Note that 0
≤
𝛾
𝑖
≤
1, so that either
𝛾
𝑚
<
1 or
𝛾
𝑚
= 1. We show that both cases lead
to a contradiction.
∙
𝛾
𝑚
<
1. Then
x
𝑚
=
1
1
−
𝛾
𝑚
𝑚
−
1
∑
𝑖
=1
(
𝛼𝛼
𝑖
+ (1
−
𝛼
)
𝛽
𝑖
)
x
𝑖
which contradicts the minimality of the set
{
x
1
,
x
2
, . . . ,
x
𝑚
}
.
∙
𝛾
𝑚
= 1. Then
𝛾
𝑖
= 0 for every
𝑖
∕
=
𝑚
. That is
𝛼𝛼
𝑖
+ (1
−
𝛼
)
𝛽
𝑖
= 0
for every
𝑖
∕
=
𝑚
which implies that
𝛼
𝑖
=
𝛽
𝑖
for every
𝑖
∕
=
𝑚
and therefore
x
=
y
.
Therefore, if
{
x
1
,
x
2
, . . . ,
x
𝑚
}
is a minimal spanning set, every point must be an extreme
point.
1.181
Assume to the contrary that one of the vertices is not an extreme point of the
simplex.
Without loss of generality, assume this is
x
1
.
Then, there exist distinct
y
,
z
∈
𝑆
and 0
< 𝛼 <
1 such that
x
1
=
𝛼
y
+ (1
−
𝛼
)
z
(1.25)
Now, since
y
∈
𝑆
, there exist
𝛽
1
, 𝛽
2
, . . . , 𝛽
𝑛
such that
y
=
𝑛
∑
𝑖
=1
𝛽
𝑖
x
𝑖
,
𝑛
∑
𝑖
=1
𝛽
𝑖
= 1
Similarly, there exist
𝛿
1
, 𝛿
2
, . . . , 𝛿
𝑛
such that
z
=
𝑛
∑
𝑖
=1
𝛿
𝑖
x
𝑖
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 Fall '10
 Dr.DuMond
 Macroeconomics, Existence, All rights reserved, ????=1, Assumption of Mary, Foundations of Mathematical Economics

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