# Proof involving numerical equivalence of sets

1. May 30, 2014

### eclayj

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

Show that for a set A$\subset$N, 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 $\approx$B

Hint: Recall the definition of A≈B and use the fact that A is numerically equivalent to N. Note that 0 $\notin$ 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 $\exists$f:A→N such that f is a bijection. Therefore, f:A→N such that Im[f] = N and $\forall$x1, x2$\in$A, x1$\neq$x2→f(x1)$\neq$f(x2). Because A $\subset$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 $\forall$n$\in$A. 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) $\neq$f(x2)→g(x1-1)$\neq$g(x2-1). Thus, we have found a bijection g:A→B, and therefore A$\approx$B. This concludes the proof.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. May 30, 2014

### 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.