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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.

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Yes, this is a counterexample. –  egreg Apr 6 at 23:35

2 Answers 2

up vote 4 down vote accepted

A lattice is a poset $P$ in which $x\lor y$ and $x\land y$ exist for all $x,y\in P$.

Your example is indeed an example of a finite poset with top and bottom which isn't a lattice, for the reasons you stated.

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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.

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