Simplest Example of a Poset that is not a Lattice

A partially ordered set $(X, \leq)$ is called a lattice if for every pair of elements $x,y \in X$ both the infimum and suprememum of the set $\{x,y\}$ exists. I'm trying to get an intuition for how a partially ordered set can fail to be a lattice. In $\mathbb{R}$, for example, once two elements are selected the completeness of the real numbers guarantees the existence of both the infimum and supremum. Now, if we restrict our attention to a nondegenerate interval $(a,b)$ it is clear that no two points in $(a,b)$ have either a suprememum or infimum in $(a,b)$.

Is this the right way to think of a poset that is not a lattice? Is there perhaps a more fundamental example that would yield further clarity?

Solution 1:

The set $\{x,y\}$ in which $x$ and $y$ are incomparable is a poset that is not a lattice, since $x$ and $y$ have neither a common lower nor common upper bound. (In fact, this is the simplest such example.)

If you want a slightly less silly example, take the collection $\{\emptyset, \{0\}, \{1\}\}$ ordered by inclusion. This is a poset, but not a lattice since $\{0\}$ and $\{1\}$ have no common upper bound.

Solution 2:

(The previous answers are perfectly fine, but it's always helpful to have a picture in mind.)

Solution 3:

In the DAG of a lattice every pair of elements must have both a common successor (and thus a successor is the sup) and predecessor (inf).

Non-examples: no common predecessor for 1 and 2:

  3
 / \
1   2

No common successor for 2 and 3:

2   3
 \ /
  1

No common successor nor predecessor for 1 and 2:

1   2

Being a lattice implies that the poset is connected, and the above is not connected.

Solution 4:

Here a simple example, consider the $\{0,1,2\}$ with the order $$\{(0,0),(1,1),(2,2),(0,1),(0,2)\}$$ this is indeed a poset (the verification is simple) but it is not a lattice because it doesn't have any supremum of $1$ and $2$, i.e $1 \lor 2$.