$ZF$ define membership by conditions demanding the existence of some constructable right-side-terms $M $ ($x \in M$). Is it meaningsful to ask for a categorical axiom system here? Shouldn't it be prooved, that any two relations $\epsilon$ and $\varepsilon$ which satisfying the axioms allways fullfill: $\forall x \forall M: x\epsilon M \Leftrightarrow x \varepsilon M$?
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$\begingroup$ Soft-question would have been a better choice, but I got the answer of my question. As I slipped and missed the point a bit I might rewrite in soft-question sometime. $\endgroup$– LehsCommented Aug 20, 2014 at 19:03
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Of course not, since you can always take a permutation of $M$ and redefine $\in$ as the transport of structure defined by the permutation.
But even if you mean "up to isomorphism", the answer is still negative. Since given a model of set theory, by usual compactness+Lowenheim-Skolem arguments we can produce a non-isomorphic model with the same cardinality (so we can assume, without loss of generality that they have the same underlying set).
If you allow second-order axioms, then models of the form $V_\kappa$ where $\kappa$ is a strongly inaccessible cardinal are exactly the models which satisfy the second-order replacement axiom, and then we can take $L_\kappa$ and it is the unique model (up to a unique isomorphism) which satisfies the second-order axioms of $\sf ZFC$ and $V=L$.
But of course, this requires large cardinals, and second-order logic has been called "set theory in wolf's clothing" before.
answered Aug 20, 2014 at 9:34
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$\begingroup$ So the first order definitions of set theory only provide a class of relations from which each mathematician make a choice $\in$ and hope that all other mathematician make the same choice (or doesn't think about it)? And it doesn't matter? $\endgroup$– LehsCommented Aug 20, 2014 at 9:57
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$\begingroup$ But a countable model of set theory can be characterized up to isomorphism by a single first-order sentence of infinite length, right? $\endgroup$– bofCommented Aug 20, 2014 at 10:07
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$\begingroup$ @bof: As is any countable model of a countable theory (I don't recall the reference for the theorem, but Scott analysis might be good keywords to start looking). $\endgroup$– Asaf Karagila ♦Commented Aug 20, 2014 at 10:59
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$\begingroup$ @Lehs: Almost, but not quite. We have axioms just so we can always ensure that certain properties are true, and then we work with these properties. This is exactly to ensure that other people who might interpret these notions differently still agree with you about core principles in a way which is relatively absolute between one person to another (meaning I can convince you why my proof work if we sit carefully enough and overview the details). $\endgroup$– Asaf Karagila ♦Commented Aug 20, 2014 at 11:03
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$\begingroup$ @Asaf Karagila: You are right. The "sets" in our minds are just mental models that stimulates the fantasy, but must be supervised and cleared by the formal rules. $\endgroup$– LehsCommented Aug 20, 2014 at 14:44