Bounded poset

Remarks

The positive divisors of $12$, ordered by divisibility, form a bounded poset with $\bot = 1$ and $\top = 12$, since $1 \mid d$ and $d \mid 12$ for each divisor $d$. The poset $(\Z, \le)$ is not bounded: it has neither a greatest nor a least element. The interval $(0, 1]$ in $\R$ has the top $1$ but no bottom, so it is not bounded. For a subset of $P$, being bounded usually means something weaker: having an upper bound and a lower bound in $P$. The elements $\top$ and $\bot$ are unique when they exist, by antisymmetry. A bounded poset with one element has $\top = \bot$.