Bijections and need of zorn's lemma

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SUMMARY

This discussion focuses on the application of Zorn's Lemma in establishing bijections between sets. It asserts that given a bijection \(\phi: X \to Y\) and subsets \(A \subset X\) and \(B \subset Y\) with a bijection \(\varphi: A \to B\), Zorn's Lemma guarantees a bijection \(\psi: X \to Y\) such that \(\psi(A) = B\). The conversation also explores whether this result holds for countable sets \(A\) and \(B\) without invoking Zorn's Lemma, concluding that a direct construction of a bijection \(\phi'\) is feasible through iterative fixing of elements.

PREREQUISITES
  • Understanding of bijections and their properties
  • Familiarity with Zorn's Lemma and its implications in set theory
  • Knowledge of countable sets and their characteristics
  • Basic concepts of mathematical induction for constructing bijections
NEXT STEPS
  • Study the formal proof of Zorn's Lemma and its applications in set theory
  • Explore the concept of countable sets and their bijections in detail
  • Learn about the construction of bijections through iterative methods
  • Investigate other mathematical principles that relate to bijections, such as the Cantor-Bernstein-Schröder theorem
USEFUL FOR

This discussion is beneficial for mathematicians, particularly those specializing in set theory, logic, and abstract algebra, as well as students seeking to deepen their understanding of bijections and Zorn's Lemma.

jostpuur
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Assume that there exists a bijection \phi:X\to Y. Also assume there exists some subsets A\subset X and B\subset Y such that a bijection \varphi:A\to B exists too. Now Zorn's lemma implies that there exists a bijection \psi:X\to Y such that \psi(A)=B.

I think I have now understood how to apply Zorn's lemma to things like this, so the above claim is clear to me (I assume).

My question is that if A and B and countable, will the result also hold without Zorn's lemma?
 
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Let A={a1, a2, ...} and B={b1, b2, ...}
This is possible as both are countable.

Now you can just loop over the elements of A and B and "fix" ϕ to get a ϕ' with ϕ'(A)=B.
 
Ok, I guess you are right. I was originally conserned that when I try to fix \phi so that \tilde{\phi}(a_n)=b_n would hold, this step could interfere with the earlier fixings at points a_1,\ldots, a_{n-1}. But it seems now that it will not happen. But it's not trivial. Very difficult to see intuitively.
 

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