The Topology of $ \mathbf{Q}_N$ (is Weird)

Definition 16.2.10 (Connected)   Let $ X$ be a topological space. A subset $ S$ of $ X$ is if there exist open subsets $ U_1, U_2\subset X$ with $ U_1\cap U_2\cap S=\emptyset$ and $ S=(S\cap U_1)\cup (S\cap U_2)$ with $ S\cap U_1$ and $ S\cap U_2$ nonempty. If $ S$ is not disconnected it is .

The topology on $ \mathbf{Q}_N$ is induced by $ d_N$, so every open set is a union of open balls

$\displaystyle B(x,r) = \{y \in \mathbf{Q}_N : d_N(x,y) < r\}.

Recall Proposition 16.2.8, which asserts that for all $ x,y,z$,

$\displaystyle d(x,z) \leq \max(d(x,y),d(y,z)).

This translates into the following shocking and bizarre lemma:

Lemma 16.2.11   Suppose $ x\in \mathbf{Q}_N$ and $ r>0$. If $ y\in\mathbf{Q}_N$ and $ d_N(x,y)\geq r$, then $ B(x,r)\cap B(y,r) = \emptyset$.

Proof. Suppose $ z\in B(x,r)$ and $ z\in B(y,r)$. Then

$\displaystyle r\leq d_N(x,y) \leq \max(d_N(x,z), d_N(z,y)) < r,

a contradiction. $ \qedsymbol$

You should draw a picture to illustrates Lemma 16.2.11.

Lemma 16.2.12   The open ball $ B(x,r)$ is also closed.

Proof. Suppose $ y\not\in B(x,r)$. Then $ r\leq d(x,y)$ so

$\displaystyle B(y,d(x,y)) \cap B(x,r)
B(y,d(x,y)) \cap B(x,d(x,y))
= \emptyset.

Thus the complement of $ B(x,r)$ is a union of open balls. $ \qedsymbol$

The lemmas imply that $ \mathbf{Q}_N$ is , in the following sense.

Proposition 16.2.13   The only connected subsets of $ \mathbf{Q}_N$ are the singleton sets $ \{x\}$ for $ x\in \mathbf{Q}_N$ and the empty set.

Proof. Suppose $ S\subset \mathbf{Q}_N$ is a nonempty connected set and $ x, y$ are distinct elements of $ S$. Let $ r=d_N(x,y)>0$. Let $ U_1=B(x,r)$ and $ U_2$ be the complement of $ U_1$, which is open by Lemma 16.2.12. Then $ U_1$ and $ U_2$ satisfies the conditions of Definition 16.2.10, so $ S$ is not connected, a contradiction. $ \qedsymbol$

William Stein 2004-05-06