A subset of $\mathbb{N}$ is finite if and only if it has a maximum element I have been stumped on both directions of the proof for quite some time now.
For the forward direction, I initially naively thought I could use the definition of finite (bijection $f: S \rightarrow  [m] $ for some $m \in \mathbb{N}$, and some subset of $S\subseteq\mathbb{N}$) and declare $f^{-1}(m)$ to be the greatest element; I can't because the bijection does not guarantee the mapping in order.
I have thought about using a recursive definition but I feel like there is another easier way to go about doing this with maybe composition of functions?
Any suggestions and advice?
 A: For the $\Rightarrow$ direction, I will use the following fact about the cardinality of finite sets: 
$|A\cup B| = |A| + |B|$ if $A\cap B =\emptyset$. 
That is the case because $A$ has a bijection with $\{1,\ldots,n\}$, $B$ has a bijection with $\{1,\ldots,m\}$, which is itself bijective with $\{1+n,\ldots,m+n\}$, and a more polished proof can be given with that $A\cup B$ is bijective with $\{1,\ldots,m+n\}$.
Now, every finite subset of 1 element has a maximum (trivial). Next, suppose that every finite subset of $k$ elements has a maximum, let's prove that any finite subset of $k+1$ elements has a maximum. Let $X$ be a finite subset of $\mathbb{N}$ with $k+1$ elements. Since $X$ is not empty, let $x \in X$. So $X = \{x\} \cup (X\setminus \{x\})$.
Since $(X\setminus \{x\})\cap \{x\} = \emptyset$ , $|X \setminus \{x\}|=k$ and the induction hypothesis apply. Then there is a maximum $y$ for $X \setminus \{x\}$, now one can easily see that $M = \max \{x,y\}$ is the maximum of $X$.
For the $\Leftarrow$ direction:
Induction on the maximum element.
