5. We have shown that
conv
𝑆
⊆
conv
𝑆
, that is conv
𝑆
is closed.
6. conv
𝑆
is a closed and bounded subset of a finite dimensional space, and hence
conv
𝑆
is compact (Proposition 1.4 and Exercise 1.215).
1.226
1.
𝑆
is bounded. Therefore, there exists some
𝑐
such that
∥
x
∥
∞
= max
𝑖
∣
𝑥
𝑖
∣
<
𝑐
for every
x
∈
𝑆
. That is
−
𝑐
≤
𝑥
𝑖
≤
𝑐
so that
x
∈
𝐶
=
{
x
= (
𝑥
1
, 𝑥
2
, . . . , 𝑥
𝑛
)
∈ ℜ
𝑛
:
−
𝑐
≤
𝑥
𝑖
≤
𝑐
for every
𝑖
}
Therefore
𝑆
⊂
𝐾
.
2. Exercise 1.177.
3.
𝐶
is the convex hull of a finite set and hence is compact (Exercise 1.225)
4.
𝑆
is a closed subset of a compact set and hence is compact (Exercise 1.110).
1.227
A polytope is the convex hull of a finite set. Any finite set is compact.
1.228
The unit simplex Δ
𝑛
−
1
in
ℜ
𝑛
is the convex hull of the unit vectors
e
1
,
e
2
, . . . ,
e
𝑛
,
that is
Δ
𝑛
−
1
= conv
{
e
1
,
e
2
, . . . ,
e
𝑛
}
=
{
(
𝑥
1
, 𝑥
2
, . . . , 𝑥
𝑛
)
∈ ℜ
𝑛
:
𝑥
𝑖
≥
0 and
∑
𝑥
𝑖
= 1
}
This simplex has a nonempty relative interior, namely
ri
𝑆
=
{
(
𝑥
1
, 𝑥
2
, . . . , 𝑥
𝑛
)
∈ ℜ
𝑛
:
𝑥
𝑖
>
0 and
∑
𝑥
𝑖
<
1
}
1.229
Let
𝑛
= dim
𝑆
. By Exercise 1.182,
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 Fall '10
 Dr.DuMond
 Macroeconomics, Topology, Empty set, Compact space, ???? ????, conv ????

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