TA: Yankun Wang
Economics 617
Problem Set 5 Solution Key
1. [Distance]
Let
d
denote the Euclidean distance function on
R
n
×
R
n
.
De
f
ne:
g
(
x, y
)=
d
(
x, y
)
/
[1 +
d
(
x, y
)]
for all
x, y
in
R
n
Show that:
(a)
g
(
x, y
)
≥
0
for all
x, y
in
R
n
(b)
g
(
x, y
)=0
if and only if
x
=
y
(c)
g
(
x, y
)=
g
(
y,x
)
(d)
g
(
x, z
)
≤
g
(
x, y
)+
g
(
y,z
)
for all
x, y, z
in
R
n
Solution:
(a)
d
(
x, y
)
≥
0
for all
x, y
in
R
n
,
and thus
1+
d
(
x, y
)
≥
1
for all
x, y
in
R
n
.
As a result,
g
(
x, y
)
≥
0
for all
x, y
in
R
n
.
(b) By the de
f
nition of
g
(
x, y
)
,g
(
x, y
)=0
if and only if
d
(
x, y
)=0
.
But
d
(
x, y
)=0
if and only if
x
=
y.
(c) Since
d
(
x, y
)=
d
(
y,x
)
,
it is straightforward to see that
g
(
x, y
)=
g
(
y,x
)
.
(d)
g
(
x, z
)
≤
g
(
x, y
)+
g
(
y,z
)
⇔
d
(
x, z
)
1+
d
(
x, z
)
≤
d
(
x, y
)
1+
d
(
x, y
)
+
d
(
y,z
)
1+
d
(
y,z
)
⇔
d
(
x, z
)[1 +
d
(
x, y
)][1 +
d
(
y,z
)]
≤
d
(
x, y
)[1 +
d
(
x, z
)][1 +
d
(
y,z
)] +
d
(
y,z
)[1 +
d
(
x, y
)][1 +
d
(
x, z
)]
⇔
d
(
x, z
)
≤
d
(
x, y
)+
d
(
y,z
)+2
d
(
x, y
)
d
(
y,z
)+
d
(
x, y
)
d
(
y,z
)
d
(
x, z
)
Since
d
(
x, y
)+
d
(
y,z
)
≥
d
(
x, z
)
,d
(
a, b
)
≥
0
for any
(
a, b
)
∈
R
n
×
R
n
,
the
last inequality is always true. Thus
g
(
x, z
)
≤
g
(
x, y
)+
g
(
y,z
)
for all
x, y, z
in
R
n
.
2. [Interior, Exterior and Boundary]
1