proof of Schroeder-Bernstein theorem using Tarski-Knaster theorem
The Tarski-Knaster theorem can be used to give a short, elegant proof of the Schroeder-Bernstein theorem.
Proof.
If X⊆Y⊆S, then S∖g(T∖f(X))⊆S∖g(T∖f(Y)), and so φ is monotone. Since P(S) is a complete lattice, we may apply the Tarski-Knaster theorem to conclude that the set of fixed points
of φ is a complete lattice and thus nonempty.
Let C be a fixed point of φ. We have
S∖C=g(T∖f(C)). |
Hence g|T∖f(C):T∖f(C)→S∖C and f|C:C→f(C) are bijections. We can therefore construct the desired bijection h:S→T by defining
h(x)={f(x)if x∈C(g|T∖f(C))-1(x)if x∉C.∎ |
The usual proof of Schroeder-Bernstein theorem explicitly constructs a fixed point of .
References
-
1
Thomas Forster, Logic, induction
and sets, Cambridge University Press, Cambridge, 2003.
- 2 M. Kolibiar, A. Legéň, T. Šalát, and Š. Znám, Algebra a príbuzné disciplíny, Alfa, Bratislava, 1992 (Slovak).
Title | proof of Schroeder-Bernstein theorem using Tarski-Knaster theorem |
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Canonical name | ProofOfSchroederBernsteinTheoremUsingTarskiKnasterTheorem |
Date of creation | 2013-03-22 15:30:24 |
Last modified on | 2013-03-22 15:30:24 |
Owner | kompik (10588) |
Last modified by | kompik (10588) |
Numerical id | 7 |
Author | kompik (10588) |
Entry type | Proof |
Classification | msc 06B99 |
Related topic | SchroederBernsteintheorem |
Related topic | TarskiKnastertheorem |
Related topic | TarskiKnasterTheorem |
Related topic | SchroederBernsteinTheorem |