The Smith-Volterra-Cantor set: the same construction removing, at stage $n\ge 1$, an open middle interval of length $4^{-n}$ from each of the $2^{n-1}$ remaining intervals

The lengths. By the recursion theorem in the index-carrying form used by Finite sums and finite products, by recursion (The recursion theorem, applied to $\mathbb{N}\times \mathbb{R}$ with starting element $(0,1)$ and the map $(n,t)\mapsto (n+1,(t-4^{-n-1})\cdot 2^{-1})$) there is a unique sequence $({\lambda }_n{)}_{n\in \mathbb{N}}$ of reals with

${\lambda }_0=1,{\lambda }_{n+1}=({\lambda }_n-4^{-n-1})\cdot 2^{-1}(n\in \mathbb{N}),$

powers being those of Integer powers $a^m$. Put $g_n:={\lambda }_n-{\lambda }_{n+1}$.

The left endpoints. Let $\mathcal{F}$ be the set of pairs $(N,\mathrm{\ell })$ with $N\in \mathbb{N}$, $N\ge 1$, and $\mathrm{\ell }$ a function from $\{j\in \mathbb{N}:j<N\}$ to $\mathbb{R}$; such a pair is a finite list of reals of length $N$. Applying The recursion theorem to $\mathbb{N}\times \mathcal{F}$, the starting element $(0,(1,{\mathrm{\ell }}^{(0)}))$ with ${\mathrm{\ell }}_0^{(0)}:=0$, and the map that sends $(n,(N,\mathrm{\ell }))$ to $(n+1,(N+N,{\mathrm{\ell }}^{\mathrm{\prime }}))$ where

${\mathrm{\ell }}_j^{\mathrm{\prime }}:={\mathrm{\ell }}_j(j<N),{\mathrm{\ell }}_j^{\mathrm{\prime }}:={\mathrm{\ell }}_{j-N}+g_n(N\le j<N+N),$

gives a unique family $(N_n,{\mathrm{\ell }}^{(n)}{)}_{n\in \mathbb{N}}$ of finite lists, with $N_0=1$, $N_{n+1}=N_n+N_n$, and ${\mathrm{\ell }}^{(n+1)}$ the concatenation of ${\mathrm{\ell }}^{(n)}$ with its translate by $g_n$. Write $e_j^{(n)}:={\mathrm{\ell }}_j^{(n)}$.

The sets. For $n\in \mathbb{N}$ put

$S_n:={\bigcup }_{j<N_n}[e_j^{(n)},e_j^{(n)}+{\lambda }_n],S:={\bigcap }_{n\in \mathbb{N}}{S}_n,$

the intervals being those of Intervals of $\mathbb{R}$: the nine order-convex forms, nondegeneracy, and length. $S$ is the Smith-Volterra-Cantor set, also called the fat Cantor set.

Counting. For every $n$ and every real $c$ one has ${\sum }_{j<N_n}c=2^{n}c$, by induction on $n$ (The principle of mathematical induction): at $n=0$ both sides are $c$; and ${\sum }_{j<N_n+N_n}c={\sum }_{j<N_n}c+{\sum }_{j<N_n}c=2^{n}c+2^{n}c=2^{n+1}c$, by the splitting law (Laws of finite sums and finite products, Finite sums and finite products, by recursion) and $2^{n+1}=2^n\cdot 2=2^n+2^n$ (Integer powers $a^m$, Ordered field). So stage $n$ has "$2^n$ intervals" in exactly this sense, and no separate arithmetic of natural-number exponents is needed.

