Prove that there is a real number $a$ such that $\frac{1}{3} \leq \{ a^n \} \leq \frac{2}{3}$ for all $n=1,2,3,...$

There exists $\alpha \in\left[\dfrac{16}{3},\dfrac{17}{3}\right]$ with the required property. To see this, we will construct an interval sequence $$\left[\dfrac{16}{3},\dfrac{17}{3}\right]=[\alpha_{1},\beta_{1}]\supset [\alpha_{2},\beta_{2}]\supset\cdots\supset[\alpha_{n},\beta_{n}],$$ where $\alpha_{n}$ and $\beta_{n}$ are such that $$\alpha^n_{n}-\dfrac{1}{3}=\beta^n_{n}-\dfrac{2}{3}=m_{n}\in \Bbb N^{+},$$ so that, for any $x\in [\alpha_{n},\beta_{n}]$, we have $$\dfrac{1}{3}\le\{x^n\}\le\dfrac{2}{3}.$$

We construct the interval sequence by induction. Assume that we have $[\alpha_{n},\beta_{n}]$. Let $$a=\alpha^{n+1}_{n},\quad\quad b=\beta^{n+1}_{n}.$$It follows that $$ b-a=(m_{n}+\dfrac{2}{3})\beta_{n}-(m_{n}+\dfrac{1}{3})\alpha_{n}>\dfrac{\alpha_{n}}{3}>\dfrac{5}{3}.$$ Then there exists $m_{n+1}\in \Bbb N^{+}$ such that $$\left[m_{n+1}+\dfrac{1}{3},m_{n+1}+\dfrac{2}{3}\right]\subset[a,b].$$ We take $$\alpha_{n+1}=\sqrt[n+1]{m_{n+1}+\dfrac{1}{3}},\qquad\beta_{n+1}=\sqrt[n+1]{m_{n+1}+\frac{2}{3}}.$$ Now $$\alpha^{n+1}_{n}=a<\alpha^{n+1}_{n+1}=m_{n+1}+\dfrac{1}{3}<\beta^{n+1}_{n+1}=m_{n+1}+\dfrac{2}{3}<b=\beta^{n+1}_{n},$$and hence $\alpha_{n}\le\alpha_{n+1}<\beta_{n+1}<\beta_{n},$ or $$[\alpha_{n},\beta_{n}]\supset[\alpha_{n+1},\beta_{n+1}].$$