Results 1 to 5 of 5

Math Help - Help with proof

  1. #1
    Senior Member
    Joined
    Sep 2009
    Posts
    299

    Exclamation Help with proof

    If S is uncountable. Show it is impossible that P({i}) > 0 for every i∈S.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    May 2010
    Posts
    1,028
    Thanks
    28
    The relevant constraint is that \sum P(i) = 1 cannot hold under these conditions.

    proposition: If S is uncountable. Show it is impossible that P({i}) > 0 for every i∈S.
    Equivalent statement: Min (P{i}) > 0

    Denote this minimum = x

    Now note that
    (P_1 + P_2 + P_3 + ...) \geq (x + x + x + x ....)

    So its sufficient to show that
    (x + x + x + x ....) > 1

    This is true, since you are adding up a strictly positive number an uncountably large number of times (ie, more than 1/x times).
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member Traveller's Avatar
    Joined
    Sep 2010
    Posts
    162
    Quote Originally Posted by SpringFan25 View Post
    Equivalent statement: Min (P{i}) > 0
    Can we always talk of a minimum when the set under consideration is countable ?

    Also, it seems that your proof does not really use uncountability.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    May 2010
    Posts
    1,028
    Thanks
    28
    Can we always talk of a minimum when the set under consideration is countable ?
    I think you mean uncountable, and I dont understand why that would be a problem. For example the subset of real numbers  [2,\infty) is uncountable but clearly has a minimum of 2.



    Also, it seems that your proof does not really use uncountability.
    it did use and require uncountability. The final step was to show that (x + x + x + x + x... ) > 1
    This is true iff there are more than \displaystyle \lceil \ 1/x \rceil terms in the summation. Every element of the set S creates 1 term in the summation. Since S is uncountable, it must have more than \displaystyle \lceil \ 1/x \rceil elements.

    (note that \displaystyle \lceil \ 1/x \rceil is well defined as the proposition was that x>0)
    Last edited by SpringFan25; September 27th 2010 at 04:17 AM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7
    Quote Originally Posted by Sneaky View Post
    If S is uncountable. Show it is impossible that P({i}) > 0 for every i∈S.
    You haven't said that P is supposed to denote a probability on the set S, so that \sum P(\{i\}) = 1. As SpringFan25 pointed out, this is crucial to the proof.

    For each value of n=1,2,3,..., there are at most n elements i of S with P({i}) > 1/n (otherwise the sum of their probabilities would be greater than 1). Call the (finite) set of all such elements A_n. If P({i}) > 0 then it must be true that P({i}) > 1/n for some n and so i\in A_n, and therefore i is in the union of all the A_n. Thus the set of all elements i such that P({i}) > 0 is a countable union of finite sets and therefore countable.

    So if S is uncountable it cannot be true that P({i}) > 0 for all i in S.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 15
    Last Post: June 8th 2011, 11:13 AM
  2. Replies: 5
    Last Post: October 19th 2010, 10:50 AM
  3. Replies: 0
    Last Post: June 29th 2010, 08:48 AM
  4. Proof with algebra, and proof by induction (problems)
    Posted in the Discrete Math Forum
    Replies: 8
    Last Post: June 8th 2008, 01:20 PM
  5. proof that the proof that .999_ = 1 is not a proof (version)
    Posted in the Advanced Applied Math Forum
    Replies: 4
    Last Post: April 14th 2008, 04:07 PM

Search Tags


/mathhelpforum @mathhelpforum