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Cantor paradox?

  1. Sep 23, 2006 #1
    Can someone explain how the Cantor set can be uncountable but also contain no intervals? I am assuming that as k goes to infinity, we are left with 0 and 1 in the final interation so the set is finite with those elements. The set of natural numbers is countable so I can bijectively map every Cantor interval to an element in N, right? So it seems countable to me with no intervals. But my notes say the opposite.
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  3. Sep 23, 2006 #2

    matt grime

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    Your intuition is wrong. Not that I know what an 'interation' is (nor does anyone else). How can you map every 'cantor interval' to an elementin N? What is a 'cantor interval', for that matter?
  4. Sep 23, 2006 #3
    Yes the word is iteration. In any case, if I call the union of sets produced after the kth iteration, ie after the middle interval is removed, Ak and bijectively map that to k in N then I am stuck thinking that the collection of intervals produced after k iterations can be mapped to N and thus the Cantor set is countable.
  5. Sep 23, 2006 #4


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    There are only finitely many intervals at any stage, but this says nothing about the infinite limit. In fact, the cantor set is uncountable, as can be proved by showing that the elements in the set are precisely those whose base 3 expansion contains only 0's and 2's. So it can be mapped bijectively into [0,1] by replacing all the 2's with 1's in the base 3 expansion of an element in the cantor set, and looking at this as a number written in base 2 (check out http://en.wikipedia.org/wiki/Cantor_set" [Broken] if that isn't clear).

    The counterintuitive thing is that the set has lebesque measure 0. I'm not going to try to explain Lebesgue measure too precisely, but basically it works like this: the interval [a,b] has measure b-a, and a countable disjoint union of intervals has as its measure the sum of the constituent intervals. All countable sets have measure 0, and it seems at first to someone learning the theory like uncountable sets could be characterized as those with positive measure, but this turns out not to be true.

    I'm not 100% sure, but I believe the resolution is that subsets of R with Hausdorff dimension greater than 0 are uncountable, but only sets with dimension 1 have positive Lebesgue measure. The cantor set has dimension ln(2)/ln(3), so it falls in this in between region.
    Last edited by a moderator: May 2, 2017
  6. Sep 24, 2006 #5

    matt grime

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    So, you're thinking is :

    A_k is countable

    A is the limit of A_k

    k is in a countable index set

    therefore A is countable.

    Cardinality jsut doesn't behave like that I'm afraid.

    The techincal reasonfor that is that you're taking an inverse limit rather than a direct limit. I hadn't thought in those terms for Cantor sets before....

    You can think of embedding A_{k+1} in A_k, this means there is a diagram


    and because all the arrows point 'down' (from a larger numbered set to a smaller one) we have an inverse limit. This means that limit, the cantor set, will be the product of the A_i modulo an equivalence relation, which means the cardinality can be uncountable, since the countable product of finite sets is uncountable.

    Had the arrows pointed the other way the limit would have been the disjoint union of the A_i modulo some equivalence relation, and that would have been countable. But they don't go that way.
  7. Sep 24, 2006 #6

    I think I am starting to make some sense of it. My problem logically was that I had just proved that a collection of open disjoint sets was countable, and was trying to treat each set of intervals produced in the cantor kth iteration as a similar collection.
  8. Sep 24, 2006 #7
    also by product of sets I assume you mean the cartesian product?
  9. Sep 24, 2006 #8

    matt grime

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    If you want to call it that, then yes, though there is no need to call it the cartesian product.
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