MATH 255
ASSIGNMENT 3, short solutions
Problems
Please justify carefully your answers.
1.
[10 points] Is it true that in any metric space (
X, d
),
cl(
D
(
x, r
)) =
{
y
∈
X
:
d
(
x, y
)
≤
r
}
.
Provide a proof or find a counterexample.
Solution.
Counterexample. Let
X
be a set that contains more then 2 points and let
d
(
x, y
) = 0 if
x
=
y
,
d
(
x, y
) = 1 if
x
6
=
y
.
We have shown in class that any subset of
X
is both open and closed. Clearly,
D
(
x,
1) = cl(
D
(
x,
1)) =
{
x
}
. On the other hand,
{
y
∈
X
:
d
(
x, y
)
≤
1
}
=
X
. If
X
has more then 2 points,
X
6
=
{
x
}
.
2.
[10 points] Let (
X, d
) be a metric space and
S
1
, S
2
⊂
X
. Prove that
cl(
S
1
∪
S
2
) = cl(
S
1
)
∪
cl(
S
2
)
.
Solution.
It is an immediate consequence of the definition of the closure that
A
⊂
B
⇒
cl(
A
)
⊂
cl(
B
) (you should write a short justification). Hence, cl(
S
1
)
⊂
cl(
S
1
∪
S
2
),
cl(
S
2
)
⊂
cl(
S
1
∪
S
2
), and so
cl(
S
1
)
∪
cl(
S
2
)
⊂
cl(
S
1
∪
S
2
)
.
To prove the opposite inclusion, note that
S
1
⊂
cl(
S
1
),
S
2
⊂
cl(
S
2
), imply
S
1
∪
S
2
⊂
cl(
S
1
)
∪
cl(
S
2
). Since cl(
S
1
)
∪
cl(
S
2
) is a closed set and cl(
S
1
∪
S
2
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 Spring '09
 Math, Topology, Metric space, Open set, Topological space, Closed set

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