# Prove sets are equipotent using Schroder - Bernstein

1. Oct 16, 2008

### fk378

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. Oct 16, 2008

### HallsofIvy

Staff Emeritus
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.