# Independence of the comprehension axiom

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

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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