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Russell's paradox: how is it resolved?

  1. Oct 22, 2006 #1
    well, i read about the russell's paradox recently. but what i was wondering is "how is the paradox resolved?"
     
  2. jcsd
  3. Oct 22, 2006 #2

    matt grime

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    By disallowing unrestricted comprehension. I.e. restricting what is allowed to be called a set. Things that are too large to be called sets are called (proper) classes.
     
  4. Oct 22, 2006 #3
    can you please give an example of what cannot be allowed to be called a set (to avoid the paradox)?
     
  5. Oct 22, 2006 #4

    Hurkyl

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    You are always allowed to construct sets like

    [tex]\{ x \in X \, | \, P(x) \}[/tex]

    (where X is some set, and P is some logical predicate) and

    [tex]\{ f(x) \, | \, x \in X\}[/tex]

    (where f is some logical function) but you aren't generally allowed to construct

    [tex]\{ x \, | \, P(x) \}[/tex]

    (And, of course, you can always do pairs, unions, and power sets)
     
    Last edited: Oct 22, 2006
  6. Oct 22, 2006 #5
    all of those went above my head. can you give some easier examples?
     
  7. Oct 22, 2006 #6

    matt grime

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    No. Nothing, a priori, is a set. All we know is that we have a notion of what we want a set to be, and the rules the collection of sets should obey. We also know what rules we can't allow sets to obey (Russell's paradox). Thus we have to sit down and write out a collection of axioms that we want sets to obey, and then think about what things can be sets *in some model* of the axioms. Thus some collection of objects might be a set in one theory and not in another. What all set theories do agree on is that there is an empty set, that is an object that satisfies x in X is false for all x. Apart from that you're on your own as to what is and what isn't a set.

    Some things cannot be sets in any model. For instance if M is some model of some set theory, the collection of sets in M is not a set in M. The class of all cardinal numbers is not a set in any theory. Indeed that is one way to show something is not a set: show it has 'cardinality' greater than kappa for any cardinal kappa.

    There are models of set theory where the real numbers as you know them are not a set.

    In short think of 'set' as a label that we can apply to some objects and not to others, depending on the axioms we have chosen.

    (Actually, this wasn't how Russell got round the issue, but his notion of higher types did not catch on as it is far too wieldy).
     
    Last edited: Oct 22, 2006
  8. Oct 22, 2006 #7

    Hurkyl

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    It's an algebraic sort of thing: ZFC explicitly posits the existence of only two sets: the empty set, and the set of natural numbers... but it provides us with a lot of ways to build new sets out of old sets.

    For example, if X and Y are sets, then we can apply the { , } operator to create the "pair set": {X, Y}.
     
  9. Oct 23, 2006 #8
    suppose we construct a set [tex]A = \{ S \ | \ S \not\in S \}[/tex]. so to avoid Russell's paradox, this kind of sets are NOT allowed. am i right?
     
  10. Oct 23, 2006 #9

    Hurkyl

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    Correct; that construction is in neither of the two forms I mentioned earlier, so it is not expressly permitted. (And the paradox proves that it cannot be permitted)
     
  11. Oct 24, 2006 #10
    Russel and Zermelo

    Two people actually solved the problem independently. Zermelo introduced an axiom called the axiom of separation and Russell introduced his theory of types.
     
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