Does for any infinite set $M \subseteq \Bbb N$ hold $|M| =|\Bbb N|$?

Question

In a proof I do a case split about a set $M \subseteq \Bbb N$. In one case $M$ is finite and it means that there’s some bijection only between n elements of $\Bbb N$ and all the elements of $M$, so $|M| \lt |\Bbb N|$. In the other case M is infinite. Can I now conclude/prove that $|M| = |\Bbb N|$? I have no theory of cardinal numbers available. I still feel it must be provable that there cannot be some set $M \subseteq \Bbb N$ that is infinite but still somehow $|M| \le |\Bbb N|$. (If indeed it was not provable I‘d like to just assume it because it‘s natural because then the proof would formally work, right?)

Answer 1

The proof is simple: let

$$a_1=\min(M)$$

$$a_2=\min(M-\{a_1\})$$

$$a_3=\min(M-\{a_1,a_2\})$$

$$\vdots$$

$$a_n=\min(M-\{a_1,a_2,...,a_{n-1}\})$$

$$\vdots$$

Since $M$ is a subset of the natural numbers, you are assured that the minimum is well defined and exists at every step. Also, since $M$ is infinite, this process will never end. Now, does this sequence $a_n$ encompass every element of $M$? Said differently, for all $b\in M$, is there $N\in\mathbb{N}$ such that $a_N=b$? This is obviously true as after $b$ steps we have that $a_b\geq b$. Thus, $b=a_N$ for some $1\leq N\leq b$. Finishing the proof, note that the sequence $a_i$ puts the elements of $M$ into a bijection between the elements of $\mathbb{N}$, which is the definition of countable.