MAT 125a, HW4 Solutions
10
Suppose that lim
n
→∞
p
n
=
p
in a given metric space. The set of points
S
=
{
p, p
1
, p
2
...
}
is closed.
lemma:
If
x
6∈ {
a
1
, a
2
, a
3
, ...
}
and for all
>
0
B
(
x
)
∩ {
a
1
, a
2
, ...
} 6
=
∅
then
∀
>
0 and
∀
n
∈
N
we have
B
(
x
)
∩ {
a
n
+1
, a
n
+2
, a
n
+3
...
} 6
=
∅
.
proof
We will show the contrapositive. Suppose there exist
>
0 and
n
∈
N
such that
B
(
x
)
∩ {
a
n
+1
, a
n
+2
, ...
}
=
∅
. Let
*
= min
{
, d
(
x, a
1
)
, d
(
x, a
2
)
, ..., d
(
x, a
n
)
}
/
2 (Why is this
>
0?)
and consider
B
*
∩ {
a
1
, a
2
, ...
}
.
a
i
6∈
B
*
∩ {
a
1
, a
2
, ...
}
for
i
∈ {
1
, ..., n
}
because
*
< d
(
x, a
i
) for
i
∈ {
1
, ..., n
}
. Also for
i
∈ {
n
+ 1
, n
+ 2
, ...
}
we know
a
i
6∈
B
*
∩ {
a
1
, a
2
, ...
}
because
*
<
and
B
(
x
)
∩ {
a
n
+1
, a
n
+2
, ...
}
=
∅
. So we have that
B
*
(
x
)
∩ {
a
1
, a
2
, ...
}
=
∅
,
proving our lemma.
To show that
S
is closed, we will show its complement is open. Consider some point
x
∈
S
{
.
We wish to show there is a ball centered at
x
and contained entirely in the complement. For
a contradiction, suppose there is no such ball, then for every
k
∈
N
, there is a point
p
n
k
such
that
p
n
k
∈
B
1
/k
(
x
). Furthermore, by our lemma, we can pick the indices such that
p
n
k
is a
subsequence of
p
n
(pick
p
n
1
from
{
p
1
, p
2
, p
3
, ...
}
, pick
p
n
2
from
{
p
n
k
+1
, p
n
k
+2
, p
n
k
+3
, ...
}
, and
continue in this manner). Now we see that by construction (
p
n
k