Does the empty set have a supremum or infimum? I actually dont need this question to be answered for university or something, it was just a discussion I had with a friend.
Does the empty set have an upper or lower bound?
Since the supremum is the minimal bound it must be $-\infty$ since  there is no element in the empty set such that it is bigger than $-\infty$. It somehow makes sense. On the other hand the infimum is $\infty$. 
Can one define something like that for the empty set? Or is it just heuristic...?
 A: Let $X$ be a set equipped with an order relation $\le$. The supremum of a subset $A$ is computed (if it exists) in two steps.


*

*Consider the set $UB(A)$ of upper bounds of $A$:
$$
 UB(A)=\{x\in X: a\le x,\text{ for all }a\in A\}
 $$

*Take the minimum $m$ of $UB(A)$ (if it exists).
When both steps succeed, $m$ is the supremum of $A$.
Now, if $A$ is the empty set, what's the set of upper bounds? Let $x\in X$; may $x$ fail to be in $UB(\emptyset)$? In order for this to happen there should be $a\in\emptyset$ such that it is false that $a\le x$. But this isn't possible, because no element can be found in $\emptyset$. Therefore no element of $x$ fails to be in $UB(\emptyset)$, that is,
$$
UB(\emptyset)=X
$$
and so the supremum of $\emptyset$ is, if it exists, the minimum element of $X$.
If your set is $\mathbb{R}$ with the usual order relation, then the supremum of $\emptyset$ doesn't exist. If you're considering $\bar{\mathbb{R}}$, the set $\mathbb{R}$ with $-\infty$ and $\infty$ added in the usual fashion, then the supremum of the empty set is indeed $-\infty$.
