Theorem: Let X and Y bet non empty sets, then there exist a surjection from X to Y and a surjection from Y to X if and only if there exist a bijection betwen X and Y.
Do anyone have an idea how to prove this?
Asked
Active
Viewed 191 times
1
-
Related: http://en.wikipedia.org/wiki/Schr%C3%B6der%E2%80%93Bernstein_theorem – Jan 22 '15 at 06:26
1 Answers
3
The proof of this is actually quite involved. First, we need the Partition Principle:
If there exists a surjection from $Y$ onto $X$ then there exists an injection from $X$ into $Y$.
The proof of this fact relies on the axiom of choice. You can refer to a previous stack exchange thread for the details. From here, we now have two injections, $h:X\to Y$ and $k: Y\to X$. The problem of finding a bijection is now reduced to the well-known Cantor-Bernstein theorem, whose proof also is quite lengthy. You can refer here for a few different methods. Taken together, this result is called the "Dual Cantor-Berstein Theorem."
