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Linear ordered dense subset.

  1. Jan 10, 2013 #1
    1. The problem statement, all variables and given/known data
    If a linear ordered set P has a countable dense subset, then
    [itex] |P| \leq 2^{\aleph_0} [/itex]
    3. The attempt at a solution
    because it has a linear order then all elements of P can be compared x<y .
    And because it is dense that mean that I can find an element f
    such that x<f<y for any x or y.
    so we can partition this set into countable many pieces.
    now either their are countably many elements between
    any two elements. And the union of countably many things
    with a countable number of objects is countable.
    Or their are an uncountable number of things in between
    any of the 2 dense elements, but this union would
    be [itex] 2^{\aleph_0} [/itex] if anything larger was in between
    then the cardinality would not work.
    This is a little informal, just want to know if this is on the right track.
  2. jcsd
  3. Jan 10, 2013 #2
    How will you define this partition??
  4. Jan 10, 2013 #3
    Can i just take an arbitrary partition will some element w.
    And because the set is linear ordered, their will be either 0 elements to the left or
    at most [itex] 2^{\aleph_0} [/itex] to the left or right. I guess im assuming that.
    does this look ok? Im thinking of using something like if I have a countable dense
    subset on an interval I can only have [itex] 2^{\aleph_0} [/itex] limit points.
    all thought I will probably need to prove that.
  5. Jan 10, 2013 #4
    I don't really get it. Can you write it out more formally?
  6. Jan 10, 2013 #5
    I guess I am going to pick a random element from the set call it x.
    now I have partitioned the set into 2 sub sets.
    I call them E and U. E is everything less than or equal to x and U is everything greater than x. Now either there is 0 elements in E or countable or 2^N,
    and U must either have finite, countable or 2^N.
    I was going to assume one of the partions have more than 2^N elements
    and then say this was a contradiction. But I dont think that really proves it and I havn't
    used the fact that the set has a countable dense subset.
    Im not really sure what to do next.
    Maybe I could assume there were more than 2^N elements
    in between and 2 of my countable elements and then someohow get a contradiction by showing that
    I couldn't embed a countable subset into a set with more than 2^N elements.
  7. Jan 10, 2013 #6
    This is the direction you want to head in. Also, make sure to look up (or recall) the definition of the cardinal ##2^{\aleph_0}##. You wouldn't want to try to prove something that's true by definition.
  8. Jan 11, 2013 #7
    so [itex] 2^{\aleph_0} [/itex] is the set of all possible combinations of the natural numbers. I still dojnt really know how to prove what I want.
    I guess beacuse we have a countable dense subsets between any 2 of our
    countable elements that their exists at most [itex] 2^{\aleph_0} [/itex]
    limits points and there are only a countable many partitions in out set.
  9. Jan 11, 2013 #8
    Yes. I would characterize ##2^{\aleph_0}## as the cardinality of the power set of the natural numbers (or, for that matter, any set of cardinality ##\aleph_0##). In general, ##2^{|X|}## is the cardinality of the power set of ##X##.

    You have some good ideas (mixed in with some not-as-good ones), so let's try to get them organized. First let's give names to the pertinent sets to make things easier to read and write; let ##P## be the linearly ordered set and ##A## be the countable subset of ##P## which is dense in ##P##, and let ##\bar{A}## be the set of limit points of ##A##.

    1) Forget about dense part for the moment. Can you show that ##|\bar{A}|\leq2^{\aleph_0}## using only the definition of limit point, the fact that ##|A|=\aleph_0##, and our definition for ##2^{\aleph_0}##? So your proof here should really depend only on the cardinality of ##A## and not at all on its relationship to the ambient space ##P##.

    2) Now for the dense part ... Can you show that ##P=\bar{A}## using the definition of dense?

    Out of curiosity, what definitions for limit point and dense are you using? Part 2 could potentially end up being very easy.
    Last edited: Jan 11, 2013
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