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This question is probably very naive but it does bother me and I am not sure where to look for an answer.

Define $A = \{ 1,2,A \}$, $A$ can not be a set (Axiom of regularity). Can $A$ be a "class" or a "collection" of elements. Suppose it is. Don't we get the Russel's paradox again? Take the collection of all collections which does not contain themselves as an element. Does it contain itself as an element?

Is there a short answer or is there a lot of knowledge involved?

Thank you

Remark: I know that $A$ cannot be a set because of axiom of regilarity. But, what prevents it from being a class or a collection?

A student
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  • The "definition" of $A$ is not good, because it is redundant: $A$ is not defined, but you're defining something ($A$ itself) in terms of $A$. However, you can ask if there is some object (class, etc...) which has the form of $A$. In the theories of classes I'm aware of (NBG, MK), the limitation of size axiom implies that no such object is a class. Wikipedia also has something about "New Foundations" and "Semisets", which I don't know. Might be interesting to you: http://en.wikipedia.org/wiki/Class_(set_theory)#Classes_in_formal_set_theories – Luiz Cordeiro Sep 16 '14 at 15:00
  • Related: http://math.stackexchange.com/questions/850986/russells-paradox-with-bounded-comprehension/ (with a very short answer). – Asaf Karagila Sep 16 '14 at 15:32

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A class in standard (ZF) set theory generally coincides with a property that any set may or may not have. For instance, the property might be "$\phi(x) \equiv x \not\in x$". Then $\{x \;:\; x\not\in x\}$, while not necessarily a set, is a well-defined class. But note that the elements of a class are sets, not classes. In your case, you want to know whether there is a class corresponding to $A$. Such a class would contain exactly $1$, $2$, and a third set equal to the entire class. This could only occur if $A$ were itself a set, which you've already noted is ruled out by the axiom of regularity.

mjqxxxx
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Luiz Cordeiro already mentioned that you can't define $A$ like this, but as you are interested in the details of axiomatic set theory let me try to explain why exactly you can't do it.

You will know that in ZFC $\{a,b\}$ is just an abbreviation. You usually introduce the pairing axiom which says that for any sets $a$ and $b$ there's a (unique) set which contains exactly the elements $a$ and $b$. This unique set is then written as $\{a,b\}$ and it is obvious how this can be extended to one or (using union) three elements.

So, when we later write something like "let $A$ be the set $\{1,2,c\}$" we are actually applying a couple of axioms under the assumption that $1$, $2$, and $c$ are sets. What we should have said is: "We know that $1$, $2$, and $c$ are sets and by the axioms of pairing, union, and extensionality we know there's a unique set which has exactly these three elements. We'll call it $A$."

Now it should be clear why this fails for the attempted "definition" of your $A$ above. The step "we know that $1$, $2$, and $A$ are sets" doesn't work because at this point $A$ hasn't been defined yet.

Frunobulax
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  • Thank you for your answer. I know that $A$ can not be a set. But, how do i know that it can not be a "class" or a "collection". I mean, are there any axiom limitations about classes or collections? – A student Sep 16 '14 at 15:32
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    My point was not that $A$ cannot be a set. My point was that $A$ cannot be defined like this. You should probably rephrase your question (or start a new one) as: Can there be a class $A$ with $A \in A$. But as @mjqxxxx just pointed out at least in ZF the elements of classes are sets, so that already settles it. – Frunobulax Sep 16 '14 at 15:40
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Classes are really just "too big of a set" (at least in the context of separation axioms for every formula). So sets are really just "small classes". In particular, a class with three elements is a set.

The issue with defining $A=\{1,2,A\}$ is that the definition is circular. Just because you wrote something which looks like it has meaning, doesn't mean that it has meaning (e.g. "The table ate lunch, while the dogs policed the dust" is even syntactical, but has no real meaning).

Sets with certain properties cannot be defined explicitly. We can at best prove their existence is consistent with some axioms, then deduce that such set exists. For example you can't define $x=\{x\}$, but you can prove that it is possible that the axiom of regularity fails and such $x$ does in fact exists, then you can just pick one.

As for "collection", this is an informal notion which means either set, or a class, or something else, depending on the context. Since you haven't specified the context, it seems that you mean "set or class". So it's not a collection.

Asaf Karagila
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  • So,.., if it can be proved that there exists an $x$, such that, $x = { x }$, can we say that $x \in x$? And, if we can say that, let's take all items $y$ such that $y \notin y$. Don't we get Russel paradox again? Which of the moves I made here is the wrong one? – A student Sep 16 '14 at 15:46
  • If you consider all the $y$ such that $y\notin y$, then $x$ is not one of them. This class simply does not include all the sets; and if you only consider ${y\in x\mid y\notin y}$ then you get $\varnothing$. Is $\varnothing\in x$? – Asaf Karagila Sep 16 '14 at 15:48