Cantor set has dimension zero Define 'zero-dimensional' as follows: for every pair of closed disjoint sets $A,B$ in a metric space $X$ there exist disjoint open sets $U$ and $V$ such that $A \subseteq U$, $B \subseteq V$ and $X = U \cup V$.
What is a direct way to prove that the Cantor set $C$ (defined by removing open middle thirds from $[0,1]$ or by considering numbers with only $0$s and $2$s in the ternary expansion) with the Euclidean metric satisfies this property? I struggle to come up with anything as I find it difficult to work with $C$ and its open/closed sets.
 A: SKETCH: Perhaps the easiest approach is to prove that your definition of zero-dimensional is equivalent to the usual one for a compact Hausdorff space $X$: $X$ is zero-dimensional if it has a base of clopen sets.1 Both directions are fairly easy to prove. Then show that $C$ has a base of clopen sets by showing that its complement is dense in $\Bbb R$ and considering open intervals whose endpoints are in the complement.
1 Your definition implies the clopen base definition in any $T_1$ space. If $U$ is an open nbhd of $x$, $\{x\}$ and $X\setminus U$ are disjoint closed sets, so there is a clopen set $V$ such that $x\in G\subseteq U$. The clopen base definition implies yours if $X$ is also compact. If $A$ and $B$ are disjoint closed sets, for each $x\in A$ let $U_x$ be a clopen nbhd of $x$ disjoint from $B$. By compactness there is a finite $F\subseteq A$ such that $A\subseteq\{U_x:x\in F\}$, and $\bigcup_{x\in F}U_x$ is then a clopen nbhd of $A$ disjoint from $B$.
The two properties really are distinct in general. A space with yours is usually said to be strongly zero-dimensional, though I’ve also seen such spaces called ultranormal.
