COT3100C01, Spring 2000
S. Lang
Solution Key to Assignment #3 (40 pts.)
3/19/2000
1.
(15 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
is symmetric, then
R
∪
{(
a
,
a
) 
a
∈
A
} is symmetric.
Proof:
In order to prove the symmetric property, let (
x
,
y
)
∈
R
∪
{(
a
,
a
) 
a
∈
A
}  (1),
we need to prove (
y
,
x
)
∈
R
∪
{(
a
,
a
) 
a
∈
A
}  (2).
From (1), (
x
,
y
)
∈
R
or (
x
,
y
)
∈
{(
a
,
a
) 
a
∈
A
}.
Thus, there are two cases:
(Case 1) Suppose (
x
,
y
)
∈
R
.
Since
R
is symmetric by assumption, so (
y
,
x
)
∈
R
⊂
R
∪
{(
a
,
a
) 
a
∈
A
}, so (2) is proved in this case.
(Case 2) Suppose (
x
,
y
)
∈
{(
a
,
a
) 
a
∈
A
}.
In this case,
x
=
y
.
Thus, (
y
,
x
) = (
x
,
y
)
∈
{(
a
,
a
) 
a
∈
A
}
⊂
R
∪
{(
a
,
a
) 
a
∈
A
}, so (2) is proved in this case.
Therefore, we proved (2) in both cases.
(b) If
R
is transitive and
R
⊂
T
, then
T
is transitive.
Disproof:
Let
A
= {
a
,
b
,
c
},
R
= {(
a
,
b
), (
b
,
c
), (
a
,
c
)},
T
= {(
a
,
b
), (
b
,
c
), (
a
,
c
), (
b
,
a
)}.
Then,
R
is transitive and
R
⊂
T
, but
T
is not transitive because both (
a
,
b
) and (
b
,
a
)
∈
T
but (
a
,
a
)
∉
T
.
(c)
(
R
∪
T
)
–1
=
R
–1
∪
T
–1
.
Proof:
We will prove (
x
,
y
)
∈
(
R
∪
T
)
–1
⇔
(
x
,
y
)
∈
R
–1
∪
T
–1
.
Note that (
x
,
y
)
∈
(
R
∪
T
)
–1
⇔
(
y
,
x
)
∈
(
R
∪
T
)
, by the definition of
(
R
∪
T
)
–1
⇔
(
y
,
x
)
∈
R
or (
y
,
x
)
∈
T
, definition of union
⇔
(
x
,
y
)
∈
R
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 Spring '09
 Binary relation, Transitive relation, Symmetric relation, brief explanation

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