# 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?

• 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-Peter 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.
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.