Unformatted text preview: 7.17 Recall the BolzanoWeierstrass theorem for sequences: Let xi , i = 1, 2, 3, . . . be a bounded
sequence in RN . Then xi contains a convergent subsequence. Let xν ∈ R, ν = 1, 2, 3, . . .. xν =
1
(−1)ν + ( 2 )ν .
1. Is xν bounded? Explain.
2. Is xν convergent? Explain.
3. If your answer to part “b” is “yes,” ﬁnd the limit. Explain
4. If your answer to part “a” is “yes” and to part “b” is “no,” then ﬁnd a convergent subsequence
and its limit. Demonstrate convergence.
Suggested Answer: 1. The sequence is completely contained in the interval [−2, 2]. It is
bounded.
2. The sequence bounces between values near 1 and greater than +1 . It is not convergent.
3. Omit
1
4. For ν odd the subsequence converges to −1: − 2 , − 7 , − 31 , · · · .
8
32
1
1
For ν even the subsequence converges to +1: 1 + 1 , 1 + 16 , 1 + 64 , · · · .
4 1 Problems 7.19  7.25: Suggested Answers
7.19 Recall the BolzanoWeierstrass theorem for sequences: Let xi , i =
1, 2, 3, . . .be a bounded sequence in RN . Then xi contains a convergent subsequence. Let xν ∈ R, ν = 1, 2, 3, . . .. xν = (−1)ν (10). That is, xν = −10 for ν
odd and xν = 10 for ν even.
1. Is xν bounded? Explain.
2. Is xν convergent? Explain.
3. If your answer to part b is “yes” ﬁnd the limit. Explain. If your answer
to part a is “yes” and to part b is “no” then ﬁnd a convergent subsequence
and its limit. Demonstrate convergence.
Suggested Answer (a) xν ∈ [−10, 10] so the sequence is bounded.
(b) No. There is no single limit point since the sequence moves between 10
and 10 indeﬁnitely.
(c) The subsequence of xν for odd values of ν converges to 10; the subsequence for even values converges to 10.
7.20 Think of a vector of purchases of N goods as a point in RN , a =
(a1 , a2 , . . . , aN ) where a1 is the amount of good 1 purchased, a2 is the amount
of good 2 purchased, and so forth. Think of prices of the N goods as represented
by a point in RN p = (p1 , p2 , . . . , pn , . . . , pN ) where p1 is the price of good 1,
p2 is the price of good 2, and so forth. Prof. Debreu writes (paraphrasing
N
slightly): “The value of an action a relative to the price system p is i=1 pi ai ,
the [dot] product p · a.”
Brieﬂy explain this deﬁnition of the value of the purchase plan a = (a1 , a2 , . . . , aN )
relative to prices p = (p1 , p2 , . . . , pn , . . . , pN ). What does it mean in words?
Why does it make sense? Your answer should run between ten and one hundred
words.
Suggested Answer: This is just adding up the cost of a shopping list. For
each item, take unit price times number of units; that gives the gross value of
the purchase of that item. Then add up the gross costs along the list of items.
7.21 Consider each of the following functions of R to R. In each case state
whether the function is continuous at 0.
1. f (x) = x + 10
2. g(x) = −1 for x ≤ 0, g(x) = 1 for x > 0
3. h(x) = x2
Suggested Answer: (a) Yes, deﬁnitely. There’s no jump at 0. The function
value progresses steadily from slightly less than 10 to slightly more than 10.
(b) No, deﬁnitely not. The function value jumps at 0 from 1 to +1.
1 (c) Yes. The function value hovers consistently about 0.
7.22 Recall the following deﬁnitions, concerning subsets of RN :
• a set is ‘closed’ if it contains all of its cluster points (limit points).
• a set is ‘open’ if, for each point in the set, there is a small ball (neighborhood) centered at the point, contained in the set.
• a set is ‘bounded’ if it can be contained in a cube of ﬁnite size, centered
at the origin.
• a set is ‘compact’ if it is both closed and bounded.
1. Is the following subset of R2 closed? open? bounded? compact? Explain
your answer. T=45o line through the origin = {(x, y)(x, y) ∈ R2 , x = y}.
2. Is the following subset of R2 closed? open? bounded? compact? Explain
your answer.
U =
= ball of radius 10 centered at the origin, including its boundary
{(x, y)(x, y) ∈ R2 , x2 + y2 ≤ 100}. Suggested Answer: (a) The set T is closed. It is not bounded, open,
or compact. Since the line extends indeﬁnitely in both directions, it is not
bounded or compact. Since at each point it does not contain the surrounding
neighborhood in R2 , it is not open.
(b) The set U is closed, bounded, compact, not open. U is closed since
it contains its limit points. It is bounded since it is of ﬁnite extent: it can
be contained in a ﬁnite cube centered at the origin. It is compact since it is
closed and bounded. It is not open, since for points on the boundary, a small
neighborhood of the point is not fully contained in U .
7.23 Let A, B ⊂ R. A = [−1, 12], the closed interval from 1 to 12. B =
[7, 18], the closed interval from 7 to 18.
