# In class universe, on what basis can a conclusion that every element of a set is set be made?

Is every element in a set set? In a set model it is obvious true. However in the class universe it is another story since it need to be shown not a proper class.

Let $A$ be a set, $a \in A$.

Firstly consider about the Axioms of Comprehension. Of course $a= \{x \in \cup A|x \in a\}$. However $x \in a$ may not a formula in the language of sets since $a$ may not a constant symbol. But if we can find a formula $\varphi_a(x)$ in the language such that $\varphi_a(x) \iff x \in a$ then the Axioms of Comprehension can be spelt. Unfortunately this kind formula does not always exist if $A$ is uncountable whereas the language is countable.

Then the Axioms of Replacement. If $a=\emptyset$ it is obviously a set; else it contains at least one element, let $m$ be one of them. The being $f=\{(x,y)|x\in \cup A \land (x\in a \rightarrow y=x)\land(x \not \in a \rightarrow y=m)\}$, if it is a function then $a$ is a set by Axioms of Replacement. But note that $a$ and $m$ may not definable in the language hence there may no formula $\psi(x,y)$ to define $f$.

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If you’re talking about ZF(C), every object is a set; it’s a fundamental assumption. – Brian M. Scott Nov 30 '12 at 9:57

In the universe of sets all the objects are sets. In particular the statement $x\in y$ is saying that $x$ and $y$ are sets and the relation $\in$ as interpreted by the universe agrees that $x \in y$ is true (note that in this case $\in$ is used as a syntactic object whereas in the first use above it meant as a semantic object).

Therefore if the universe knows that $x\in y$ at all then it knows about $x$, therefore $x$ is a set.

Ah, but you could retort with the following question "But why can we write $\alpha\in\mathsf{Ord}$ if the ordinals do not form a set?", and indeed this is the right question to ask. The answer is that when we write that we actually write that $\varphi(\alpha)$ is true, where $\varphi$ is the formula describing the class of ordinals (i.e. transitive and well-ordered by $\in$). This is a step in the proof so we simply use the fact that $\exists x\varphi(x)$ and then we take $\alpha$ to be an instance of such $x$.

There is no semantic object corresponding to the ordinals, though. At least in ZFC. If one talks about class-set theories (e.g. NBG or MK) then this is different story. In those theories it is common to define sets as elements of classes anyway, so the question itself becomes trivial.

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I see, ordinal number class is definable on any set model of ZF but not in it(thus not a set). It sounds like nominalism... – Popopo Dec 1 '12 at 3:26
No, the class of ordinals is internally definable. However as the various paradoxes show us, not every definable collection is a set. – Asaf Karagila Dec 1 '12 at 8:00
What I meant is the class of ordinals is a definable subclass of the set universe, but not an element of the set universe. – Popopo Dec 1 '12 at 12:57
Your post seems really informative; unfortunately, I don't have the background to fully understand your comments. I'm in my fourth year of undergraduate math work, so in some sense I speak the language, but I lack exposure to this set theory/axiom stuff. Do you know of any good resources or texts that cover this material at an introductory (graduate) level? I'm not sure if I should be looking for a text in set theory or logic (or even how the two topics interact). – Josh Keneda Dec 31 '12 at 9:22
@Josh: There are several threads on the site about intro and advanced books on set theory and logic. I can recommend Kunen's book, and perhaps Just-Weese Discovering Modern Set Theory. Other viable options are Levy's Basic Set Theory and Jech's Set Theory. These are all intermediate texts, but seeing how you have some mathematical experience you might manage to wade through them. – Asaf Karagila Dec 31 '12 at 12:41