According to the definition, generally speaking, a structure $\langle A;R;F,C\rangle$ is such that $A$ is a non-empty set, $R$ is the set of relations, $F$ is the set of functions, and $C$ is a set of constants. For example $\langle\mathbb{R};; +,\cdot, ^{-1};0,1\rangle$ would be the field of the real numbers. Now, in the construction of the set theory (ZF), we need to make a structure. But then $A$ would be the set of sets, which is impossible... What am I missing?
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2I changed $<\mathbb{R};; +,\cdot, ^{-1};0,1>$ to $\langle\mathbb{R};; +,\cdot, ^{-1};0,1\rangle$. That is standard usage. – Michael Hardy May 25 '13 at 21:39
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1One note about the formulation of your question, ZF is not a structure. It's a list of axioms. – Asaf Karagila May 25 '13 at 21:50
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This is a very delicate point in set theory.
First of all we need to point out that the concept of "set" is not an absolute one. Different models of set theory, will think that different mathematical objects are indeed sets.
This is an external point of view. We consider the universes of set theory as sets, from the outside. The Russell paradox, and indeed all the classical paradoxes of set theory, state that the universe cannot be a set from its own point of view. This is where internal and external points of view are important.
Furthermore, if in a universe of set theory which satisfies the axioms of $\sf ZF$ and we can find a set $M$ and a relation $E$ such that $\langle M,E\rangle$ is a model of $\sf ZF$ then by the completeness theorem we have proved that $\sf ZF$ is consistent - in that particular universe. But from the incompleteness theorem we know that a theory like $\sf ZF$ cannot prove its own consistency, therefore we can never even prove that such set structure exists.
If $V$ is the universe of all sets that mathematics have, and suppose that it satisfies all the axioms of $\sf ZF$, then we know that (1) it is not a set itself; (2) we cannot prove that there is a set which is a structure satisfying the axioms of $\sf ZF$; (3) if there is such set, then we can talk about objects which are in that structure, which that particular structure thinks of as sets, and we can contrast them to actual sets.
This is close to the Skolem paradox, by the way.
In any case, as I said, this is a very delicate point and one has to study quite a lot of logic and set theory in order to understand it completely and become comfortable with it.
Asaf Karagila
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Very nice answer. Now, please tell me if I'm wrong. this doesn't mean that the "system" that generate the set theory in ZF is not a structutre, rigth? You only say that we cannot prove that such a structure is a set. But, more importantly, how do I call it if it is not a structure (Like when we are trying to generate a theory we begin saying that we are making a structure, what do we say instead if we are trying to generate the theory of sets)? – Daniela Diaz May 25 '13 at 22:29
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@Daniela: Internally the result of the generating "system" is never a set. That is, if we construct a universe of $\sf ZF$ using the von Neumann hierarchy (or other similar hierarchies), then the result is not a set. But if you begin with a countable model of set theory, and use forcing to add a new set to that model, the result is in fact a countable model of set theory. Generally, however, when we work internally to $\sf ZF$, we consider it as the whole universe of sets. So we call it "the universe". – Asaf Karagila May 25 '13 at 22:33
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@AsafKaragila When we are building classical logic (first and second order) we speak about "sets", right. Can we talk about such "sets" (that are not properly speaking the same as the ones generated by the "system" of $ZF$) as the "universe". Like saying I'm going to build a structure that we will call ZF:=$<A;;\in;\emptyset>$ along with the axioms...? In this case $A$ is a "set" in the "universe" that contain the new sets generated by the structure that we are defining. – Daniela Diaz May 25 '13 at 22:48
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@Daniela: This is directly related to my comment to your question. $\sf ZF$ is a list of axioms. It's not a structure (and also, $\varnothing$ is not part of the language). Mathematically speaking, something exists if it is in the universe. If we accept, for a moment, that the universe of mathematics is made of sets which obey the axioms of $\sf ZF$, then the universe itself is not a set as we know. We can prove however, that this universe can be generated as an increasing union of sets. But there is no actual object of the form $\langle V,\in\rangle$. It doesn't exist. – Asaf Karagila May 25 '13 at 23:01
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@AsafKaragila Thank you so much. Just one las thing. Could you suggest a good book, or books, to study this concepts. – Daniela Diaz May 25 '13 at 23:37
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@Daniela: That very much depends on how much do you already know about set theory (naive and axiomatic) and logic (first-order logic, including completeness and incompleteness). There has been several books requests threads about set theory on the site. – Asaf Karagila May 25 '13 at 23:55
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ok. I'll look for them. Thank you sooooo much for your answer. It make me ralize about many things that I was using wrongly. – Daniela Diaz May 25 '13 at 23:59
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