This question is somewhat related to my previous question here.
Here, $\lbrace{ \cdot \rbrace}$ means fractional part.
Is it true that if $x > 1\ $ then either $0$ or $1$ (or both) are accumulation points of the set $ A_x:= \lbrace{\ \lbrace{\ x^n \rbrace}: n\in\mathbb{N}\ \rbrace}\ ?$
I think this is true for Pisot Numbers at least, because, for example, $(3+\sqrt{5})^n + (3-\sqrt{5})^n \in\mathbb{Z}.\ $ and $\ \vert 3-\sqrt{5}\vert < 1.$
Also relevant is here, where it says,
The sequence $\displaystyle (\alpha ,\alpha ^{2},\alpha ^{3},\dots )$ is completely uniformly distributed for almost all $\alpha >1.$
Unfortunately this does not decide whether or not $A_x$ contains either $0$ or $1$ as a limit point for every $x>1.$ And also the above fact does not tell us that it is not known whether or not $0$ or $1$ is a limit point of $A_x$ for every $x>1.$ So the question remains.
I have some vague idea about considering the interval $(\liminf A_x, \limsup A_x)$ and coming to a contradiction, but I don't have anything concrete yet.