Set theory, functions, bijectivity

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SUMMARY

The discussion focuses on the implications of bijectivity in set theory, specifically examining the function f:X→Y and the subset A⊆X. It establishes that if f(Ac)=[f(A)]c, then the function f is bijective. The participants analyze the definitions of injective and bijective functions, emphasizing that f(x)=f(y) implies x=y, and for every y in Y, there exists an x in X such that f(x)=y. The equivalence of these conditions is critical for proving bijectivity.

PREREQUISITES
  • Understanding of set theory concepts, including subsets and complements.
  • Familiarity with functions, particularly the definitions of injective and bijective functions.
  • Knowledge of mathematical notation and implications in function theory.
  • Basic skills in logical reasoning and proof techniques.
NEXT STEPS
  • Study the definitions and properties of injective, surjective, and bijective functions in detail.
  • Explore examples of bijective functions and their applications in set theory.
  • Learn about the implications of function composition and its relation to bijectivity.
  • Investigate the role of set complements in function mapping and their impact on bijective proofs.
USEFUL FOR

Mathematicians, students of mathematics, and educators looking to deepen their understanding of set theory and function properties, particularly in the context of bijectivity and its implications.

Daveyboy
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f:X\rightarrowY, A\subsetX
f(Ac)=[f(A)]c implies f bijective.

Just trying to apply the definitions of injective and bijective. The equivalence makes sense but I am having a hard time showing it.

f(x)=f(y) implies x=y and for every y in Y there exists a x in X s.t. f(x)=y.

I mean all I have is if f(x)=f(y)
then f(X\x)=Y\f(x\x)=f(X\y)-Y-f(X\y)...
 
Last edited:
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what's the question?
 
Daveyboy said:
f(Ac)=[f(A)]c implies f bijective.
this implication
 

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