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Measurable subset of [0,1].

  1. Mar 4, 2008 #1

    Tom Mattson

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    1. The problem statement, all variables and given/known data
    Construct a measurable set [itex]E\subset [0,1][/itex] with [itex]0<m(E)<1[/itex] such that both [itex]E[/itex] and [itex][0,1]-E[/itex] (that's a set difference) are dense in [itex][0,1][/itex].


    2. Relevant equations
    None.


    3. The attempt at a solution
    Well, the obvious dichotomy here is rationals vs. irrationals, but of course the rationals are countable and hence have measure zero, so that's no good. So now I'm thinking about including [itex]\mathbb{Q}\cap [0,1][/itex] in [itex]E[/itex], plus a selection of irrationals. Unfortunately, I can't think of how to pick the irrationals to go in [itex]E[/itex] without screwing up the denseness of the irrationals.

    I have the feeling that this is one of those questions where you either see the trick, or you don't.

    I don't. :cry:
     
    Last edited: Mar 4, 2008
  2. jcsd
  3. Mar 4, 2008 #2

    NateTG

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    I don't understand why you've got "...and [0,1] are dense in [0,1]" is there an extra condition?

    That said, in response to:
    "Unfortunately, I can't think of how to pick the irrationals to go in [itex]E[/itex]. Reload this page in a moment. without screwing up the denseness of the irrationals."

    I have to ask, "Is it ever possible to add points to a dense set, and make that set not dense?"
     
  4. Mar 4, 2008 #3

    Tom Mattson

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    Sorry, that was a typo. It's fixed now. The condition is that both [itex]E[/itex] and its complement in [itex][0,1][/itex] must be dense.

    I wasn't thinking of adding points to the irrationals, I was thinking of taking points away from them and adding them to the rationals, in such a way that I get two sets with nonzero measure. Am I totally off base in thinking that way?
     
  5. Mar 4, 2008 #4

    quasar987

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    I remember seeing this question asked here before. If you get desperate, you can always browse the forums.
     
  6. Mar 4, 2008 #5

    NateTG

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    Not necessarily, since this is a rather open-ended question. That said, it's usually easier to operate in terms of intervals since it can be tricky to prove that a particular point set is measureable.

    Hint: One way to construct an example is to separate [itex][0,1][/itex] into two intervals.
     
  7. Mar 4, 2008 #6

    Tom Mattson

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    That doesn't sound right. Both subsets of [itex][0,1][/itex] have to be dense, and the conditions are rigged so as to exclude the trivial examples (such as choosing [itex][0,1)[/itex] and [itex]\{1\}[/itex], for instance). So if I choose two intervals, say [itex][0,a][/itex] and [itex](a,1][/itex] with [itex]a\in (0,1)[/itex], there will be neighborhoods in [itex][0,1][/itex] that do not contain points in one interval or the other (hence, neither will be dense in [itex][0,1][/itex]).
     
    Last edited: Mar 4, 2008
  8. Mar 4, 2008 #7

    NateTG

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    I'm sorry, that wasn't phrased as well as it could have been. I meant that you think of [itex][0,1][/itex] as two intervals and treat each interval differently.
     
  9. Mar 4, 2008 #8

    Tom Mattson

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    I'm still not following you. What does, "think of [0,1] as two intervals" mean, if not splitting it up into two intervals such that [0,1] is covered by them?
     
  10. Mar 4, 2008 #9

    quasar987

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    Oh, I see it! :x

    you were on the right track with those rationals and irrationals... now add nate's hint about "splitting [0,1] in two.. as in "[0,½) and (½,1]"
     
  11. Mar 4, 2008 #10

    Tom Mattson

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    I assume you mean [0,1/2) and [1/2,1], since one subset is to be the complement of the other, right?

    Even so, I must be missing something that is obvious to you guys. If I split up the interval like that, then I run into the problem that I described a few posts ago: Neither of those is dense in [0,1], as there are neighborhoods of [0,1] that don't contain any points of one or the other of those intervals.
     
  12. Mar 4, 2008 #11

    NateTG

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    Let's say that I take each of [itex][0,\frac{1}{2}) [\frac{1}{2},1][/itex] and split them into their rational and irrational components....
     
  13. Mar 4, 2008 #12

    Tom Mattson

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    Ah, I think I see. The following would be a valid construction for [itex]E[/itex] then.

    * Take [itex]\mathbb{Q} \cap [0,1/2)[/itex]

    * Take [itex](\mathbb{R}-\mathbb{Q}) \cap [1/2,1][/itex].

    * Take the union of the above.
     
  14. Mar 4, 2008 #13

    NateTG

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    Right, and you should be able to show that all the specified criteria are met.
     
  15. Mar 4, 2008 #14

    Tom Mattson

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    Thank you both.
     
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