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Ross-Littlewood Paradox

  1. May 15, 2013 #1

    BDV

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    Ross-Littlewood vase filling paradox (from Wikipedia):

    To complete an infinite number of steps, it is assumed that the vase is empty at one minute before noon, and that the following steps are performed:

    The first step is performed at 30 seconds before noon.
    The second step is performed at 15 seconds before noon.
    Each subsequent step is performed in half the time of the previous step, i.e., step n is performed at 2−n minutes before noon.
    This guarantees that a countably infinite number of steps is performed by noon. Since each subsequent step takes half as much time as the previous step, an infinite number of steps is performed by the time one minute has passed.

    At each step, ten balls are added to the vase, and one ball is removed from the vase. The question is then: How many balls are in the vase at noon?



    To me, it is somewhat obvious that it is ω+ω+ω+ω+ω+ω+ω+ω+ω. (9ω's).

    Where am I wrong/what am I missing?
     
    Last edited: May 15, 2013
  2. jcsd
  3. May 15, 2013 #2
    If your answer is "correct" then I would say that the real correct answer is ω, since |ω| = |ω+ω+ω+ω+ω+ω+ω+ω+ω|.

    Also I don't understand the part that says step n is performed at 2−n minutes before noon. That would mean step 2 was performed at noon which would make the anwer 19.
     
  4. May 18, 2013 #3
    It's not 2-n, it's 2 to the power of -n.
     
  5. May 18, 2013 #4
    how is this possible that there is infinte steps.before noon??
     
  6. May 18, 2013 #5
    What if you number each ball and take out the n-1 ball on step n. Then which balls are left?
     
  7. May 21, 2013 #6

    BDV

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    Why should I be able to name even one of the remaining elements?

    The elimination procedure in the RL paradox simply ensures I have removed the entirety of the vocabulary for naming elements, so the remaining elements are unnamed.

    The RL pradox procedure yields a completely unorderd set of N0 power, (which can be ordered solely by whacking it with the AC sledgehammer).
     
  8. May 21, 2013 #7

    micromass

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    You have countably many elements. So you have a bijection with ##\mathbb{N}##. So you can name the elements.
     
  9. May 21, 2013 #8
    There are no unnamed elements because every ball you put in was labeled with a natural number.
     
  10. May 21, 2013 #9

    BDV

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    Ergo, having removed all named balls (ie balls covered under the natural notation) you are left with N0 of unnamed ones (not covered under natural notation).

    One cannot take the intuition from the completion of a finite number of steps and apply it to the totality of the infinite procedure.

    The set of all steps prescribed by the RL "paradox" procedure exists only by axiom. Once we're contemplating the result of the TOTALITY of those steps (set exant strictly by axiom) to apply rules derived from finite steps is silly. And incorrect. The rules of the game are those of the infite sets, not those of the finite sets.
     
  11. May 21, 2013 #10
    You say you are left with N_0. Where do u get that? You put in omega balls and you took out omega balls. I'm not saying that alone means there are no balls left but it surely does not mean there has to be N_0 left. (Although there could be, it depends on how you do your steps).
     
  12. May 21, 2013 #11

    BDV

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    But the procedure is very clear that at each step the number of balls increases by 9. N0 steps of 9 balls each.

    However, the RL procedure ensures that the resulting set is not effectively denumerable (cf. Sierpinski's usage of the word, e.g in "Cardinal and Ordinals"). Actually we cannot name even one single ball!
     
  13. May 21, 2013 #12
    The procedure is not clear about which balls are added and removed. Even if they are not numbered, they must be different objects (or the answer will be one). Therefore we can define a well ordering on the balls by the order that they were put in. So there is always this concept of "which ball". I do however agree that if we remove the balls in an arbitrary order the probability of being left with N_0 balls is 1. We can intentionally choose to remove balls such that we are left with 0 balls though.
     
  14. May 21, 2013 #13
    Here's what I think is the essence of the Ross-Littlewood Paradox. (I'm happy to simplify this if people can't understand the technicalities.) Let us define an equivalence relation on the collection of infinite subsets of ##\mathbb{N}##, as follows: M1~M2 if there exists a reordering f of ##\mathbb{N}## which maintains the order type ω (i.e. a bijection from ##\mathbb{N}## to ##\mathbb{N}##), such that f(M1) = M2. Then I think that M1~M2 if and only if ##\mathbb{N}## - M1 has the same cardinality as ##\mathbb{N}## - M2. So we can define the degree of an equivalence class [M1] to be the cardinality of ##\mathbb{N}## - M1.

    And then it turns out that some equivalence classes have finite degree, while other equivalence classes have infinite degree. So long story short: it doesn't actually matter how you label the elements. The paradox remains regardless of how you reorder ##\mathbb{N}##. The paradox is that, independent of labeling, some infinite subsets have complements with finite cardinality, while other infinite subsets have complements with infinite cardinality.
     
  15. May 25, 2013 #14
    Can anyone verify that what I said in post #13 makes sense, or should I clarify it?
     
    Last edited: May 25, 2013
  16. Nov 4, 2015 #15

    BDV

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    I agree!

    Thank you for the clarification, it reinforces that intuition derived from examining finite phenomena and collections is a weak guide when dealing with infinities.
     
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