Order Relations Explained: Partial vs Total, Weak vs Strict (With Examples)
- Tyler Buffone
- Oct 12
- 5 min read
Updated: Oct 17
Read this first (prerequisites)
This post is best experienced after these ones:
Why this matters
Relations are the language for comparing mathematical objects. Two big families appear everywhere:
Equivalence relations: group objects that are “the same in essence,” even if they look different.
Order relations: arrange objects from “smaller” to “larger” in a consistent, logical way.
This post builds a clear, example-ready framework for understanding order relations.
Weak Order and Strong Order Relations
Order relations have two standard flavours:
Weak (non-strict) order ⪯ on A:
Reflexive, transitive, and antisymmetric.
Strong (strict) order ≺ on A:
Irreflexive and transitive.
They are interconvertible under mild conditions:
From weak to strict: a ≺ b ⟺ a ⪯ b and a ≠ b.
From strict to weak: a ⪯ b ⟺ a ≺ b or a = b.
Total Order and Partial Order Relations
Total order relation:
Any two elements are comparable. For all a, b, either a ⪯ b or b ⪯ a.
Partial order relation:
Some pairs might be incomparable. You cannot always decide which should come first.
Proving a Relation is a Partial Order
Given a proposed definition of a ⪯ b:
Reflexive: Show a ⪯ a holds for every a.
Antisymmetric: Assume a ⪯ b and b ⪯ a. Prove a = b using the definition.
Transitive: Assume a ⪯ b and b ⪯ c. Prove a ⪯ c.
Close the loop with a clear statement: "Thus ⪯ is a weak partial order."
Maximal and Minimal vs Maximum and Minimum
Let (A, ⪯) be a partially ordered set (poset).
Maximal element m: there is no x with m ≺ x inside A. Formally, if m ⪯ x then x = m.
Minimal element u: there is no x with x ≺ u inside A. Formally, if x ⪯ u then x = u.
Maximum M: x ⪯ M for every x ∈ A. At most one can exist, and if it exists it is also maximal.
Minimum m: m ⪯ x for every x ∈ A. At most one can exist, and if it exists it is also minimal.
A poset can have several maximal elements and no maximum at all. Same story for minimal vs minimum.
Upper and Lower Bounds, Supremum, and Infimum
For S ⊆ A:
Upper bound of S: an element U with s ⪯ U for all s ∈ S.
Lower bound of S: an element L with L ⪯ s for all s ∈ S.
Supremum sup S: the least upper bound. It is an upper bound U such that U ⪯ U' for every other upper bound U'.
Infimum inf S: the greatest lower bound. It is a lower bound L such that L' ⪯ L for every other lower bound L'.
The Supremum or infimum may fail to exist in a general poset. When they exist and happen to lie in S, they coincide with maximum or minimum respectively.
Order Relation Examples
Subset inclusion on a power set is a weak partial order
Let X be any set. Consider the power set P(X) (the set of all subsets of X). Define a relation R on P(X) by: ARB if A ⊆ B.
Prove that R is a weak (non-strict) partial order on P(X).
Solution
Reflexivity
Take any A ∈ P(X). Every element of A is of course in A itself, so A ⊆ A. Hence ARA. Reflexivity holds.
Antisymmetry
Assume ARB and BRA. By the definition of R, this means A ⊆ B and B ⊆ A. Mutual inclusion gives equality, so A = B. Antisymmetry holds.
Transitivity
Assume ARB and BRC. That is, A ⊆ B and B ⊆ C. Let x ∈ A. Since A ⊆ B, we get x ∈ B. Since B ⊆ C, we get x ∈ C. Therefore every element of A lies in C, so A ⊆ C. Hence ARC. Transitivity holds.
The relation is reflexive, antisymmetric, and transitive on P(X). Therefore R is a weak partial order.
Helpful Remarks
This relation is not total in general:
For instance, on P({1, 2}), the sets {1} and {2} are incomparable since neither is a subset of the other. So this is a partial order relation, not a total order relation.
Least and greatest elements:
Under ⊆ on P(X):
The least element (minimum) is ∅ since ∅ ⊆ A for every A ⊆ X.
The greatest element (maximum) is X since A ⊆ X for every A ⊆ X.
Bounds, infimum, and supremum (subset order on P(X)):
Let F ⊆ P(X) be a collection of subsets of X.
Upper bound: a set U ⊆ X such that A ⊆ U for every A ∈ F.
Least upper bound (supremum): ∪F, which is the union of all sets in F.
Lower bound: a set L ⊆ X such that L ⊆ A for every A ∈ F.
Greatest lower bound (infimum): ∩F, which is the intersection of all sets in F.
Divisibility on a finite set is a weak partial order
Let:
A = {1, 2, 3, 5, 6, 10, 15}
On A, define a relation ⪯ by:
a ⪯ b if ∃n ∈ ℤ with b = na
Since every element of A is a positive integer, this is equivalent to divisibility:
a ⪯ b ⟺ a divides b
(a) Show that ⪯ is a weak (non-strict) partial order on A.
(b) Determine the maximal elements of A with respect to ⪯.
Solution a)
Reflexivity
For any a ∈ A, a = 1 ⋅ a with 1 ∈ ℤ. Hence:
a ⪯ a
Reflexivity holds.
Antisymmetry
Suppose a ⪯ b and b ⪯ a. Then b = na and a = mb for some integers m, n. Substituting gives:
a = m(na) = (mn)a
Since a > 0, we have mn = 1. Over the integers this forces m = n = 1. Hence:
a = mb
a = 1 ⋅ b
a = b
Antisymmetry holds.
Transitivity
Suppose a ⪯ b and b ⪯ c. Then b = na and c = mb for some integers m, n. Hence:
c = m(na) = (mn)a
So a ⪯ c.
Transitivity holds.
Partial order
This order relation is partial since not every pair of elements is comparable.
For instance, we cannot say 2 ⪯ 15 since there is no integer n such that 15 = n ⋅ 2.
Then, since reflexivity, antisymmetry, and transitivity hold, this relation is a weak partial order on A.
Solution b)
Recall that x ∈ A is maximal if there is no y ∈ A with x ⪯ y and x ≠ y.
In divisibility language: x is not a proper divisor of any other element of A.
Check each element:
1 divides everything → not maximal.
2 divides 6, 10 → not maximal.
3 divides 6, 15 → not maximal.
5 divides 10, 15 → not maximal.
6 divides only 6 in A → maximal.
10 divides only 10 in A → maximal.
15 divides only 15 in A → maximal.
Thus, the maximal elements are 6, 10, and 15.
There is no single maximum element, since no member of A is divisible by all the others; the least element exists and is 1.
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