Proof involving numerical equivalence of sets

1. The problem statement, all variables and given/known data



Show that for a set A[itex]\subset[/itex]N, which is numerically equivalent to N=Z₊, and the set B = A [itex]\cup[/itex]{0}, it holds that A and B are numerically equivalent, i.e., that A [itex]\approx[/itex]B



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





2. Relevant equations







3. 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 [itex]\exists[/itex]f:A→N such that f is a bijection. Therefore, f:A→N such that Im[f] = N and [itex]\forall[/itex]x₁, x₂[itex]\in[/itex]A, x₁[itex]\neq[/itex]x₂→f(x₁)[itex]\neq[/itex]f(x₂). Because A [itex]\subset[/itex]N, 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 [itex]\forall[/itex]n[itex]\in[/itex]A. Then we can define a function h(n) = g(n-1). Because B = A[itex]\cup[/itex]{0}, g:A→B is a bijective function. This is true b/c Im[g] = B and f(x₁) [itex]\neq[/itex]f(x₂)→g(x₁-1)[itex]\neq[/itex]g(x₂-1). Thus, we have found a bijection g:A→B, and therefore A[itex]\approx[/itex]B. This concludes the proof.

1. The problem statement, all variables and given/known data







2. Relevant equations







3. The attempt at a solution