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What's wrong with the following line of reasoning, if so?

The comprehension axiom of Zermelo's set theory would be provable by the other axioms, if the following was provable:

($*$) All subclasses of a set are sets.

Then, especially all definable subclasses of a set were sets, which is essentially what the comprehension axiom says.

Since I assume that the comprehension axiom is not provable by the other axioms, ($*$) must not be provable. What does this mean? Are there models of set theory with subclasses of sets that are not sets?

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Now when you say the axioms of $ZF$, do you include Replacement? –  Asaf Karagila Jun 18 '11 at 16:27
    
Sorry, I didn't want to include Replacement, so I changed ZF to Zermelo's. –  Hans Stricker Jun 18 '11 at 16:46

2 Answers 2

up vote 1 down vote accepted

You can't translate your axiom "all subclasses of a set are sets" into a logical sentence of ZFC (with fist-order quantifiers, logical connectives, and $\in$). Since this sentence doesn't exist, you can't discuss its provability from the other axioms.

However you can probably translate it if you work with a richer language than the one of ZFC, like von Neumann–Bernays–Gödel set theory

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What would you say: Would there be only one model of set theory, if it was possible to translate "all subclasses of a set are sets" into the language of ZFC? But how could that be? Could "second order" help - like in the case of the induction axiom in PA? –  Hans Stricker Jun 19 '11 at 9:15
    
I accepted this answer for the very first sentence which made everything clear. –  Hans Stricker Jun 20 '11 at 19:16

At the informal level, sure. If ZFC has a model, it has a countable model. That countable model, from the external point of view, has only countably many sets.

Now look at the picture of $\mathbb{N}$ in this model. The class of all "real world" subsets of this is "really" uncountable. So most of these subsets are not sets in our countable model of ZFC.

Technical remarks: In connection with the countable model assertion, I should have mentioned the Lowenheim-Skolem Theorem. And one needs in addition to show that the countable model can be chosen so that its elements are sets, and the $\in$ relation of the model is the ordinary $\in$ relation. This can be done.

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