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Cantor sets (poof needed)

  1. Oct 11, 2011 #1
    1. The problem statement, all variables and given/known data

    Let C be a Cantor set and let x in C be given
    prove that
    a) Every neighborhood of x contains points in C, different from x.
    b) Every neighborhood of x contains points not in C

    2. Relevant equations

    How can I start to prove?

    3. The attempt at a solution

    n/a
     
  2. jcsd
  3. Oct 11, 2011 #2

    Dick

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    The usual way would be to think about what the Cantor set means in terms of an expansion in base 3, i.e. ternary.
     
  4. Oct 12, 2011 #3
    I need some kind of initiation. Prof. had just defined the Cantor set and assigned this problem.
    I do not have more info about this. I looked the internet they are little bit complicated.
     
  5. Oct 12, 2011 #4

    HallsofIvy

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    Okay, what definition did your teacher give you? And are you talking about a Cantor set or the Cantor set?
     
  6. Oct 12, 2011 #5
    a cantor set
     
  7. Oct 12, 2011 #6

    HallsofIvy

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    Okay, but once again, what definition of "Cantor set" are you using?
     
  8. Oct 12, 2011 #7
    A1={[0,1/3],[2/3,1]}
    A2={[0,1/9],[2/9,3/9],[6/9,7/9],[8/9,9/9]}
    :
    :
    :

    intersection of all Ai is the Cantor set.

    This is definition that Prof. defined.
     
  9. Oct 12, 2011 #8

    Dick

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    The hint in post 2 remains valid. You might also think about what the measure (length) of the Cantor set might be.
     
  10. Oct 12, 2011 #9

    HallsofIvy

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    Do you see that, in base 3 notation, every number in [0, 1/3] starts "0.0..." and every number in [2/3, 1] starts "0.2..". That is, A1 contains all real numbers between 0 and 1 whose base 3 representation does NOT have a "1" in the first place.

    Do you see that, again in base 3 notation, every number in [0, 1/9] starts 0.00..., every number in [2/9, 3/9] starts 0.02..., every number in [6/9, 7/9] starts 0.20..., and every number in [8/9, 1] starts 0.22... That is, A2 contains all real numbers between 0 and 1 whose base 3 representation does NOT have a "1" in the first two places.

    Ai is the set of all numbers between 0 and 1 whose base three representation does not have a "1" in the first i places.

    So the Cantor set is the set of all numbers between 0 and 1 whose base 3 representation does not contain any "1" in its base three representation. Now look at intervals around a number "x" in that set.
     
  11. Oct 12, 2011 #10

    Dick

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    You could also think about endpoints. Did you notice all of the endpoints of the intervals in A_n are also in A_n+1?
     
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