Math 2200
Exercises on Relations
27 February, 2014
For each statement below, assume that R is a relation on a set A. Determine whether
each statement is true or false. If it is true, prove it. If it is false, give a counterexample,
add a necessary assumption, and the prove your new theorem.
1. If R is transitive and symmetric, then R is reflexive.
This statement is false.
Counterexamples:
• Empty Relation: A = {1, 2, 3}, R = ∅
• A = {1, 2, 3}, R = {(1, 1), (1, 2), (2, 1), (2, 2)}
Theorem: Let R be a relation on A. If R is transitive and symmetric, and every element
of A is related to at least one element, then R is reflexive.
Proof: Let a ∈ A. By the last assumption, there is some b ∈ A such that aRb.
Because R is symmetric, bRa. Combining these with transitivity gives us aRa. Thus R
is reflexive. 2. If R is symmetric and antisymmetric, then R is a function.
This statement is false.
Counterexamples:
• Empty Relation: A = {1, 2, 3}, R = ∅
• A = {1, 2, 3}, R = {(1, 1), (2, 2)}
Theorem: Let R be a relation on A. If R is symmetric and antisymmetric, and every
element of A is related to at least one element, then R is a function.
Proof: Let a ∈ A. By the last assumption, there is some b ∈ A such that aRb. Because
R is symmetric, bRa. Since R is antisymmetric, this implies that a = b. Thus every
element of A is related to exactly one element. (In particular, each element is related
only to itself.) This proves that R is a function. 3. If R is symmetric and antisymmetric, then R is a function.
This statement is false.
Counterexamples:
• A = {1, 2, 3}, R = {(1, 1), (2, 1), (3, 1)}
• A = Z, R = {(z, 1)|z ∈ Z}
Theorem: Let R be a relation on A. If R is an onto function and transitive, then R is
reflexive.
Proof: Let b ∈ A. Since R is onto, there is some a ∈ A such that aRb. Because R is a
function, there is some c ∈ A such that bRc. By transitivity, aRc. Since R is a function,
a can be related to only one element; therefore b = c. Thus bRb and R is reflexive.