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Instead of writing "$r\in\mathbb{R}$", we can write "$r$ is a real number". In the latter statement we are asserting $P(r)$ where $P(x)$ is the predicate "$x$ is a real number". It seems like sets give rise to predicates. How far can this be taken? Can I replace every mention of a set with a predicate and have an equivalent theory (to set theory)? If not, why not?

The reason for the question is this: the notion of "is in a set", to me, suggests something different than "is a". "is a" seems to just tell me that a certain object has a certain property or type (and we don't have to think about size). "is in a set" tells me that we have to think of all of those objects together in a certain place (and we do have to think about size).

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  • $\begingroup$ You might wish to Google "axiom of subsets" in ZFC. $\endgroup$ – GFauxPas Sep 14 '16 at 21:18
  • $\begingroup$ can you explain that comment? $\endgroup$ – fdshfdsk Sep 14 '16 at 21:30
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    $\begingroup$ I mean that Googling "axiom of subsets zfc" will give you websites that are relevant to your line of questions. $\endgroup$ – GFauxPas Sep 14 '16 at 21:40
  • $\begingroup$ It's a subtle and surprising fact that not every predicate is a set. I had a bit more to say about it in math.stackexchange.com/a/820785/25554 $\endgroup$ – MJD Sep 14 '16 at 21:46
  • $\begingroup$ But can every set be thought of as a predicate? $\endgroup$ – fdshfdsk Sep 14 '16 at 22:03
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While every set gives rise to a predicate, the reverse is not true. In particular, the predicate "$x$ is a set" does not correspond to a set (which would then be the set of all sets).

If you try to identify predicates with sets, you get naive set theory. But naive set theory is not consistent. For example, the power set of a set is provably larger that the set, but the power set of the set of all sets would consist only of sets, so it cannot be larger than the set of all sets.

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    $\begingroup$ If we don't use the notion of a set and instead use the notion of a predicate, then "$x$ is a set" is not even meaningful (it's not a predicate). $\endgroup$ – fdshfdsk Sep 14 '16 at 21:29
  • $\begingroup$ @fdshfdsk: Sure, if you just completely get rid of sets, you'll get no contradictions. But you'll also not get anything set theory gives you. Try for example to define a group without using sets. $\endgroup$ – celtschk Sep 14 '16 at 21:40
  • $\begingroup$ @celtschk Defining groups without sets: i.snipboard.io/alzQiG.jpg. 'range' and 'domain' can be likewise defined as predicates. The series of predicates can be combined into one by and-ing. $\endgroup$ – Joe Borysko Mar 10 at 7:14
  • $\begingroup$ @JoeBorysko: In that “set-free” definition, I see lots of $\in$. For example, “is a function” is given as $x\in\mathbb F$. I don't see how I should interpret that, if not as “x is an element of the set $\mathbb F$”. $\endgroup$ – celtschk Mar 10 at 7:59
  • $\begingroup$ There's a long logical tradition of trading in talk about objects, plural, for talk about a set of objects, singular. But you don't have to. You can keep plural talk, use a plural logic, and -- in particular -- treat a group as some objects (oe or more) equipped with a binary operation etc. etc. $\endgroup$ – Peter Smith Mar 10 at 18:13
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I'm not sure exactly what you mean by "predicate," but here's one potential pitfall: there are sets which are not "definable" in any nice way. Do these correspond to predicates? If by "predicate" you have in mind some description in a certainl language, I'd argue the answer should be no; in which case most sets do not correspond to predicates.


Re: celtschk's point that some predicates don't define sets, note that some "predicate"-like things can't even be predicates: consider the property "Is a predicate which does not apply to itself" (this is Russell's paradox). Basically, as soon as I have a notion of "set" or "predicate" or "property" or . . . , I'll be able to "diagonalize out of it": there will be some thing, definable in terms of that notion, which is not one of those.

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