Dual Question: Uniqueness of Suprema and Definition of Bounded Set In my book on elementary real analysis, there is a definition promptly on the third page of the book (in particular, before defining any topological concepts):
''Suppose $S$ is an ordered set and $E \subset S$. If there exists a $\beta\in S$ such that $x \leq \beta$ for every $x \in E$, we say that $E$ is bounded above
and call $\beta$ an upper bound of $E$.''
My question first question is: is it correct to read this as 
"$E$ is bounded above if both the following are true


*

*There exists an ordered superset $S$ of $E$, and 

*There exists a $\beta\in S$ such that $x \leq \beta$ for every $x \in E$ ?
My second question is: Would this then imply that the supremum of $E$ is only unique up to choice of ordered superset? Is it possible that the supremum of $E$ in $S_1$ could be distrinct from the supremum of $E$ in $S_2$ for two ordered sets $S_1$ and $S_2$ which both satisfy $S_i \supset E$, $i=1,2$ ?
Thank you.
 A: About the first question.
No, the ordered set is given, and it doesn't make sense to say that a set is bounded if it is not ordered.
About the second question.
If you endow $S$ with two different orders, it is indeed possible that the supremum of $E \subseteq S$ is different for one order or the other.
It is even possible the $E$ has a supremum for one order but not for the other, or even that it has a supremum for one order and it is not bounded above for the other.
As an example, take $E=S=\{1,2,3\}$, and define $\leq$ to be the natural order in $S$ (inherited from the order in the natural numbers: $x \leq y$ if $x + z = y$, for some $z$), that is $1 \leq 2 \leq 3$, and define $\lesssim$ by $x \lesssim y$ if $x$ divides $y$ (this gives a partial order relation).
So $1 \lesssim 2$ and $1 \lesssim 3$ and that's all, that is,
$$\lesssim \;= \{(1,1),(1,2),(1,3),(2,2),(3,3)\}.$$
Now $3$ is the supremum of $E$, with respect to $\leq$, but $E$ is not bounded above, with respect to $\lesssim$ and so it doesn't have a supremum.
