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Set M of all elements not in M

  1. Jul 14, 2010 #1
    Is this set a variation of Russell's paradox?

    [tex]M = \{x : x \notin M \}[/tex]

    I understand this formulation a lot more than

    [tex]R = \{S : S \notin S\}[/tex]

    because I don't understand how, for example, 1 is a member of itself. Is 1 a set? Are all numbers sets?
     
  2. jcsd
  3. Jul 14, 2010 #2

    tiny-tim

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    Hi Unit! :smile:

    Your first definition is not a definition …

    you can't have the thing you're defining on both sides of the equation.

    (and 1 is not in 1 … 1 is the set {φ,{φ}}, so the only elements in 1 are φ and {φ})
     
  4. Jul 14, 2010 #3
    Hi tiny-tim! Thanks for your reply!

    If it is not a definition, then what is it? I'm not trying to be sarcastic here; I actually don't know. Would it qualify as a recursive definition?

    Also, is φ the empty set? I have only seen it written as Ø.

    I have never heard of defining 1 as {φ, {φ}}! I'm assuming 0 = φ, so 1 is {φ} U {{φ}} = {φ, {φ}}. This is the empty set and the set containing the empty set. Now, there are 2 elements in total ... so would 2 = {φ, {φ}, {φ, {φ}}}?
     
  5. Jul 14, 2010 #4

    tiny-tim

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    Hi Unit! :smile:
    I suppose you could call it an implicit definition, but the Russell paradox really presupposes that everything is constructed explicitly.
    ah, I only had a φ to hand (and not a Ø), so I used that. :wink:

    yes, 2 would be {φ, {φ}, {φ, {φ}}} … that's the standard Peano construction for numbers (except I might be one out … maybe that's 3, and maybe 1 is just {φ}? :redface:)
     
  6. Jul 14, 2010 #5

    Office_Shredder

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    You are off by one. n should have n elements in it. So 0 is the empty set, 1 is the set containing the empty set, etc.
     
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