# 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?

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In $\mathbb{R}$, any two elements are comparable; so the supremum of $\{a,b\}$ is $\max\{a,b\}$, and the infimum of $\{a,b\}$ is $\min\{a,b\}$. In particular, any nonempty subset of $\mathbb{R}$ is a lattice (any nonempty subset of a totally ordered set is always a lattice). – Arturo Magidin Feb 22 '12 at 18:35

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.

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

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True about the picture, for those of use who are not math geniuses ;-) +1 – joshmcode Nov 23 '15 at 4:13

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 $1 \lor 2$.

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I don't understand your example. In any total order, the infimum exists and is equal to the smaller of the two elements, and the supremum exists and is equal to the larger of the two elements. This is true in both $\mathbb{R}$ and $(a, b)$, and completeness plays no part in the discussion. (Completeness is about the existence of suprema and infima of infinite sets.)

Since the two notions are dual, I'll restrict myself to talking about the supremum. The supremum can fail to exist in two ways:

• There may be no upper bounds. For example, no antichain (with at least two elements) has suprema (unless $x = y$).
• The set of upper bounds may not have a least element. For example, in the four-element poset $\{ a_1, a_2, b_1, b_2 \}$ with $a_i \ge b_j$ for all $i, j$, the supremum of $\{ b_1, b_2 \}$ doesn't exist.
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So, every poset whose order is total is a lattice, correct? – ItsNotObvious Feb 22 '12 at 18:36
@3Sphere: Yes: total orders are always lattices. – Arturo Magidin Feb 22 '12 at 18:36

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.

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Take a lattice which is not totally ordered and has good number of elements. Most of the times you can delete an element from the lattice and the subset remaining would be a poset and not a lattice. (Easy to see from the Hasse diagram)

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