The Fat Cantor Set

Consider the fat Cantor set: at stage $n$ , from each of the $2^{n-1}$ surviving closed intervals remove the middle open interval of absolute length $4^{-n}$ . Prove the total removed length over all stages is $\tfrac12$ via a geometric series, and check that the length remaining after $N$ stages is $\dfrac{1 + 2^{-N}}{2}$ .

\sum_{n=1}^{\infty} 2^{n-1} \cdot 4^{-n} = \sum_{n=1}^{\infty} 2^{-n-1} = \frac{1/4}{1 - 1/2} = \frac12

1 - \sum_{n=1}^{N} 2^{-n-1} = 1 - \tfrac12(1 - 2^{-N}) = \tfrac12(1 + 2^{-N}) \;\xrightarrow[N\to\infty]{}\; \tfrac12

Stage

n = 4

: 16 intervals, remaining length

\tfrac{17}{32} = 0.53125 \to \tfrac12

— the standard Cantor set has 16 intervals with

\left(\tfrac23\right)^4 = \tfrac{16}{81} \approx 0.197531 \to 0

.

Show proof

At stage $n$ there are $2^{n-1}$ surviving closed intervals from stage $n-1$ , and exactly one middle open interval of length $4^{-n}$ is removed from each. Removed open intervals from different stages and different parent intervals are all pairwise disjoint (each is carved out of the interior of a closed interval, and the surviving sets are nested). So the total length removed through stage $N$ is the sum of the per-stage contributions.
The stage- $n$ contribution is $2^{n-1} \cdot 4^{-n}$ . Simplifying the exponent, $\dfrac{2^{n-1}}{4^n} = \dfrac{2^{n-1}}{2^{2n}} = 2^{-n-1}$ , so the total removed length is the geometric series $\sum_{n=1}^{\infty} 2^{-n-1}$ with ratio $\tfrac12$ . Its first term is $\tfrac14$ , giving $\dfrac{1/4}{1 - 1/2} = \tfrac12$ .
The finite-stage version: through stage $N$ the total removed is the partial sum $\sum_{n=1}^{N} 2^{-n-1} = \tfrac14 \cdot \dfrac{1 - 2^{-N}}{1 - 1/2} = \tfrac12(1 - 2^{-N})$ . Starting from $[0,1]$ , the length remaining after $N$ stages is $1 - \tfrac12(1 - 2^{-N}) = \tfrac12(1 + 2^{-N})$ , tending to $\tfrac12$ as $N \to \infty$ . For $N = 1$ this is $1 - \tfrac14 = \tfrac34 = \tfrac12(1 + \tfrac12)$ — the surviving intervals $[\,0, \tfrac38\,]$ and $[\,\tfrac58, 1\,]$ with the middle $(\tfrac38, \tfrac58)$ carved out, exactly as the scene shows.
Hence the pairwise-disjoint union $G$ of the removed open intervals has Lebesgue measure $m(G) = \tfrac12$ , and since $C = [0,1] \setminus G$ , we get $m(C) = 1 - \tfrac12 = \tfrac12 > 0$ . This is where the fat Cantor set parts company with the standard Cantor set (middle third removed each stage, remaining length $(2/3)^n \to 0$ ).

Let the fat Cantor set be $C = \bigcap_{n \ge 0} F_n$ , where $F_0 = [0,1]$ and $F_n$ removes the middle open interval (length $4^{-n}$ ) from each closed interval of $F_{n-1}$ . Prove $C$ is compact, nowhere dense, and perfect (noting that its interior is empty yet $m(C) = \tfrac12$ ).

\ell_n = \frac{\ell_{n-1} - 4^{-n}}{2} < 2^{-n}

C = \bigcap_{n\ge 0} F_n\ \text{closed} \subset [0,1]\ \text{compact}

\operatorname{int}(C) = \varnothing\ \text{(every open interval meets a removed piece)}

x \in C \Rightarrow x = \lim z_n,\ z_n \in C \setminus \{x\}

Stage

n = 4

: 16 intervals with remaining length

0.53125

— sixteen thin slivers, yet topologically the set is "hollow inside" (proof below: every open interval meets a removed piece).

