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My question is to find a sequence $x_n$ such that $\forall r \in\mathbb Q$, there is a subsequence of $x_n$ which converges to $r$. I'll be trying to do the same with $\mathbb R$.

I decided to simplify it first by finding a sequence $y_n$ such that $\forall n \in\mathbb N$, there is a subsequence of $x_n$ which converges to $n$. Namely, I constructed the sequence $1,1,2,1,2,3,1,2,3,4,1,2,3,4,5....$ because you can easily construct a subsequence that converges to $n$ by taking the subsequence $n$ recurring.

I'm trying to think similarly with $\mathbb Q$. Any help is appreciated.

Florian Suess
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  • I'm currently trying to find some sort of onto map for $\mathbb Q$, I think that's the best way to do this. So checking out things like these; https://math.stackexchange.com/questions/7643/produce-an-explicit-bijection-between-rationals-and-naturals – Florian Suess Sep 08 '18 at 01:22

3 Answers3

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You can use the same trick for $\mathbb Q$ that you used for $\mathbb N$: if $(q_1,q_2,q_3,\ldots)$ is any enumeration of $\mathbb Q$, then the sequence $S=(q_1,q_1,q_2,q_1,q_2,q_3,q_1,q_2,q_3,q_4,q_1,q_2,q_3,q_4,q_5,\ldots)$ does the job.

In fact, given any sequence $T=(r_1,r_2,r_3,\ldots)$ of rational numbers, the sequence $S$ contains the sequence $T$ as a subsequence. So given $x\in\mathbb R$, choose a rational sequence $T$ converging to $r$, and then $S$ contains $T$ as a subsequence that converges to $x$.

TonyK
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HINT: A version for $[0,1]$. Let $a_1=0$, $a_2=1$. Then every next $a_i$ is the centre of obtaining intervals, i.e., $a_3=1/2$, $a_4=1/4$, $a_5=3/4$, $a_6=1/8$ and so on.

Przemysław Scherwentke
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  • It's not acceptable to call on a sequence like $0,1, -1, 1/2, -1/2, 1/4, -1/4, ..., 2, -2, 2/3, -2/3, 2/6, -2/6...$ and so on ay? – Florian Suess Sep 08 '18 at 01:50
  • To elaborate why I think it's not, suppose $a_n$ is the sequence I mentioned, there is no precise choice of $x \in \mathbb N$ such that the sequence $a_x = 2$ for example. – Florian Suess Sep 08 '18 at 01:53
  • @FlorianSuess There are many methods of extending it to the whole line, e.g, adding the next interval from time to time and going through all of them, e.g. $[0,1]$, then add $[-1,0]$, then $[1,2]$, then $[-2,1]$ and so on. – Przemysław Scherwentke Sep 08 '18 at 01:56
  • Oh. Like the sequence $0, 1/1, 2/1, 1/2, 3/1, 3/2, 1/3, 2/3$, but then take negative of each (right after it's occurrence) and then use the trick I used in my question. – Florian Suess Sep 08 '18 at 02:09
  • Sweet! Yeah you're answer for sure pushed me to think in a way that answers the question in the end. Thanks @PrzemysławScherwentke – Florian Suess Sep 08 '18 at 02:11
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Any enumeration of $\mathbb Q$ will work. Proof: Let $q_1,q_2,\dots $ be such an enumeration. Let $x\in \mathbb R.$ Denote by $n_1$ the smallest $n$ such that $q_n\in (x,x+1).$ Then let $n_2$ be the smallest $n>n_1$ such that $q_n\in (x,x+1/2).$ Then let $n_3$ be the smallest $n>n_2$ such that $q_n\in (x,x+1/3),$ etc. The subsequence $q_{n_k}$ then converges to $x.$

zhw.
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