Cantor–Bernstein–Schroeder theorem

In set theory, the Cantor–Bernstein–Schroeder theorem, named after Georg Cantor, Felix Bernstein, and Ernst Schröder, states that, if there exist injective functions f : A → B and g : B → A between the sets A and B, then there exists a bijective function h : A → B. In terms of the cardinality of the two sets, this means that if |A| ≤ |B| and |B| ≤ |A|, then |A| = |B|; that is, A and B are equipollent. This is a useful feature in the ordering of cardinal numbers.

The theorem is also known as the Schroeder–Bernstein theorem, the Cantor–Bernstein theorem, or the Cantor–Schroeder–Bernstein theorem.

An important feature of this theorem is that it does not rely on the axiom of choice. However, its various proofs are non-constructive, as they depend on the law of excluded middle, and therefore rejected by intuitionists.

König's definition of a bijection h:A→B from given example injections f:A→B and g:B→A. An element in A and B is denoted by a number and a letter, respectively. The sequence 3→e→6→… is an A-stopper, leading to the defintitions h(3)=f(3)=e, h(6)=f(6), …. The sequence d→5→f→… is a B-stopper, leading to h(5)=g^-1(5)=d, …. The sequence …→a→1→c→4→… is doubly infinite, leading to h(1)=f(1)=a, h(4)=f(4)=c, …. The sequence b→2→b is cyclic, leading to h(2)=f(2)=b.

Proof

This proof is attributed to Julius König.

Assume without loss of generality that A and B are disjoint. For any a in A or b in B we can form a unique two-sided sequence of elements that are alternately in A and B, by repeatedly applying $f$ and $g$ to go right and $g^{-1}$ and $f^{-1}$ to go left (where defined).

$\cdots \rightarrow f^{-1}(g^{-1}(a))\rightarrow g^{-1}(a)\rightarrow a\rightarrow f(a)\rightarrow g(f(a))\rightarrow \cdots$

For any particular a, this sequence may terminate to the left or not, at a point where $f^{-1}$ or $g^{-1}$ is not defined.

Call such a sequence (and all its elements) an A-stopper, if it stops at an element of A, or a B-stopper if it stops at an element of B. Otherwise, call it doubly infinite if all the elements are distinct or cyclic if it repeats. See the picture for examples.

By the fact that $f$ and $g$ are injective functions, each a in A and b in B is in exactly one such sequence to within identity, (as if an element occurs in two sequences, all elements to the left and to the right must be the same in both, by definition). Therefore, the sequences form a partition of the (disjoint) union of A and B. Hence it suffices to produce a bijection between the elements of A and B in each of the sequences separately, as follows:

• For an A-stopper, the function $f$ is a bijection between its elements in A and its elements in B.
• For a B-stopper, the function $g$ is a bijection between its elements in B and its elements in A.
• For a doubly infinite sequence or a cyclic sequence, either $f$ or $g$ will do ($f$ is used in the picture).

Another proof

Below follows an alternative proof.{{ safesubst:#invoke:Unsubst||date=__DATE__ |$B= {{#invoke:Category handler|main}}{{#invoke:Category handler|main}} }}

Idea of the proof: Redefine f in certain points to make it surjective. At first, redefine it on the image of g for it to be the inverse function of g. However, this might destroy injectivity, so correct this problem iteratively, by making the amount of points redefined smaller, up to a minimum possible, shifting the problem "to infinity" and therefore out of sight.

More precisely, this means to leave f unchanged initially on C₀ := A \ g[B]. However, then every element of f[C₀] has two preimages, one under f and one under g ^–1. Therefore, leave f unchanged on the union of C₀ and C₁ := g[f[C₀]]. However, then every element of f[C₁] has two preimages, correct this by leaving f unchanged on the union of C₀, C₁, and C₂ := g[f[C₁]] and so on. Leaving f unchanged on the countable union C of C₀ and all these C_n+1 = g[f[C_n]] solves the problem, because g[f[C]] is a subset of C and no additional union is necessary.

In the alternate proof, C_n can be interpreted as the set of n-th elements of A-stoppers (starting from 0).

Indeed, C₀ is the set of elements for which g⁻¹ is not defined, which is the set of starting elements of A-stoppers, C₁ is the set of elements for which $f^{-1}\circ g^{-1}$ is defined but $g^{-1}\circ f^{-1}\circ g^{-1}$ is not, i.e. the set of second elements of A-stoppers, and so on.

The bijection h is defined as f on C and g⁻¹ everywhere else, which means f on A-stoppers and g⁻¹ everywhere else, consistently with the proof below.

