Let's consider the "axiom 0" of ZFC: $\exists x (x=x)$

I "think to" it as "there exist something which is equal to itself, in other words there exist at least something".

But on the notes where I am studying I find: "There exists at least a set".

My question is: what is the correct interpretation? If the latter, why $x$ must be a set and can't be a proper class, for example?

Thanks in advance.

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    $\begingroup$ Proper class is not an object of ZFC. Rather, proper classes are predicates of ZFC. $\endgroup$ – Hanul Jeon Jan 29 '15 at 7:36
  • $\begingroup$ In ZFC everything is a set. The $x=x$ is just a dummy for any true statement, afaik. $\endgroup$ – Stefan Perko Jan 29 '15 at 8:33

In $\sf ZFC$ objects of the universe are called sets. Something exists if it is an object of the universe, so if an axiom of set theory says that something exists, we say that there is a set with such and such properties.

(The same it true for other theories, $\sqrt 2$ is a real number since it exists in the real numbers, where as $\sqrt{-1}$ is not a real number for similar considerations. The difference is that we have some "absolute" understanding of what is a real number, but we don't have such understanding regarding sets.)

So what does the axiom tells us? Really just that some set exists. The universe is not empty.

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    $\begingroup$ Does this not also follow from the axiom of infinity? $\endgroup$ – mrp Jan 29 '15 at 15:27
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    $\begingroup$ Yes. Also from many formulations of first-order logic that require the universe of discourse is non-empty. But repetition is not necessarily a bad thing when writing axioms. $\endgroup$ – Asaf Karagila Jan 29 '15 at 15:29
  • $\begingroup$ Presumably, the axiom coud be written $\exists x: true$ ? $\endgroup$ – Yves Daoust Jan 29 '15 at 15:31
  • $\begingroup$ @Yves: Yes, but in a language with just equality and $\in$ you don't have many options... :-) $\endgroup$ – Asaf Karagila Jan 29 '15 at 15:33
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    $\begingroup$ @Biagio: Yes, but you can say something like "There is no set which is equal exactly to the elements satisfying $\varphi$", in which case you say that $\varphi$ defines a proper class. Then when you say that $x$ is in that proper class, you actually say that $\varphi(x)$ is true. $\endgroup$ – Asaf Karagila Jan 30 '15 at 11:33

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