COT3100.01, Fall 2002
S. Lang
Solution
Keys to Assignment #6
11/26/2002
1.
(12 pts.) Let
A
denote an arbitrary nonempty set, and
R
and
T
denote arbitrary binary
relations defined over
A
.
Determine if each of the following statements is
true
or
false
, and
give a brief
explanation
of your answer.
In the cases of a false statement, use
A
= {
a
,
b
,
c
}
and appropriate relations over
A
for your
counterexample
.
(a)
If
R
⊆
T
, then
s
(
R
)
⊆
s
(
T
).
Proof:
Note that
R
⊆
T
implies
R
–1
⊆
T
–1
, by the definition of inverse
Thus,
s
(
R
) =
R
∪
R
–1
, by the definition of
s
(
R
)
⊆
T
∪
T
–1
=
s
(
T
).
(b)
If
R
is transitive and
T
is transitive, then
R
∪
T
is transitive.
Disproof:
We give a counterexample as follows.
Let
A
= {1, 2, 3}.
Define two relations
R
= {(1, 2)}, and
T
= {(2, 3)}.
Then both
R
and
T
are transitive (since there are no pairs to be composed together in both relations).
However,
R
∪
T
={(1, 2), (2, 3)}; it is not transitive since the pair (1, 3)
∉
R
∪
T
.
(c)
If
R
=
T
–1
, then
T
=
R
–1
.
Proof:
We prove
T
⊆
R
–1
and
R
–1
⊆
T
simultaneously as follows.
Note that (
a
,
b
)
∈
T
⇔
(
b
,
a
)
∈
T
–1
(definition of inverse)
⇔
(
b
,
a
)
∈
R
(since
R
=
T
–1
by assumption)
⇔
(
a
,
b
)
∈
R
–1
(definition of inverse)
(d)
If
R
∪
T
is irreflexive, then
R
is irreflexive.
Proof:
We prove by contradiction.
That is, we assume
R
is not irreflexive.
Thus, there
exists some (
a
,
a
)
∈
R
.
Since
R
⊆
R
∪
T
, so (
a
,
a
)
∈
R
∪
T
.
Thus,
R
∪
T
is not
irreflexive, a contradiction to the assumption.
2.
(4 pts.) Let
A
,
B
, and
C
denote 3 sets.
Prove that
A
×
(
B
∩
C
) = (
A
×
B
)
∩
(
A
×
C
).
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 Spring '09
 Binary relation, Transitive relation, Transitive closure, Reflexive relation

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