# 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?

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What do you know about the set of subsequential limits of a sequence $(a_{n})_{n \in \mathbb N}$? – t.b. Dec 22 '11 at 6:37
@t.b. Can you provide a little more please? – mathmath8128 Dec 22 '11 at 6:40
Okay, I'd say there isn't such a sequence if $Q\cap[0,1]$ are taken to be exclusive (e.g. the only possible subsequence limits). Assume that $a_n$ fulfills this property, and let $b_n$ be a sequence of rational numbers in $[0,1]$ whose limit is an irrational number in $[0,1]$. For each $b_i$ we can find infinite members of $a_n$ that're close to $b_i$ as much as we like, so we can create a subsequence of $a_n$ whose limit is $b_i$'s limit. Is this correct? – ro44 Dec 22 '11 at 6:49
Yes, that's correct -- you could write that up as an answer. – joriki Dec 22 '11 at 6:53
@joriki I had to wait 8 hours (!) before I could type it up, so I selected Kannappan's answer instead :). – ro44 Dec 22 '11 at 7:03

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.

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What's $E$? I think it should say "whenever $x\in\overline{\mathcal{S}}$, there is $\delta\gt0$ such that $B_\delta(x)\subset\overline{\mathcal{S}}$? – joriki Dec 22 '11 at 7:14
@joriki Fixed, Thanks for the pointer. But, I would keep it $\mathcal{S}$, because $\mathcal{S}$ is a closed set, and will coincide with its closure, $\bar{\mathcal{S}}$. – user21436 Dec 22 '11 at 7:25
Sorry, I should have defined my notation -- that was supposed to be the complement, not the closure. What you've written is the definition of $\mathcal S$ being open rather than closed. (BTW you can get the bar to cover the entire $\mathcal S$ by using \overline instead of \bar.) – joriki Dec 22 '11 at 7:28
@joriki I probably must have gone completely off my brains. I'll fix it. Thanks for your LaTeX help in this direction. – user21436 Dec 22 '11 at 7:32
@Kannapan: I don't see the need to edit it but using standard notation in a non-standard way is a very bad idea, even if it's explained. (By the way: given that you're so eager to edit so many things so often elsewhere, I'm a bit surprised by this statement...) – t.b. Dec 22 '11 at 8:19

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.

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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.

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Such sequences (having subsequences that converge to every point in $[0,1]$) do exist: one example is the van der Corput sequence, obtained by reversing the base-$n$ representation of consecutive integers. – Ilmari Karonen Dec 22 '11 at 8:14