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Simplest Example of a Poset that is not a Lattice

Question is simple

Lattices

A Poset $(S,\leq)$ can be a Lattice if every pair of elements $a,b \in S$ in the Poset has a meet $\exists a \wedge b$ and a join $\exists a \vee b$.

Question 1

Is my definition of lattice correct?

Question2

Is every finite Poset a Lattice?

I guess no, but cannot find an example. Could you please make an example of Poset which is not a Lattice.

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marked as duplicate by Qiaochu Yuan Feb 23 '12 at 3:01

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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A poset with two elements and no nontrivial relations (a two-element antichain) is the minimal counterexample. –  Qiaochu Yuan Feb 23 '12 at 3:02
    
Really sorry for creating a duplicate, actually I searched under the lattice-orders label finding no other similar questions. –  Andry Feb 23 '12 at 4:24

1 Answer 1

Most posets are not lattices, including the following. A discrete poset, meaning a poset such that x ≤ y implies x = y, is a lattice if and only if it has at most one element. In particular the two-element discrete poset is not a lattice. Although the set $\{1,2,3,6\}$ partially ordered by divisibility is a lattice, the set $\{1,2,3\}$ so ordered is not a lattice because the pair 2,3 lacks a join, and it lacks a meet in $\{2,3,6\}$. The set $\{1,2,3,12,18,36\}$ partially ordered by divisibility is not a lattice. Every pair of elements has an upper bound and a lower bound, but the pair 2,3 has three upper bounds, namely 12, 18, and 36, none of which is the least of those three under divisibility (12 and 18 do not divide each other). Likewise the pair 12,18 has three lower bounds, namely 1, 2, and 3, none of which is the greatest of those three under divisibility (2 and 3 do not divide each other).

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