3

I recalled this interesting proposition from my real analysis course, for which the answer is true but I forgot the construction of such a sequence. I remember that's a point to get to the answer:

theorem: every real number has a strictly increasing infinite sequence of rational numbers whose limit is that real number. i.e. $\{1, 1.4, 1.414, \ldots\} \to \sqrt{2}$ (could be generalized to $\mathbb{Q}'$), $\{1,1.5,1.75,1.875,\ldots\}\to 2$ (could be generalized to $\mathbb{Q}$).

I know there is a way to include all rational numbers into a sequence using the numerator/denominator table construction. Although this sequence contains all distinct rational numbers, they are not in an increasing order so I can't utilize the theorem above. I'm stuck here then, do you have any hint to it?

shouya
  • 145
  • 1
    Why do they need to be increasing? – Thomas Andrews Oct 03 '15 at 23:02
  • Thomas, no they don't have to if they don't utilize the theorem I posed there. I'm just wondering how if that theorem can be used. – shouya Oct 03 '15 at 23:04
  • 1
    Enumerate the rationals as $ q_0, q_1,\dots $ Consider, for instance, $\pi $. Find the first $ n $ with $ q_n <\pi $. Then find the first $ m> n $ with $q_n <q_m <\pi $. Then find the first $ k> m $ with $ q_m <q_k <\pi $. Etc. – Andrés E. Caicedo Oct 03 '15 at 23:18
  • 1
    Have a look at these older questions:http://math.stackexchange.com/questions/1158521/a-sequence-for-which-the-set-of-limits-points-is-the-interval-0-1 and http://math.stackexchange.com/questions/50992/sequence-with-all-rationals-as-limit-points They are not exactly the same, but you could use the same approach to get a sequence with all real numbers as limit points. – Martin Sleziak Oct 04 '15 at 06:41

3 Answers3

2

The 'theorem above' (that 'every real number has a strictly increasing infinite sequence of rational numbers tending to it') offers a little help with a question given in the title: Is there a sequence that has all real numbers as cluster points?

The set $\Bbb Q$ is countably infinite, so there is a sequence $S$ containing all members of $\Bbb Q$. The set is dense in $\Bbb R$ — every open interval contains infinitely many rational numbers, so for every real number $y$ and arbitrarily small positive $\varepsilon$ there is infinitely many terms of $S$ in a neighborhood $(y-\varepsilon, y+\varepsilon)$. So every real $y$ is a cluster point of $S$,
Q.E.D.

Of course, for any $y$ you can find an infinite monotonic subsequence of $S$, which has a limit $y$ – but you don't need a 'subsequence', let alone 'monotonic'. Of course having 'infinitely many terms' in a neghborhood implies some infinite subsequence, but that subsequence is not necessary, 'infinitely many terms' is enough.

CiaPan
  • 13,049
0

By the monotone convergence theorem (on real numbers), if an increasing sequence is bounded, it has a unique limit. (If it's unbounded, it diverges instead)

Nate 8
  • 514
0

I was looking for a construction to arrange rational numbers into a sequence so every real number is a cluster point of the sequence. The problem turns out to be pretty easy.

My basic idea of decimal approximation indeed works, and we can see it intuitively. The point is how to describe the arrangement.

To make the construction cleaner, we restrict 'all real numbers' to 'real numbers in $[0,1)$'.

Let $\{a_n\}$ be the sequence. $a_0=0$, $a_{10^b+i}=\frac{i}{10^{-b}}$ where $b$ takes all integers and $i$ takes all integers in $[1, 10^b)$.

The terms in the sequence looks like:

0, 0.1, 0.2, 0.3, ..., 0.9, 0.01, 0.02, ..., 0.99, 0.001, 0.002, ... 0.999, ...

It's now intuitive to see this sequence contains all subsequence approximation to all real number in $[0,1)$.

shouya
  • 145