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!

Prove sets are equipotent using Schroder - Bernstein

  1. Oct 16, 2008 #1
    1. The problem statement, all variables and given/known data
    Using the Schroeder-Bernstein Theorem, prove the following sets must have the same cardinality by producing explicit one-to-one (but not necessarily surjective) maps f: A-->B and g: B-->A.

    (a) The unit disc in the plane A={(x,y) in R2: x^2 + y^2 <1} and the unit square B={(x,y) in R2: x,y in [-1,1]}.

    (b) The unit disc in the plane A={(x,y) in R2: x^2 + y^2 <1} and the entire plane B=R2.



    3. The attempt at a solution
    I'm having trouble with this section we're working on because I don't know how to come up with an injective function at the top of my head.

    For (a) the map g, can I use the map (x,y)-->(x^2 + y^2)/2?

    For (b) the map g, can I use the map (x,y)-->(x^2+y^2)/(x^2+y^2)+1???

    You can probably tell I am lost.
     
  2. jcsd
  3. Oct 16, 2008 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Re: Schroeder-Bernstein

    No, of course those don't work:
    1) because of the squares, neither of those is one to one.
    2) you need two functions, f and g, for each.
    3) those are functions from R2 to R, not R2 to R2.

    My first thought. for (a), since the unit disk is already inside the unit square, you can take g to be the identity map: g((x,y))= (x,y). To go the other way "reduce" the size of the unit square so it is inside the unit dist: first map (x,y) to (x/2, y/2), then use the identity map: in other words, use f(x,y)= (x/2, y/2).


    For (b) you can again use g((x,y))= (x,y). For f, dividing by x2+ y2+ 1 is good but you must have 2[/sup] components and the function must be one to one. There is an almost trivial choice for the numerators.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Prove sets are equipotent using Schroder - Bernstein
  1. Prove using sets (Replies: 2)

Loading...