Prove sets are equipotent using Schroder - Bernstein

[fk378]

Homework Statement

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.

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.

 

[HallsofIvy]

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.