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.
Best Answer
$S=\{0,\frac1n,\frac2n,...,1\}$ 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.