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One more question about the cantor set.

  1. Jan 7, 2012 #1
    Lets start with a line segment from zero to 1 and instead of removing like the middle 1/3 can we remove an infinitesimal amount, and then keep doing this forever. It seems like this set would still have measure 1. Unless I don't understand measure or infinitesimals. And if we looked at the line would it look like a line or dust on the line?
     
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  3. Jan 7, 2012 #2

    lavinia

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    you need to better define what you mean by removing an infinitesimal amount.


    If you mean what would be the limit of the sets obtained say by removing middle fifths then middle sevenths then middle 11'ths and so on it is clear that it will have measure one. What do you think the set would look like?
     
  4. Jan 7, 2012 #3
    I wanted to remove an infinitesimal amount from the start, like as close as I can get to zero. But on your example, it seems like the set would look like scattered points,
     
  5. Jan 8, 2012 #4

    lavinia

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    What do you mean by an infinitesimal amount?
     
  6. Jan 8, 2012 #5
    can I define it as 1/x and x goes to infinity?
     
  7. Jan 8, 2012 #6

    lavinia

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    to me that is zero. I do not know what an infinitesimal amount is.
     
  8. Jan 8, 2012 #7
    close to zero but not zero. could I define it as multiplying 1/2 to itself forever.
     
  9. Jan 8, 2012 #8

    jgens

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    Nope. Notice that limn→∞2-n = 0. In fact, since the ordinary real number system is archimedean, it has no non-zero infinitesimal elements.

    One way to get infinitesimal elements involves using the compactness theorem to construct a non-standard model of the reals. I am not familiar with the measure theory of non-standard models of R so I cannot give you any more information than this.
     
  10. Jan 8, 2012 #9
    ok, thanks for your responses. so I can have stuff like (0)*(Infinity)=1
     
  11. Jan 8, 2012 #10

    jgens

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    No. Even in non-standard models of the reals, you still do not have anything like 0 * ∞ = 1.
     
  12. Jan 8, 2012 #11
    why couldn't I just have [itex] \frac{1}{2^x}(2^x) [/itex] and have x go to infinity
     
  13. Jan 8, 2012 #12

    jgens

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    Notice that limx→∞2-x2x+1 = 2. Do you see any problem with this? If you want to include the term +∞ in your number system, then you have to leave things like 0 * ∞ as undefined.
     
  14. Jan 9, 2012 #13
    ok. I thought we could do some 0*infinity limits with L'Hôpital's rule,but maybe im wrong.
    And yes I do see something wrong with what you said.
     
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