The Topology of
(is Weird)
Definition 16.2.10 (Connected)
Let
![$ X$](img23.png)
be a topological space. A subset
![$ S$](img121.png)
of
![$ X$](img23.png)
is
if there exist open subsets
![$ U_1, U_2\subset X$](img1827.png)
with
![$ U_1\cap U_2\cap S=\emptyset$](img1828.png)
and
![$ S=(S\cap U_1)\cup (S\cap U_2)$](img1829.png)
with
![$ S\cap U_1$](img1830.png)
and
![$ S\cap U_2$](img1831.png)
nonempty.
If
![$ S$](img121.png)
is not disconnected it is
.
The topology on
is induced by
, so every open set is a union
of open balls
Recall Proposition 16.2.8, which asserts that for
all
,
This translates into the following shocking and bizarre lemma:
Proof.
Suppose
![$ z\in B(x,r)$](img1838.png)
and
![$ z\in B(y,r)$](img1839.png)
. Then
a contradiction.
You should draw a picture to illustrates Lemma 16.2.11.
Proof.
Suppose
![$ y\not\in B(x,r)$](img1842.png)
. Then
![$ r\leq d(x,y)$](img1843.png)
so
Thus the complement of
![$ B(x,r)$](img1841.png)
is a union of open balls.
The lemmas imply that
is ,
in the following sense.
Proposition 16.2.13
The only connected subsets of
are the singleton sets
for
and the empty set.
Proof.
Suppose
![$ S\subset \mathbf{Q}_N$](img1846.png)
is a nonempty connected set and
![$ x, y$](img1847.png)
are distinct
elements of
![$ S$](img121.png)
. Let
![$ r=d_N(x,y)>0$](img1848.png)
. Let
![$ U_1=B(x,r)$](img1849.png)
and
![$ U_2$](img1850.png)
be
the complement of
![$ U_1$](img1851.png)
, which is open by Lemma
16.2.12.
Then
![$ U_1$](img1851.png)
and
![$ U_2$](img1850.png)
satisfies the conditions of Definition
16.2.10,
so
![$ S$](img121.png)
is not connected, a contradiction.
William Stein
2004-05-06