Can A be Countable if f is 1-1 and Onto?

[cragar]

Homework Statement

Let n_1=min(n\in\mathbb{N}:f(n){\in}A)
As a start to a defintion of g:N→A, set g(1)=f(n_1)
Show how to inductively continue this process to produce a 1-1 function g from
N onto A.

The Attempt at a Solution

g(1)=f(n_1) so this is our base case for induction.
so g(2)=f(n_1+1)
If I understand this correctly g is a function that has input values of natural numbers and maps these to the set A.
So I guess I need to show that f(n) is in A and f(n+1) is in A
By definition f(n) is in A for all n, so f(n+1) is in A for all n.
Could I maybe do a proof by contradiction and assume that f(n+1) was not in A and show that it was because n+1 is in the Naturals, therefore it works for f(n), and f(n+1)

 

[jbunniii]

Are there some additional assumptions that you haven't listed? It is obviously not going to be possible to construct a bijection between \mathbb{N} and A unless A is countable. Also, there must be some assumption about f. In general, the set \{n \in \mathbb{N} : f(n) \in A\} could be empty, in which case the minimum isn't even defined.

 

[cragar]

ok, you sorry, It says there exists f:\mathbb{N} --->B which is 1-1 and onto.
Let A\subseteq B be an infinite subset of B.We must show that A is countable. The question was phrased in 2 parts and I thought they were separate.
Thanks for looking at that.