In Rudin's "Principles of Mathematical Analysis", there is the following definition of bounded.
Definition: Suppose $S$ is an ordered set, and $E \subset S$. If there exists a $\beta \in S$ such that $x \leq \beta$ for every $x \in E$, we say that $E$ is bounded above, and call $\beta$ an upper bound of $E$. Lower bounds are defined in the same way(with $\geq$ in place of $\leq$).
Let $S = \emptyset$.
Let $E = \emptyset$.
Then, there exists no $\beta \in S$ such that $x \leq \beta$ for every $x \in E$.
So, in this case, I think $E (\subset S)$ is not bounded above.
Am I correct or not?
Isn't $\emptyset$ an orderd set?
Being bounded is relative to a given ordered set. If $S$ is empty, then indeed it is not bounded, which is just one more reason to agree that in general the empty set shouldn't be counted as a partial order. But if $S$ is non-empty, then the empty set is bounded, and every element is an upper and lower bound.
There is an analogy here with $(0,1)$. As a subset of itself, it is not bounded. There is no maximal element. But as a subset of $[0,1]$ or $\Bbb R$ it is certainly bounded by $0$ and $1$.