Definition3.2.1Reflexive
A relation \(R\) on \(A\) is said to be reflexive if \(x\,R\,x\) for all \(x \in A\text{.}\) Analogously, it is said to be irreflexive if \(x\,\not R\,x\) for all \(x \in A\text{.}\)
Section3.2Equivalence relations
¶Definition3.2.2Symmetric
A relation \(R\) on \(A\) is said to be symmetric if for all \(x,y \in A\) \(x\,R\,y\) if and only if \(y\,R\,x\text{.}\)
Definition3.2.3Transitive
A relation \(R\) on \(A\) is said to be transitive if for all \(x,y,z \in A\text{,}\) if \(x\,R\,y\) and \(y\,R\,z\text{,}\) then \(x\,R\,z\)
Definition3.2.4Antisymmetric
A relation \(R\) on a set \(A\) is said to be antisymmetric if for all \(x,y \in A\text{,}\) if \(x\,R\,y\) and \(y\,R\,x\text{,}\) then \(x = y\text{.}\)
Let us consider a few examples:
- The less than relation on \(\mathbb{R}\) is irreflexive, antisymmetric, and transitive.
- The divides relations on \(\mathbb{N}\) is reflexive, antisymmetric, and transitive.
- Equality is a relation which is reflexive, symmetric, and transitive.
- The subset relation is reflexive, antisymmetric, and transitive.
- Trust is a relation which is reflexive (probably?), neither symmetric nor antisymmetric, and not transitive.
- For the relation from Example 3.1.2 we cannot even talk about it being reflexive, irreflexive, symmetric, antisymmetric, or transitive because it is a relation between two different sets.
- The relation \(R\) on \(\mathbb{R}\) given by \(x\,R\,y\) if and only if \(\sin x = \sin y\) is clearly reflexive, symmetric and transitive.
Prove that the only relations \(R\) on \(A\) which are both symmetric and antisymmetric are subsets of the identity relation \(I_A\text{.}\)
Solution
Recall that \(I_A = \{ (a,a) \in A^2 \mid a \in A \}\text{.}\) Let \(R\) be a symmetric and antisymmetric relation on \(A\text{.}\) Then for all \(x,y \in A\text{,}\) if \(x\,R\,y\) then \(y\,R\,x\) by symmetry, and hence by antisymmetry, \(x=y\text{.}\) Thus, a pair can only be in \(R\) if it is of the form \((x,x)\) for some \(x \in A\text{.}\) Hence \(R \subseteq I_A\text{.}\)
A few corollaries of this fact are that the only symmetric, antisymmetric, reflexive relation is the identity, and the only symmetric, antisymmetric, irreflexive relation is the empty relation.
Prove that a relation \(R\) on a set \(A\) is symmetric if and only if \(R = R^{-1}\text{.}\)
Solution
Suppose that \(R\) is a symmetric relation on \(A\text{.}\) Then for all \(x,y \in A\text{,}\) we have \(x\,R\,y\) if and only if \(y\,R\,x\) (by symmetry) if and only if \(x\,R^{-1}\,y\) (by definition of inverse). This proves that symmetric relations are equal to their own inverses.
Now suppose that \(R = R^{-1}\text{,}\) then all we do is rearrange the if and only if statements from above. Then for all \(x,y \in A\text{,}\) we have \(x\,R\,y\) if and only if \(y\,R^{-1}\,x\) (by definition of inverse) if and only if \(y\,R\,x\) (by \(R = R^{-1}\)). This proves that relations which are equal to their own inverses are symmetric.
A relation \(R\) on \(\mathbb{R}\) can be viewed as a subset of the plane \(\mathbb{R}^2\) as we have already disussed. For each of the properties on the relation \(R\) given below, provide its geometric equivalent.
- reflexive
- irreflexive
- symmetric
- antisymmetric
- transitive
Solution
- reflexive: contains the diagonal line \(y=x\text{.}\)
- irreflexive: does not contain the diagonal line \(y=x\text{.}\)
- symmetric: symmetric about the line \(y=x\text{,}\) i.e., when reflected about this line, the set doesn't change.
- antisymmetric: the only points that lie both in this set and its reflection about the line \(y=x\) are perhaps some points on the diagonal.
- transitive: this one is harder to see. If points \(P\) and \(Q\) lie in the relation, and if \(Q\) lies on the vertical line through the reflection of the point \(P\) across the line \(y=x\text{,}\) then the point of intersection of the vertical line through \(P\) and the horizontal line through \(Q\) also lies in the relation.
