When doesn't a supremum exist? Other than ∞, is there another case where a supremum (or an infimum for that matter) doesn't exist?
 A: 
Supremum axiom: Any nonempty subset $A\subset\mathbb{R}$ which is bounded above has a supremum $s\in\mathbb{R}$.

This is only valid in $\mathbb{R}$. For example, the set $\{x\in\mathbb{Q}:x\geq 0, \ x^{2}<2\}$ does not admit supremum, since $\sqrt{2}\notin\mathbb{Q}$. 
A: Yes, depends on the partial ordering you define. A partial ordering for which a bounded set have sup and inf is called complete. 
An example of an incomplete ordering would be on set $\lbrace a, b, c \rbrace$, $a < b$, $a < c$, but $b$ and $c$ are not comparable. Then the subset $\lbrace a, b, c \rbrace$ do not have a supremum
Note that the real numbers are complete, therefore supremum exists for all bounded sets
A: Within the extended line $[-\infty,+\infty] = \mathbb R\cup \{\pm\infty\}$ every subset has a supremum and an infimum.  Within the line $(-\infty,+\infty) = \mathbb R$ every subset has a supremum and and infimum except when the supremum or infimum within the extended line is $-\infty$ or $+\infty$.  For example
$$
\sup \{1,2,3,\ldots\} = +\infty
$$
and
$$
\sup\varnothing = -\infty.
$$
