Is there a sequence whose set of subsequence limits is $\mathbb{Q}\cap [0,1]$? Is there a sequence whose set of subsequential limits are $\mathbb Q \cap [0,1]$?
I was told that there wasn't such a sequence, is this true?
 A: Let $\alpha$ be your favourite irrational number in the interval $[0,1]$. For definiteness, let $\alpha=\sqrt{2}/2$.
Suppose that the set of subsequential limits of the sequence $a_1,a_2,a_3,\dots$
includes all rationals in the interval $[0,1]$.  We will show that $\alpha$ is also a subsequential limit of the sequence $(a_n)$.
The rationals are dense in the reals. So there is a sequence $r_1,r_2,r_3,\dots$ of rationals in $[0,1]$ such that $(r_n)$ has limit $\alpha$. Such a sequence, can, for example, be obtained by truncating the decimal expansion of $\alpha$ further and further along.
We now construct our subsequence of $(a_n)$ that has limit $\alpha$.
Let $n_1$ be the smallest index $i$ such that $|a_i-r_1|<\frac{1}{2}$.  There is such an $i$ because the sequence $(a_n)$ has, by assumption, a subsequence that converges to $r_1$.
Let $n_2$ be the smallest index $i$ such that $i>n_1$, and  $|a_i-r_2|<\frac{1}{2^2}$.  There is such an $i$ because $(a_n)$ has a subsequence that converges to $r_2$.
Let $n_3$ be the smallest index $i$ such that $i>n_2$, and  $|a_i-r_3|<\frac{1}{2^3}$.  There is such an $i$ because $(a_n)$ has a subsequence that converges to $r_3$.
Continue in the obvious way. The sequence $(a_{n_k})$ is a subsequence of $(a_n)$ and converges to $\alpha$.
Comment: We have proved that every real number in $[0,1]$ is a subsequential limit of $(a_n)$. More generally, let $(a_n)$ be any sequence. The same argument shows that the set of subsequential limits of $(a_n)$ is closed.
A: You have been told the right answer, if you want $\mathbb{Q} \cap [0,1]$ to be the exclusive limit points!
As t.b. points out, the set of subsequential limits, $\mathcal{S}$ is a closed set. In the 'epsilonics', it would mean, 
 Whenever $x \in \mathcal S^c$, there exists a $\delta > 0$ such that, $B_\delta(x) \subseteq \mathcal{S}^c$, where $ \mathcal{S}^c$ stands for complement of $\mathcal {S}$. A special thanks to @joriki, (also @t.b.) in this connection!! (Read the comments, in case you are wondering why.)
But as irrational numbers are dense in $[0,1]$, there does not exist any open ball $B_\delta(x)$ with property as mentioned. 
A: Suppose $\{a_j\}_{j=0}^{\infty}$ is a sequence that has $\mathbb{Q}\cap [0,1]$ as limit points. Let $\alpha$ be an irrational number in $[0,1]$ and let $\{b_j\}_{j=0}^{\infty}$ be a sequence of rationals in $[0,1]$ that converges to $\alpha$.  Set $d_i = |\alpha-b_i|$; by construction $d_i\rightarrow 0$.  Since there is a subsequence of $\{a_i\}_{i=0}^{\infty}$ converging to $b_i$ there must be some $a_i$ within $d_i$ of $b_i$.  Thus $a_i$ is within $2d_i$ of $\alpha$.  Doing this for every $i$ yields a subsequence of $\{a_j\}_{j=0}^{\infty}$ converging to $\alpha$.  
Consequently, any sequence that has $\mathbb{Q}\cap [0,1]$ as limit points (and I make no claims about the existence of such a sequence) actually has all of $[0,1]$ as limit points.
