Proof involving functions and intersection

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SUMMARY

The proof that f(∩Tα) = ∩f(Tα) holds for all choices of (Tα) where α ∈ λ if and only if the function f is one-to-one. The discussion emphasizes the necessity of assuming f is one-to-one to demonstrate the inclusion of elements in both sets. Additionally, it suggests using the contrapositive approach to establish a counterexample when f is not one-to-one. This logical structure is crucial for understanding the relationship between functions and their intersections.

PREREQUISITES
  • Understanding of set theory and intersections
  • Familiarity with functions and their properties, particularly one-to-one functions
  • Knowledge of logical proofs, including direct proof and contrapositive
  • Basic concepts of order isomorphisms
NEXT STEPS
  • Study the properties of one-to-one functions in depth
  • Learn about set intersections and their implications in function mappings
  • Explore logical proof techniques, focusing on direct and contrapositive methods
  • Review resources on order isomorphisms and their applications in mathematics
USEFUL FOR

Mathematicians, students studying advanced mathematics, and anyone interested in the properties of functions and set theory will benefit from this discussion.

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Prove that [tex]f(\cap T_\alpha)=\cap f(T_\alpha)[/tex] for all choices of [tex](T_\alpha) \alpha \in \lambda[/tex] if and only if f is one-to-one.

I've been working on this on and off for a day and have nothing to show for it... Any help pointing me in the right direction would be appreciated.

Also, more important then this question, can any of you recommend additional resources to review this type of stuff, mostly functions, order isomorphisms, etc. The book I am using does not give as many examples as I'd like.
 
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It's not really that hard. First assume f is one-to-one. Now can you show that if y is an element of [itex]f(\cap T_\alpha)[/itex] then it is an element of [itex]\cap f(T_\alpha)[/itex] and vice versa? For the opposite direction I'd do the contrapositive. Show if f is not one-to-one then you can find a counterexample.
 

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