1. Describe A
suﬃcient. B. Representing A B as one or more intervals is 2. Describe A
suﬃcient. B. Representing A B as one or more intervals is 3. Describe A\B . Representing A\B as one or more intervals is suﬃcient.
Suggested Answer: (a) A ∪ B = [−1, 18]
(b)A ∩ B = [7, 12]
(c) A \ B = [−1, 7)
7.24 Let S ⊂ R2 , S compact (closed and bounded).
ν = 1, 2, 3, ..., be a sequence in S . xν ∈ S , all ν .
2 Let xν = (xν , xν ),
1
2 Show that A B is convex.
Let x, y A BThen x, y A and B. Then by convexity of A and B we have that
x+(1)y A and B. Then x+(1)y A B.
Show that A+B is convex.
x, y A+B means that there are xa, ya A, and xb, yb B so that xa + xb = x A+B and
ya + yb y A+B. Then
x + (1)y = xa + xb + (1)ya +(1)yb = xa+(1)ya+xb+(1)yb. But
a
x +(1)ya A, xb+(1)yb B, by convexity of A and B. So x + (1)y A+B.
Show that A is convex.
Let x, y A . We wish to show that x+(1)y A . The only difficulty arises if
x or y A. Suppose x A. But then x is the limit of a sequence x A . Then the sequence
x+(1)y A and approaches x+(1)y as its limit, so x+(1)y A .
8.1 1 9.1, 9.2 Suggested Answers
9.1
The Brouwer FixedPoint Theorem can be stated in the following way:
Let S ⊂ RN be compact and convex. Let f : S → S be a continuous function.
Then there is x∗ ∈ S so that f (x∗ ) = x∗.
Show how a ﬁxed point would fail to exist when the assumptions of the Brouwer
FixedPoint theorem are not fulﬁlled, as speciﬁed in the following cases:
i. Suppose S is not convex. Let S = [1, 2] ∪ [3, 4]; S ⊂ R. That is, S is the union
of two disjoint closed intervals in R. Find continuous f : S → S so that there is no
ﬁxed point x∗ fulﬁlling the theorem.
ii. Suppose f is not continuous. Let S = [1, 4]; S ⊂ R. Let
f (x ) = 4 − x for x < 2,
x − 1 for x ≥ 2. Show that although f : S → S there is no ﬁxed point of f in S .
iii. Suppose S is not compact. Let S = R and f (x) = x + 1. Note that f : S → S
and f is continuous. Show that there is no ﬁxed point of f in S .
Suggested Answer :(i) Suppose S is not convex. Let S = [1, 2] ∪ [3, 4]; S ⊂ R.
That is, S is the union of two disjoint closed intervals in R. Find continuous f : S → S
so that there is no ﬁxed point x* fulﬁlling the theorem.
Let f(x) = 5  x. f : [1, 2] → [3, 4]; f : [3, 4] → [1, 2]. f : S → S and f continuous
but there is no ﬁxed point.
(ii) Suppose f is not continuous. Let S = [1, 4]; S ⊂ R. Let f (x) = 4 − x for x < 2,
f (x) = x − 1 for x ≥ 2. Show that f : S → S but there is no ﬁxed point of f in S.
For 1 ≤ x < 2, we have 3 ≥ f (x) > 2,
2 ≤ x ≤ 4, we have 1 ≥ f (x) ≤ 3, f (x) = x. So f : S → S, but there is no ﬁxed
point.
(iii) Suppose S is not compact. Let S = R, f (x) = x + 1. Show that f : S → S
but there is no ﬁxed point of f in S.
To have a ﬁxed point would require that f (x) = x + 1 = x, or equivalently that
0 = 1. This is impossible so there is no ﬁxed point. 1 9.2
Recall the Intermediate Value Theorem:
Let [a, b] be a closed interval in R and h a continuous realvalued function on [a, b]
so that h(a) < h(b). Then for any real k so that h(a) < k < h(b) there is x [a, b] so
that h(x) = k .
Recall the Brouwer FixedPoint Theorem:
Let S ⊂ RN be compact and convex. Let f : S → S be a continuous function.
Then there is x∗ ∈ S so that f (x∗ ) = x∗.
Consider the special case S = [0, 1], the unit interval in R, and let f be a continuous function from S into itself. Using the Intermediate Value Theorem, prove the
Brouwer FixedPoint Theorem for this case. You may ﬁnd the function g (x) = x−f (x)
useful.
Suggested Answer : Deﬁne g (x) = x − f (x) . Then g (0) ≤ 0 and g (1) ≥ 0.
Then by the Intermediate Value Theorem there is x∗ ∈ [0, 1] so that g (x∗) = 0. But
then g (x∗ ) = 0 = x∗ − f (x∗) so x∗ = f (x∗ ) and x∗ is the required ﬁxed point. 2 ...
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Full Document
 Fall '08
 ZAMBRANO
 Microeconomics, Topology, Continuous function, Brouwer, Brouwer FixedPoint Theorem

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