. Firstly, suppose that it does correspond to an object in the domain of discourse, well what do we gain by proving that this is the case? And secondly, if the class of all such objects does not correspond to an object in our domain of discourse, what are we NOT allowed to do with this class which we would perhaps like to be able to do?

Now firstly, let me apologize in advance if my conceptualization above is either imprecise, or founded on false assumptions. My experience is limited to naive set theory, and I have no model theory under my belt.

Secondly, let me provide some background context for my question.

Suppose I have a unary function $f$ and I want to speak of the set of all sets that are closed with respect to $f$. We might write this as $\mathrm{clo}(f)$. My understanding is that even if $\mathrm{clo}(f)$ is defined only on 'small' functions $f$ (that is, those functions that are defined by a set of ordered pairs), nonetheless $\mathrm{clo}$ is going to be 'big'. Is this going to be a problem?

Now things get messier. I want to define a binary function $\curvearrowright$ such that for any object in the domain of discourse $x$ and any function $f$, we may write that $x \curvearrowright f = f(x)$. However, I don't want to limit this to small functions $f$. For instance, letting the object $x$ be replaced by a small function $g$, and letting $f$ be replaced by the large function $\mathrm{clo}$, I want to be able to write $g \curvearrowright \mathrm{clo}$ for the set of all sets that are closed with respect to $g$. Can this be done without introducing contradiction?

Yianni Mitropoulos
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    If a class is a set, then there are a lot more things you can do to it: you can form its powerset, take subsets of it, etc. As for your unary "function" $f$ – if $f$ is a "small" function, then $f$ has a domain of definition, so the set of all sets closed under $f$ is going to be a set. – Zhen Lin Aug 14 '12 at 02:09

2 Answers2

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Well, let us assume we are working in ZFC. In this case:

  • We are not allowed to quantify over classes which are not sets.
  • We cannot collect proper classes into other classes.
  • We cannot say that a class "exists".
  • We cannot use classes as parameters internally.

One of the main uses of set theory is to give a foundation to the rest of mathematics. If you cannot say that an object exists in your foundation, how can you delegate it to other theories? You can't. You can only talk about it as a syntactical construct, without any semantical meaning.

If you want to define the class $\operatorname{clo}(f)$ and $f$ itself is a class then your definition is not within the theory. Of course if you are not a set theorist it will probably interest you as much as it would interest you to know that I'm hungry right now. You should know, and you probably do, that there are ways around it like class-set theories or large cardinals.

Furthermore, if you define the operation for class functions and you want to talk about the collection of "all functions which have such and such closure" then you can only do that if these functions form a set. You cannot talk about a collection of classes, not in ZFC and not in NBG. You need to go another step to allow $2$-classes (classes of classes).

This form of iteration can go on for a while before it settles down and you find yourself with a large cardinal. It is somewhat necessary for some constructs in categories, but it is not always necessary otherwise.

Your $\curvearrowright$ function takes input of a function and a set. If the function is not a set, then $\curvearrowright$ cannot take it as input.

Asaf Karagila
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  • So would the best solution involve simply defining things as abbreviations? For example: If $f$ is a map (not necessarily set-like) with domain $X$ and if $x \in X$, we will use $x \curvearrowright f$ as shorthand for $f(x)$. If $f$ is a set-like map with domain $X$, we will write $f \curvearrowright \mathrm{clo}$ as shorthand for ${U \subseteq X ,|,\forall u(u \in U \rightarrow f(u)\in U)}$. – Yianni Mitropoulos Aug 18 '12 at 03:47
  • @Yianni: You'll notice I changed your "answer" to a comment. I also combined the account you used to write the question with the account you used to write the "answer" (did you know they were different?). – davidlowryduda Aug 18 '12 at 06:15
  • @Yianni: Yes, that is one solution. Another is to use some trick which allows you to have classes as sets (large cardinals, for example). – Asaf Karagila Aug 18 '12 at 07:16
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One thing that you can do with a set is put a poset structure on it and use Zorn's Lemma, which is a very powerful technique in proving the existence of certain objects. The following is a false application of Zorn's Lemma that illustrates its usefulness and the necessity of proving set-hood.

