Schroeder-Bernstein theorem, proof of
Inductively define a sequence of subsets of by and . Now let , and define by
If , then . But if , then , and so . Hence is well-defined; is injective by construction. Let . If , then . Otherwise, for some , and so there is some such that . Thus is bijective; in particular, if , then is simply the identity map on .
To prove the theorem, suppose and are injective. Then the composition is also injective. By the lemma, there is a bijection . The injectivity of implies that exists and is bijective. Define by ; this map is a bijection, and so and have the same cardinality.
|Title||Schroeder-Bernstein theorem, proof of|
|Date of creation||2013-03-22 12:49:56|
|Last modified on||2013-03-22 12:49:56|
|Last modified by||mps (409)|
|Synonym||proof of Cantor-Bernstein theorem|
|Synonym||proof of Cantor-Schroeder-Bernstein theorem|