Proof: Define

$C_{0}=A\setminus g[B],\qquad C_{n+1}=g[f[C_{n}]]\quad {\mbox{ for all }}n\geq 0,$

and

$C=\bigcup _{n=0}^{\infty }C_{n}.$

Then, for every a ∈ A define

$h(a)={\begin{cases}f(a)&{\mbox{if }}a\in C,\\g^{-1}(a)&{\mbox{if }}a\notin C.\end{cases}}$

If a is not in C, then, in particular, a is not in C₀. Hence a ∈ g[B] by the definition of C₀. Since g is injective, its preimage g ^–1(a) is therefore well defined.

It remains to check the following properties of the map h : A → B to verify that it is the desired bijection:

• Surjectivity: Consider any b ∈ B. If b ∈ f[C], then there is an a ∈ C with b = f(a). Hence b = h(a) by the definition of h. If b is not in f[C], define a = g(b). By definition of C₀, this a cannot be in C₀. Since f[C_n] is a subset of f[C], it follows that b is not in any f[C_n], hence a = g(b) is not in any C_n+1 = g[f[C_n]] by the recursive definition of these sets. Therefore, a is not in C. Then b = g ^–1(a) = h(a) by the definition of h.

• Injectivity: Since f is injective on A, which comprises C, and g ^–1 is injective on g[B], which comprises the complement of C, it suffices to show that the assumption f(c) = g ^–1(a) for c ∈ C and a ∈ A \ C leads to a contradiction (this means the original problem, the lack of injectivity mentioned in the idea of the proof above, is solved by the clever definition of h). Since c ∈ C, there exists an integer n ≥ 0 such that c ∈ C_n. Hence g(f(c)) is in C_n+1 and therefore in C, too. However, g(f(c)) = g(g ^–1(a)) = a is not in C — contradiction.

Note that the above definition of h is nonconstructive, in the sense that there exists no general method to decide in a finite number of steps, for any given sets A and B and injections f and g, whether an element a of A does not lie in C. For special sets and maps this might, of course, be possible.

Original proof

An earlier proof by Cantor relied, in effect, on the axiom of choice by inferring the result as a corollary of the well-ordering theorem. The argument given aboveTemplate:Clarify shows that the result can be proved without using the axiom of choice.

Furthermore, there is a simple proof which uses Tarski's fixed point theorem.

History

As it is often the case in mathematics, the name of this theorem does not truly reflect its history. The traditional name "Schröder-Bernstein" is based on two proofs published independently in 1898. Cantor is often added because he first stated the theorem in 1895, while Schröder's name is often omitted because his proof turned out to be flawed while the name of the mathematician who first proved it is not connected with the theorem.

Georg Cantor (1895) states the theorem (B.)

In reality, the history was more complicated:

• 1887 Richard Dedekind proves the theorem but does not publish it.
• 1895 Georg Cantor states the theorem in his first paper on set theory and transfinite numbers (as an easy consequence of the linear order of cardinal numbers which he was going to prove later).
• 1896 Ernst Schröder announces a proof (as a corollary of a more general statement).
• 1897 Felix Bernstein, a young student in Cantor's Seminar, presents his proof.
• 1897 After a visit by Bernstein, Dedekind independently proves it a second time.
• 1898 Bernstein's proof is published by Émile Borel in his book on functions. (Communicated by Cantor at the 1897 congress in Zürich.)

Both proofs of Dedekind are based on his famous memoir Was sind und was sollen die Zahlen? and derive it as a corollary of a proposition equivalent to statement C in Cantor's paper:

$A\subset B\subset C\quad {\textrm {and}}\quad |A|=|C|\qquad \Rightarrow \qquad |A|=|B|=|C|$

Cantor observed this property as early as 1882/83 during his studies in set theory and transfinite numbers and therefore (implicitly) relying on the Axiom of Choice.

See also

• Myhill isomorphism theorem
• Schröder–Bernstein theorem for measurable spaces
• Schröder–Bernstein theorems for operator algebras
• Schröder–Bernstein property

References

Template:Citizendium Peter Schmitt contributed the History section to Citizendium which TakuyaMurata copied into this article.

• Proofs from THE BOOK, p. 90. ISBN 3-540-40460-0
• Template:Planetmath reference
• MathPath – Explanation of and remarks on the proof of Cantor–Bernstein Theorem
• Papers on the history of the Cantor–Bernstein theorem
• Template:Nlab

External links

• Citizendium: Schröder-Bernstein theorem and Schröder–Bernstein property
• Weisstein, Eric W., "Schröder-Bernstein Theorem", MathWorld.
• Template:PlanetMath