Why is Russell's Paradox not just wordplay?

  • #1
Suppose I have a set that contains all sets that it does not contain. Obviously no such set exists because the thing I just said makes no sense. I might as well say consider the natural number which is greater than zero and less than one. No such thing exists. So I don't understand why Russell's Paradox isn't just considered wordplay too - it defines a set that doesn't make sense, so obviously that set doesn't exist. It's like a self-contained reductio ad absurdum.

Now, here's another way of looking at things, which is sort of facetious. Say you have a shelf in a library that is just filled with indexes. Some of those indexes list themselves (for whatever reason). If you wanted to, you could make an index which lists every other index that does not list itself, and then you could either add the index to its own list or choose to leave it out. The index is then an actual existing object which contains the things it contains, regardless of how you choose to describe it. You could talk about a hypothetical index which lists every index that does not list itself, but that's just talk, because no such thing exists or could exist, and the fact that you could talk about such a thing does not undermine the existence or consistency of your shelf full of indexes in any way.
 

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  • #2
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It is indeed true that

[tex]\{x~\vert~x\notin x\}[/tex]

does not exist. But the question is, what sets do exist??

Before Russels paradox, it was said that every "collection of objects" was a good set. That means, if [itex]\varphi[/itex] is a property, then

[tex]\{x~\vert~\varphi(x)\}[/tex]

is a set. This is the comprehension principle. Of course, this leads to a contradiction with Russel's paradox. So apparently, not every collection satisfying some property is a set. But the question becomes: what is a set?? Can we make a definition/axioms of a set such that all of mathematics still holds but such that things like Russel's paradox not come up.

One such axiom system are the ZFC axioms. Here, the comprehension axiom is replaced by the seperation axiom. That is: for every set A, the set

[tex]\{x\in A~\vert~\varphi(x)\}[/tex]

is a set. Then Russels paradox does not occur. But does this absolve us from other paradoxes or contradictions?? We don't know and we will likely never know. Godels incompleteness theorem states that it is impossible to prove that ZFC does not have contradictions. But as of now, no contradiction has been found. So most mathematicians believe that ZFC is consistent (but we are unable to prove it).
 
  • #3
Hurkyl
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Suppose I have a set that contains all sets that it does not contain. Obviously no such set exists because the thing I just said makes no sense.
There is nothing nonsensical about the predicate "X contains all sets that X does not contain".

It is also true that there do not exist any sets satisfying the predicate, so it cannot be used to implicitly define a set.

I might as well say consider the natural number which is greater than zero and less than one. No such thing exists.
Again, there is no problem considering the notion of a natural number between zero and one. And as before, we discover there is no natural number satisfying that property.

So I don't understand why Russell's Paradox isn't just considered wordplay too - it defines a set that doesn't make sense, so obviously that set doesn't exist. It's like a self-contained reductio ad absurdum.
As before, there is nothing nonsensical. We can consider the notion of a set that contains only those sets not containing themselves. And by an easy proof by contradiction, we can show no such set exists.

The thing that is different, however, is that 'naive set theory' proves this set does exist, by invoking unrestricted comprehension. Thus, paradox.

Happily, Zermelo developed a set theory that lets us 'construct' most of the objects we would like to be able to construct, and restricts comprehension to be merely the axiom of subsets (aka the axiom of restricted comprehension), and is free of all of the known paradoxes of naive set theory. Fraenkel added in a couple other useful constructions, and ZFC (or other ZFC-like theories) has become the standard set theory.
 
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  • #4
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There is nothing nonsensical about the predicate "X contains all sets that X does not contain".
That's one point-of-view. Other people think it is nonsensical and explicitely forbid sentences like that. This is the approach in positive set theory or New Foundations.
 
  • #5
Hurkyl
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That's one point-of-view. Other people think it is nonsensical and explicitely forbid sentences like that. This is the approach in positive set theory or New Foundations.
Looking over the wiki page, NF doesn't forbid it: [itex]x \notin x[/itex] is a perfectly good predicate, as is [itex]\forall y: ((y \notin y) \Leftrightarrow y \in x)[/itex].

But the point holds in other sorts of contexts: e.g. multi-sorted theories where there are multiple 'types' of sets, or a more category-theoretic style of set theory where set membership isn't a globally defined relation.
 
  • #6
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Looking over the wiki page, NF doesn't forbid it: [itex]x \notin x[/itex] is a perfectly good predicate, as is [itex]\forall y: ((y \notin y) \Leftrightarrow y \in x)[/itex].

But the point holds in other sorts of contexts: e.g. multi-sorted theories where there are multiple 'types' of sets, or a more category-theoretic style of set theory where set membership isn't a globally defined relation.
In NF [itex]x\notin x[/itex] is not a stratified formula and cannot be used in comprehension. So NF does forbid that formula to be used in comprehension.
 
  • #7
Hurkyl
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In NF [itex]x\notin x[/itex] is not a stratified formula and cannot be used in comprehension. So NF does forbid that formula to be used in comprehension.
Saying that such a predicate cannot be used in the comprehension schema is a very different claim than saying isn't even in the language of NF.
 
  • #8
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Saying that such a predicate cannot be used in the comprehension schema is a very different statement than saying isn't even in part of the language of NF.
Of course I meant that it was forbidden in comprehension. It is comprehension that this entire topic is about. Surely you did not think that I said the language forbids the formula?? That would of course be false.
 

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