Is My Proof of the Cantor Bernstein Theorem Correct?

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    Cantor Theorem
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The discussion focuses on the Cantor-Bernstein theorem, which states that if there are injective functions between two sets A and B, then a bijective function exists between them. A user presents their proof of an equivalent statement claiming that if such injective functions exist, then one of them must be surjective. However, this assertion is challenged, with a counterexample provided using the inclusion of natural numbers into integers and a specific mapping from integers to natural numbers. The counterexample demonstrates that the user's proof is incorrect, as the functions can be injective without being surjective. The conversation highlights the importance of understanding the distinctions between injective and surjective functions in relation to the theorem.
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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 \in B, such that b \notin f(A).
Let C be the set of g({ f(A) \bigcup b }). Then C \subseteq A, and f(A) \subset g^-1(C). Because both C and g^-1(C) contain the same amount of elements (because g is injective), from f(A) \subset g^-1(C) we get A \subset C, which is a contradiction with C \subseteq A.
 
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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 f:\mathbb{N} \rightarrow \mathbb{Z} be the inclusion and define the map g:\mathbb{Z} \rightarrow \mathbb{N} by setting g(n) = 3|n| + \mathrm{sgn}(n). Both of these maps are injections, but neither of them is a surjection.
 
Thanks!
 
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