Homework Week Eight
Part 1 Mathematics
You can find the solutions for last weeks mathematics part at this weeks homework.
Def. 1 Cauchy sequence
A sequence (xn )n∈N in R is called Cauchy sequence, if:
∀ > 0∃N ∈ N∀n, m ≥ N :
|xn − xm | < .
Remark 2 We will proof that every convergent sequence in R is a Cauchy
sequence.
Proof:
Suppose (xn )n∈N is a convergent sequence in R with limit x. Thus we have
∀ > 0∃N ∈ N∀n ≥ N :
|xn − x| < .
2
Therefore:
∀ > 0∃N ∈ N∀n, m ≥ N :
and |xm − x| <
2
2
|xn − x| + |xm − x| < .
|xn − x| <
∀ > 0∃N ∈ N∀n, m ≥ N :
The triangle inequality gives us
∀ > 0∃N ∈ N∀n, m ≥ N :
|xn − xm | < .
Def. 3 Bounded sequence A sequence (xn )n∈N in R is bounded from above
(below), if there exist a constant C ∈ R, such that for all n ∈ N
xn ≤ C
(xn ≥ C).
A sequence is bounded if it is bounded from below and from above.
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1.1 Prove the following propositions:
Prop. 1 A Cauchy sequence in R is bounded. Particularly, every convergent
sequence is bounded.
Prop. 2 Suppose the sequences (xn )n∈N and (yn )n∈N in R converge to x and y.
The we have
1. (xn + yn )n∈N converges to x + y
2. (xn · yn )n∈N converges to x · y
Hint: You will need the inverse triangle inequality in Prop. 1 and the triangle
inequality in Prop. 2. For Prop. 2.2 you will also want to use Prop. 1.
Part 2 Economics
Read Chapter 1 of Mas-Colell
2.1 Prove Proposition 1.B.1.
If - is rational then:
i) ≺ is both irreflexive and transitive.
ii) ∼ is reflexive, transitive and symmetric.
iii) If x ≺ y z, then x ≺ z.
2.2 Consider the choice structure (B, C(·)) with B = {{x, y}, {x, y, z}} and
C({x, y}) = {x}. Show that if (B, C(·)) satisfies the weak axiom, then we must
have C({x, y, z}) = {x}, = {z}, or = {x, z}.
2.3 Show that the weak axiom is equivalent to the following property:
Suppose that B, B 0 ∈ B, that x, y ∈ B and that x, y ∈ B 0 . Then if x ∈ C(B)
and y ∈ C(B 0 ), we must have {x, y} ⊆ C(B) ∩ C(B 0 ).
2.4 Give an example of a choice structure that can be rationalized by several
preference relations. Note that if the family of budgets (B) includes all the twoelement subsets of X, then there can be at most one rationalzing preference
relation.
Please hand this in on Thursday (14.01.2015).
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Part 3 Solutions
Ex. 1 The sequence
1−
1
n
converges to 1.
n∈N
Proof: Suppose > 0. The fundamental lemma gives us the existence of a
N ∈ N such that
1
N
≤ . For n ≥ N we have
1 − 1 − 1 = 1 ≤ 1 ≤ n
n
N
Thus the sequence converges
to 1.
Ex. 2 The sequence 5n22+6
converges to 0.
n∈N
Proof: Suppose > 0. The fundamental lemma gives us the existence of a
N ∈ N such that
1
N
≤ . For n ≥ N we have
2
2
1
1
1
2
=
−
0
5n2 + 6
5n2 + 6 ≤ 5n2 ≤ n2 ≤ n ≤ N ≤ Thus the sequence converges
to 0.
2n n!
Ex. 3 The sequence nn+1
converges to 0.
n∈N
Proof: Suppose > 0. The fundamental lemma gives us the existence of a
N ∈ N such that
1
N
≤ . For n ≥ N = max{N , 6} we have
n
n
n
2 n!
= 2 n! ≤ n n! ≤ 1 ≤ 1 ≤ −
0
nn+1
nn+1
n!nn+1
n
N
Thus the sequence converges to 0.
Proposition 1.6
A convergent sequence in R has a unique limit.
Proof: Suppose there exists a sequence (xn )n∈N with two distinct limits x and
y. The we have
∀ > 0∃N ∈ N∀n ≥ N :
∀ > 0∃N ∈ N∀n ≥ N 0 :
2
|xn − y| < .
2
|xn − x| <
For Ñ := max{N, N 0 } we have
∀ > 0∃Ñ ∈ N∀n ≥ Ñ :
|xn − x| + |xn − y| <
The triangle inequality leaves us with
∀ > 0∃Ñ ∈ N∀n ≥ Ñ :
3
|x − y| ≤ .
+ = .
2 2
In particular
∀ > 0 :
|x − y| ≤ .
Form this we get x = y, because if x 6= y we’d get |x − y| > 0. For = |x − y|/2
we get the contradiction |x − y| ≤ |x − y|/2.
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