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The version of Cantor's notion of sets that I've come across goes something like this:

"...collection of well defined, distinguishable objects of our intuition or of our thought to be conceived of as a whole. The objects are called the members of the set..."

With Russell's paradox $B = \{x:x \notin x\}$, I understand the mistake is assuming the collection $B$ is a set (i.e. if by sets we mean a collection which has the membership relation with its elements). The paradox shows not all collections are sets.

So far I haven't seen any paradox phrased like this: $B = \{x:x \notin B\}$? It seems slightly different from Russell's paradox in that the question isn't so much about whether $B$ is a collection which is also a set, but whether $B$ is a collection at all. Is this formulation allowed in Cantor's notion of sets, where it must satisfy the criterion of being well defined?

Thanks!

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    $\begingroup$ The basic problem here is: what does "is a collection" mean? How much must a collection be defined to be a collection? For $B$, for instance, it is impossible to consistently answer the question of whether $B \in B$; in what sense is such an object a collection? $\endgroup$
    – Ian
    Jan 3, 2015 at 3:51
  • $\begingroup$ Why $B = \{x:x \notin B\}$? Did you mean $B^c$? $\endgroup$
    – Gaston Burrull
    Jan 3, 2015 at 3:53
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    $\begingroup$ @VijayKonnur: Naive set theory has no problem with unions, or complements, or with an universal set. Indeed there are (widely assumed to be non-paradoxical) axiomatic set theories that have all three of these, such as Quine's theory NF. Naive set theory explodes when it encounters Russell's paradox, but writing down $\{x\mid x\notin x\}$ does not in particular depend on having a universal set. In ZF(C) in particular, the set builder notation is restricted such that you need to write $\{x\in A\mid x\notin x\}$, and then you reach a paradox if $A$ can be a universal set. (contd.) $\endgroup$
    – hmakholm left over Monica
    Jan 4, 2015 at 1:54
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    $\begingroup$ (..cont'd) But once you require $\{x\in A\mid\cdots\}$ rather than $\{x\mid\cdots\}$ you have definitely left naive set theory behind -- then you're doing some variant of Zermeloan set theory. $\endgroup$
    – hmakholm left over Monica
    Jan 4, 2015 at 1:56
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    $\begingroup$ @VijayKonnur: The meaning of "naive set theory" I assume we're speaking of here is inconsistent due to the paradoxes of Cantor, Russel, and more. As such it "has a problem", as you say, but this doesn't mean that you can point to any particular construction and say that naive set theory disallows it. Its basic characteristic is that it allows everything, and thereby it is inconsistent. Once you start adding restrictions (such as "no universal set") in order to avoid the paradoxes, the set theory you have is not naive anymore. $\endgroup$
    – hmakholm left over Monica
    Jan 4, 2015 at 18:32

2 Answers 2

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The basic problem here is that even in naive set theory you have no right to assume that $$ B = \{ x\mid x\notin B\} $$ works as the definition of $B$. Since the letter $B$ appears in the expression that supposedly defines its meaning, what you have is not a definition, but merely an equation that you want $B$ to be a solution of. And there's nothing paradoxical about the fact that this equation happens not to have any solution, any more than it is paradoxical that $$ x=x+1 $$ fails to define a number $x$.

Even if we look at the non-contradictory case, $$ C = \{ x \mid x \in C \} $$ that too fails to define anything because it just says that $C$ equals itself, which does not single out any particular set among all the other sets that also equal themselves. Again, this is not mysterious, but the same as the fact that $$ y = 1\cdot y $$ fails to define a number $y$.

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  • $\begingroup$ Based on this answer and comments, I can see I was confused between what's a paradox and what's a definitional issue (namely, that paradoxes aren't ones emerging from illegal definitions). Russell's paradox is based on a proper definition. Thanks! $\endgroup$
    – Stranger
    Jan 3, 2015 at 5:03
  • $\begingroup$ @GastónBurrull: No, the empty set is not a solution either. $\endgroup$
    – hmakholm left over Monica
    Jan 3, 2015 at 18:54
  • $\begingroup$ oops youre right $\endgroup$
    – Gaston Burrull
    Jan 4, 2015 at 1:43
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$B = \{x:x \notin B\}$ is an equation rather than a definition, and an equation without solutions.

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