I had the following idea but it doesn't look like a proper proof:
Let's say $\mathbb N$ is finite. Now consider an arbitrary number $k$ in $\mathbb N$ such that it is the biggest number in the set $\mathbb N$. Now let's consider the number $k+1$. $k+1$ is in $\mathbb N$ and is greater than $k$. Hence we can conclude that $\mathbb N$ has no greatest element because I can always add $1$, and thus is infinite.
EDIT 1: I just want to prove it using basic set theory , without using cardinality or other complicated stuff.
EDIT 2: I want to know if this is a formal proof that proves that the natural number set is infinite, supposed we do not know anything about the natural number. In other words i want to know if this can work as stand alone proof.
EDIT 3: I can use the following : If the size of the natural numbers set $\mathbb{N}$ is less than or equal to the size of a set $F$ then $F$ is infinite. $\mathbb{N}$ is less than or equal to $F$ iff there is an injective function from $\mathbb{N}$ to $F$
Best Answer
This argument works, though depending on what facts you are allowed to assume there may be some details that need to be filled in. In particular, you should probably say more about how you know that any finite subset of $\mathbb{N}$ has a greatest element (so if $\mathbb{N}$ were finite, then it would have a greatest element). For instance, if your definition of "finite" is "in bijection with $\{m\in\mathbb{N}:m<n\}$ for some $n\in \mathbb{N}$", then you could prove this by induction on $n$. That is, prove by induction on $n$ that if $A\subseteq\mathbb{N}$ and there is a bijection from $A$ to $\{m\in\mathbb{N}:m<n\}$, then there is some $k\in A$ such that $k\geq a$ for all $a\in A$.