Totally disconnected implies nowhere dense Why is it true that a totally disconnected space implies it is nowhere dense in the reals?
I know that totally disconnected implies the component is a singleton, but how do we construct a nowhere dense set such that their union equals the set itself?
 A: As stated, the given proposition is false.
The rationals $\mathbb Q$ is a totally disconnected subset of $\mathbb R$ (essentially because between any two rationals there is an irrational, and if $r$ is irrational, then $( - \infty , r ), (r , + \infty )$ partitions $\mathbb{Q}$ into disjoint nonempty (relatively-)open sets). But far from being nowhere dense, it is actually dense in $\mathbb{R}$.

However, a closed totally disconnected subset of $\mathbb R$ is nowhere dense. This follows because totally disconnected subsets of $\mathbb R$ cannot include any non-degenerate intervals, and so given $a < b$ as $A \not\supseteq (a,b)$, then $(a,b) \setminus A$ is nonempty open set disjoint from $A$. (Remembering that a set $A$ is nowhere dense if given any nonempty open $U$ there is a nonempty open $V \subseteq U$ disjoint from $A$.)
A: It is as in the proof in the book. The set is totally disconnected, hence it doesn't contain any open interval (since an interval is connected). Since it is closed, it is nowhere dense, dense because open intervals form a base of open sets of reals and a set is nowhere dense iff its closure doesn't contain any nonempty open set.
