Is My Proof of the Cantor Bernstein Theorem Correct?

In summary, the conversation discusses the Cantor-Bernstein theorem and the speaker's attempt at providing a proof for an equivalent statement. The theorem states that if there exist injective functions between two sets, then there exists a bijective function between them. However, the speaker's proof for the equivalent statement is incorrect, as shown by a counterexample provided by one of the participants in the conversation.
  • #1
pk1234
11
0
Hi guys, I've got some problems with the cantor bernstein theorem. I'm having a hard time with all the proofs I've found, but I've actually come up with a proof myself... it will be no doubt wrong in some part though, so it would be great if you could check it for me and tell me what's wrong with the simple idea I came up with.

The theorem:
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.


My proof of an equivalent statement-
If there exist injective functions f : A → B and g : B → A between the sets A and B, then f is surjective.

By contradiction:
f is not surjective. Then there exists a b [itex]\in[/itex] B, such that b [itex]\notin[/itex] f(A).
Let C be the set of g({ f(A) [itex]\bigcup[/itex] b }). Then C [itex]\subseteq[/itex] A, and f(A) [itex]\subset[/itex] g^-1(C). Because both C and g^-1(C) contain the same amount of elements (because g is injective), from f(A) [itex]\subset[/itex] g^-1(C) we get A [itex]\subset[/itex] C, which is a contradiction with C [itex]\subseteq[/itex] A.
 
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  • #2
pk1234 said:
My proof of an equivalent statement-
If there exist injective functions f : A → B and g : B → A between the sets A and B, then f is surjective.

That is not an equivalent statement. Let [itex]f:\mathbb{N} \rightarrow \mathbb{Z}[/itex] be the inclusion and define the map [itex]g:\mathbb{Z} \rightarrow \mathbb{N}[/itex] by setting [itex]g(n) = 3|n| + \mathrm{sgn}(n)[/itex]. Both of these maps are injections, but neither of them is a surjection.
 
  • #3
Thanks!
 

What is the Cantor-Bernstein theorem?

The Cantor-Bernstein theorem, also known as the Schröder-Bernstein theorem, is a mathematical theorem that states that if there exist injective functions from set A to set B and from set B to set A, then there exists a bijective function between the two sets. This means that if there is a one-to-one mapping from set A to set B and also from set B to set A, then these two sets have the same cardinality (number of elements).

Why is the Cantor-Bernstein theorem important?

The Cantor-Bernstein theorem is important because it allows us to compare the sizes of infinite sets, which was previously thought to be impossible. It also has many applications in various fields of mathematics, including topology, measure theory, and set theory.

Can the Cantor-Bernstein theorem be extended to more than two sets?

Yes, the Cantor-Bernstein theorem can be extended to any finite number of sets. This is known as the Cantor-Schröder-Bernstein theorem, which states that if there exist injective functions from set A to set B, from set B to set C, and so on up to the last set, and also an injective function from the last set back to set A, then there exists a bijective function between all the sets.

What is the difference between the Cantor-Bernstein theorem and the Cantor-Schröder-Bernstein theorem?

The Cantor-Bernstein theorem only considers two sets, while the Cantor-Schröder-Bernstein theorem can be applied to any finite number of sets. Additionally, the Cantor-Schröder-Bernstein theorem requires an additional injective function from the last set back to the first set, which is not needed in the Cantor-Bernstein theorem.

Are there any limitations to the Cantor-Bernstein theorem?

Yes, the Cantor-Bernstein theorem cannot be applied to infinite sets if there is no injective function from one set to the other. This is known as the failure of the Cantor-Bernstein theorem. It also cannot be used to compare the sizes of uncountable sets, as these sets may have the same cardinality but still be incomparable by the theorem.

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