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Russell, with his paradox, proves that the set $\{x:x\notin x\}$ of all sets that are not members of themselves doesn't exist. So, he demonstrates that the set $\{x:p(x)\}$ doesn't exist necessary (it will not exist if $p(x)$ leads to a contradiction).

But Zermelo answers (axiom of subsets):

If the set $A$ exists, and if $p$ is a predicate, then the set $\{x \in A : p(x)\}$ of all elements in $A$ satisfying $p$ also exists.

According to axiom of subsets, the set $W = \{x \in \mathbb{R} : x \notin x\}$ exists. But I don't see what this set is. For instance, do we have $\pi \in W$ (i.e $\pi \notin \pi$)?

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If you define the reals by Dedekind cuts or equivalence classes of rational Cauchy sequences, you'll always have $x\notin y$ for any real numbers $x,y$. –  Olivier Bégassat May 9 '12 at 11:02

3 Answers 3

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You need to remember that in a universe of set theory everything is a set. In particular real numbers are represented by sets.

Furthermore in the Zermelo-Frenkael set theory, $x\notin x$ for every set $x$, so $W$ as you defined it is simply $\mathbb R$.

If you take Zermelo's original set theory which allows sets of the form $x=\{x\}$, it is possible to construct the real numbers in such way which allows for some (perhaps all) real numbers to be of this form. In this case $W$ might be a proper subset of $\mathbb R$ - even empty.

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Thanks for the answer. –  Wondrous May 9 '12 at 11:15

This depends on how you define $\mathbb{R}$. You can say that real numbers are not sets and so $W$ will be $\mathbb{R}$ trivially, since a number can't contain itself; but even if you define the real numbers using sets you'll end up finding that $x\notin x$ for every real number. This is not a paradox since we don't expect $W$ to be a real number, so it has no reason to belong to itself in the first place.

What the "axiom of subsets" (usually called the separation axiom schema) actually proves with Russell's paradox is this: That there can be no "universal" set containing all sets, since using the separation axiom we can build from this universal set the paradoxical set of Russell.

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Thanks for the answer. –  Wondrous May 9 '12 at 11:16

This doesn't answer your question directly, but you should note that for any sets $r$ and $w$, if $w = \{x \in r : x \notin x \}$, then it is easy prove $w \notin w$ and $w \notin r$ using only the rules of first-order predicate logic. That is, these ideas are independent of any set theory or interpretation of $\in$ as anything other than an arbitrary binary relation.

See my formal proof at: http://www.dcproof.com/informant.htm

So, however you may define $\in$ on $R$, whether some, all or no real number is an element of itself, $w = \{x \in R : x \notin x \}$ would exist. Furthermore, we would also have $w \notin w$ and $w \notin R$.

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First, note that you really did not answer the question here which was essentially asking why can we even write $\pi\notin\sqrt 2$, as "real numbers are not sets" (although in set theory they might be represented as sets); the second and more important point is that you already decide the semantics of $\in$ in your answer which is exactly what the axioms of ZF tell us about. Remember that $\in$ is just a symbol, we can interpret it differently than set membership if we want. Of course we would never want that... :-) –  Asaf Karagila May 20 '12 at 7:02
    
Yes, it was more a comment than an answer. I just wanted to point Oout these ideas are independent of any set theory. In FOPL You can easily prove: ALL(w):ALL(r):[ALL(a):[a @ w <=> a @ r & ~a @ a] => ~w @ w & ~w @ r] –  Dan Christensen May 20 '12 at 13:01
    
Yes, it was more a comment than an answer. I just wanted to point out that these ideas are independent of any set theory or interpretation of $\in$. –  Dan Christensen May 20 '12 at 13:13
    
But if I have no axioms, and I interpret $\in$ as the "entire" binary relation, that is for all $x,y$ we have $x\in y$. In such case no matter how you want to twist and turn, $w\in w, r\in r, w\in r$, etc. so it does depend on the interpretation of $\in$. –  Asaf Karagila May 20 '12 at 13:25
    
I have fixed up my original reply. I hope this addresses your concern. –  Dan Christensen May 20 '12 at 13:44

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