4.3 KiB
Orders
Orders and posets
Definition: a relation
\sqsubseteq
on a setP
is called an order if it is reflexive, antisymmetric and transitive.
- The pair
(P, \sqsubseteq)
is called a partially ordered set or for short poset.- Two elements
x
andy
in a poset(P, \sqsubseteq)
are called comparable ifx \sqsubseteq y
ory \sqsubseteq x
. The elements are incomparable ifx \not\sqsubseteq y
andy \not\sqsubseteq x
.- If any two elements are comparable then the relation is called a linear order.
For example on the set of real numbers \mathbb{R}
the relation \leq
is an order relation. For any two numbers x,y \in \mathbb{R}
we have x \leq y
or y \leq x
. This makes \leq
into a linear order.
Definition - Hasse diagram: let
(P, \sqsubseteq)
be a poset. The graph with vertex setP
and two verticesx,y \in P
adjacent if and only ifx \sqsubseteq y
and there is noz \in P
different fromx
andy
withx \sqsubseteq z
andz \sqsubseteq y
.
Maximal and minimal elements
Definition: let
(P, \sqsubseteq)
be a partially ordered set andA \subseteq P
. An elementa \in A
is called the maximum (\top
) ofA
, if for alla' \in A
we havea' \sqsubseteq a
. An elementa \in A
is called maximal if for alla' \in A
we have that eithera' \sqsubseteq a
ora
anda'
are incomparable.Similarly we can define the notion of minimum (
\bot
) and minimal element.
If we consider the poset of all subsets of a set S
then the empty set \varnothing
is the minimum of the poset, whereas the whole set S
is the maximum. The atoms are the subsets of S
containing just a single element.
Definition: if a poset
(P, \sqsubseteq)
has a minimum\bot
, then the minimal elements ofP\backslash \{\bot\}
are called the atoms ofP
.
Lemma: let
(P, \sqsubseteq)
be a partially ordered set. ThenP
contains at most one maximum and one minimum.
??? note "Proof:"
Suppose $p,q \in P$ are maxima. Then $p \sqsubseteq q$ as $q$ is a maximum. Similarly $q \sqsubseteq p$ as $p$ is a maximum. By antisymmetry of $\sqsubseteq$ we have $p = q$.
Lemma: let
(P, \sqsubseteq)
be a finite poset thenP
contains a minimal and a maximal element.
??? note "Proof:"
Consider the directed graph associated to $(P, \sqsubseteq)$ and pick a vertex in this graph. If the vertex is not maximal, then there is an edge leaving it. Move along this edge to the neighbour. Repeat this as long as no maximal element is found. Since the graph contains no cycles, a vertex will never be met twice. Hence, as $P$ is finite, the procedure has to stop. Implying a maximal element has been found. A minimal element of $(P, \sqsubseteq)$ is a maximal element of $(P, \sqsupseteq)$ and thus also exists.
Definition: if
(P, \sqsubseteq)
is a poset andA \subseteq P
then an upperbound forA
is an elementu
witha \sqsubseteq u
for alla \in A
. A lowerbound forA
is an elementu
withu \sqsubseteq a
for alla \in A
.If the set of all upperbounds of
A
has a minimal element then this element is called the least upperbound or supremum ofA
. Such an element is denoted by\mathrm{sup} A
.If the set of all lowerbounds of
A
has a maximal element then this element is called the largest lowerbound or infimum ofA
. Such an element is denoted by\mathrm{inf} A
.
For example let S
be a set. In (\wp(S), \subseteq)
any set A
of subsets of S
has a least upperbound and an largest lowerbound. Indeed
\mathrm{sup} A = \bigcup_{X \in A} X ;\text{ and }; \mathrm{inf} A = \bigcap_{X \in A} X.
If (P, \sqsubseteq)
is a finite poset then the elements from P
can be ordered as p_1, p_2, \dots, p_n
such that p_i \sqsubseteq p_j
implies i < j
. This implies that the adjacency matrix of \sqsubseteq
is uppertriangular, which means that it has only nonzero entries on or above the main diagonal.
Definition: an ascending chain in a poset
(P, \sqsubseteq)
is a sequencep_1 \sqsubseteq p_2 \sqsubseteq \dots
of elementsp_i \in P,i \in \mathbb{N}
. A descending chain in(P, \sqsubseteq)
is a sequencep_1 \sqsupseteq p_2 \sqsupseteq \dots
of elementsp_i \in P, i \in \mathbb{N}
.The poset
(P, \sqsubseteq)
is called well founded if any descending chain is finite.