If $X$ is an infinite set and there exists an injection $X \to \mathbb{N}$, is there also a bijection?

Question

Full question is in the title. It seems to me the answer is yes, because we can just order the numbers in the image of the function from starting from the least, and then there's 1-1 (is there?) correspondence with $\{1, 2, 3 , \dots \}$, but obviously this isn't a rigorous argument. Help is appreciated.

Answer 1

Yes. If there is an injection $f\colon X\to\mathbb N$ we can define the following:

$$g(x) = |\{y\in X\mid f(y)<f(x)\}|$$

Where $|A|$ denotes the cardinality of $A$. This is a well-defined function because $g(x)$ is the cardinality of a bounded set of natural numbers, and therefore it is a natural number itself.

We can show by induction that $g$ is injective, and that its range is an initial segment of $\mathbb N$. If $X$ is infinite then $g$ is also surjective, because there is only one initial segment of $\mathbb N$ which is infinite... $\mathbb N$.