My background in set theory is that of a casual acquaintance that I would like to know become friends with (I am not sure set theory feels the same way). For my question, I would like to stay within the hospices of ZFC or something similarly strong (ETCS for instance). I can imagine a situation where we have a class (defined as a logical formula in set theory Let us call it $P$), we can define an equivalence relation on $P$ as follows as a new logical formula, $Q$ such that $$P(x)\wedge P(y)\wedge P(z)\implies \{Q(x,x)\wedge (Q(x,y)\Leftrightarrow Q(y,x))\wedge (Q(x,y)\wedge Q(y,z)\implies Q(x,z))\}.$$

Now I would like to say that "the collection of equivalence classes is a set" (not literally true since a set cannot contain proper classes) if $$\exists S \forall x P(x)\implies \exists s\in S: Q(x,s).$$ My question is "is my definition correct"? So the type of example (simplified a bit) we declare two sets to be equivalent if they are both finite or both infinite. The collection of equivalence classes is of size two, the class of finite sets and the class of infinite sets. But the collection of these two classes is not a set in the literal sense. My interest in this is to be able to phrase the statement " the localization of a category exists". Any references would be welcome.

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    $\begingroup$ Let's be friends. -- Set Theory $\hspace{4in}$ $\endgroup$ – Trevor Wilson Mar 8 '13 at 3:53

Your definition says, in simple words, that there is a set which is a system of representatives.

Indeed it is true that if a system of representatives forms a set then you can regard the equivalence classes as "small".

Some [possibly] relevant links:

  1. homeomorphism of topological spaces is an equivalence relation ?
  2. What can I do with proper classes?
  • $\begingroup$ This is essentially what I want. I wanted to make sure that this notion makes sense within ZFC or something weaker, perhaps. Thank you. $\endgroup$ – Baby Dragon Mar 8 '13 at 2:57
  • $\begingroup$ It looks like Scott's trick is what is needed to form the collection of equivalence classes as a class. $\endgroup$ – Baby Dragon Mar 8 '13 at 3:08
  • $\begingroup$ @BabyDragon: Yes. Although it can be helpful in the other case as well. You can simply claim that the collection of all the "Scott sets" is a set. $\endgroup$ – Asaf Karagila Mar 8 '13 at 3:10

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