TITLE: If a finite poset has greatest and least elements, is it a lattice?
QUESTION [7 upvotes]: Let $P$ be a finite partially ordered set with elements $0$ and $1$ such that $0 \le x \le 1$ for any $x \in P$. Does it follow that $P$ is a lattice? If not, what is a counterexample?
I believe this is a counterexample, but I'm not entirely clear on the definition of a lattice, so I was hoping someone would confirm or deny:
  1
 / \
a   b
|\ /|
| / |
|/ \|
c   d
 \ /
  0

In this poset, $a$, $b$, and $1$ are upper bounds of the set $\{c,d\}$, but $a$ and $b$ are incomparable, so $\{c,d\}$ has no least upper bound.

REPLY [2 votes]: You can take the set of divisors of of a positive integer $n$ partially  ordered by divisibility condition.  Clearly $n$ and $1$ will be global maximum and minimum. Do this for some $n$ with at leat two prime divisors, say $n=24$. Draw the Hasse diagram of this poset and you can remove an element suitably to get more examples.