Prove a set is totally disconnected

Question

Let $I=[0,1]$ be a set with the topology induced from the usual topology of $\mathbb{R}$ and $D \subset I$ an open and dense subset. Prove that $I-D$ is totally disconnected.

Answer 1

You want to prove that all the connected subsets of $I \setminus D$ are singletons. By hypothesis $I\setminus D$ is closed with empty interior. Recall that every connected subset of $I$ is an interval. Let $x \in I \setminus D$: any connected subset of $I\setminus D$ containing $x$ has thus to be an interval with empty interior, hence it is the point $\{x\}$ itself.

Answer 2

Let $a, b \in I - D$ such that $a < b$.

Then there exists a point $d \in D$ such that $a < d < b$, because $D$ is dense in $I$.

And, then the sets $(I-D) \cap [0, d)$ and $(I-D) \cap (d, 1]$ are non-empty disjoint open sets in $I-D$ whose union is $I-D$.

Note that the subspace topology inherited by $I = [0, 1]$ from the usual topology on $\mathbb{R}$ has as a basis all sets of the form $(a, b)$, where $0 \leq a < b \leq 1$; $[0, b) = (-1, b) \cap I$, where $0 < b \leq 1$; and $(a, 1] = (a, 2) \cap I$, where $0 \leq a < 1$.

Thus we have shown that any two-point subset of $I - D$ is disconnected. Thus we can show that any subset of $I-D$ having more than one-point is also disconnected. Hence $I-D$ is totally disconnected.

Here we have not even required set $D$ to be open in $I$.