HW 3
1
1. (a) Let
(
A;
°
)
be a wellordered set,
B
±
A
and
f
:
A
onto
!
B
be an
order isomorphism of
(
A;
°
)
onto
(
B;
°
)
. Prove that for every
a
2
A
,
a
°
f
(
a
)
.
(b) Let
(
A;
°
)
be a wellordered set and
a
2
A
. Recall that the set
I
a
=
f
x
2
A
j
x
²
a
g
is called an
initial segment
of
A
.
Use part (a) to prove that
(
A;
°
)
cannot be order isomorphic to one of its initial segments.
Proof.
(a) Contrary to the claim, suppose that there is
a
0
2
A
such
that
f
(
a
0
)
²
a
0
. Then the set
A
=
f
a
2
A
j
f
(
a
)
²
a
g
is not empty because
a
0
2
A
. Because
(
A;
°
)
is a wellordered set and
A
6
=
?
, the set
A
contains the smallest element
a
0
2
A
, i.e.,
a
0
°
a
8
a
2
A
:
Now set
b
0
=
f
(
a
0
)
. Because
b
=
f
(
a
0
)
²
a
0
and because
f
preserves
the order, we have
f
(
b
)
²
f
(
a
0
) =
b
0
, i.e.,
b
0
2
A
and
b
0
²
a
0
, a
contradiction because
a
0
is the smallest element of
A
.
(b) Contrary to the claim, suppose that there is
a
2
A
such that
(
A;
°
)
is order isomorphic to the initial segment
I
a
. Then
8
x
2
A
f
(
x
)
²
a;
in particular,
f
(
a
)
²
a
, a contradiction with the result of part (a).
2. An ideal in a partial ordered set
(
X;
°
)
is a set
I
±
X
such that
(
x
2
I
)
^
[(
y
2
X
)
^
(
y
°
x
)]
)
y
2
I
. Prove that every ideal
I
in a
wellordered set
(
X;
°
)
is either an initial segment of
(
X;
°
)
or
X
.
Proof.
If
I
=
X
, then we are done. Let
I
6
=
X
. Let
X
n
I
6
=
?
. Set
b
= inf (
X
n
I
)
. We claim that
I
=
I
b
. Indeed, if
x
2
I
, then
x
°
b
;
1
Note to the grader.
Please grade the following problems:
#1, 2, #3, 4, 5
(20 pts for each problem)
1
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otherwise,
b
²
x
.
Because
I
is an ideal,
b
2
I
, a contradiction.
So,
I
±
I
b
. Now we show that
I
b
n
I
=
?
. Contrary to the claim, suppose
that there is
y
2
I
b
n
I
. Then
y
²
b
because
y
2
I
b
and
y
<
b
because
y
2
X
n
I
and
b
is the smallest element in
X
n
A
. Hence, because
y
²
b
and
y
<
b
, we arrive at a contradiction. So,
I
b
n
I
=
?
.
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 Spring '08
 Ruan
 Addition, Order theory, Natural number, Total order, initial segment

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