Proof involving numerical equivalence of sets

[eclayj]

Homework Statement

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.

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 [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.

 

[haruspex]

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.