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Cantor Set?

  1. Jan 6, 2012 #1
    How can the cantor set be uncountable and have zero measure. Couldn't I map the cantor set to another uncountable set that did not have zero measure. I probably don't understand measure or the cantor set very well. Any input will be much appreciated.
  2. jcsd
  3. Jan 6, 2012 #2
    Yeah, you could map the Cantor set to another set that has nonzero measure. You could even map it bijectively onto the real numbers!!

    However, there is no reason at all to believe that this would preserve measure. Why would it be true that [itex]\mu(A)=\mu(f(A))[/itex]?? It isn't even true for nice maps like [itex]f(x)=2x[/itex]. Indeed: if A=[0,1], then f(A)=[0,2], but [itex]\mu([0,1])\neq \mu([0,2])[/itex].
  4. Jan 6, 2012 #3
    For measure to work do my numbers have to be defined like in decimal form, or can I just talk about an uncountable number of mathematical objects? In your last post does mu stand for measure and what is A
  5. Jan 6, 2012 #4
    We just need them to be real numbers. We don't care whether they are in decimal form or not.

    We're just talking about a measure on the real numbers here. But actually, you can define a measure on any set you like. You just need a set, a sigma-algebra and something that satisfies the axioms of a measure. IT don't need to be the real numbers.

    A is a measurable set. For example, A= the cantor set.
  6. Jan 6, 2012 #5
    so then if my set can be uncountable objects, how would I know if it had measure zero.
    I guess I need to learn more about measure. If its not too hard to explain, why does the cantor set have zero measure. If its hard to explain, are there any books you would recommend on measure?
  7. Jan 6, 2012 #6


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    You know how the Cantor set is defined right?
    Start with [0,1]. It has measure 1.
    Remove the middle third. [itex][0,1/3]\cup[2/3,1][/itex] has measure 2/3.
    And so on.
    At each stage the measure keeps getting smaller and converges to zero. The measure of the Cantor set is that limit, so it must be zero.
  8. Jan 6, 2012 #7
    ok, this might sound crazy but can I start with the whole real axis and then keep removing segments of it forever like what is done to the cantor set? And If I could do this what would its measure be.
  9. Jan 6, 2012 #8
    A very easy but good book on measure theory is "Lebesgue integration on Euclidean space" by Jones. It won't touch much on abstract measure theory, but it'll give you enough information to get comfortable with the topic.
  10. Jan 6, 2012 #9
    Very good remark!!!

    Basically, what pwsnafu uses is the following theorem, sometimes called "continuity from above":

    The question you raise doesn't apply here since the real numbers do not have finite measure.
  11. Jan 6, 2012 #10
    thanks everyone for your posts. When you say it doesn't apply because the reals don't have finite measure, is this because I used all the reals at once? What If I just did it in intervals? Like the cantor set from 0 to 1, then 1 to 2 and did this out to infinity and considered them one at a time and then added them up at the end. Or does this still not work
  12. Jan 6, 2012 #11
    So you mean that you want to take an infinite union of Cantor sets?? That will have measure 0 because

    [tex]\mu\left(\bigcup A_n\right)\leq \sum \mu(A_n)=0[/tex]

    for all measurable [itex]A_n[/itex] of measure 0.
  13. Jan 6, 2012 #12
    ya. What did you mean when you said the reals didn't have finite measure, were you referring to all the reals?
  14. Jan 6, 2012 #13
    I was referring to the set [itex]\mathbb{R}[/itex].
  15. Jan 6, 2012 #14
    ok, I wanted to add an infinite amount of cantor sets together to cover the real axis from the bottom up. And was wondering if this would be equivalent to starting with the real axis and removing segments. But this is probably very ill defined and not possible. And we would need to change our segments instead of starting with segments of length 1, we might need to adjust that to make it work.
  16. Jan 6, 2012 #15
    This can be made to work. That is: you can probably find a construction that indeed gives you the same two sets. The infinite amount of Cantor sets will have measure 0.

    However, if you say that this is equivalent with "starting with the real axis and removing segments", then you are correct. But you won't find the measure this way. Continuity from above only holds with spaces of finite measure.

    That is: if you would apply the same reasoning and naively apply continuity from above, then you would find that the set has measure infinity. This would be the wrong value. The only way to find the value of the infinite amount of Cantor sets would be by applying the inequality

    [tex]\mu \left(\bigcup A_n\right)\leq \sum \mu (A_n)[/tex]
  17. Jan 6, 2012 #16
    Interesting, alot of new stuff to think about, thanks for your posts
  18. Jan 6, 2012 #17


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    One can also create Cantor sets of positive measure. Instead of removing middle thirds, remove middle fifths. This is also uncountable.

    Uncountability tells you nothing about measure. Not only can an uncountable set have any measure at all, but it can also be unmeasurable - i.e. not have any measure.
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