example of a sequence $x_n$ such that the set of limit points of $x_n$ is $[0,1]$. Give an example of a sequence $x_n$ such that the set of limit points of $x_n$ is $[0,1]$.
I know the answer is $\mathbb{Q}$ intersection $[0,1]$, but how can I prove that? 
Can you help me please? 
 A: The rationals in $[0,1]$ are countable, so they can be enumerated as a sequence $\{q_n\}$. Since the rationals in $[0,1]$ are dense in $[0,1]$, its limit point set is the whole $[0,1]$.
The only step you need to finish is the denseness property. Now observe that any point $x$ in $[0,1]$ has a decimal representation $x=0.a_1a_2\dots$. Then each $x_n=0.a_1a_2\dots a_n$ is rational, and $\lim_{n\to\infty}x_n=x$. So $x$ is a limit point of $\mathbb{Q}\cap [0,1]$.
A: Let $X$ be a set. A sequence in $X$ is  function from $f:\mathbb{N}\rightarrow X$ 
Take $X=\mathbb{Q}\cap [0,1]$ 
$X$ is countable and hence we have a bijection from $f:\mathbb{N}\rightarrow X$
Now use the fact that $\mathbb{Q}$ is dense.
A: In fact, any countable dense sequence $\{x_n\}$ in $[0,1]$ can satisfy your requirement.
A: Another sequence is $\{ \sin^2(n)\}_n$ whose limit point set is $[0,1]$. The set $D:=\{m+2\pi n : m,n\ are\ integrs \}$ is dense in $R$. Hence for suitabilly chosen sequence $\{a_n\}_n$ from $D$ one can converge the sequence $\{\sin^2(a_n)\}_n$ to $\sin(x)$ for any $x\in R$.
