Math 131A  Section 2
Spring 2010
Homework 2
Graded problems: 4.8 (3 points), 4.11 (2 points), 4.12 (2 points), 4.14 (3 points), 5.6 (3 points), 7.4 (2 points)
4.4.(e). 0; (i). 0; (j). 2
/
3; (l). No inf; (v).

1
/
2; (w).

√
3
/
2
4.8.(a). Since
T
is nonempty, there exists
t
0
∈
T
. Since
s
≤
t
0
for all
s
∈
S
, we have
S
is bounded above. Similarly,
since
S
is nonempty, there exists
s
0
∈
S
. Since
s
0
≤
t
for all
t
∈
T
, we have
T
is bounded below. Thus, sup
S
and inf
T
exists and are ﬁnite.
(b). Fix
t
0
∈
T
. Since
s
≤
t
0
for all
s
∈
S
, we have that
t
0
is an upper bound for
S
. Thus, sup
S
≤
t
0
(since the
supremum is the least upper bound). Our choice of
t
0
∈
T
was arbitrary, we have that sup
S
≤
t
for all
t
∈
T
. Thus,
sup
S
is a lower bound for
T
, and since the inﬁmum is the greatest lower bound, we have that sup
S
≤
inf
T
.
(c). Take
S
=
T
=
{
0
}
. Then,
S
∩
T
=
{
0
} 6
=
∅
, and the condition
s
≤
t
for all
s
∈
S
and
t
∈
T
holds (since 0
≤
0).
(d). Let
S
= (0
,
1) and
T
= (1
,
2). Then,
S
∩
T
=
∅
, but sup
S
= 1 = inf
T
.
Notes: The most confusing part of this problem seemed to have been what to do when the book says “observe that.
..”
in the beginning of a problem. In general, this should mean you prove the result they state, and your proof should be
rather short and pretty much follow immediately from the assumptions in the problem.
4.11. Let