Does Cantor's set contain a copy of each finite set?

Question

Let $C$ denote the (usual) Cantor's set in the interval $[0,1]$. If $S=\{x_1,x_2,\cdots,x_n\}$ is a finite set of points in $\mathbb R$, is it true that $aS+b\subset C$ for some $a\neq 0$ and $b\in\mathbb R$? (i.e. $C$ contains a copy of $S$)

For the fat Cantor-type sets the answer is yes (from Steinhaus Theorem), since in this case the fat Cantor-type set has positive Lebesgue measure. Also it is known that there exists Lebesgue measure zero sets that contain a copy of each finite set.

Answer 1

Here is a detailed answer. Let us start with some definitions.

Let $S$ be a subset of $\Bbb R$ and $p,q \in S$ such that $p<q$. We say that a interval $(c,d)$ is a gap in $S\cap [p,q]$ if $(c, d) \subseteq [p,q]$ and $(c,d)\cap S=\emptyset$.

We also define, for any gap $(c,d)$ in $S\cap [a,b]$, the relative size of the gap to be $\frac{d-c}{b-a}$.

Given any finite set $F=\{a_0, .... , a_n\}$ where $a_0 < ... < a_n$, we define $g(F)= \max \left\{\frac{a_i-a_{i-1}}{a_n-a_0} \;|\; i=1,...,n \right\}$.

Our first result is easy to prove.

1. Any gap $(c,d)$ in $F \cap [a_0,a_n]$ will have relative size less or equal to $g(F)$.

Proof:

Any open interval $(c,d) \subseteq [a_0,a_n]$ such that $d-c \geqslant g(F) (a_n-a_0)$ contains a point of $F$. So $(c,d)$ is not a gap in $F \cap [a_0,a_n]$.

$\square$

Now, let $a,b \in \Bbb R$ where $a\ne 0$ and let $T_{a,b}$ the function defined by $T_{a,b}(x) = ax+b$, for all $x \in \Bbb R$.

Our second result is also easy to prove.

2. $g(T_{a,b}(F))=g(F)$

Proof:

For all i=1,...,n, in the case $a>0$ we have $\frac{T_{a,b}(a_i)-T_{a,b}(a_{i-1})}{T_{a,b}(a_n)-T_{a,b}(a_0)}= \frac{a_i-a_{i-1}}{a_n-a_0}$. In the case $a<0$ we have $\frac{T_{a,b}(a_{i-1})-T_{a,b}(a_i)}{T_{a,b}(a_0)-T_{a,b}(a_n)}= \frac{a_i-a_{i-1}}{a_n-a_0}$. So, in both case it follows immediately that $g(T_{a,b}(F))=g(F)$.

$\square$

Our next result is a little more subtle to prove.

3. Let $C$ be the (usual) Cantor set. Given any two points $r,s \in C$, $r<s$, there is a a gap $(c,d)$ in $C\cap[r,s]$ whose relative size is greater or equal to $\frac{1}{3}$

Proof:

Since $r,s \in C$, we know that they have expressions (finite or infinite) in base 3 not containing de digit 1. So

$r=0.r_1r_2r_3...$ where for all $i$, $r_i$ is the digit 0 or 2.

$s=0.s_1s_2s_3...$ where for all $i$, $s_i$ is the digit 0 or 2.

Since $r \ne s$, let $n$ be the smallest index such that $r_n \ne s_n$. Since $r<s$, we must have $r_n=0$ and $s_n=2$. It follows that $0< s-r \leqslant \frac{1}{3^{n-1}}$

Now, note that, defining

$r'= 0.r_1r_2r_3...r_{n-1}011111111111...$ e $s'= 0.s_1s_2s_3...s_{n-1}2000000...$,

we have that $(r',s')$ is a gap in $C\cap [r,s]$. Note that $s'-r'= \frac{1}{3^n}$. So the relative size of the gap is $(r',s')$ in $C\cap [r,s]$ is greater or equal to $\frac{1}{3}$.