Suppose you want to prove that any field $F$ has an algebraic closure. Consider the collection of all algebraic extensions of $F$ and order this collection by inclusion. It is easily seen that all of the conditions for Zorn's Lemma hold and that a maximal element is an algebraic closure. The only problem is that the collection is not a set and thus this proof is wrong. It can be corrected but it requires a bit more work to actually produce a set.

Now, the following example shows that a very similar variant of the 'proof' above can be construed that 'proves' nonsense. Consider now the collection of all sets and order it by inclusion. Again, the conditions of Zorn's Lemma are easily satisfied and a maximal element now gives us a set with the property that it is not properly contained in any other set, obviously nonsense.

Another thing one can do with a set is use it to index other sets and then take the cartesian product of the indexed family. If one instead uses a proper class to index some sets then the cartesian product may or may not exist and determining if exists or not can be very difficult. This type of importance in distinguishing between sets and proper classes is very general. Indeed, given any small category it is well known that if the category admits arbitrary (i.e., class indexed and not just set indexed) categorical products (of which the cartesian product of sets is a special case) then the category degenerates to a poset. This argument thus shows that in any category which is not a poset (the category of sets is an example of a non-poset category) there must exist a proper class indexed family of sets which has no product in the category. In contrast, many categories (that of sets included) have the property that any set indexed family of objects has a product in the category (more generally, all small limits and colimits exist in many categories but if all limits, or all colimits, exist then the category must be a poset).

