Proving countable set using a function that is one-to-one My problem reads: 

Prove that if there is a function $f\colon A\rightarrow \mathbb{N}$ that is one-to-one, then $A$ is countable

Assuming $\mathbb{N}$ to be the set of natural numbers. 
I am not too sure how to go about proving this. Would I need the definition of denumerable in this case? or can I use the cardinality of A < or equal to N?
 A: If $f\colon A\rightarrow \mathbb{N}$ is one-to-one, then $f\colon A\rightarrow f(A)$ is bijective. Now, since $f(A)\subseteq \mathbb{N}$, then by this result $f(A)$ is countable, which gives us that $A$ is countable too.
A: After much thought, I've decided this isn't entirely a triviality.  If we assume the following definitions.
1) countable means either finite or denumerable.  Finite means there is a bijection from some {1,....., n} to the set.  Denumerable means there is a bijection from $\mathbb N$.  (Frankly I never use the term "denumerable" and use "countably infinite" instead.  Furthermore I assume in context that "countable" should be assumed infinite if not explicitly stated to be finite.)
2) 1-1 means injective but not surjective.  I.e. for every $x \in A$ there is exactly one and only one $z \in f(A)$ so that $f(x) = z$.  (Frankly, I never use 1-1 to mean injective and I always mean it to mean bijective.  $f:A \rightarrow f(A)$ is always surjective and we can always for $f(A) \subsetneq X$ to make $f:A \rightarrow X$ not surjective if we wanted to so to say something misleading like $f:\mathbb R \rightarrow \mathbb R: f(x) = e^x$ is one to one because it is injective [but not surjective] is pointless and ... unsporting.)
So if we interpret the statement to be:
Prove: if  an injection $f:A \rightarrow \mathbb N$ exists, then $A$ is countable.
That's not quite so trivial after all.
A: Let A be a non-empty set. Then the following are equivalent.
(a) A is countable.
(b) There exists a surjection $f :\mathbb N → A $.
(c) There exists an injection $g : A → \mathbb N$.
Proof. $(a) \Rightarrow (b)$ If $A$ is countably infinite, then there exists a bijection $f :\mathbb N → A $ and then (b) follows. If $A$ is finite, then there is bijection
$h : ${$1,... ,n$} $→ A$ for some $n$. Then the function $f :\mathbb N → A $ defined by
$$f(i) = 
\left\{
   \begin{array}{ll}
h(i) 1 \leq i \leq n\\
h(n) i > n. 
   \end{array}
\right.$$
is a surjection.  
$(b)\Rightarrow (c)$. Assume that $f : \mathbb N → A $ is a surjection. We claim that there is an injection $g: A →\mathbb N$. To define $g$ note that if $a \in A$, then $f^{-1}$({a})$6=∅$.
Hence we set $g(a) = \min f^{-1}$({a}). Now note that if $a \neq a'$then the sets $f^{-1}$({a})$\cap f^{-1}$({a′})= $∅$ which implies $\min^{−1}$({a})$\neq \min^{−1}$({a′}). Hence $g(a)\neq g(a')$ and $g : A → \mathbb N $ is an injective.  
$(c)\Rightarrow (a)$. Assume that $g : A → \mathbb N$ is an injection. We want to show that $A$ is countable. Since $g : A → g(A)$ is a bijection and $g(A) \subset N$, and as we know that any subset of a countable set is countable, it implies that $A$ is countable.
