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Section 2.2
35
In this case,
b
E
A
EEl
B,
so
bE
A
EEl
C
and since
b
fJ.
A
it follows that
bE
C.
Case 2:
b
E
A.
Here we have
b
E
B
n
A
and, hence,
b
fJ.
A
EEl
B,
so
b
fJ.
A
EEl
C.
Since
b
E
A,
we
must have
bE
C
(otherwise,
bE
A, C
<;;;
A
EEl
C).
In either case, we obtain
bE
C.
It
follows that
B
<;;;
C.
A
similar argument shows
C
<;;;
Band,
hence,
C
=
B.
(d) This is false since for
A
= 0,
A
x
B
=
A
x
C
=
0
regardless of
Band C.
28.
(a) True. Let
(a, b)
E
A
x
B.
Since
a
E
A
and
A
<;;;
C,
we have
a
E
C.
Since
b
E
Band B
<;;;
D,
bED.
Thus, (
a, b)
E
C
x
D
and
A
x
B
<;;;
C
x
D.
(b) False: Consider
A
=
{I},
B
=
{2, 3},
C
=
{I, 2, 3}.
(c) False. Let
A
=
{I},
B
= 0,
C
=
{2},
D
=
{3}. Then
A
x
B
=
0
<;;; {(2,3)}
=
C
x
D,
but
A~C.
(d) False since, by (b), the implication + is false. .
(e)
[BB]
True. Let
x
E
A.
Then
x
E
AU B,
so
x
E
An B
and, in particular,
x
E
B.
Thus,
A
<;;;
B.
Similarly, we have
B
<;;;
A,
so
A
=
B.
29.
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 Summer '10
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 Graph Theory

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