3
Example 12.
In Example 8, we saw that
C
1
= [0
,
1]
in the metric space
[0
,
1]
∪
(2
,
3)
. Fol
lowing Lemma 9, this means that for any point
x
∈
[0
,
1]
,
C
x
= [0
,
1]
(indeed, this is
easy to show directly). On the other hand, if
y
∈
(2
,
3)
, the reader can readily verify that
C
y
= (2
,
3)
. Hence, the metric space
[0
,
1]
∪
(2
,
3)
has two connected components:
[0
,
1]
and
(2
,
3)
.
Example 13.
Consider the metric space
Q
equipped with the usual metric
d
(
q, p
) =

q

p

.
Let
q
0
be any nonzero rational number. By the Archimedean property of
Q
, there exists a
natural number
n
so that
n
·
q
2
0
>
1
. Since
n
2
+ 1
> n
, we therefore have
(
n
2
+ 1)
·
q
2
0
>
1
.
Now, define
A
=
{
p
∈
Q
;
p
2
>
1
1+
n
2
}
and
B
=
{
p
∈
Q
;
p
2
<
1
1+
n
2
}
. By definition,
q
0
∈
A
.
Note,
1 +
n
2
is not a perfect square (since
n >
0
here), and therefore (following standard
proofs like the one in the first lecture)
1
1+
n
2
is not the square of any rational number. It
follows that
Q
=
A
∪
B
. What’s more, if
p
∈
A
, then there is a small rational distance
ω
so
that the interval
(
p

ω
, p
+
ω
)
∩
Q
is contained in
A
=
B
c
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 Spring '09
 Koskesh
 Topology, Metric space, Topological space

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