Problem Set 1 – Solutions
Economics 204  August 2006
1. Set Theory
(a) Determine whether this formula is always right or sometimes wrong. Prove it if
it is right. Otherwise give both an example and a counterexample and state (but
don’t prove) an additional neccesary and sufficient condition for it to always be
right:
A
\
(
B
\
C
) = (
A
\
B
)
∪
C.
Sometimes wrong. Let
A
be the set of firstyear econ grad students at Berkeley,
B
be the set of international econ grad students at Berkeley, and
C
be the set
of male econ grad students at Berkeley. Then your GSI, Zack, is an element of
(
A
\
B
)
∪
C
, but not of
A
\
(
B
\
C
). He
is
a member of the set
{
domestic firstyears
}∪
{
male econ grads
}
but not of the set
{
firstyear international female econ grads
}
.
It can be correct though. Let
A
be the set of firstyears,
B
be the set of interna
tional econ students, and
C
be the set
{
Econ 204 students with econ.berkeley.edu email addresses
}
.
In this case,both the left and righthand sides consist of
{
firstyear econ grad students except international students who are not in Econ 204
}
.
An additional necessary and sufficient condition for the formula to be always cor
rect would be that
C
⊆
A
.
While not a proof, Figure 1 illustrates why this is
true. Observe that the lefthand side is contained in the righthand side. The set
C
\
A
is included on the right, but not the left.
C
⊆
A
is precisely the condition
needed for this to be empty and render the two sides equal.
(b) Certain subsets of a given set
S
are called
A
sets and others are called
B
sets.
Suppose that these subsets are chosen in such a way that the following properties
are satisfied:
•
The union of any collection of
A
sets is and
A
set.
•
The intersection of any finite number of
A
sets is an
A
set.
•
The complement of an
A
set is a
B
set and the complement of a
B
set is an
A
set.
Prove the following:
1. The intersection of any collection of
B
sets is a
B
set.
Proof:
Let
I
index
B
I
, a collection of
B
sets, and let
x
∈
i
∈
I
B
i
.
Now
x
∈
i
∈
I
B
i
if and only if
x
∈
B
i
∀
i
∈
I
, which is true if and only if
x /
∈
B
C
i
for all
i
∈
I
, which is true if and only if
x /
∈
i
∈
I
B
C
i
.
This means that
i
∈
I
B
i
= (
i
∈
I
B
C
I
)
C
.
Because
B
i
is a
B
set for all
i
, the complement
B
C
i
is
1
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an
A
set. The union of these
A
sets is an
A
set, and its complement is, in
turn, a
B
set. Thus
i
∈
I
B
i
is a
B
set.
2. The union of any finite number of
B
sets is a
B
set.
Proof:
Now let
I
be finite.
x
∈
i
∈
I
B
i
⇐⇒
x
∈
B
i
for some
i
∈
I
. This
is true if and only if
x /
∈
B
C
i
for some
i
∈
I
which is true if and only if
x /
∈
i
∈
I
B
C
i
. Thus,
i
∈
I
B
i
= (
i
∈
I
B
C
i
)
C
.
B
C
i
is an
A
set for all
i
so the
intersection of
B
i
is also an
A
set and the complement of this intersection is
a
B
set. Thus,
i
∈
I
B
i
is a
B
set.
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 Summer '08
 Staff
 Economics, Vector Space, Natural number, Equivalence relation, Metric space, i∈I Bi

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