I tried to prove the well-ordering principle but I’m not sure whether or not it’s rigorous or even valid. Feedback would be greatly appreciated:
Let S be a nonempty set of positive integers.
S is bounded below by 0 and therefore, $\inf S$ exists.
But we know that for any $\phi > \inf S$, there must exist some “a” such that
$a \in S$ and $a < \phi$
Let
$ \phi = \inf S + \varepsilon$
where $\varepsilon > 0$
$$a < \inf S + \varepsilon $$
but since $a \in S$
$$a \geq \inf S$$
And I ended up with the inequality,
$ \inf S \leq a < \inf S + \varepsilon $ (1)
Meaning there must exist some element $a \in S$ that satisfies the inequality.
if $\inf S \notin Z^{+}$, there exists $\varepsilon$ such that no element of S could satisfy the inequality:
set $\varepsilon = k – \inf S$, where k is a positive integer.
$\inf S = k – \varepsilon$
Plugging that into (1)
$$k – \varepsilon \leq a < k$$
For $0 < \varepsilon < 1$, there exists no integer a that satisfies the inequality which is a contradiction.
Thus, $\inf S \in Z^{+}$ (2)
By definition,
$\inf S \geq 0$, since $0$ is a lower bound of S (3)
From (2) and (3), $\inf S \in S$ which means that $min S$ exists.
EDIT: I replaced $\frac{1}{n}$ with $\varepsilon$ to generalize the problem further without making the assumption that $\varepsilon$ is rational and also because $\frac{1}{n}$ basically served no purpose.
EDIT 2: I was advised against defining the set to make the proof more general.
Best Answer
You seem to want to prove the well-ordering of the natural numbers from the completeness of the reals. OK, although this looks like a petitio principii.
Let $S$ be a nonempty set of natural numbers. It is certainly bounded below by $0$, so it has an infimum $a$.
You want to prove that $a$ is actually the minimum of $S$. By the properties of the infimum, there exists $s\in S$ such that $a\le s<a+\frac{1}{2}$.
If $a\notin S$, then $a<s$, so there exists $t\in S$ such that $a<t<s$. However, this is a contradiction, because $s-a>s-t\ge1$, whereas $s-a<1/2$.
Basically, we can choose an interval starting from the infimum that can contain no more than one natural numbers. Because of the properties of the infimum, such interval must contain one natural number, but if the infimum is not the minimum, each interval starting from it must contain infinitely many members of the set. Now you have to prove the lemma: the difference between two distinct natural numbers is at least $1$.
This requires induction, but not the full strength of the well-ordering principle.