Boğaziçi University
Department of Economics
Fall 2012
EC 521 MATHEMATICAL METHODS FOR ECONOMICS
Problem Set 5 - Answer Key
1. (a) (i) d1 (x, x) = d(x, x)/(1 + d(x, x)) = 0 for all x ∈ X since d(x, x) = 0.
(ii) d1 (x, y) = d(x, y)/(1 + d(x, y)) = d(y, x)/(1 + d(y, x)) = d1 (y, x) for all x, y ∈ X
since d(x, y) = d(y, x).
(iii) d1 (x, y) = d(x, y)/(1 + d(x, y)) ≤ (d(x, z) + d(z, y))/(1 + d(x, z) + d(z, y)).
Above inequality follows from d(x, y) ≤ d(x, z) + d(z, y) and from the fact that
f (x) = x/(1 + x) is increasing in x.
Thus d1 (x, y) ≤ d(x, z)/(1 + d(x, z) + d(z, y)) + d(z, y)/(1 + d(x, z) + d(z, y)) ≤
d(x, z)/(1 + d(x, z)) + d(z, y)/(1 + d(z, y)) = d1 (x, z) + d1 (z, y) for all x, y, z ∈ X.
(b) Let S ⊆ X be open in (X, d). Then for all x ∈ S there exists > 0 such
that N,d (x) ⊆ S. Now note that N,d (x) = {y ∈ X : d(x, y) < } = {y ∈ X :
d(x, y)/(1 + d(x, y)) < /(1 + )}.
To see the second equality above, note that d(x, y) < if and only if d(x, y) +
d(x, y) < + d(x, y) if and only if d(x, y)(1 + ) < (1 + d(x, y)) if and only if
d(x, y)/(1 + d(x, y)) < /(1 + ).
Thus N0,d1 (x) = {y ∈ X : d1 (x, y) < 0} ⊆ S where 0 = /(1 + ). Thus S is open
in (X, d1 ).
2. (i) (f ◦ d)(x, x) = f (d(x, x)) = f (0) = 0. for all x ∈ X.
(ii) (f ◦ d)(x, y) = f (d(x, y)) = f (d(y, x)) = (f ◦ d)(y, x) for all x, y ∈ X.
(iii) (f ◦ d)(x, y) = f (d(x, y)) ≤ f (d(x, z) + d(z, y)) ≤ f (d(x, z)) + f (d(z, y)) =
(f ◦ d)(x, z) + (f ◦ d)(z, y) since f is strictly increasing and concave for all x, y, z ∈ X.
3. Since limxn = x, for all > 0 there exists Nx such that d(xn , x) < /2 for all n > Nx .
And since limyn = y, for all > 0 there exists Ny such that d(yn , y) < /2 for all
n > Ny . Then;
|d(xn , yn ) − d(x, y)| ≤ |d(xn , x) + d(x, yn ) − d(y, x)| ≤ |d(xn , x) + d(x, y) + d(y, yn ) −
d(x, y)| = |d(xn , x) + d(y, yn )| < /2 + /2 = for all n > max{Nx , Ny } = N .
Therefore for all > 0 there exists N such that |d(xn , yn )−d(x, y)| < for all n > N.
Thus limd(xn , yn ) = d(x, y).
1
4. (a) {(1/n, 1/n2 ) : n ∈ N } ∪ (0, 0) ⊆ R2 is closed its complement in R2 is open in
R2 .
(b) {(x, y, x2 y 2 ) : x2 + y 2 < 1} ⊆ R2 is closed because its complement in R3 is open
in R2 .
S
(c) [−n, (n − 1)/n] = (−∞, 1) ⊆ R is open in R.
T
(d) (0, 1/n] = ∅ ⊆ R is clopen in R.
5. (a) Let x ∈ Int(A) ∩ Int(B). Then x ∈ Int(A) and x ∈ Int(B). Then there exists
A > 0 s.t. NA (x) ⊆ A and there exists B > 0 s.t NB (x) ⊆ B. Then defining
= min{A , B }, we get N (x) ⊆ A ∩ B. Thus x ∈ Int(A ∩ B).
Now let x ∈ Int(A ∩ B). Then there exists > 0 s.t N (x) ⊆ A ∩ B. So, N (x) ⊆ A
and N (x) ⊆ B. Therefore x ∈ Int(A) and x ∈ Int(B). Thus x ∈ Int(A) ∩ Int(B).
