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Which one is more fundamental, Set theory or Axiomatic system? Which one can be defined without the other?


marked as duplicate by TravisJ, s.harp, Asaf Karagila set-theory Jan 24 '17 at 19:42

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When begining any mathematical theory, you need axioms, same is for sets. The very basics of Mathematics is true-false accounts and tautologies, and predicate counting of fist order (forall and exists) and after that you could do something.

  • $\begingroup$ But isn't predicate counting of first order somehow part of the set theory itself? Doesn't it look like a circular definition? $\endgroup$ – user650585 Jan 24 '17 at 17:29
  • $\begingroup$ No, it does not. Predicate counting are how exists and forall behave without sets let us say $(\forall x) \Longrightarrow (\exists x)$ and this x has some property P, which does not need to be enough to form a set, you can search that in elementar logic, there are counterexamples that not every property is coming to form a set. $\endgroup$ – nikola Jan 24 '17 at 21:01
  • $\begingroup$ Thanks @nikola for the reply. Would you please elaborate a little bit further on what you mentioned in your last comment? $\endgroup$ – user650585 Jan 26 '17 at 2:10
  • $\begingroup$ Forall and exists address to, well, some property P of given element $x$, if you tell that something holds for all $x$, then you have said that all of those $x$ have that property P. Now, let us have the sentence $(\forall x \in X) x \in x$, that is totally correct if you are relying on the predicate account exactly, but if you form a set with that property, you will end up in contradiction. $\endgroup$ – nikola Jan 26 '17 at 15:29
  • $\begingroup$ But when we say (∀x∈X) , aren't we talking about a set named X which already exist? I mean as soon as we talk about (∀x∈X) or (∃x∈X) we are already assuming the set X exists. Am I right? $\endgroup$ – user650585 Feb 6 '17 at 17:26

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