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
fi
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
fi
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
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Let
A
be a set in
R
2
,
de
fi
ned by:
A
=
{
(
x
1
, x
2
)
∈
R
2
: (
x
1
)
2
−
(
x
2
)
2
>
1
}
(a) Describe the interior, exterior and boundary of
A
(algebraically).
(b) Graph the sets in (a) in an appropriate diagram.
Solution:
(a) Algebraically,
Int
(
A
)
=
{
(
x
1
, x
2
)
∈
R
2
: (
x
1
)
2
−
(
x
2
)
2
>
1
}
;
Ext
(
A
)
=
{
(
x
1
, x
2
)
∈
R
2
: (
x
1
)
2
−
(
x
2
)
2
<
1
}
;
Bond
(
A
)
=
{
(
x
1
, x
2
)
∈
R
2
: (
x
1
)
2
−
(
x
2
)
2
= 1
}
.
(b)
The graph is as follows:
3. [Limits of Sequences and Continuity]
Let
{
a
n
}
be a sequence, with
a
n
∈
S
⊂
R
m
for each
n
∈
{
1
,
2
,
3
,
....
}
,
and
lim
n
→∞
a
n
=
p.
Suppose
f
:
S
→
R
is continuous on
S,
and
p
∈
S,
show
that:
lim
n
→∞
f
(
a
n
) =
f
(
p
)
Intuition:
The intuition behind this problem is as follows: We know
that as
n
becomes very large,
a
n
becomes very close to
p.
We also know that
as
a
n
gets very close to
p, f
(
a
n
)
gets very close to
f
(
p
)
.
Now we are asked to
prove that as
n
gets large,
f
(
a
n
)
gets very close to
f
(
p
)
.
Proof:
Given any
ε >
0
, by the de
fi
nition of continuous functions, we know that
there exists
δ >
0
,
which might depend on
ε,
such that
k
a
n
−
p
k
< δ
implies
k
f
(
a
n
)
−
f
(
p
)
k
< ε.
(We have used the fact that if a function is continuous
on its domain, it is continuous at each point on its domain by the de
fi
nition
of continuous functions.
)
Now, for this speci
fi
c
δ >
0
, by the fact that
lim
n
→∞
a
n
=
p,
we know that there exists
N
∈
N
, where
N
might depend
on
δ,
such that for every
n > N
,
k
a
n
−
p
k
< δ.
Essentially we have proved that for any
ε >
0
, there exists
N
∈
N
,
where
N
might depend on
ε,
such that for every
n > N
,
k
f
(
a
n
)
−
f
(
p
)
k
< ε.
Therefore,
lim
n
→∞
f
(
a
n
) =
f
(
p
)
.
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