Selected items from the Introduction to set theory and to methodology and philosophy of mathematics and computer programming.

1. Introduction to set theory and to methodology and philosophy of
mathematics and computer programming
Reflexivity, symmetry and transitivity
An overview
by Jan Plaza
c 2017 Jan Plaza
Use under the Creative Commons Attribution 4.0 International License
Version of December 4, 2017

2. Definition. Let R be a binary relation over X.
1. R is reflexive on X if ∀x∈X xRx.
2. R is symmetric on X if ∀x,y∈X (xRy → yRx).
3. R is transitive on X if ∀x,y,z∈X (xRy ∧ yRz → xRz).
Example. Let X be the set of all straight lines on a plane.
1. Relation isParallelTo is:
reflexive on X,
symmetric on X,
transitive on X.
2. Relation isPerpendicularTo is:
not reflexive on X,
symmetric on X,
not transitive on X.

3. Informal Example
Let X be a set of people.
1. Relation isASiblingOf on X is:
not reflexive on X,
symmetric on X,
not transitive on X.
It is not transitive, because
x isASiblingOf y ∧ y isASiblingOf x → x isASiblingOf x does not hold.
2. Relation isAnAncestorOf on X is:
not reflexive,
not symmetric,
transitive.

4. Exercise
Consider binary relations =Z, =Z, <Z, >Z, Z, Z, the total relation 1Z2 , the empty
relation ∅, and the relation | of divisibility.
1. For each of these relations, check if it is reflexive on Z, if it is symmetric on Z,
and if it is transitive on Z.
2. Graph these relations on the Cartesian plane. Observe that it is reflexivity and
symmetry of the relation is easy to recognize in a graph.
Exercise
Recall congruence modulo k: x ≡k y iff k|(y − x), for any fixed integer k > 1.
Make a discrete Cartesian graph of ≡3 which shows only those ordered pairs x, y
where x ≡3 y and x, y ∈ {−2..6}.

5. Proposition
≡k, is reflexive on Z, symmetric on Z and transitive on Z.
Proof. We will prove transitivity.
Take any x, y, z such that x ≡k y and y ≡k z.
Goal: x ≡k z.
By the definition of congruence mod k we have:
y − x = m · k, for some m ∈ Z and
z − y = n · k, for some n ∈ Z.
By adding these equations we obtain:
z − x = (m + n) · k, for some m, n ∈ Z.
So, z − x = l · k, for some l ∈ Z.
So, x ≡k z, by the definition of congruence mod k.
Exercise. Prove that ≡k is reflexive on Z and symmetric on Z.

6. Proposition
Let X is a set with n elements. Then, there are 2n(n−1) relations over X that are
reflexive on X.
Proof
Instead of counting relations reflexive on X it is enough to count subsets of the
Cartesian product X × X without the diagonal.
(Such relations and such subsets are in 1-1 correspondence – there is a bijection.)
The number of such relations is the same as the number of subsets of
(X × X) − { a, a : a ∈ X}.
This set contains n2 − n = n(n − 1) elements.
So the number of its subsets, and the number of relations in question is 2n(n−1).

7. Exercise. Assume that X is a set with n elements.
1. How many binary relations over X are there?
2. How many binary relations over X are reflexive on X?
3. How many binary relations over X are not reflexive on X?
4. How many binary relations over X are symmetric on X?
5. How many binary relations over X are not symmetric on X?
6. How many binary relations over X are reflexive on X and symmetric on X?
7. How many binary relations over X are reflexive on X and not symmetric on X?
8. How many binary relations over X are not reflexive on X and symmetric X?
9. How many binary relations over X are not reflexive on X and not symmetric on
X?
Hint: in order to answer the next question you may need to use answers to some of the
previous questions.

8. Claim. If R is symmetric on X and transitive on X then it is reflexive on X.
We will attempt to prove it.
Assume R is symmetric on X and transitive on X.
Take any x, y such that xRy.
By symmetry, yRx.
By transitivity, xRy ∧ yRx → xRx.
So, xRx.
We conclude that R is reflexive on X.
Is this conclusion correct? No. We proved only this:
If R is symmetric on X and transitive on X then it is reflexive on domain(R).
Exercise. Find R and X that disprove the Claim above.
Proposition
Let X = field(R). If R is symmetric on X and transitive on X then it is reflexive on X.
Exercise. Prove this proposition.