Set $A$ is countably infinite if and only if there exists a bijection $f: \mathbb{N} \rightarrow A$

Question

Using the fact that for any $A$, $A$ is countably infinite if there exists a bijection $f: A \rightarrow \mathbb{N}$, how do I prove the statement:

$A$ is countably infinite if and only if $\exists$ bijection $f: \mathbb{N} \rightarrow A$?

Answer 1

Hints:

1. If $f: A \rightarrow \mathbb N$ is a bijection , then $f^{-1}: \mathbb N \to A$ is a bijection.

2. If $g: \mathbb N \to A$ is a bijection, then $g^{-1}: A \rightarrow \mathbb N$ is a bijection

Answer 2

If by $N$, you mean $\mathbb{N}$, then just remember that $|\mathbb{N}| = \aleph_0$. Now since, the set of natural numbers, $\mathbb{N}$, is countably infinite(this is exactly what $|\mathbb{N}| = \aleph_0$ means) and if there is such a bijection, $f$, for the set $A$ such that $f: A \mapsto \mathbb{N}$ (you are already given one-to-one correspondence), then by the property of transitivity of cardinality, $|A| = |\mathbb{N}|$, which further implies $|A|= \aleph_0$. Which just simply means that the set $A$ is countably infinite.