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

Definition 14.2.10 (Connected)   Let $X$ be a topological space. A subset $S$ of $X$ is disconnected 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 connected.

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

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

Recall Proposition 14.2.8, which asserts that for all $x,y,z$,

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

This translates into the following shocking and bizarre lemma:

Lemma 14.2.11   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

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

a contradiction. $\qedsymbol$

You should draw a picture to illustrates Lemma 14.2.11.

Lemma 14.2.12   The open ball $B(x,r)$ is also closed.

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

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

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

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

Proposition 14.2.13   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 14.2.12. Then $U_1$ and $U_2$ satisfies the conditions of Definition 14.2.10, so $S$ is not connected, a contradiction. $\qedsymbol$