Infinite intersection between a arbitrary set of integers and a set of floor powers Let $E$ be an infinite set of positive integers, proves that there is a $\alpha \in \mathbb{R}$ such that $\{\left \lfloor \alpha^k \right \rfloor  ;k \in \mathbb{N} \}\cap E$ is infinite.
I have two ideas in mind, first I can try to use $\lfloor \alpha^k \rfloor=n \iff \log(\alpha)\in [\frac{log(n)}{k},\frac{log(n+1)}{k})$ in order to prove that some infinite intersection of the type $\bigcap_{n\in E,k\in A} [\frac{log(n)}{k},\frac{log(n+1)}{k})$ is not empty, and my other idea is to prove that the set $A_m=\{\alpha; | \{\left \lfloor \alpha^k \right \rfloor  ;k \in \mathbb{N} \}\cap E |>m \}$ contains an open dense set (I don't know if that is true) for later use the baire theorem.
can anyone give me a hint or show me a solution?
 A: Your idea of using Baire's theorem is good. 
Let $\alpha>1$ and $k$ a positive integer. Call the pair $(\alpha,k)$ good if for some $n\in E$ we have $n<\alpha^k<n+1$. 
Define $A_m:=\{\alpha>1:\exists k_1<\dots<k_m\text{ s.t. }(\alpha,k_i)\text{ are good}\}$. This is almost your definition, except that now $A_m$ is an open set (why?). 
Let's now prove that $A_m$ is dense in $(1,\infty)$: fix a real $\alpha>1$, $\alpha\not\in A_m$. Call $s:=\max\{j:\alpha\in A_j\}$, so that there exist $k_1<\dots<k_s$ such that $(\alpha,k_i)$ are good pairs. (Here $s$ could be $0$ as well!) 
We have
$$E\cap [\alpha^{k_s+1},\infty)=\bigcup_{k>k_s}E\cap [\alpha^k,\alpha^{k+1})$$
and since $E$ is infinite we have $|E\cap [\alpha^{k_s+1},\infty)|=\infty$; thus for infinitely many values of $k>k_s$ there is some $u\in E$ s.t. $\alpha^k\le u<\alpha^{k+1}$. 
Put $\beta:=\sqrt[k]{u}$: we have $\alpha\le\beta<\alpha^{1+\frac{1}{k}}$, so we can choose $\beta$ arbitrarily close to $\alpha$. In particular we can choose $\beta$ s.t. $(\beta,k_i)$ are good pairs (for $1\le i\le s$). But there is an arbitrarily small $\epsilon$ s.t. $u<(\beta+\epsilon)^k<u+1$ and $(\beta+\epsilon,k_i)$ are still good,
thus $(\beta+\epsilon,k)$ is a new good pair and we obtain $\beta+\epsilon\in A_{s+1}$. 
Iterating the argument with $\beta+\epsilon$, we will eventually obtain a real number in $A_m$. This number can be chosen arbitrarily close to $\alpha$ (why?), so $A_m$ is dense.
