Find a set A (subset of R,set of real numbers) and an element a of R

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SUMMARY

The discussion focuses on finding a set A, a subset of real numbers R, and an element a of R such that there is no bijection between the set A and the set A + a (where a is added to each element of A). Participants confirm that both the empty set and finite sets like A = {1, 2, 3} with a = 0 serve as valid examples. The empty set is established as a subset of R, and it is noted that A + a results in a different cardinality, preventing a bijection.

PREREQUISITES
  • Understanding of bijections in set theory
  • Familiarity with subsets and cardinality
  • Basic knowledge of real numbers (R)
  • Concept of set addition (A + a)
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  • Explore examples of finite and infinite sets
  • Learn about cardinality and its implications in mathematics
  • Investigate the concept of set operations, particularly addition of elements to sets
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Mathematicians, students studying set theory, and anyone interested in understanding the properties of sets and bijections in real numbers.

julia89
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Find a set A (subset of R,set of real numbers) and an element a of R
such that there is no bijecton from a+A(we add a to the set A)to A.

I can't find a good example. Can someone help
Are we done if we choose the empty set? (And is the empty set a subset of R?)

Thank you
 
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?? Any finite set will do. Take A= {1, 2, 3} and a= 0. Then A has 3 members while A+ a has 4. There cannot be a bijection from one to the other.

Yes, the empty set is a subset of R (it is a subset of any set). Choosing A= {} and a any real number so that A+ a= {a} will also work.
 

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