$\sup(S)$ does not belong to $S$? Problem:
can anyone come up with an ordering of $\mathbb{N}$ different than the standard one we know, where we can find a subset $S\subset \mathbb{N}$ in a way such that $\sup(S)$ exists in $\mathbb{N}$, but $\sup(S)$ does not belong to the set $S$? 
$\mathbb{N}$: the set of natural numbers.
 A: The simplest way is to maintain the standard order on $\mathbb{N} \setminus \{1\}$
and make $1 > n$ for all $n \ne 1$. That way, for $S = \mathbb{N} \setminus \{1\}$
you have $\sup S = 1 \notin S$. 
A: Let $A$ consist of the rationals together with $\sqrt{2}$. Then $A$ is countable, so there is a bijection $\phi$ from $\mathbb{N}$ to $A$. For any pair of natural numbers $a$ and $b$, we say that $a\prec b$  iff $\phi(a)\lt \phi(b)$ in the ordinary ordering of $A$.
Then $\prec$ is a "non-standard" ordering of $\mathbb{N}$.
Let $K\subset A$ consist of all numbers $x$ in $A$ such that $x\lt \sqrt{2}$, and let $S=\phi^{-1}(K)$.
Remark: As long as we do not ask for any kind of interaction between the ordering and the ordinary structure of $\mathbb{N}$, anything that can happen in a countably infinite set can be replicated in $\mathbb{N}$.
A: An example with applications in dynamical systems is Sharkovski's order. In it
$$
\sup\{2\,k+1:k\ge1\}=6.
$$
A: HINT: There is a bijection between $\Bbb N$ and $\Bbb Q$, and $\Bbb Q\cap(0,1)$ is such a set in $\Bbb Q$.
