Solutions to homework # 2.
1. For arbitrary
z
=
a
+
b
i,
w
=
c
+
d
i, define
z < w
if
a < c
or
a
=
c
and
b < d
. Prove that
this turns C
into an ordered set (”dictionary” or lexicographic” order). Does this ordered
set have the least upper bound property?
Solution. To check that this defines an ordered set, we must establish that any two elements
are comparable. Indeed, if
w
=
z
, then
a < c
,
a
=
c
or
a > c
. If
a < c
, then
z < w
; if
a > c
,
then
z > w
. If
a
=
c
, then either
b < d
(and then
z < w
) or
b > d
(and then
z > w
). So,
any two elements of C
are comparable using the lexicographic order.
Next, we need to check that the lexicographic order is transitive. Indeed, if
z
1
< z
2
< z
3
,
then Re
z
1
≤
Re
z
2
≤
Re
z
3
. If Re
z
1
<
Re
z
3
, then
z
1
< z
3
and we are done. If Re
z
1
= Re
z
3
,
then both inequalities must be equalities, i.e., Re
z
1
= Re
z
2
= Re
z
3
, but then Im
z
1
<
Im
z
2
<
Im
z
3
, hence
z
1
< z
3
. Thus C
is ordered by the lexicographic order.
This order does not have the least upper bound property. Consider the set
S
:=
{
z
: Re
z <
0
}
. This set is nonempty and bounded above by say 0. Suppose that
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 Spring '09
 #, Order theory, Complex number, upper bound, Partially ordered set

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