Solutions  Problem Set 1
Problem 1
(a)
F
must contain all intersections, unions, and complements the two sets, so
{
2
}
,
{
1
,
3
,
4
}
,
{
1
,
2
,
4
}
,
{
3
}
are elements. Now, intersections and unions of these taken together with the original
two sets, and their complements, yield
{
2
,
3
}
,
{
1
,
4
}
,
{
1
,
2
,
3
}
,
{
4
}
,
{
1
}
At this point we have all elements (
{
1
}
,
{
2
}
,
{
3
}
,
{
4
}
), so
F
will be the power set
of Ω.
(b)
F
must contain all intersections, unions, and complements the two sets, so
F
=
{{
1
,
2
}
,
{
3
,
4
}
,
{
1
,
2
,
3
,
4
}
, φ
}
Problem 2
(a) One possible choice is Ω =
ZZ
+
∪ {
0
}
.
(b)
F
can be the power set of Ω.
(c)
A
1
=
{
t <
1
}
and
A
2
=
{
t >
2
}
(d)
B
1
=
{
t <
2
}
and
B
2
=
{
t >
1
}
Problem 3
(a)
E
k
&
since
{
N
1
>
k
} ⊃ {
N
1
>
k
0
}
for
k
0
> k
.
(b) None.
(c)
E
k
%
since
{
min
m
6
k
N
m
= 0
}
implies the occurance of
{
min
m
6
k
0
N
m
= 0
}
for
k
0
> k
and also if
{
min
m
6
k
N
m
= 0
}
does not occur,
{
min
m
6
k
0
N
m
= 0
}
can still occur for
some
k
0
> k
.
(d) None. Consider the sequence
N
1
= 1
, N
2
= 0
, N
3
= 3 for which the sequence of
events goes from occuring (
E
1
occurs) to not occuring (
E
2
does not occur) and back
to occuring again (
E
3
occurs).
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 Spring '05
 UYSALBIYIKOGLU
 Trigraph, i=1

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