(b) Let x ∈ Cl(A) ∪ Cl(B). Then x ∈ Cl(A) or x ∈ Cl(B). Say x ∈ Cl(A). Then for
all > 0 N (x) ∩ A 6= ∅. Thus N (x) ∩ (A ∪ B) 6= ∅. Thus x ∈ Cl(A ∪ B).
Now let x ∈ Cl(A ∪ B). Then for all > 0 N (x) ∩ A 6= ∅. Then N (x) ∩ A 6= ∅ or
N (x) ∩ A 6= ∅. Thus we get x ∈ Cl(A) and x ∈ Cl(B), that is, x ∈ Cl(A) ∪ Cl(B).
6. (→) let x ∈ Bd(S). Then for all > 0 N ∩ S 6= ∅ and N ∩ (X/S) 6= ∅. Then
for any n ∈ N, we have N(1/n)(x) ∩ S 6= ∅ and N(1/n)(x) ∩ (X/S) 6= ∅. Pick xn ∈
N(1/n)(x) ∩ S and xn 0 ∈ N(1/n)(x) ∩ (X/S). Thus d(xn x) < 1/n < for any n > 1/
and d((xn )0, x) < 1/n < for any n > 1/. Therefore (xn ) is in S with limxn = x
and (xn )0 is in X/S with lim(xn )0 = x.
(←) Suppose there exists (xn ) ∈ S, there exists (xn )0 ∈ X/S such that limxn = x =
limxn 0. Then x ∈ Cl(S) and x ∈ Cl(X/S). Thus x ∈ S ∪ Bd(S) and x ∈ (X/S) ∪
Bd(X/S). Suppose x ∈
/ Bd(S). Then x ∈ S. Thus x ∈
/ X/S, so x ∈ Bd(X/S).
But this contradicts x ∈
/ Bd(S) since Bd(S) = Bd(X/S). To see this note that
Bd(S) = X/(Int(S) ∪ Ext(S)) = X/(Ext(X/S) ∪ Int(X/S)) = Bd(X/S).
7. a) Let x ∈ S. Then N( 1/2)(x) = x ⊆ S. Thus S is open.
b) Let x ∈ X.Then for all > 0 picking δ = 1/2 we get if d(x, y) < δ = 1/2 then
x = y which implies dY (f (x), f (y)) = dY (f (x), f (x)) = 0 < .Thus f is continuous
at x. Since x was arbitrary, f is continuous.
8. Let y ∈ f (Cl(A)). Then there exists x ∈ Cl(A) such that f (x) = y. Then for
all > 0 N (x) ∩ A 6= ∅. Now since f is continuous, there exists δ > 0 such that
f (Nδ (x)) ⊆ N (f (x)). Since Nδ (x) ∩ A 6= ∅, we have f (Nδ (x) ∩ A) 6= ∅. Now since
2
∅=
6 f (Nδ (x) ∩ A) ⊆ f (Nδ (x)) ∩ f (A) ⊆ N (f (x)) ∩ f (A), we get N (y) ∩ f (A) 6= ∅.
Thus y ∈ Cl(A).
9. To see f (x) = 1/x is continuous, first observe that |f (x) − f (y)| = |1/x − 1/y| =
|x − y|/xy. Given > 0, we need to find δ > 0 such that |x − y|/xy < whenever
|x − y| < δ. Note that |x − y|/xy < δ/x(x − δ) when |x − y| < δ for all 0 < δ < x.
But δ/x(x − δ) < if delta < x2 /(1 + x).
So by choosing any δ > 0 such that δ < x2 /(1 + x), we get d(f (x), f (y)) < whenever d(x, y) < δ.
To show f (x) = 1/x is not uniformly continuous, choose = 1. Now by choosing
y = x + δ/2, we have |x − y| = δ/2 < δ. Then for |1/x − 1/y| < 1 to hold one needs
|x − y|xy, that is, δ/2 < x(x + δ/2) for any x ∈ X, that is, δ/2 < x2 + xδ/2 or
δ/2 < x2 /(1 − x) for any x ∈ X. However for x > 1 there does not exist δ > 0 that
satisfies δ/2 < x2 /(1 − x).
10. For any > 0, pick δ =
K.
Then, if dX (x, y) < δ then,
dY (f (x), f (y)) ≤ KdX (x, y) < Kδ = K
3
=
K