A finite collection of natural numbers has a largest element. Let $A$ be a nonempty finite collection of natural numbers. I want to prove that $A$ has a largest element, that is $A$ contains a number $m'$ such that $m'\ge a$ for every element $a$ in $A$.
I defined the set
$$M:=\{n\in\mathbb N:n+1>a,\forall a\in A\}$$
so by Well-ordering principle of $\mathbb N$, there is $m'\in M$ such that $m'\le m$ for all $m\in M$ which implies that $m'+1>a$ or $m'\ge a$ for all $a\in A$. Now while it seems easy that $m'\in A$, I can't show that. Can you help me, please?
 A: Consider ordering the elements in any which way and running a find max algorithm.  Just keep track of the max-so-far as you go, and after a finite number of steps you've checked them all, so your max-so-far is the global max.
A: One small correction is needed: the set $A$ (Edit and M) had better be non-empty. But I suppose this is implicitly assumed here.
Edit: In light of @Henning Makholm's comment on the boundedness, let me add that argument here for completeness. Since $A$ is finite, we can
let $a_1,\ldots, a_n$ be the elements of $A$. Then $a_1+a_2+\ldots+a_n+1>a_i$
for each $i=1,\ldots, n$, so $a_1+\ldots+a_n\in M$ hence $M\neq \emptyset$.
Anyway, you have concluded that the is an $m'\in M$, hence $m'+1>a$ for all $a\in A$.
In particular, $m'\geq a$ for all $a\in A$. Suppose $m'\neq a$ for any $a\in A$.
Then $m'>a$ for all $a\in A$ (since $m'\geq a$ but $m\neq a$). Furthermore, note
that since $A\neq \emptyset$, there is at least one $a\in A$ with $m'>a\geq 0$, so $m'-1\in \mathbb N$ and $m'>a$ for all $a\in A$. Thus by definition,
$m'-1\in M$. But then $m'-1<m'$, which contradicts that $m'$ was least in $M$.
A: Following the logic you already have.
$M$ is a subset of $\mathbb N$
Every subset of non-empty $\mathbb N$ has a smallest element.
The smallest element in M equals the largest element in $A.$
I suppose it is a still an open question that $M$ is non-empty.
A: If $A$ is finite and nonempty. It's elements can be listed as $x_1,.... x_n$.  And by Well ordering principal, $A$ has a minimum.  Call it $a$.
The sum $X = \sum x_i$ is finite.  As each $x_i \ge 0$ $X \ge x_j$ for all $j$.  (Let's include $0$ as a natural number.)
So the set $A^- = \{n -X|n \in A\}$ is a set of non-positive numbers and has a minimum number (namely $a - X$).
The set $-A^- = \{-v|v  \in A^-\}$ is a set of non-negative numbers and by the well ordering principal has a minimum element.  Call it $v$. And and it has a maximum element $-\min A^- = X - a$.
So $A^-$ has a maximum element in $-v$ and a minimum in $a -X$.
An $A$ has a maximum element in $X -v$ as well as the minimum in $a$.
Of course, this assumes any finite sum is finite.  Which may be circular.  The basic induction axioms should show you can exhaust anything finite.
A: What you're doing doesn't really work, because you can't apply the well-ordering principle without first showing that your $M$ is non-empty -- so you need to show that $A$ is bounded, and that itself seems to be roughly as involved as showing that it has a maximal element.
Instead, a way forward would be to forget that $A\subseteq\mathbb N$ specifically and prove generally

A finite (but nonempty) set $A$ of elements from any totally ordered set contains a largest element.

Badam Baplan's answer gives an informal sketch for how to do that. More formally, you can use induction on $|A|$.
