What is wrong with the following proof that $\{ (\frac{3}{2})^n\mod 1: n\in\mathbb{N} \} $ is dense in $\ [0,1]\ $?

Question

What is wrong with the following straightforward proof that $\ \lbrace{ \left(\frac{3}{2}\right)^n \mod 1: n\in\mathbb{N} \rbrace}\ $ is dense in $\ [0,1]\ $ ?

In fact, let $\ f(x) = \lbrace{\ x^n \mod 1: n\in\mathbb{N} \rbrace}.\ $ Then the following proves that $\ f(x)\ $ is dense in $\ [0,1]\ $ for all $\ x>1,\ x\notin \mathbb{N}.\ $ However, I will show my attempted proof for $\ x=\frac{3}{2}\ $ and then you can just replace $\ \frac{3}{2}\ $ with any $\ x>1,\ x\notin \mathbb{N}, $ throughout.

Firstly, wikipedia suggests that $\ (\alpha \mod 1 ,\alpha ^{2} \mod 1,\alpha ^{3} \mod 1,\dots )\ $ is uniformly distributed for all $\ \alpha >1\ $ except for a set of measure $0.$ Let $\ Y\ $ be the set of all $\ \alpha>1\ $ such that $\ (\alpha \mod 1 ,\alpha ^{2} \mod 1,\alpha ^{3} \mod 1,\dots )\ $ is uniformly distributed. Then $Y\cap[1,2]\ $ must be dense in $[1,2].$

Therefore, there are (we can choose) two sequences, $\ a_n\subset Y\ $ and $\ b_n\subset Y\ $, such that $a_n\to\frac{3}{2}^-\ $ and $\ b_n\to\frac{3}{2}^+.$ From what wikipedia says about the set $Y$, $\ f(a_n)\ $ and $f(b_n)\ $ are dense subsets of $[0,1]\ $ for every $\ n\in\mathbb{N}.$

Fix $\ \gamma \in (0,1)\ $ and let $ \delta\ $ be any number with $ \ 0<\delta<\min\{\gamma, 1-\gamma \},\ $ and $\ \delta\ $ can be as small as you like.

Then there exist $\ n_1,\ n_2,\ m,\ N,\ $ such that

$$ N +\gamma -\delta < {a_{n_1}}^m < N + \gamma < {b_{n_2}}^m < N + \gamma + \delta.\quad (1)$$

Since $\ \frac{3}{2}\ $ is sandwiched between $\ a_{n_1}\ $ and $\ b_{n_2},\ $ we must also have that

$$N+\gamma-\delta < \left(\frac{3}{2}\right)^m < N+\gamma + \delta.\quad (2)$$

Since $\ \delta\ $ was arbitrary, we can always find $\ N,\ m\ $ so that $\ (2)\ $ holds. This completes the proof.

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Note: I didn't prove $\ (1)\ $ because it seems elementary and would distract from the overall proof. However, I can provide a proof if anyone isn't sure or requests it. But for example, the real function $\ g(x) = x^n\ $ is continuous, so that ought to do it [and this "continuity" proof highlights that the proof won't work for $\ x\in \mathbb{N}.$]

Anyway, my attempt to answer my own question is that either Wikipedia is wrong in it's claim, or I misunderstand Wikipedia's claim. Basically my logic is this: Wikipedia claims that the sequence $ (\alpha\mod 1 ,\alpha ^{2}\mod 1,\alpha ^{3}\mod 1,\dots )$ is completely uniformly distributed for all $\ \alpha >1\ $ except for a set of measure $0$, which implies that it is $k-uniformly$ distributed for each natural number $k\geq 1$ which implies it is $1-uniformly$ distributed, which is the same as saying it is uniformly distributed...

Answer 1

As pointed out in the comments by Zerox, $\ (1),\ $

There exist $\ n_1,\ n_2,\ m,\ N,\ $ such that $ N +\gamma -\delta < {a_{n_1}}^m < N + \gamma < {b_{n_2}}^m < N + \gamma + \delta.\quad (1)$

has not been justified.

My idea when writing the proof attempt was that for any $\ \delta>0,\ $ no matter how small, $\ \exists\ \varepsilon>0, m,N\in\mathbb{N}\ $ such that $\ y\in \left(\frac{3}{2}-\varepsilon, \frac{3}{2}-\varepsilon\right) \implies N +\gamma -\delta < \left(\frac{3}{2}-\varepsilon\right)^m\ < N+\gamma < \left(\frac{3}{2} +\varepsilon\right)^m\ < N +\gamma +\delta. $

However, this only makes sense if we assume that $\ \lbrace{ \left(\frac{3}{2}\right)^n \mod 1: n\in\mathbb{N} \rbrace}\ $ is dense in $\ [0,1],\ $ which is the result we are trying to prove. [I think I did this without realising it was what I was doing].

I have tried to salvage $(1)$ but I don't see a way to do this.