MAT237Y – Multivariable Calculus
Summer 2009
Assignment # 3 Solutions
1.
(a) Show that if
A
⊂
R
n
is convex, then the closure of
A
is convex.
(b) Show that if
A
⊂
R
n
is convex, then the interior of
A
is convex.
Hint:
If
r >
0 and
x
and
y
are such that
B
r
(
x
) and
B
r
(
y
) are
subsets of
A
, show that
B
r
(
z
)
⊂
A
for any
z
on the line between
x
and
y
.
Solution.
(a) Let
A
be convex. Two show that the closure
A
is convex, we must
show that for any two points
a, b
∈
A
, and any
t
∈
[0
,
1], the
interpolated point
c
=
ta
+ (1
−
t
)
b
is also in
A
.
Since
a, b
are in the closure of
A
, we may find sequences (
a
n
)
∞
n
=1
,
(
b
n
)
∞
n
=1
of elements in
A
which converge to
a, b
respectively. Let
c
n
=
ta
n
+ (1
−
t
)
b
n
.
Since
A
is convex, we see that
c
n
∈
A
. Moreover, we have
lim
n
→∞
c
n
= lim
n
→∞
ta
n
+ (1
−
t
)
b
n
=
t
( lim
n
→∞
a
n
) + (1
−
t
)( lim
n
→∞
b
n
)
=
ta
+ (1
−
t
)
b
=
c.
Thus,
c
is the limit of a sequence of points in
A
, so that
c
∈
A
.
(b) Let
A
be convex. To show that the interior int
A
of
A
is convex, we
must show that for any two points
a, b
∈
int
A
, and any
t
∈
[0
,
1],
the interpolated point
c
=
ta
+ (1
−
t
)
b
is also in int
A
.
Since
a
∈
int
A
, we have for some
r
1
>
0 that
B
r
1
(
a
)
⊂
A
. Simi
larly, for some
r
2
>
0,
B
r
2
(
b
)
⊂
A
. By taking
r
= min
{
r
1
, r
2
}
, we
thus have that
B
r
(
a
)
⊂
A
and
B
r
(
b
)
⊂
A.
Claim.
B
r
(
c
)
⊂
A
.
Proof of claim.
For
d
∈
B
r
(
c
), let
h
=
d
−
c
, so that
bardbl
h
bardbl
< r
. We
thus have
a
+
h
∈
B
r
(
a
)
⊂
A
and similarly,
b
+
h
∈
B
r
(
b
)
⊂
A
.
Also,
d
=
c
+
h
=
ta
+ (1
−
t
)
b
+
h
=
ta
+ (1
−
t
)
b
+
th
+ (1
−
t
)
h
=
t
(
a
+
h
) + (1
−
t
)(
b
+
h
)
,
and since
A
is convex, this shows that
d
∈
A
, as required.
1
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2. A function
f
:
U
→
V
is
invertible
if there is
g
:
V
→
U
such that
g
(
f
(
x
)) =
x
for all
x
∈
U
, and
f
(
g
(
y
)) =
y
for all
y
∈
V
.
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 Fall '09
 RomauldStanczak
 Calculus, Multivariable Calculus, SEPTA Regional Rail, Continuous function, Invertible matrix, Inverse function, ∂v

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