The closure of a complex Set of Powers. I came across this question while doing some exercises in complex analysis:

For fixed $x\in[0,1)\setminus \mathbb{Q}$ let $a=e^{2\pi ix} \in\mathbb{C}_{|z|=1}$ and define: 
  $$X_a:=\{a^n:n\in\mathbb{N}\}=\{e^{(2\pi i)·x·n}:n\in\mathbb{N}\}.$$
  What is $\overline{X_a}$?

I believe I've solved the exercise: $$\overline{X_a}=\mathbb{C}_{|z|=1},$$ but I have two related questions...
Solution:
First, I know that $xn$ is always irrational and always different for each $n\in\mathbb{N}$. $X_a$ is therefore infinite and its elements never have a rational multiple of $\pi$ as argument. Then one realizes that all: 
$$\xi\in\mathbb{C}_{|z|=1}:\quad \xi=e^{2\pi i \frac{p}{q}},\quad p<q\in\mathbb{N}$$
are the limit points of $X_a$. Thus:
$$ X_a \subset X_a \cup \{\,\xi\in\mathbb{C}_{|z|=1}:\, \xi=e^{2\pi i\frac{p}{q}},\,p<q\in\mathbb{N}\}\subseteq\overline{X_a}=\mathbb{C}_{|z|=1}$$
"by density".
My questions is the following:

Could I have had simplified the solution by focusing from the beginning on: 
  $$x\in\overline{[0,1)\setminus\mathbb{Q}}\leadsto \overline{X_a}?$$ 
  In general does:
  $$ \overline{\{e^{(2\pi i)x}:x\in A\}}=\{e^{(2\pi i)x}:x\in\bar{A}\}?$$

 A: Your last claim would be true if you demand that $A$ be invariant under integer translations.
The points of $A=\{mx+n| m,n\in\mathbb Z\}$, form a lattice in $\mathbb R$. It is a result of Kronecker (or Minkowski?) that there are only two possibilities, 


*

*either each point of $A$, and thus of $\exp(i2\pi\,A)$ is isolated 

*or $A$ is dense in $\mathbb R$, and in consequence $\exp(i2\pi\,A)$ is dense in $S^1$.

A: For your second question, let $A$ be the set $\{n+\frac{1}{n}\}_{n\in\mathbb{Z}^+}$. Then $\bar{A} = A$, because elements of $A$ are always separated by distance at least $1$. However, the points $\{e^{2\pi i / n}\}_{n\in \mathbb{Z}^+}$ are all in the image of $A$, because $e^{2\pi i n} = 1$ for $n\in\mathbb{Z}$, so $e^0=1$ is in the closure of the image of $A$, but $0 \not\in A$.
A weaker form of this does hold, however: the first set always contains the second as a subset, as can be seen by taking a sequence $\{x_i\}_{x\in\mathbb{N}}$ converging to $x\in\bar{A}$, and noting that $e^{2\pi i x_i}$ converges to $e^{2\pi i x}.$
A: First, the solution of the excercise is indeed:
$$\overline{X_a}=\mathbb{C}_{|z|=1}.$$
Respondig your first question and using your notation, it is true that:
$$ \overline{\{e^{(2\pi i)x}:x\in [0,1)\setminus\mathbb{Q}\}}=\{e^{(2\pi i)x}:x\in\overline{[0,1)\setminus\mathbb{Q}}\}=\{e^{(2\pi i)x}:x\in\overline{[0,1)\cap\mathbb{Q}}\}.$$
Note that:
$$\overline{[0,1)\setminus\mathbb{Q}}=\overline{[0,1)\cap\mathbb{Q}}=[0,1].$$
For the general case, your second question, refer to LutzL's answer. 
