Find all cluster points for the sequence $x_{n}$ = The $n$-th rational number Find all cluster points for the sequence $x_{n}$ = The $n$-th rational number
Note: In this problem a labeling of rational numbers by positive integers is used. Such labellings do exist because $\mathbb{Q}$ is countable and we choose any of them. The answer does not depend on the choice.
Def: $c$ is a cluster point for $x_{n}$ if $\forall ε > 0$, and $\forall N$, $\exists$ an $n>N$ such that $|x_{n} - c| < ε$.
We also know that $\mathbb{Q}$ is dense in $\mathbb{R}$. So $\forall r \in \mathbb{R}$, and $\forall ε > 0$, $\exists q \in \mathbb{Q}$ such that $|r - q| < ε$.   
It seems like all real numbers are cluster points, but I'm not sure how to prove this. 
Thanks for your help!
 A: Let $c$ be an arbitrary real number, let $N$ be an arbitrary natural number and let $\varepsilon>0$ be arbitrary. If none of the rationals $x_{N+1},x_{N+2},x_{N+3},\ldots$ belongs to the open interval $(c-\varepsilon,c+\epsilon)$ then $(c-\varepsilon,c+\varepsilon)\cap\mathbb Q$ is finite, which cannot happen since $\mathbb Q$ is dense in $\mathbb R.$ Thus $|r_n-c|<\varepsilon$ for some $n>N$ and hence $c$ is a cluster point of $\mathbb Q.$ Since $c$ is arbitrary, we conclude that every real number is a cluster point of $\mathbb Q.$
A: Let $c$ be any real number.  What you want to show is that there is some subsequence that converges to $c$.  Let $\epsilon_{n}=\frac{1}{n}$ (really, anything that is positive and converges to $0$).  Now, pick $n_1$ so that $x_{n_1}\in (c-\epsilon_1, c+\epsilon_1)$.  Then, pick $n_2$ so that $x_{n_2}\in (c-\epsilon_2, c+\epsilon_2)$ AND so that $n_2>n_1$.  How do we know such an $n_2$ exists?  By excluding the rationals with index smaller than $n_1$ we've only excluded finitely many rationals, and there are infinitely many rationals in that interval.
Inductively, suppose we've chosen $n_1, ..., n_{k}$ with $n_1 < n_2 < ... < n_k$ and $x_{n_i}\in (c-\epsilon_i, c+\epsilon_i)$.  Then, we choose the next element $n_{k+1}$ so that $x_{n_{k+1}}\in (c-\epsilon_{k+1}, c+\epsilon_{k+1})$ and $n_{k+1}>n_{k}$.  Again, this can be done because excluding all the rationals with index smaller than $n_k$ only excludes finitely many rationals and there are infinitely many rationals in the interval.
Then, it is fairly clear that $(x_{n_k})_{k=1}^{\infty}\to c$ and by construction $(x_{n_k})$ is a subsequence of $x_n$.  Hence $c$ is a limit point of the sequence.
A: As you said Q is dense in R meaning that for any real number there exists a sequence of rationals converging toward that real number, here the sequence converging toward any real number is a sub-sequence of Xn.Now use the definition of the limit of a sequence to conclude 
