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In my (limited) experience, it is usually easy to see when something is large enough to be a proper class, by constructing an element of the class for every set.

However, sometimes such a proper class has lots of redundant information, so we consider it modulo some equivalence relation. This way, in some cases, we end up with something small enough to be (represented by) an actual set!

Motivating example: The Witt group $W(F)$

(actually, we don't care about the group structure, I'm sorry)

Let $F$ be a field. Let $$ W(F) := \{(V, q)\ \text{quadratic}\ F\!-\!\text{vector spaces}\}/\sim $$ Where $(V, q)\sim (W, r)$ if there exist metabolic (i.e., hyperbolic plus degenerate) spaces $E_1, E_2$ such that $(V, q)\perp E_1 \simeq (W, r)\perp E_2$, where $(V, q)\perp (W,r) = (V\oplus W, (v,w)\mapsto q(v)+r(w))$.

Then $W(F)$ is a set has a set of representatives.

When I first learned this, this was not at all obvious to me.

(To be honest, I only vaguely recall this, and am not too sure of the technical details. The gist of this was however, that we only consider finite-dimensional vector spaces in the first place, which allow only for a finite number of non-equivalent quadratic forms, and even after that, most of those can be seen to be equivalent to a quadratic space of lower dimension by adding elements to make a large portion look hyperbolic)

Question

So, in a similar spirit to this question:

What are other non-obvious examples of sets, e.g. proper classes being “quotiented” to a set?

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    $\begingroup$ The big one is localization in homotopy theory (or more broadly, (model) category theory). Entire structures are invented just to make sure that hom-sets are truly sets and not classes. $\endgroup$
    – Randall
    Jul 29 '19 at 14:57
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    $\begingroup$ Neat question! $\;\;$ $\endgroup$ Jul 29 '19 at 15:02
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    $\begingroup$ I don't know if this counts, since it is not an example of a class being quotiented, but in set theory one defines, for an infinite cardinal $\kappa$, $H(\kappa)=\{x\mid |\mathrm{trcl}(x)|<\kappa\}$ and it is not immediately obvious that this is a set. (One usually proves $H(\kappa)\subseteq V_\kappa$ to establish that) $\endgroup$ Jul 29 '19 at 15:02
  • $\begingroup$ @AlessandroCodenotti what is $\operatorname{trcl}(x)$? $\endgroup$ Jul 29 '19 at 15:07
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    $\begingroup$ You are confusing the quotient with "there is a set of representatives". $\endgroup$
    – Asaf Karagila
    Jul 29 '19 at 17:13
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Consider compact metrizable topological spaces, modulo homeomorphism there's actually set many such spaces.

One way to show this is to show that for every compact metrizable space $X$ there is an embedding $X\to [0,1]^\Bbb N$, so there are at most $|\mathcal P([0,1]^\Bbb N)|=2^{|[0,1]^\Bbb N|}=2^\mathfrak c$ such spaces.

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    $\begingroup$ You're confusing the class with its class of representatives. $\endgroup$
    – Asaf Karagila
    Jul 29 '19 at 17:11
  • $\begingroup$ That's why I'm taking spaces up to isomorphism @Asaf $\endgroup$ Jul 29 '19 at 17:18
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    $\begingroup$ Which means that every equivalence class is itself a proper class... $\endgroup$
    – Asaf Karagila
    Jul 29 '19 at 17:19
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    $\begingroup$ I see your point, all I want to say is that there is set many compact metrizable spaces up to homeomorphism, which is somewhat surprising $\endgroup$ Jul 29 '19 at 17:21
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    $\begingroup$ @FrancescoBilotta: Because there is a universal compact metrizable space. And in general, when you can easily bound the cardinality of an object, then there are only "set many options" to have that sort of object, up to the relevant isomorphisms. And in the case of compact metric spaces, it's not hard to show they are separable, and therefore have size of at most $2^{\aleph_0}$. $\endgroup$
    – Asaf Karagila
    Jul 30 '19 at 7:29
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Based on the comments (in particular, when I say "isomorphism classes of ... form a set", what I really mean is "there is a set such that any ... is isomorphic to one element of this set"; or I could argue that what I say makes sense using Scott's trick : this all should be equivalent anyway), here are some examples :

1- Given a cardinal $\kappa$, a first order language $L$ and a theory $T$ in $L$, the isomorphism classes of models of $T$ form a set. An interesting special case is when you take $L$ to consist entirely of function symbols and $T$ a list of universally quantified equations; then knowing that this is a set actually helps in one of the usual proofs of the Birkhoff variety theorem.

It's important to note that I restricted to a first order language, but actually there's no need for symbols in $L$ to be finitary, as long as $L$ is a set (and so its symbols have bounded (possibly infinite) arity), this works and the proof is the same.

2- If you include "second countable" in the definition of manifold (I don't know how consensual that is, my teachers did it that way), then diffeomorphism classes of manifolds form a set. This can be interesting to know in some considerations where we have functors defined on $\mathrm{Diff}$, and so to deal with functor categories, knowing that this category is essentially small is interesting (e.g. to see that prestacks on $\mathrm{Dif}$ actually form a category, or cobordism categories)

3- Given a cardinal $\kappa$, homeomorphism classes of compact Hausdorff spaces of cardinality $\leq \kappa$ form a set. This follows because the topology on a compact Hausdorff space is completely determined by the $\lim$ function $\beta X\to X$ ($\beta X$ is the set of ultrafilters on $X$). I don't know how useful that is, but I used it (rather a trivial variation of it) in a bachelor project to prove the existence of a universal minimal flow of a topological group.

4- (part of ) 1- and 3- generalize cleanly to monadic categories : whenever a category of objects is equivalent to $\mathbf{Set}^T$, the Eilenberg-Moore category of $T$-algebras for a certain monad $T$, then for any cardinal $\kappa$ there is a set of isomorphism classes of objects of this category of cardinality $\leq \kappa$. In 1-, $T$ is the monad associated to an algebraic theory, in 4- $T$ is the ultrafilter monad.

One can modify slightly 4 (or 1) to get that isomorphism classes of "finitely generated stuff" usually form a set : $R$-modules for any ring $R$ for instance.

5- Homeomorphism classes of separable metric spaces (in fact, isometry classes of separable metric spaces) form a set. In fact, with the ultrafilter trick as above, given a cardinal $\kappa$, homeomorphism classes of Hausdorff spaces that have a dense subset of size $\leq \kappa$ form a set.

6- As Randall pointed out in the comments, given a model category $C$ and objects $X,Y\in C$ it is not clear at all that $\hom_{\mathbf{Ho}(C)}(X,Y)$ should be a set. In fact if we don't take a model category but just a category with weak equivalences and localize, in general these don't form sets. But in a model category, fibrant and cofibrant replacements and various lemmas ensure that $\hom_{\mathbf{Ho}(C)}(X,Y)$ is (isomorphic to) a quotient of a hom-set in $C$ and is therefore essentially a set.

7- An example that is not "isomorphism classes of", that is also a bit weird, but that has some interesting (philosophical) implications : the class of all sets $x$ such that $\neg\mathrm{(Fermat's\, last\, theorem)}$ is a set (known since the 90's only !)

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