Solutions to Assignment 1
Sophia Knight
September 29, 2006
Question 1
Consider the following relation on finite sequences of natural numbers:
n
1
n
2
n
3
...n
k
<
m
1
m
2
m
3
...m
l
if and only if
∃
i
such that
n
i
< m
i
and
∀
j < i n
j
=
m
j
. The case where
k < l
and
∀
i
≤
k n
i
=
m
i
is regarded as a special case of this definition. We define
n
1
n
2
n
3
...n
k
≤
m
1
m
2
m
3
...m
l
as
n
1
n
2
n
3
...n
k
< m
1
m
2
m
3
...m
l
or
n
1
n
2
n
3
...n
k
=
m
1
m
2
m
3
...m
l
.
Show that
≤
is a partial order.
Solution: We must show that
≤
is antisymmetric, transitive and reflexive. In order to include the
special case of the definition, we consider a blank space to be an element of a sequence which is
smaller than any number.
Antisymmetric
: If
a
=
a
1
a
2
...a
m
< b
=
b
1
b
2
...b
n
, then at the first position
i
where
a
and
b
differ,
a
i
< b
i
. If we also have
b < a
, then at the first position
i
where
b
and
a
differ,
b
i
< a
i
. Clearly,
the
i
=
i
, so both cannot be true. Thus, if
a < b
, then it is not the case that
b < a
, so if
a
≤
b
and
b
≤
a
, then
a
=
b
. So
≤
is antisymmetric.
Transitive
: Assume that
a
=
a
1
a
2
...a
l
< b
=
b
1
b
2
...b
m
< c
=
c
1
c
2
...c
n
. So,
∃
i
1
such that
a
i
1
< b
i
1
and for all
j < i
1
,
a
j
=
b
j
.
And
∃
i
2
such that
b
i
2
< c
i
2
and for all
j < i
2
,
b
j
=
c
j
.
So, let
i
= min(
i
1
, i
2
). Then for all
j < i
,
a
j
=
b
j
=
c
j
. Furthermore, if
i
=
i
1
=
i
2
, then
a
i
< b
i
< c
i
.
Otherwise,
i
=
i
1
or
i
=
i
2
. If
i
=
i
1
, then
i < i
2
and so
a
i
< b
i
=
c
i
. If
i
=
i
2
, then
a
i
=
b
i
< c
i
.
In any case,
a
i
< c
i
, so
a < c
. Finally, if
a
=
b
and
b < c
or if
a < b
and
b
=
c
, or if
a
=
b
=
c
then
it is still the case that
a
≤
c
. So, if
a
≤
b
≤
c
, then
a
≤
c
, and the relation is transitive.
Reflexive
This is obvious from the definition.
Show that
≤
is a wellfounded order.
Solution: It turns out that it’s not a wellfounded order! Here’s an infinite descending sequence:
((1);(0,1);(0,0,1);(0,0,0,1);...). We thank Artour Tomberg for this counterexample.
It is wellfounded if we bound the lengths of the sequences.
Here is the proof: We must show that for every nonempty subset
S
of finite sequences of natural
numbers, there is a minimal element with respect to our relation, that is an element
x
with no
other element in the set less than
x
.
Let
S
=
{
s
1
, s
2
, ...
}
=
{
(
a
1
,
1
a
1
,
2
a
1
,
3
...a
1
,n
1
); (
a
2
,
1
a
2
,
2
...a
2
,n
2
); (
a
3
,
1
a
3
,
2
...a
3
,n
3
);
...
}
Now, let
a
1
= min
{
a
1
,
1
, a
2
,
1
, a
3
,
1
, ...
}
. We know that
a
1
exists because
N
is a wellordered set. Let
S
1
=
{
s
m
∈
S

a
m,
1
=
a
1
}
. If
S
1
has an element of length one, then it is clear from the definition of
1