1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    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

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    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
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook