Let $S$ be an ordered set. Let $A$ subset of $B$ be a nonempty subset that is bounded above. Let $S$ be an ordered set. Let $A \subset S$ be a nonempty subset that is bounded above. Suppose $\sup A$ exists and $\sup A$ is not in $A$. Show that $A$ contains a countably infinite subset. In particular, $A$ is infinite.
Let $x_0 ∈ A $
then $x_0$ is not equal to $\sup(A)$ since $x_0 ∈ A$
Then, by definition of $\sup (A)$, there has got to be another element $x_1 ∈ A$ such that $x_1 > x_0 $ 
I don't understand why there has got to be another element $x_1$ 
I know the definition of $ \sup (A)$. 
$\sup (A)$ is greater than or equal to all elements of $A$. 
But in problem $\sup A$ is not in $A$. From that I can conclude $x_0 < \sup A$ 
My question is: Why there has to be another element? 
 A: Otherwise $\forall x \in A$, $x \leq x_0$, then $x_0$ is upper bound of $A$, let $y$ another upper bound of $A$, then $y \geq x_0$  then $x_0$ is the smaller upper bound, therefore $x_0 = sup(A)$.
A: Let $x_0 \in A \subset S$. Suppose that there does not exist another $x_1 \in A$ such that $x_1 > x_0$. This implies that $x_0$ is an upper bound of $A$
$$\forall x \in A \implies x < x_0$$
Because $S$ is ordered we have that $A$ is ordered and then we can have the above implication. Now this is the least upper bound possible. If there is another upper bound, say $u$, we must have $u \geq x_0$ because if it is such that $u < x_0$ then, by definition, $u$ is not an upper bound, because $x_0 \in A$. The upper bound must be greater (or equal) than all numbers of the set. So, with this, we get that $x_0$ is the supremum.  
But this is not possible because the supremum cannot be an element of $A$ so there must exist $x_1$. Because $x_0$ was arbitrary we must have that $A$ is infinite. 
Note that you use two absurd arguments: If does not exist and element grater I get an upper bound, using absurd argument this must be supremum, using absurd argument this is impossible. Tried to make it clear. Comment for EDIT's such that you can understand.
A: Importantly, $\sup{A}$ is the least upper bound on $A$. If there isn't a member of $A$ larger than $x_0$, then $x_0$ is itself an upper bound on $A$ - that's just what "upper bound" means. Since $\sup{A}$ is the least upper bound, it must be that $\sup{A} \leq x_0$. But $\sup{A}$ is certainly an upper bound, so $x_0 \leq \sup{A}$. So we would have $x_0 = \sup{A}$, and since $x_0 \in A$ we would have $\sup{A} \in A$. Since that can't be the case, it must be that there is a member of $A$ larger than $x_0$.
