Showing $\{ 0, 1 \}^{\mathbb{N}} $ is totally disconnected with given metric

Question

I would like to show $(\{ 0, 1 \}^{\mathbb{N}}, d) = (X,d)$ is totally disconnected where $d(v,w) = 2^{-n}$ where the $n$th term is the first for which the entries of $v,w$ differ.

Suppose $E \subseteq X$ has more than 1 element. Then I'd like to construct open sets $U, V$ such that $E \subseteq U \cup V$ and $U \cap V \cap E = \emptyset $.

The only thing I can see how to potentially exploit is that max distance of points in $E$ is achieved, since the distances take discrete values in $[0,1]$. So a sensible thing to try is pick some $x_1, y_1 \in E$ of maximal distance $d_1$ apart, then try isolated these points in a cover of $E$. Then take $B_{d_1 /2} ( x_1), B_{d_1 /2} ( y_1)$ and repeat for the remaining elements. However I feel this won't work as the balls may overlap.

How can I proceed?

Answer 1

Let $v,w\in X$, $v\neq w$. Let $d(v,w)=2^{-n}$. Consider

$$V=\{x\in X\ |\ d(v,x)<d(v,w)\}$$

$V$ is obviously open and nonempty. It's an open ball.

Now consider

$$W=\{x\in X\ |\ d(v,x)\geq d(v,w)\}$$

and note that $d(v,x)\geq d(v,w)=2^{-n}$ which is if and only if $d(v,x)>2^{-n-1}$ (due to the discrete nature of the metric), i.e.

$$W=\{x\in X\ |\ d(v,x)>2^{-n-1}\}$$

which is again open and nonempty. Obviously $V\cap W=\emptyset$ and $V\cup W=X$. Now since $v\in V$ and $w\in W$ then the partitioning shows that $v,w$ cannot belong to a common connected subset.

Answer 2

Let $S=\Bbb R^+\setminus \{2^{-n}:n\in \Bbb N\}.$ Every open ball $B(x,r)$ with $r\in S$ is closed because $\overline B(x,r)\setminus B(x,r)\subset \{y: d(x,y)=r\}=\emptyset.$

The set of open balls with radii in $S$ is a base (basis) for the topology. A space with more than one point and a base whose members are open-and-closed is called totally disconnected.