Proving $\mathbb N$ isn't bounded In this video (@ ~33 mins) we're trying to prove a part of the Archimedean property:
(i) For any $x \in \mathbb R$, there is an $n \in \mathbb N$ such that $ x < n$.
To do so the lecturer uses an "equivalent formulation of a supremum", video here (@ ~1:18:30):
Proposition: suppose $A \subseteq \mathbb R$ with upper bound $l$, denoted $A \leq l$. Then $l= \sup A$ if and only if for all $\epsilon > 0$, there is an $a \in A$ in the interval $(l-\epsilon, l]$.
Now, back in the proof of (i), the video proceeds as follows:
Proof of (i)
Suppose for the sake of contradiction that $\mathbb N \leq x$ for some $x \in \mathbb R$. Then, by the Axiom of Completeness, there is a least upper bound $l = \sup \mathbb N$. Then we know that there must be an $n \in \mathbb N$ such that $n \in (l-1, l]$. The proof then goes on but we now arrive at...
The confusing part/my question: in the above to get $n \in (l-1, l]$ we are taking $\epsilon = 1$ from the Proposition. However, the lecturer says in voice over that we could take $\epsilon$ to be any value, since the Proposition holds for any $\epsilon > 0$. But how can that be since we know that the elements of $\mathbb N$ are spaced out s.t. $|m-n|=1$ for any $m,n \in \mathbb N$ and $m \neq n$?
Phrased more loosely in words: I don't understand how we can take $\epsilon$ to be small and still expect to find a natural number in the interval $n \in (l-\epsilon, l]$.
 A: It might help to look at the case of a subset that is bounded, for instance $S=\{1,2,3\}$. The supremum of $S$ is $3$, and it is indeed true that for any $\epsilon > 0$, there is an element of $S$ in $(3-\epsilon,3]$: namely, $3$ itself.
Now back to your proof: the argument is that if $\mathbb{N}$ were bounded with supremum $l$, then there would be a natural number in $(l-\epsilon, l]$ for any $\epsilon>0$. Note that this is not claiming that there is a natural number in $(x-\epsilon, x]$ for any arbitrary real number $x$: only when $x=l$, the hypothesized supremum of $\mathbb{N}$.
A: Taking $\epsilon=1$ is convenient and just for the proof. With that he can do the trick to get the $l<n+1$. About finding a natural in $(l-\epsilon,l]$, since you are assuming that for every $\epsilon >0$ you can find a natural in that interval and, as you said naturals are spaced for at least one, if you take $\epsilon=\frac{1}{2}$ you get that there is a natural in $(l-\frac{1}{2},l]$. If you take $\epsilon=\frac{1}{3}$ you get a natural in $(1-\frac{1}{3},l]$. As you take $\epsilon$ smaller you get a smaller interval near $l$, the coclussion is that the natural in $(l-\epsilon,l]$ is $l$ itself.
