# $\sup(S)$ 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.

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An example with applications in dynamical systems is Sharkovski's order. In it $$\sup\{2\,k+1:k\ge1\}=6.$$

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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 $\sup S = 1 \notin S$.

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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 $\mathbb{N}$.

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:Literally. what do you mean by the notation $R(a,b)$? –  C. Lambda Oct 10 '12 at 3:25
@C.Lambda: Yes, sort of. The relation $R$, as defined, is an order relation on $\mathbb{N}$. It is not the ordinary order on $N$, so I did not want to confuse things by using the ordinary order symbol. But maybe I will search my table of symbols for something that looks more like $\lt$. –  André Nicolas Oct 10 '12 at 3:29
It is $\phi^{-1}(\sqrt{2})$. The function $\phi^{-1}$ is just a way of giving the elements of $A$ natural number names. –  André Nicolas Oct 10 '12 at 3:35
I think it is not worthwhile for you to write out the details. If you really want to, first write out a proof that $\sqrt{2}$ is the sup of the set of rationals $\lt \sqrt{2}$. Certainly is upper bound, and nothing cheaper will do, since if we pick any element $b\lt \sqrt{2}$ of our set, there is always another rational $c$ such that $b\lt c\lt \sqrt{2}$. Then translate that back into our convoluted $\phi^{-1}$, which codes ordinary facts into facts about $\mathbb{N}$ and the funny order. The translation task is so mechanical that once you see it can be done, it becomes not worth doing. –  André Nicolas Oct 10 '12 at 5:20

HINT: There is a bijection between $\Bbb N$ and $\Bbb Q$, and $\Bbb Q\cap(0,1)$ is such a set in $\Bbb Q$.

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