If a subset of the real numbers has a lowest upper bound, then that lowest upper bound in unique. (The same is, of course, also true for greatest lower bounds.) This is because the real numbers are totally ordered, so that, given any two distinct real numbers, one of them has to be greater than the other. Thus, in particular, there can be only (at most) one lowest or greatest anything in the real numbers.
The empty set has no lowest upper bound in the real numbers: every real number is, trivially, an upper bound of the empty set, and there is no lowest real number. However, on the extended real number line, $-\infty$ is the lowest of all numbers, and thus the lowest upper bound of the empty set.
On a partially ordered set, it can happen that some subsets might have several lowest upper bounds (or greatest lower bounds). For example, consider the set $\lbrace \mathrm A, \mathrm B, \mathrm X, \mathrm Y \rbrace$, where $\mathrm A$ and $\mathrm B$ are both less than $\mathrm X$ and $\mathrm Y$ (that is, $\mathrm A < \mathrm X$, $\mathrm A < \mathrm Y$, $\mathrm B < \mathrm X$ and $\mathrm B < \mathrm Y$), but $\mathrm A$ and $\mathrm B$ are incomparable, and so are $\mathrm X$ and $\mathrm Y$ (i.e. $\mathrm A \nleq \mathrm B$, $\mathrm B \nleq \mathrm A$, $\mathrm X \nleq \mathrm Y$ and $\mathrm Y \nleq \mathrm X$). Then $\mathrm X$ and $\mathrm Y$ are both lowest upper bounds of the subset $\lbrace \mathrm A, \mathrm B \rbrace$ (which has no greatest lower bound), and $\mathrm A$ and $\mathrm B$ are both greatest lower bounds of $\lbrace \mathrm X, \mathrm Y \rbrace$ (which has no lowest upper bound).