Show proof

Compact: each $F_n$ is a finite union of closed intervals, hence closed, and $C = \bigcap_n F_n$ is an intersection of closed sets, hence closed. Since $C \subset [0,1]$ is bounded, Heine–Borel makes $C$ compact.
Length lemma: by symmetry every stage- $n$ interval has the same length. Calling it $\ell_n$ , we have $\ell_0 = 1$ and $\ell_n = (\ell_{n-1} - 4^{-n})/2$ , so $\ell_n < \ell_{n-1}/2$ . By induction $\ell_n < 2^{-n}$ , and in particular $\ell_n \to 0$ .
Nowhere dense: since $C$ is closed, it suffices to show its closure — itself — has empty interior. Take a nonempty open interval $I$ of length $\delta > 0$ , shrinking it into $(0,1)$ if needed (otherwise $I \not\subset C$ trivially). Choose $n$ with $2^{-n} < \delta$ , so $\ell_n < \delta$ . Suppose $I$ meets no removed open interval; then $I \subset F_n$ . But $F_n$ is a union of pairwise-disjoint closed intervals and $I$ is connected, so $I$ lies entirely inside exactly one of them, say $J$ . Then $\delta = \lvert I \rvert \le \lvert J \rvert = \ell_n < \delta$ — contradiction.
Hence $I$ always meets a removed open interval, and the intersection is a nonempty open subset of $I \setminus C$ . No nonempty open interval sits inside $C$ , so $\operatorname{int}(C) = \varnothing$ — $C$ is nowhere dense.
Perfect: let $x \in C$ . At each stage $n$ , $x$ lies in some stage- $n$ interval $J_n$ . At stage $n+1$ , $J_n$ splits into two closed children $J', J''$ separated by the removed open middle, and $x$ belongs to one of them, say $J'$ . Choosing a nested chain of children, the nested-interval principle gives $C \cap J'' \ne \varnothing$ , so pick $z_n \in C \cap J''$ . The middle gap separates $x$ from $z_n$ , so $z_n \ne x$ , and $\lvert z_n - x \rvert \le \ell_n < 2^{-n} \to 0$ . Thus $x$ is a limit point of $C$ . This holds for every $x \in C$ , so $C$ is perfect.

Let $U = [0,1] \setminus C$ be the complement of the fat Cantor set. Prove $U$ is (relatively) open, dense in $[0,1]$ , and has $m(U) = \tfrac12$ . Then finish by completing this table: "empty interior, yet half the length" — measure and topological size are independent notions.

U = \bigcup_{\text{stages}} \text{(removed open intervals)}

m(U) = \tfrac12,\quad m(C) = \tfrac12

\begin{array}{c|cc} & m = 0 & m > 0 \\ \hline \text{nowhere dense} & \text{standard Cantor} & \textbf{fat Cantor} \\ \text{dense} & \mathbb{Q} \cap [0,1] & \text{irrationals in } [0,1] \end{array}

Stage

n = 4

: the removed open intervals total

\tfrac12(1 - 2^{-4}) = 0.46875

in length, leaving

0.53125

— the complement is already nearly half, and in the limit exactly

\tfrac12

.

Show proof

$U$ is the complement of the closed set $C$ , hence relatively open in $[0,1]$ . Concretely, $U$ is the union of the removed open intervals over all stages — a countable union of pairwise-disjoint open intervals.
Dense: the previous problem showed that every nonempty open interval meets a removed open interval. Hence every open interval contains a point of $U$ , so $U$ is dense in $[0,1]$ .
Measure: the removed open intervals are pairwise disjoint, so $m(U) = \sum_{n=1}^{\infty} 2^{n-1} 4^{-n} = \tfrac12$ (the geometric series from the first problem). Hence $m(C) = m([0,1]) - m(U) = 1 - \tfrac12 = \tfrac12$ .
To summarize: $C$ is closed, has empty interior (nowhere dense), is perfect, and has length $\tfrac12$ ; its complement $U$ is open, dense, and has length $\tfrac12$ . The table above fills all four cells — the standard Cantor set (nowhere dense, measure 0), the fat Cantor set (nowhere dense, measure $\tfrac12$ ), the rationals in $[0,1]$ (dense, measure 0), the irrationals (dense, measure 1). There are even sets that are open and dense with arbitrarily small measure (around each rational $q_k$ take an open interval of radius $\varepsilon/2^k$ : total measure $\le 2\varepsilon$ ). The punchline: the topological statement "has empty interior" and the measure-theoretic statement "has such-and-such length" impose nothing on each other — and the fat Cantor set is the most dramatic witness to their independence.

Part of these learning paths

Examples & Exercises