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Math 486 Homework 1 Notes
Paul Shafer
pshafer@math.cornell.edu
February 5, 2008
Problem I.1.8a
We are to prove that (
N
n
,<
L
) is a well order. It is reasonably clear that (
N
n
,<
L
) is a linear order,
so we need to prove the following: every nonempty
A
⊆
N
n
has a
<
L
least element. This is stronger
than saying
N
n
has a
<
L
least element. It is easy to give an example of a linear order with a least
element that is not a well order. Consider the set
{∞} ∪
Z
with
Z
ordered in the usual way and
∞
< z
for all
z
∈
Z
. Then
∞
is least, but the order is not a well order because the subset
Z
has no least element.
Given
A
⊆
N
n
nonempty, we can ﬁnd the least element of
A
exactly as you would in the dictionary.
Let
A
1
=
{
x
1
 ∃
x
2
,x
3
,...,x
n
< x
1
,x
2
,x
3
,...,x
n
>
∈
A
}
= the set of 1
st
coordinates of
A
a
1
= the least element of
A
1
A
2
=
{
x
2
 ∃
x
3
,...,x
n
< a
1
,x
2
,x
3
,...,x
n
>
∈
A
1
}
= the set of 2
nd
coordinates of
A
1
a
2
= the least element of
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 Spring '08
 DORAIS
 Math, Logic

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