Results 1 to 3 of 3

Math Help - Show that every denumerable set is equivalent to a proper subset of itself

  1. #1
    Junior Member
    Joined
    Sep 2009
    Posts
    42

    Show that every denumerable set is equivalent to a proper subset of itself

    I'm not sure how to do this. Help would be appreciated, thanks.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Oct 2009
    Posts
    5,418
    Thanks
    718
    If A is denumerable, there is a bijection f between A and \mathbb{N}. Take any a\in A. Show that \displaystyle\mathbb{N} and \mathbb{N}\setminus\{f(a)\} are equinumerous. Without loss of generality, one may assume that f(a)=0.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Quote Originally Posted by dynas7y View Post
    I'm not sure how to do this. Help would be appreciated, thanks.

    If a set A is denumerable (meaning: INFINITE denumerable), then we can write A=\{a_1,a_2,a_3,\ldots\}, so define

    f:A\rightarrow A-\{a_1\}\,,\,\,f(a_i):=a_{i+1} . Now just prove f is a bijection...

    Tonio
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. If no proper subset of X is dense..
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: September 10th 2011, 06:03 AM
  2. Replies: 1
    Last Post: October 2nd 2010, 11:21 AM
  3. Replies: 8
    Last Post: April 14th 2010, 09:36 AM
  4. Replies: 4
    Last Post: June 3rd 2009, 10:15 PM
  5. show has no proper subgroups of finite index
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: October 26th 2008, 07:36 PM

Search Tags


/mathhelpforum @mathhelpforum