Sequence in uncountable set Given an uncountable subset $A$ of $(0,1)$, does there always exist $a,r>0$ such that $a+r,a+r^2,a+r^3,\dots$ are all in $A$? For example, if $A$ contains an interval, it is easy to find such $a,r$. 
To try to show this, we can assume that no such $a,r$ exist (meaning that for any $a,r$, there exists $k$ with $a+r^k\not\in A$) and show that the set must be countable.
 A: Let $A$ be uncountable set of algebraically independent (over $\mathbb{Q}$) elements of $(0,1)$.  Then it can contain no such sequence.  Indeed, if $x=a+r$, $y=a+r^2$, and $z=a+r^3$, then $x,y,$ and $z$ are all in the subfield $\mathbb{Q}(a,r)\subset\mathbb{R}$ which has transcendence degree at most $2$ over $\mathbb{Q}$.  Thus $x,y,$ and $z$ must be algebraically dependent, so they cannot all be in $A$.
(Such a set $A$ exists, for instance, because you can take a transcendence basis for $\mathbb{R}$ over $\mathbb{Q}$ and then multiply each element by an appropriate rational number to get an element of $(0,1)$.  Less obviously, such an $A$ can be constructed without the axiom of choice; see this answer on MO, for instance.)
A: Theorem. There is a nonempty perfect set $P\subseteq(0,1)$ (so $|P|=|\mathbb R|=2^{\aleph_0}$) such that, for any $c\in\mathbb R$ and $r\in(0,1),$ we have $\{c+r^n:n\in\mathbb N\}\not\subseteq P.$
Proof. We construct $P$ by imitating the construction of the Cantor set, with the following differences: instead of starting with the closed unit interval $[0,1],$ we start with a closed subinterval of $(0,1);$ at the first step, instead of removing the open middle third, we remove the open middle half; at the second step we remove the open middle two-thirds of each of the two remaining intervals; at the third step we remove the open middle three-fourths of each of the four remaining intervals; and so on. At the end we are left with a nonempty perfect set $P\subseteq(0,1)$ with the property that, for each $\varepsilon\gt0,$ there exist $n\in\mathbb N$ and $0\lt a_1\lt b_1\lt a_2\lt b_2\lt\cdots\lt a_m\lt b_m$ such that:
$$P\subseteq[a_1,b_1]\cup[a_2,b_2]\cup\cdots\cup[a_m,b_m];\tag1$$
$$b_i-a_i\lt\varepsilon\text{ for each }i\in\{1,2,\dots,m\};\tag2$$
$$\frac{b_i-a_i}{a_{i+1}-b_i}\lt\varepsilon\text{ for each }i\in\{1,2,\dots,m-1\}.\tag3$$
Let $c\in\mathbb R$ and $r\in(0,1),$ and assume for a contradiction that $\{c+r^n:n\in\mathbb N\}\subseteq P.$ Let $a_1\lt b_1\lt\cdots\lt a_m\lt b_m$ satisfy (1)–(3) for some $\varepsilon\in(0,r).$
Now $c=\lim_{n\to\infty}c+r^n\in P,$ so $c\in[a_i,b_i]$ for some $i\in\{1,\dots,m\};$ and $b_i\lt c+r\in P,$ so $c+r\ge a_{i+1}.$ Choose $n\in\mathbb N$ so that $c+r^{n+1}\le b_i\lt a_{i+1}\le c+r^n,$ i.e.,
$$a_i\le c\lt c+r^{n+1}\le b_i\lt a_{i+1}\le c+r^n;$$
then
$$\varepsilon\gt\frac{b_i-a_i}{a_{i+1}-b_i}\ge\frac{r^{n+1}}{r^n-r^{n+1}}=\frac r{1-r}\gt r,$$
contradicting our choice of $\varepsilon.$
