$\emptyset$ does not have a supremum. Ghorpade-Limaye, A couse in Calculus and Real Analysis, p.5) says that $\emptyset$ does not have a supremum but there is not any explanation.
My question is: Why?
 A: Well, a supremum of a set $A$ is a number $a$ such that 


*

*$a$ is $\ge$ every element of $A$ (such an $a$ is called an upper bound of $A$), and

*$a$ is $\le$ any number $b$ with the previous property.
Now, here's the problem:

What isn't an upper bound of the empty set?

For example, $4$ is an upper bound of $\emptyset$. Why? Well, $4$ is $\ge$ every element of the emptyset. You don't believe me? OK, find me an element of the emptyset which is not $\le 4$.
(Similarly, every element of the emptyset is an elephant.)
So the problem is that every real number is an upper bound of $\emptyset$. But there is no least real number, so the emptyset doesn't have a least upper bound - that is, the emptyset has no supremum.
A: Every real number is an upper bound for $\emptyset$. So there is no smallest real number. The same reasoning applies for the infimum. What is interesting, however, is that you can use this reasoning to find $\sup\emptyset = -\infty$ and $\inf\emptyset = \infty$ in the extended reals.
A: Every real number is an upper bound of $\emptyset$. That is to say, the set of all upper bounds of $\emptyset$ is $\Bbb{R}$, which has no least element.
A: The supremum is the least upper bound. What's an upper bound of $\emptyset$? It is $a\in\mathbb R$ such that $x\leq a$ for all $x\in \emptyset$. Since no such $x$ exists, either you interpret that all $a$ are upper bounds, so there is no least upper bound, or that no $a$ is an upper bound, in which case there is no least upper bound either. 