The lengths are positive and shrink. By induction on $n$: $0<{\lambda }_{n+1}\le {\lambda }_n\cdot 2^{-1}$ and $2^n{\lambda }_n\ge 2^{-1}$. Indeed $2^{n+1}{\lambda }_{n+1}=2^n({\lambda }_n-4^{-n-1})=2^n{\lambda }_n-4^{-1}\cdot 2^{-n}$ by Laws of integer exponents, so by induction $2^n{\lambda }_n=1-4^{-1}{\sum }_{i<n}{2}^{-i}\ge 1-4^{-1}\cdot 2=2^{-1}$, using ${\sum }_{i<n}{2}^{-i}\le {\sum }_{i=0}^{\mathrm{\infty }}{2}^{-i}=2$ (For $\mathrm{\mid }r\mathrm{\mid }<1$, ${\sum }_{k\ge 0}{r}^k=1\mathrm{/}(1-r)$, and for $\mathrm{\mid }r\mathrm{\mid }\ge 1$ the series diverges, A series of nonnegative terms converges iff its partial sums are bounded, and then the sum is their supremum, Series, partial sums, convergence and the sum, divergence, and the tail series). Hence ${\lambda }_n\ge 2^{-n-1}>0$; and ${\lambda }_{n+1}=({\lambda }_n-4^{-n-1})\cdot 2^{-1}\le {\lambda }_n\cdot 2^{-1}$ gives ${\lambda }_n\le 2^{-n}$ by a second induction, so the lengths tend to $0$.

Each stage removes an open middle interval of length $4^{-n-1}$. From the recursion, the two sub-intervals of $[e,e+{\lambda }_n]$ retained at stage $n+1$ are $[e,e+{\lambda }_{n+1}]$ and $[e+g_n,e+g_n+{\lambda }_{n+1}]=[e+g_n,e+{\lambda }_n]$, so what is dropped from that piece is the open interval

$M=(e+{\lambda }_{n+1},e+g_n),\text{of length }{g}_n-{\lambda }_{n+1}={\lambda }_n-2{\lambda }_{n+1}=4^{-n-1}\mathrm{.}$

In particular ${\lambda }_{n+1}<g_n$, so $M$ is nonempty, and $g_n>0$, so $[e+g_n,e+{\lambda }_n]\subseteq [e,e+{\lambda }_n]$. Counting from $1$ as in the title: at stage $n\ge 1$ an open interval of length $4^{-n}$ is removed from each of the $2^{n-1}$ intervals then present.

The family is nested and lies in $[0,1]$. Each retained sub-interval is contained in the piece it came from, by the previous paragraph, so $S_{n+1}\subseteq S_n$; and $S_0=[0,1]$ since $N_0=1$, $e_0^{(0)}=0$ and ${\lambda }_0=1$. Hence $S\subseteq S_m\subseteq [0,1]$ for every $m$.

• What is different from The Cantor middle-thirds set as the intersection of the sets $C_n$ obtained by removing open middle thirds. There the removed middle is a fixed proportion of each piece, so the construction is self-similar and the total removed length is $1$. Here the removed middle has a fixed length $4^{-n-1}$, chosen to shrink faster than the pieces multiply, and the total removed length is only $2^{-1}$. Everything topological survives the change: the set is still compact, perfect and nowhere dense (The Smith-Volterra-Cantor set is compact, perfect and nowhere dense, and does not have measure zero). Everything metric fails: $S$ is not of measure zero.

• Why the construction is written with explicit lists. The set $S_n$ is a union of $2^n$ intervals, and both the estimate of the removed length and the finite covers used later need those intervals as a list, indexed by naturals below $N_n$. Building the list by recursion, rather than asserting its existence at each stage, is also what keeps the construction free of any choice: $(N_n,{\mathrm{\ell }}^{(n)})$ is a single function of $n$.

• The name. The set was described by Smith in 1875, by Volterra in 1881 and by Cantor in 1883; "fat Cantor set" is the informal name, and the two names are used interchangeably below.

• $0$ and $1$ belong to $S$. Both are instances of the general fact that every $e_j^{(n)}$ and every $e_j^{(n)}+{\lambda }_n$ lies in $S$, proved where it is used, in The Smith-Volterra-Cantor set is compact, perfect and nowhere dense, and does not have measure zero: take $n=0$ and $j=0$, where $e_0^{(0)}=0$ and $e_0^{(0)}+{\lambda }_0=1$.