I am trying to prove that if there is an injective mapping $f:X\to \mathbb{N}$, then $X$ is finite or countable.
My proof goes like this:
Suppose that $X$ is infinite and uncountable.
Since $X$ is uncountable, there is no bijection for $X\to \mathbb{N}$, nor for $\mathbb{N} \to X$.
We know that since $f$ is injective, then in order to follow the definition that $X$ is uncountable, $f$ must not be surjective.
However, since $X$ is uncountable and infinite, it is clear that $|X| >> |\mathbb{N}|$.
Since $f$ is injective, each and every fibre of $X$ must map to some infinitely countable element $n \in \mathbb{N}$ uniquely.
This forces $f$ to be surjective. Which causes a contradiction because $f$ is now a bijection while there cannot be one for $X \to \mathbb{N}$.
Therefore, $X$ must be finite or countable if $f:X \to \mathbb{N}$ is injective.
I am sceptical about this proof because of the parts in italics. I feel that I have made some incorrect assumptions/statements.
To solidify my conceptual knowledge of the problem, could someone help me?
Best Answer
An easy approach
Since $f:X\rightarrow \mathbb N$ is injective hence $\lvert X\rvert \le \lvert\mathbb N\rvert$ which proves that cardinality of $X$ cannot be greater than that of $\mathbb N$.