$\square$

Note that the result 3 above does not hold for fat Cantor sets.

Now the main result (the counter-example):

Let $F=\{1,2,3,4,5\}$. There is no $a,b \in \Bbb R$ where $a\ne 0$, such that $T_{a,b}(F) \subseteq C$.

Proof:

Now, consider $F=\{1,2,3,4,5\}$, it is easy to see that $g(F) =1/4$. Suppose there is $a,b \in \Bbb R$ where $a\ne 0$, such that $T_{a,b}(F) \subseteq C$. Then, by result 2 above, $g(T_{a,b}(F))=g(F)=1/4$.

Case 1: $a >0$. By result 1, any gap in $T_{a,b}(F) \cap [T_{a,b}(a_0),T_{a,b}(a_n)]$ will have relative size less or equal to $g(T_{a,b}(F))$, that is $1/4$. But, by result 3, since $T_{a,b}(a_0),T_{a,b}(a_n) \in C$ and $T_{a,b}(a_0)<T_{a,b}(a_n)$, there is a gap $(c,d)$ in $C \cap [T_{a,b}(a_0),T_{a,b}(a_n)]$ whose relative size is greater or equal to $1/3$. Since $T_{a,b}(F) \subseteq C$, $(c,d)$ is also a gap in $T_{a,b}(F) \cap [T_{a,b}(a_0),T_{a,b}(a_n)]$ with the same relative size greater or equal to $1/3$. Contradiction.

Case 2: $a<0$. It is completely analogous to case 1. By result 1, any gap in $T_{a,b}(F) \cap [T_{a,b}(a_n),T_{a,b}(a_0)]$ will have relative size less or equal to $g(T_{a,b}(F))$, that is $1/4$. But, by result 3, since $T_{a,b}(a_0),T_{a,b}(a_n) \in C$ and $T_{a,b}(a_n)<T_{a,b}(a_0)$, there is a gap $(c,d)$ in $C \cap [T_{a,b}(a_n),T_{a,b}(a_0)]$ whose relative size is greater or equal to $1/3$. Since $T_{a,b}(F) \subseteq C$, $(c,d)$ is also a gap in $T_{a,b}(F) \cap [T_{a,b}(a_n),T_{a,b}(a_0)]$ with the same relative size greater or equal to $1/3$. Contradiction.

So, there is no $a,b \in \Bbb R$ where $a\ne 0$, such that $T_{a,b}(F) \subseteq C$.

$\square$

Remark regarding fat Cantor sets: For the argument to work it is critical to have, in any "segment" of the (usual) Cantor set, gaps whose relative size have a positive lower bound (1/3), so we can start the counter-example by getting a set whose gaps have relative size smaller than such lower bound. For fat Cantor set, we don't have such property, so the argument above does not apply to fat Cantor sets.

Answer 2

$S=\{0,\frac1n,\frac2n,...,1\}$ for any $n\geq5$ is a counterexample.

To see this, take any set $\{0,1\}\subseteq S\subseteq[0,1]$ and assume there were constants $a,b\in\mathbb R$ as you described. Then $b$ and $b+a$ are in $aS+b$ and thus $C$; due to the symmetries of the Cantor set, we can choose $a$ and $b$ such that $b$ is in the left half of $C$ and $b+a$ is in the right, i.e. the interval $(\frac13,\frac23)\subset\mathbb R\setminus C$ lies between them. Since $aS+b$ is a subset of $C$, this means that $aS+b$ and $(\frac13,\frac23)$ are disjoint, and so are then $S$ and $\frac1a(\frac13,\frac23)-\frac ba\subseteq[0,1]$. In other words, since $a$ must obviously be $\leq1$, this means that $S$ has a hole at least $\frac13$ in length in it; for sets $S$ like the one mentioned above that's not the case, so they can't be linearly mapped to subsets of $C$ as you described.