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In the ZFC set theory the "Axiom schema of specification" states that given an infinite set z, for any formula $\phi$ the subset $\{x\in z:\phi (x)\}$ always exists. My question is: Are there any other means for assuring the existence of subsets not specified this way? (See the Wikipedia article) https://en.wikipedia.org/wiki/Zermelo–Fraenkel_set_theory#3._Axiom_schema_of_specification_.28also_called_the_axiom_schema_of_separation_or_of_restricted_comprehension.29

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    Not really a problem with the question, but I would like to point out that the axiom schema does not require $z$ to be infinite, nor is the produced set necessarily a subset of $z$. – 2017-02-10
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    @Paul: the produced set _is_ necessarily a subset of _z_ . – 2017-02-10
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    Sorry, you are right. I was thinking of the existence of $\{\phi(x) : x \in z\}$, which is a different schema. At least I was right about $z$ not needing to be infinite. – 2017-02-10

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The well-ordering theorem or its equivalent the axiom of choice can do so. A classic is a Vitali set You define an equivalence relation on the reals in $[0,1]$ by $x \sim y \iff |x-y| \in \Bbb Q.$ This gives you uncountably many equivalence classes, each countable. Now the axiom of choice says there is a set having one member of each equivalence class. We can't write a formula $\phi$ to separate out that set.

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    Dear Ross, Thank you for your valuable answer and for giving a concrete example. The construction based on the Vitali set works nice on the reals. Now, could you show a construction on a countable infinite set _z_ ? – 2017-02-09
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    That is not quite correct. Separation *allows* parameters to be used. So in fact, every set can be obtained using separation by allowing a parameter to be used as the set itself. – 2017-02-09
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    Not to mention that obtaining a specific set happens in a model. And certainly there are models where there is a parameter free definition of a Vitali set (e.g. in $L$ there is one). – 2017-02-09
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    Dear Asaf, When you wish to specify the subset _s_ you can not use _s_ as a parameter. – 2017-02-09
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    @Zoltan: That is not true. You are working *inside* a given model. The set *already exists*. – 2017-02-09
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    @Asaf: What is the case, if we do not consider any given model, just the axioms of ZFC? The ZFC system assures the existence of an infinite set _z_ . What about the subsets of this set? – 2017-02-09
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    @Zoltan: It could still prove the existence of subsets, sure. And probably not all of them, although this isn't necessarily the case. Your question isn't new, it's a natural question most people run into when learning about set theory. But the answer to this question is far harder to formulate to someone who cannot answer it on their own. – 2017-02-09
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    @Asaf: Now, I formulate a concrete question: What is the guarantee for that all the (intuitively conceivable) subsets of the integers really exist? – 2017-02-09
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    @Zoltan: Exists *where*? In what model? – 2017-02-09
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    @Asaf: In any model of ZFC. – 2017-02-09
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    @Zoltan: Again, your question makes a "naive sense", but not much more. And answering it is impossible without getting technical about models, existence, and otherwise results related to the fine lines of theory and metatheory. – 2017-02-09
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    @Asaf: There exist only countable infinite $\phi$ formulae with any free variables. Using theese formulae it seems to be possible to create only a countable infinite number of subsets of the integers. How can the other subsets be created? – 2017-02-09
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    @Zoltan: And there are countable models of set theory. In those only countably many sets exist. – 2017-02-09
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The separation schema is not just for a single-variable formulas. Parameters are allowed. So given any subset $A\subseteq X$, it is an eligible parameter for defining a subset. Namely, $\phi(x,A)$ defined as $x\in A$ is a valid formula for a separation axiom. And of course that $\{x\in X\mid x\in A\}=A$.

You might want to ask whether or not every set can be obtained using a parameter free-formula, and now we can prove that it is consistent that there are sets which cannot be obtained that way. Of course, things are even worse, since the same set can have different definitions which depend on the ambient universe of sets in which we work. So it is possible that some sets (e.g. a Vitali set mentioned by Ross) does have a parameter free-definition in some universe of set theory, but not in another.

The question, ultimately, is what are we trying to do. If we want a definition which provably does something, then the answer is most likely negative; whereas given a set in a universe, we want to know whether or not its subsets are all definable without parameters (from a meta-theoretic point of view, at least), then this becomes a somewhat more concrete question, but now there are additional constraints as to whether or not the answer is positive or negative.

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    You are not allowed to define the subset _A_ using itself. – 2017-02-09