Answer:
Yes.
Consider an eigenvector associated with
λ
= 1
/
2.
Set
x
0
= (1
,

2)
T
.
Then
A
t
x
0
= (1
/
2)
t
x
0
. Since (1
/
2)
t
→
0,
A
t
x
0
→
0
.
3. Two embedding problems:
a
) Consider
A
= (0
,
1)
⊂
R
, and
B
=
{
(
x,
0)
∈
R
2
: 0
< x <
1
} ⊂
R
2
. Is
A
open? Is
B
open? If the
two differ, explain why embedding
A
in
R
2
changed the result.
Answer:
The set
A
is open. Let
x
∈
A
and set
= min(
x,
1

x
). Then
B
(
x
)
⊂
A
. Since
A
contains an open ball about each of its points, it is an open set.
The set
B
is not open. Let (
x,
0)
∈
B
and consider
B
(
x,
0). The point (
x, /
2)
∈
B
(
x,
0),
but is not in
B
. Thus there is a point in
B
so that no open ball about that point is contained
in
B
. (In fact, this is true of all points in
B
). Therefore
B
is not open.
The difference occurs because of the extra dimension. Balls in
R
2
contain points not on the
horizontal axis, and so cannot be contained in sets restricted to the horizontal axis. With balls
in
R
1
, this is not an issue.
b
) Now suppose
C
is a closed set in
R
2
. Show that
D
=
{
(
x, y,
0) : (
x, y
)
∈
C
}
is closed in
R
3
.
Answer:
Let (
x
n
, y
n
, z
n
)
∈
D
with (
x
n
, y
n
, z
n
)
→
(
x, y, z
). Since
z
n
= 0,
z
= 0. Because
C
is closed
and (
x
n
, y
n
)
∈
C
and (
x
n
, y
n
)
→
(
x, y
), we find (
x, y
)
∈
C
, so (
x, y, z
) = (
x, y,
0)
∈
D
. Thus
D
is closed because it contains all of its limit points.