Ittay Weiss
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  • There are class-forms of Zorn's lemma which are equivalent to Global Choice. Global choice is the most non-issue in set theory (you can always force it without adding new sets). Furthermore, if you are using large categories you need a better solution anyway. – Asaf Karagila Aug 18 '12 at 07:20
  • @ Asaf I'm not sure how that relates to what I said. Global Choice doesn't make any difference to the examples I gave on how to misuse Zorn's Lemma if one does not pay attention to the set-hood of the class. As for big categories, what do you mean by 'better solution'? Certain big categories have all small limits but not all limits, showing again where the distinction between set and class is important, which is, I think, what OP asked about. – Ittay Weiss Aug 19 '12 at 06:03
  • Ittay, the class version of Zorn's Lemma asserts that if every chain is bounded then there is a maximal element. Note that if you consider the algebraic closure example, then the class has the same properties: every chain is bounded. Yes, there are class-many members but in fact every chain is of set-length, there are not even chains which are proper classes, so you can even give a medium-sized argument and not require the full blown class-form of Zorn's lemma which allows class-chains... – Asaf Karagila Aug 19 '12 at 13:43
  • But wouldn't the same be true about the class of all sets ordered by inclusion? Every chain of set-length is bounded, thus (???) there exists a set that is not properly contained in any other set (in ZF)? – Ittay Weiss Aug 21 '12 at 06:58
  • You are talking about $(V,\supsetneq)$ as an ordered class, or about $(V,\subsetneq)$? If the former, then note that a minimal element is $\varnothing$ which has no proper subsets; if the latter than it is not true, consider just the ordinals -- they form a class-length chain which has no upper bound. – Asaf Karagila Aug 21 '12 at 07:01
  • I'm talking about (V,\subseteq). You said that in the case of the algebraic closure, having set-length chains bounded suffices to establish an algebraic closure, I'm asking about that same for (V, \subseteq). There too set-length chains are bounded, but there is no set which is not a proper subset of any other set. – Ittay Weiss Aug 21 '12 at 11:22
  • I'm not sure why you are trying to use a class-form of Zorn's lemma to prove the existence of a minimal element. Note that a maximal element corresponds to "largest set". There is no largest set, but there are chains without an upper bound (yes, those are not set-length, but that is not important). The requirement is that every chain is bounded. Note that in $(V,\subseteq)$ this is clearly false. Whereas in the case of an algebraic closure every chain is bounded, and there are no chains of class-length at all. – Asaf Karagila Aug 21 '12 at 11:27
  • I'm not trying to prove the existence of a minimal element. I'm pointing to the fact that you can't deduce, in an ordered class, from the existence of bounds for set-length chains that there exists a maximal element. Now, what you claim about the case of algebraic closure that the only chains are set-length is true but I don't see how to prove it without already knowing that the algebraic closure exists. – Ittay Weiss Aug 21 '12 at 11:48
  • First, I claimed that you need the assertion that every chain in bounded. I merely remarked that in the alg. closure case it is sufficient to require every set chain is bounded. Secondly, your argument can be reduced to claiming "how can Zorn's lemma be used to prove the existence of an upper bound? We need to know there is an alg. closure to assert that". Note that in this aspect the proof is essentially the same: we just prove that the increasing union of alg. extensions is contained in an alg. extension. (cont...) – Asaf Karagila Aug 21 '12 at 11:51
  • (...) The only difference from the usual case is that we do not limit the underlying sets. This still doesn't matter because we require the chains to be increasing in inclusion, and since they are of set-length we can use the axiom of union to produce an upper bound. – Asaf Karagila Aug 21 '12 at 11:52
  • that will be fine if you can prove that in the class of algebraic extensions all chains are set-length. How do you do that? – Ittay Weiss Aug 21 '12 at 11:55
  • let us continue this discussion in chat – Ittay Weiss Aug 21 '12 at 11:59
  • First prove: algebraic extension over an algebraic extension is algebraic; there are only set many polynomials over your field. If you had a class-length chain, you would be able to find a proper class of polynomials. Now all chains are set-length; each bounded by its union. Class-form of Zorn yields a maximal element, this is a set by definition (otherwise it wouldn't be an element), and we can easily prove it is algebraically closed. – Asaf Karagila Aug 21 '12 at 12:22
  • Of course there is only a set of polynomials over a fixed field $F$. But what stops you from having a class-long chain of alg. extensions of $F$? With each extension there are more polynomials to split. So you might as well end up with a proper class of polynomials, just not all with coefficients in the same field. Of course this doesn't happen but how do you see that? – Ittay Weiss Aug 21 '12 at 12:31
  • You are aware that we are fixing both $F$ and the polynomials over $F$, right? Proving that algebraic extension of an algebraic extension of $F$ is itself an algebraic extension of $F$ suffices. It ensures us that even if the underlying set is different, we cannot be "that" far from $F$ in this aspect. There are only set many polynomials and only set many roots to each polynomial (actually finitely many), so there can only be set many new elements. Recall that for a collection here to be a chain, each is a proper extension of the other, so we have a proper class of roots, and polynomials. – Asaf Karagila Aug 21 '12 at 12:47
  • yes, of course F is fixed. But a priori it is not sufficient to prove that you can split all polynomials over F in some algebraic extensions. An alg. closure for F is an alg. extensions K such that every polynomial over K has a root in K. Not just every polynomial over F has a root in K. So you fix F indeed, but then with each extension you need to consider, potentially, more and more polynomials to split. – Ittay Weiss Aug 21 '12 at 12:52
  • Yes, but splitting new polynomial can be done by splitting old polynomial. That is the essence of the claim: algebraic extension of an algebraic extension is itself an algebraic extension. – Asaf Karagila Aug 21 '12 at 13:06
  • shall we continue in chat? – Ittay Weiss Aug 21 '12 at 13:11
  • I rather not. I have placed a self-imposed exile from the chat system here. I dislike it, a lot. If you're fed up with this argument via comments, we can remove it and call it a day. – Asaf Karagila Aug 21 '12 at 13:18