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Construct uncountably dense and holes everywhere

  1. May 8, 2013 #1
    I want to define a set [itex]X\subset [0,1][/itex], without using the axiom of choice, with the following property: Both [itex][a,b]\cap X[/itex] and [itex][a,b]\backslash X[/itex] are uncountable for all [itex][a,b]\subset [0,1][/itex] where [itex]a<b[/itex].

    I don't know how to define [itex]X[/itex] so that [itex][a,b]\cap X[/itex] would always be uncountable without [itex]X[/itex] containing some interval. But if it contains some interval, then [itex][a,b]\backslash X[/itex] will be empty sometimes.
     
  2. jcsd
  3. May 8, 2013 #2
    Never mind. I invented a solution now.

    Now the problem is left as a challenge to the rest.
     
  4. May 8, 2013 #3
    My solution has the property [itex]m^*(X)=0[/itex]. I'm still interested to know if the [itex]X[/itex] can be defined in such way that [itex]0<m^*(X)< 1[/itex].
     
  5. May 11, 2013 #4
    Consider [itex]X:=\{x\in[0,1]\backslash \mathbb{Q}: \text{the decimal expansion of } x \text{ contains only finitely many, say, }0\text{'s and }9\text{'s}\}[/itex]. This set is uncountably dense in [0,1] and so is its complement, though I'm not sure about their measures.
     
  6. May 12, 2013 #5

    Bacle2

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    I think you can calculate its measure like this:

    Start with 0: if there are no 0's in the expansion, then you remove from [0,1]:

    i.1)The segment from 0 to 0.1 , with measure 0.1

    ii.1)The segment from 0.1 to 0.11 , with measure 0.01

    n.1)The segment from 0.11111 ( n 1's) to 0.111111 (n+1-ones)

    i.2) The segment from 0.2 to 0.21

    ...........

    And you do the same thing by removing the segments :

    i.1') From 0.9 to 1

    ii.1') From 0.09 to 0.1

    etc.
     
  7. May 13, 2013 #6
    I'm confused. If X is dense in [0,1], wouldn't removing segments of [0,1] also be removing chunks of X? ... or maybe i'm the one being dense.
     
  8. May 15, 2013 #7

    Bacle2

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    Actually, you are right in that I should throw the rationals back-in after removing all the
    other parts. If this is what you meant, you're right --but remember that the Rationals
    have measure zero. If not, please read-on:

    Modulo the above, I don't think so; you are removing chunks of [0,1] itself, and X is what remains after all these parts are removed and you throw Q back-in. After you remove all the chunks between a_n=0 and a_(n+1) for n≥1 , same for a_n=9 , then removing all the points in
    the segment/interval [an,an+1 ], you will remove all the Irrationals
    with either 0-or-9 in their decimal expansion, BUT you will also remove the Rationals in the segment. Then you can throw the Rationals back-in. But remember that the Rationals have
    Lebesgue measure zero, so removing the whole chunk with Rationals included will not affect
    the measure of the remaining set.
     
  9. May 15, 2013 #8
    But there are elements of X in all those segments you're removing. Let me repeat: X is the set of all irrationals whose decimal expansion contains at most finitely many 0s and 9s. So, for example, 0.00125225222522225 ... (nonrepeating sequence) is an element of X in [0, 0.1].
     
  10. May 15, 2013 #9

    Bacle2

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    Sorry, I misread the definition of your set; I thought it was the collection of irrationals that contained neither a zero nor a nine in their expansion.
     
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