ARS DIGITA UNIVERSITY
MONTH 2: DISCRETE MATHEMATICS
PROFESSOR SHAI SIMONSON
PROBLEM SET 2 SOLUTIONS — SET, FUNCTIONS, BIGO, RATES OF
GROWTH
(1)
Prove by formal logic:
(a)
The complement of the union of two sets equals the intersection of
the complements.
Let
x
∈
(
A
∪
B
). Then
x
is not in
A
, and
x
is not in
B
. Therefore
x
∈
A
∩
B
.
Since
x
was an arbitrary element of
(
A
∪
B
), we conclude that
(
A
∪
B
)
⊆
A
∩
B
.
Now, suppose that
x
∈
A
∩
B
. We can conclude that
x
∈
A
, and that
x
∈
B
. This
impiles that
x /
∈
A
, and
x /
∈
B
, which implies
x /
∈
(
A
∪
B
), which implies that
x
∈
(
A
∪
B
). Again, since
x
was arbitrary, we conclude that
(
A
∪
B
)
⊆
A
∩
B
.
Two sets which are subsets of each other are equal (see below), so we can conclude
that
(
A
∪
B
) =
A
∩
B
(b)
The complement of the interesection of two sets equals the union of
the complements.
Let
x
∈
(
A
∩
B
).
Then
x
cannot be in both
A
and
B
— at least one of the
statements “
x
∈
A
”, “
x
∈
B
” is true. Therefore
x
∈
A
∪
B
, and
(
A
∩
B
)
⊆
A
∪
B
.
Now let
x
∈
A
∪
B
. Again,
x
cannot be in both
A
and
B
, so
x /
∈
A
∩
B
, so
x
∈
(
A
∩
B
), and
A
∪
B
⊆
(
A
∩
B
). The two sets are equal.
(c) (
B

A
)
∪
(
C

A
) = (
B
∪
C
)

A
.
Let
x
∈
(
B

A
)
∪
(
C

A
). Then
x
∈
(
B

A
)
∨
x
∈
(
C

A
). Assume without loss of
generality that
x
∈
(
B

A
). This implies that
x
∈
B
and
x /
∈
A
. Which implies
that
x
∈
(
B
∪
C
), and, combining this with the fact that
x /
∈
A
, we conclude
that
x
∈
(
B
∪
C
)

A
, and (
B

A
)
∪
(
C

A
)
⊆
(
B
∪
C
)

A
. If
x
∈
(
B
∪
C
)

A
,
then
x /
∈
A
, and
x
∈
B
∨
x
∈
C
.
Therefore
x
∈
(
B

A
)
∨
x
∈
(
C

A
),
x
∈
(
B

A
)
∪
(
C

A
), and (
B
∪
C
)

A
⊆
(
B

A
)
∪
(
C

A
), completing the
proof.
(d)
If two sets are subsets of each other than they are equal.
If two sets
A
and
B
are not equal, then there exists some element that is in one
set that is not in the other. Without loss of generality, denote such an element
x
and assume that
x
∈
A
. Since
x
∈
A
and
x /
∈
B
, we conclude that
A
⊂
B
.
(2)
Generalize De Morgan’s laws for
n
sets and prove the laws by induction.
1
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PROBLEM SET 2
The
n
version generalization of De Morgan’s laws are as follows:
n
i
=1
A
i
=
n
i
=1
A
i
n
i
=1
A
i
=
n
i
=1
A
i
To prove these laws by induction, we use
n
= 2 as our base cases: these are the
traditional twoset De Morgan’s laws. Now, we assume that the laws hold for
n
sets,
and show that they hold for
n
+ 1 sets as well, using both our inductive hypothesis
and the twoset De Morgan’s laws:
n
+1
i
=1
A
i
=
(
n
i
=1
A
i
)
∪
A
n
+1
=
(
n
i
=1
A
i
)
∩
A
n
+1
=
(
n
i
=1
A
i
)
∩
A
n
+1
=
n
+1
i
=1
A
i
n
+1
i
=1
A
i
=
(
n
i
=1
A
i
)
∩
A
n
+1
=
(
n
i
=1
A
i
)
∪
A
n
+1
=
(
n
i
=1
A
i
)
∪
A
n
+1
=
n
+1
i
=1
A
i
(3)
Prove by induction on the size of the set, that the power set
P
(
A
)
has
cardinality
2

A

.
For a set S containing a single element
x
, the power set is
{∅
,
{
x
}}
, which is of
size 2. This is our base case. Now assume that a set
A
of size
n
has power set with
cardinality 2

A

= 2
n
. Consider adding a new element
a
to
A
to make the set
A
, of
size
n
+ 1. The power set of
A
will consist of all the sets in the power set of
A
, plus
all those sets taken again, with the element
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 Spring '09
 Sets, Natural number

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