Proof involving numerical equivalence of sets

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SUMMARY

The proof demonstrates that for a set A, which is a subset of natural numbers (A ⊆ ℕ) and is numerically equivalent to ℕ (denoted as A ≈ ℕ), the set B = A ∪ {0} is also numerically equivalent to A (A ≈ B). The proof constructs a bijective function g: A → B, leveraging the properties of bijections and the inclusion of 0 in set B. The function h(n) = g(n-1) is defined to establish the bijection, confirming that A and B are numerically equivalent.

PREREQUISITES
  • Understanding of set theory and numerical equivalence
  • Familiarity with bijective functions and their properties
  • Knowledge of the natural numbers and their notation
  • Basic concepts of mathematical proofs and logic
NEXT STEPS
  • Study the definition and properties of bijections in set theory
  • Learn about the concept of numerical equivalence in more depth
  • Explore examples of proofs involving set operations and equivalence
  • Investigate the implications of adding elements to sets in terms of equivalence
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Mathematics students, particularly those studying set theory and proofs, as well as educators looking for examples of numerical equivalence in mathematical discussions.

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Homework Statement



Show that for a set A\subsetN, which is numerically equivalent to N=Z+, and the set B = A \cup{0}, it holds that A and B are numerically equivalent, i.e., that A \approxB

Hint: Recall the definition of A≈B and use the fact that A is numerically equivalent to N. Note that 0 \notin N.


Homework Equations





The Attempt at a Solution



I really have little clue of how to complete this proof, this is sort of a wild guess, any help appreciated:

It is given that A≈N. This means \existsf:A→N such that f is a bijection. Therefore, f:A→N such that Im[f] = N and \forallx1, x2\inA, x1\neqx2→f(x1)\neqf(x2). Because A \subsetN, and A≈A by definition, then there is a function g:A→A such that g is a bijection. This describes the function g(n) = n for \foralln\inA. Then we can define a function h(n) = g(n-1). Because B = A\cup{0}, g:A→B is a bijective function. This is true b/c Im[g] = B and f(x1) \neqf(x2)→g(x1-1)\neqg(x2-1). Thus, we have found a bijection g:A→B, and therefore A\approxB. This concludes the proof.
 
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What are the domain and range of h? For a given n in the domain of h, how do you know n-1 is in the domain of g?
Thinking in terms of the function f was a good start, but I don't see where you made use of it. Try combining f with the n-1 idea.
 
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