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Math Help - R has an uncountable number of distinct subsets.

  1. #1
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    R has an uncountable number of distinct subsets.

    It's a five part question leading up to the point that, if S is a subset of T, then if S is dense then so is T. That much was easy but part four took me some time and I have no idea what part five is asking. I wrote up a proof for four but I wanted to make sure it was correct before I moved on.

    I also have the problem of getting the next element in countable set not necessarily \mathbb{Z} so I've assumed a method for the object of a member of such a set named previous which returns the value of the next element. This makes sense since every member of an infinite set has a next element.

    With the latex partially down this will be difficult to describe but here we go.

    \mathbb{R} has an uncountable number of distinct subsets.

    Let i be an element of I (an infinite countable set) and j be an element of J (an uncoubtable set) and S_i be a countable dense subset. There exists an element x_j_i in R such that x_j_i is not in S_i

    Let S_i.next() = S_i U {x_j_i}

    Because the S_i is a subset of S_i.next() there are a infinite countable number of dense subsets. However note that any combination of x_j_i can be used which means that, since it is of an uncountable index, there are an uncountable number of possible S_i.

    Unfortunately I can't render this in latex either but the symbol in part five is a 2 to the power of a gothic looking C

    The question goes:

    Again using (iii) prove that \mathbb{R} has an uncountable number of distinct subsets and 2^C distinct uncountable subsets
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  2. #2
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    Quote Originally Posted by magus View Post
    It's a five part question leading up to the point that, if S is a subset of T, then if S is dense then so is T. That much was easy but part four took me some time and I have no idea what part five is asking. I wrote up a proof for four but I wanted to make sure it was correct before I moved on.

    I also have the problem of getting the next element in countable set not necessarily \mathbb{Z} so I've assumed a method for the object of a member of such a set named previous which returns the value of the next element. This makes sense since every member of an infinite set has a next element.

    With the latex partially down this will be difficult to describe but here we go.

    \mathbb{R} has an uncountable number of distinct subsets.

    Let i be an element of I (an infinite countable set) and j be an element of J (an uncoubtable set) and S_i be a countable dense subset. There exists an element x_j_i in R such that x_j_i is not in S_i

    Let S_i.next() = S_i U {x_j_i}

    Because the S_i is a subset of S_i.next() there are a infinite countable number of dense subsets. However note that any combination of x_j_i can be used which means that, since it is of an uncountable index, there are an uncountable number of possible S_i.


    What do you mean by "distinct subsets"? Do you mean "disjoint subsets"? Then it's easy: the set { {x} / x in R } is an uncountable

    set of pairwise disjoint subsets of R. Q.E.D.

    Tonio



    Unfortunately I can't render this in latex either but the symbol in part five is a 2 to the power of a gothic looking C

    The question goes:

    Again using (iii) prove that \mathbb{R} has an uncountable number of distinct subsets and 2^C distinct uncountable subsets
    .
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  3. #3
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    Sorry, that should read distinct dense subsets and I have no clue what he means by distinct. That was never defined.

    I think the 2^C means of cardinality of the power set of \mathbb{R} though because I can prove that.

    Informally, take a uncountable dense subset in \mathbb{R} and remove a countable number of points. The set will remain dense (I prove this.) Then the number of sets that can be created is equal to the power set of R because the set of removed points will form the power set.

    Does this sound right?
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  4. #4
    Senior Member Tinyboss's Avatar
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    That's not quite right. For instance, the set Q U (0,1) is an uncountable dense set, but if you remove the rationals (countably many points), it's no longer dense. The sets { Q U {x} | x in R-Q } are all dense, and there are uncountably many of them, but that's pretty trivial.

    I wonder if the problem was meant to be about pairwise-disjoint dense subsets? I think that should be true. For a fixed x in R, let Q+x denote { q+x | q in Q }. Obviously every such set is dense, and the sets Q+x and Q+y are either equal or disjoint, depending on whether x-y is rational. No countable collection of these can cover R, since a countable union of countable sets is countable, and R is uncountable. So there must be uncountably many distinct ones, and for these sets, distinct is equivalent to disjoint.
    Last edited by Tinyboss; May 3rd 2011 at 10:19 PM.
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  5. #5
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    hmm. OK Then what about a finite amount of points removed? That would work because the x removed would become a limit point of the new set.

    I would agree that pairwise disjoint would make the most sense. Your method proves that uncountability but I need to show it's cardinality.
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  6. #6
    Senior Member Tinyboss's Avatar
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    Yes, removing a finite number of points from a dense set will leave it dense (in R--other spaces need not have this property, but I think we're strictly working with R here, right?).

    There's no way there are 2^C pairwise-disjoint subsets of R. Suppose there was such a collection, and use the Axiom of Choice to pick one point from each. Now you have 2^C distinct real numbers, which is absurd. Surely "distinct" is what is meant in that part of the question. Or we've guessed wrong about something else.
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  7. #7
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    Well if you remove the "distinct" part of it I think I've proven it. So from there there are 2^C uncountable dense subsets of R. If we were to remove the ones that were not "distinct" we can remove up to at least C of them since 2^C-C=2^C. I say at least C because I don't know if there's a cardinality between C and 2^C. I'm not sure of a criteria that would remove aleph0 or C of them though.
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