Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am trying to understand the axioms of ZFC. The axiom schema of specification or the comprehension axioms says:

If A is a set and $\phi(x)$ formalizes a property of sets then there exists a set C whose elements are the elements of A with this property.

I read Is the Subset Axiom Schema in ZF necessary? and understand that ZFC axioms are defined in the language first-order logic. I do not understand the part 'phi(x) formalizes a property of sets'. If we allowed any property, e.g. $\phi(x) = x \notin x$, then we would get a Russell's paradox. But saying any property except $\phi(x) = x \notin x$ would not eliminate all the paradoxes. Thus we have to have some way to say which properties $\phi(x)$ do not make ZFC inconsistent.

Is my understanding correct? What does it mean to 'formalize a property of a set'? How do we know that a property $\phi(x)$ does not make ZFC inconsistent?

share|improve this question
    
if by consistent with ZFC you mean that it can be used to specify a subset of a set then yes. However, it's not common to use 'consistent with ZFC' here. Usually one says about a sentence in the language of sets that it is consistent with ZFC to mean that ZFC + that sentence does not imply a contradiction. –  Ittay Weiss Dec 31 '12 at 8:21
    
Or more precisely, if ZFC is consistent, then ZFC + that sentence does not imply a contradiction. –  David Toth Dec 31 '12 at 8:30

3 Answers 3

up vote 7 down vote accepted

To formalize a property of a set simply means to give a formula in the language of set theory. That is $\phi(x)$ is nothing but a formula in the language.

Now, for each such property of sets $\phi(x)$ there is an axiom that says that given any set $A$ the set $\{x\in A\mid \phi(x)\}$ exists. This is the axiom schema of specification. Russele's Paradox does not creep in here since you still can't form the set $\{x\mid x\notin x\}$ not since you can't formulate the property $\phi(x)=x\notin x$ (you certainly can) but simply since using the axiom schema of specification will only allow you to create such sets as $\{x\in A\mid x\notin x \}$. By the axiom of foundation this set is just the empty set.

share|improve this answer
1  
Regularity tells you that $\{x\in A:x\notin x\}=A$, not $\varnothing$. –  Brian M. Scott Dec 31 '12 at 15:36

As mentioned above, by "$\varphi(u)$ formalizes a property of sets" we just mean that $\varphi(u)$ is a formula in the language of set theory. Fortunately, virtually all of mathematics can be formalised within the language of set theory, and so as long as you stick with mathematically meaningful properties, you can use Comprehension to construct subsets of any given set. Unfortunately, Comprehension could not allow you to construct the set of all happy horses (until you have given a mathematically meaningful definition of a happy horse).

The ZFC axiomatization side-steps the issue of Russell's paradoxical set (as opposed to simply outlawing its formation as was done in the Russell theory of types, and Quine's New Foundations).

It is done in the following manner: Naive set theory may be thought of as essentially axiomatizaed by the Axiom Schema of Unrestricted Comprehension:

For any formula $\varphi (u)$ (with no free occurrences of the variable $y$), the universal closure of $$( \exists y ) ( \forall u ) ( u \in y \leftrightarrow \varphi (u) )$$ is an instance of Unrestricted Comprehension.

(We disallow free occurrences of $y$ in $\varphi (x)$ so that this new set isn't defined as the collection of all objects in some relation to this heretofore undetermined set; otherwise we could take $\varphi (x) \equiv x \notin y$, and get a different paradoxical set: $y = \{ x : x \notin y \}$.) That is, for any set theoretic property $\varphi(u)$ that you can think of, there is a set whose elements are precisely all those objects $u$ satisfying $\varphi (u)$. As an example, there is a universal set, given by taking $\varphi(u) \equiv u = u$: $$V = \{ u : u = u \}.$$ Russell's paradoxical set is given by the formula $\varphi (u) \equiv u \notin u$: $$R = \{ u : u \notin u \}.$$

ZF(C) takes the stance that we cannot use Comprehension to demonstrate the existence of inherently new sets, but only the existence of sets which are definable subsets of already given sets:

For any formula $\varphi (u)$ (with no free occurrences of the variable $y$), the universal closure of $$( \forall x ) ( \exists y ) ( \forall u ) ( u \in y \leftrightarrow ( u \in x \wedge \varphi (u) ))$$ is an instance of Comprehension.

Let us naively try to use Comprehension to get Russell's paradoxical set. Given any set $X$ we may form the set $$X^\prime = \{ u \in X : u \notin u \}.$$ However, if it happens that $X \notin X$, then $X \notin X^\prime$ (since $X^\prime$ only contains elements of $X$), but $X$ itself is a set which would be an element of Russell's paradoxical set $R$. So for any set $X$ which does not contain itself as an element, the associated $X^\prime$ cannot be Russell's paradoxical set. As luck would happen, another axiom of ZFC (the Axiom of Regularity) immediately implies that no set contains itself as an element, which then implies that we cannot use the Axiom of Comprehension with the formula $\varphi(u) \equiv u \notin u$ to prove the existence of Russell's paradoxical set.

Note that because of Regularity under ZF(C) we have $R = V$, and it can easily be shown that ZF(C) proves that $V$ is not a set: if $V$ were a set, then by definition $V \in V$, contradicting Regularity!

Of course, ZF(C) could turn out to be inconsistent, and would then prove the existence of Russell's paradoxical set (and still also disprove its existence). Life would be easier if we could determine exactly which formulas are unproblematic in applications of Comprehension (or Replacement), but life isn't always easy.

share|improve this answer
    
There is nothing inherently paradoxical about your set $X'$. You can prove by contradiction that $X\notin X'$, but I don't think you can obtain any further contradictions. You can obtain only $X\in X \vee X\notin X$. You don't need to ban self-membership, as ZFC does, or any axioms of set theory to prove the non-existence of the Russell set. You need only first-order logic. –  Dan Christensen Jan 2 '13 at 7:14

As far as I know (as an amateur mathematician), if you allow set self-membership in $\phi(x)$ the only contradiction you will get is perhaps a contradiction of other axioms of ZFC, e.g. Regularity. Funny things can happen if you allow references in $\phi(x)$ to $A$ -- e.g. all sets must be empty -- so you want to avoid that. With only this restriction on $\phi(x)$, I think you should be able to avoid the known contradictions of naive set theory without citing Regularity or somehow banning self-membership.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.