Intuition Regarding Boundary of Open Set & Measure

Question

Open subsets of $\mathbb{R}$ are countable disjoint unions of intervals. I naively suspected that the boundary of such a countable union is countable, and hence measure zero. This is not true since Fat Cantor sets are positive measure and closed, so their complement is an open set with boundary that is positive measure. Why is the boundary of a countable disjoint union of intervals not just the union of the endpoints, and hence countable? Could someone provide a counterexample? I cannot quite think of one.

Answer 1

The boundary of a union of pairwise-disjoint open real intervals must contain all the limits of all convergent sequences of endpoints of those intervals, and there can be uncountably many such limits.

The open intervals that comprise the complement, in $[0,1]$, of the Cantor set $C$, are all the numbers in $[0,1]$ that $cannot$ be written in base $3$ without using the digit $1$. E.g. $1/3$ in base $3$ can be written $0.0\overline 2$ so $1/3$ is $not$ in the open complement. The members of $C$ are all the numbers of the form $\sum_{n=1}^{\infty}f(n)3^{-n}$ where every $f(n)\in \{0,2\}$. There are uncountably many of them. Indeed the function $\psi:C\to \Bbb R$ where $\psi(\sum_{n=1}^{\infty}f(n)3^{-n})=\sum_{n=1}^{\infty}\frac 1 2 f(n)3^{-n}$ maps $C$ onto the whole interval $[0,1].$

Now consider taking the open middle $1/3$ of $[0,1],$ then taking open middle $1/9$ths of what's left, taking open middle $1/27$ths of what's still left, and so on. This gives us an open set $S$ with $\overline S=[0,1]$ but the boundary $\partial S=[0,1]\setminus S$ has measure $\prod_{n=1}^{\infty}(1-3^{-n})>0.$ A subset of $[0,1]$ which is closed and nowhere-dense but has positive measure is called a fat Cantor set. Some authors may require in their definition, that a fat Cantor set must also be a perfect set, i.e. a non-empty compact subspace with no isolated points.

Footnote: An elementary theorem is that if $0\le x_n<1$ for every $n$ then $\prod_{n=1}^{\infty}(1-x_n)=0\iff\sum_{n=1}^{\infty}x_n=\infty.$ Therefore $\prod_{n=1}^{\infty}(1-3^{-n})>0$ because $\sum_{n=1}^{\infty}3^{-n}\ne\infty.$ A companion result is that if $y_n\ge 0$ for every $n$ then $\prod_{n=1}^{\infty}(1+y_n)=\infty\iff\sum_{n=1}^{\infty}y_n=\infty.$

Answer 2

Consider the boundary of the middle thirds of the Cantor set: $(1/3,2/3)$, $(1/9,2/9)\cup(7/9,8/9)$, etc. This boundary is the Cantor set, well-known to be uncountable, despite being the boundary of the union of countably-many pair-wise disjoint open sets.