Prove the fact discussed in the example above: the divides relation on \(\mathbb{N}\) is reflexive, antisymmetric and transitive.
The divides relation on \(\mathbb{Z}\) is also reflexive and transitive, but it is not quite antisymmetric. Why isn't it antisymmetric on \(\mathbb{Z}\text{?}\) Once you have proven it is antisymmetric on \(\mathbb{N}\text{,}\) ask yourself what does wrong in the proof when you switch from \(\mathbb{N}\) to \(\mathbb{Z}\text{.}\)
Solution
Let \(x,y,z \in \mathbb{N}\) be arbitrary. Clearly, \(x \mid x\) since \(x = 1 x\text{,}\) so divides is reflexive.
If \(x \mid y\) and \(y \mid x\text{,}\) then there exist positive integers \(m,n\) so that \(y = mx\) and \(x = ny\text{.}\) Then \(y = mx = m(ny) = (mn)y\) and hence by cancellation \(1 = mn\) and so \(m = n = 1\) since \(m,n \in \mathbb{N}\text{.}\) Therefore, \(x = y\) and so divides is antisymmetric.
If \(x \mid y\) and \(y \mid z\text{,}\) then there exist positive integers \(p,q\) so that \(y = px\) and \(z = qy\text{.}\) Then \(z = qy = q(px) = (qp)x\) so \(x \mid z\text{.}\)
Hopefully, from the previous examples, you saw that a bunch of relations have nice properties. Actually, it is more accurate to say that relations we find interesting have certain collections of nice properties. These are summarized in the next two definitions.
Definition3.2.10Equivalence relation
A relation \(R\) on a set \(A\) is said to be an equivalence relation if it is reflexive, symmetric and transitive.
Definition3.2.11Modular congruence
Let \(n\) be a fixed positive integer called the modulus of congruence. For \(x,y \in \mathbb{Z}\text{,}\) we say \(x\) is congruent modulo \(n\) to \(y\) if \(n\) divides \(x-y\text{.}\) In this case we write \(x \cong_n y\) or \(x = y (\mod n)\text{.}\)
Show that congruence modulo \(n\) (\(\cong_n\)) is an equivalence relation.
Definition3.2.13Equivalence class
Let \(\sim\) be an equivalence relation on a set \(A\text{.}\) For \(x \in A\text{,}\) the equivalence class of \(x\) is
\begin{equation*} [x] := \{ y \in A \mid x \sim y \}. \end{equation*}The set of equivalence classes of elements of \(A\) is denoted by \(A / \sim\)
Definition3.2.14Partition
A partition of a set \(A\) is a collection \(\mathcal{P}\) of subsets of \(A\) which is pairwise disjoint and whose union is \(A\text{.}\)
Theorem3.2.15
Let \(\sim\) be an equivalence relation on the set \(A\text{.}\) Then the set of equivalences class \(A / \sim\) partitions \(A\text{.}\)
Notice that since \(\sim\) is reflexive, for every \(x \in A\text{,}\) \(x \sim x\) and therefore \(x \in [x] \in A / \sim\text{.}\) Therefore, the union over \(A / \sim\) is \(A\text{.}\)
Now suppose \([x], [y] \in A / \sim\) and \([x] \cap [y] \not\emptyset\text{.}\) Then there is some \(z \in [x] \cap [y]\text{,}\) so \(x \sim z\) and \(z \sim y\) (by definition and symmetry of \(\sim\text{.}\) Therefore \(x \sim y\) by transitivity. Now let \(w \in [x]\text{,}\) then \(w \sim x\text{,}\) so \(w \sim y\) by transitivity, and hence \(w \in [y]\text{.}\) Thus \([x] \subseteq [y]\text{.}\) A symmetric argument proves the reverse inclusion, and thus \([x] = [y]\text{.}\) So we have proven that equivalence classes are either disjoint, or they are the same set, and hence \(A / \sim\) is a pairwise disjoint collection, and hence also forms a partition of \(A\text{.}\)
Theorem3.2.16
Let \(\mathcal{A}\) be a partition of \(A\text{.}\) Define a relation \(\sim\) on \(A\) by \(x \sim y\) if and only if there is some \([z] \in A / \sim\) with \(x,y \in [z]\text{.}\) Prove that \(\sim\) is an equivalence relation on \(A\text{.}\)
Exercise on the review sheet with a solution there.