The Fat Cantor Set

Feasibility check: let $\ell_n$ be the common length of the stage-$n$ intervals; then $\ell_0 = 1$ and $\ell_n = (\ell_{n-1} - 4^{-n})/2$.

Induction gives the exact formula $\ell_n = \frac{2^{-n} + 4^{-n}}{2}$ ($n = 0$: $\frac{1 + 1}{2} = 1$; the step is a direct substitution).

Hence $\ell_{n-1} = \frac{2^{-(n-1)} + 4^{-(n-1)}}{2} > \frac{2^{-(n-1)}}{2} = 2^{-n} > 4^{-n}$ for $n \ge 1$, so a middle interval of length $4^{-n}$ genuinely fits — the removal is possible at every stage.

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.

Passing from the finite partial sums to the infinite total uses countable additivity of Lebesgue measure (continuity from below): for the pairwise-disjoint union $G$ of the removed intervals, $m(G) = \sum_{n=1}^{\infty} 2^{n-1} 4^{-n} = \lim_{N \to \infty} \sum_{n=1}^{N} 2^{-n-1} = \tfrac12$.

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 $(2/3)^n \to 0$).

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 $\ell_n < \ell_{n-1}/2$.

The bound $\ell_n < 2^{-n}$ is valid only for $n \ge 1$ (at $n = 0$ it would read $1 < 1$, which is false).

The base case is $n = 1$, verified directly from the construction: $\ell_1 = (1 - 4^{-1})/2 = 3/8 < 1/2 = 2^{-1}$.

Inductive step: for $n \ge 2$, if $\ell_{n-1} < 2^{-(n-1)}$ then $\ell_n < \ell_{n-1}/2 < 2^{-(n-1)}/2 = 2^{-n}$.

Hence $\ell_n < 2^{-n}$ for every $n \ge 1$, 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 $J'$.

Build a point of the sibling $J''$ explicitly: starting from $J''$, choose one child at each stage (e.g. always the left child) to form a nested chain $J'' = K_{n+1} \supset K_{n+2} \supset \cdots$; since $\lvert K_j \rvert = \ell_j \to 0$, the nested-interval principle says $\bigcap_j K_j$ is a single point $z_n$.

Each $K_j$ is a closed interval of some $F_j$ and the $F_j$ are nested, so $z_n \in \bigcap_j F_j = C$, and by construction $z_n \in J''$.

The middle gap separates $x$ from $z_n$, so $z_n \ne x$, and $\lvert z_n - x \rvert \le \ell_n < 2^{-n} \to 0$ (for $n \ge 1$).

Hence every neighborhood of $x$ contains a point of $C \setminus \{x\}$, so $x$ is a limit point of $C$.

This holds for every $x \in C$, so $C$ is perfect.

$U$ 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.

Handle the relative topology as well: a nonempty relatively open subset of $[0,1]$ contains either an open interval or an endpoint-touching piece $[0, a)$ with $a > 0$ or $(b, 1]$ with $b < 1$.

In the endpoint cases it still contains a nonempty open interval, namely $(0, a)$ or $(b, 1)$, so the previous argument applies verbatim.

Hence every relatively open set 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 (enumerate the rationals as $q_1, q_2, \dots$ — the index starts at $k \ge 1$ — and around each $q_k$ take an open interval of radius $\varepsilon/2^k$: each interval has length $2\varepsilon/2^k$, and $\sum_{k=1}^{\infty} 2\varepsilon/2^k = 2\varepsilon$, so the total measure is $\le 2\varepsilon$).

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.