Does the empty set have a supremum? Is it $-\infty$?

Question

I need to give an example of this, or provide an argument for why it's impossible.

Two sets $A$ and $B$ with A intersection $B = \emptyset$ $\sup A = \sup B$, $\sup A$ is not an element of $A$ and $\sup B$ is not an element of $B$.

I was thinking if both $A$ and $B$ are the empty set and their supremum is negative infinity that this would satisfy the question. Are we allowed to say the empty set is bounded and are we allowed to set negative infinity as a bound?

If it's not allowed I'm just going to say something like $A =\{x:\;x\text{ is a rational number and }2 < x < 3\}$ and $B =\{x:\; x \text{ is irrational and }2 < x < 3\}$. This doesn't seem nearly as interesting though.

Answer 1

If you are working in the reals the empty set does not have a $\sup$. If you are working in the extended reals you do have $\sup \emptyset=-\infty$. Your second example is fine.

Answer 2

Your examples looking good to me. More generaly, taking $A = (x,y)\cap \mathbb{Q}$ and $B = (x,y)\cap (\mathbb{R}-\mathbb{Q})$ we have $A \cap B = \emptyset$, $\sup A = \sup B = y$, $\inf A = \inf B = x$ and $x,y \not\in A;\; x,y \not\in B$.

PS. $x, y \in \mathbb{Z}$.

Adding to Ross's answer. If $X = \emptyset$ so for any $r \in \mathbb{R}$, $r$ bounded $X$. So, if there is $c = \sup X$, then $c \leq r$ for any $r \in \mathbb{R}$, an absurd.