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Does there exist a sequence $a_n$, such that, for all $a \in [0,1]$, there is some subsequence $\{a_{n_k}\}$ whose subsequential limit is $\lim\limits_{k\to+\infty}a_{n_k}=a$.

I suppose $a_n=|\sin n|$ is satisfied the property ,but i cann't prove it. if you take a better example ,i will appreciate you .

CiaPan
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123321lll
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    You could take a sequence which enumerates $[0, 1] ∩ ℚ$. – Jean Abou Samra Dec 03 '24 at 10:46
  • Sorry , I don't understand your meaning.How to take a sequence which enumerates $[0,1]\cap \mathbb{Q}$? – 123321lll Dec 03 '24 at 11:12
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    a sequence that enumerates the rational numbers between 0 and 1 is not sufficient to ensure every real between 0 and 1 is a subsequential limit if each rational appears only once. – Colin Tan Dec 03 '24 at 11:51
  • I think that it can make a circle that number on it such that each rational appers infinity times . I 'm not sure it is right . And how to write it strictly? – 123321lll Dec 03 '24 at 12:12
  • I know what you mean. Of you go along the circumference of a unit circle by moving one unit at a time, then, in principle, you can get as close you want to any given point on the circumference after sufficiently many steps, right? – Colin Tan Dec 03 '24 at 12:34
  • Yes,but I found it is may not a good idea in this question – 123321lll Dec 03 '24 at 13:20
  • FYI, the possible sets of subsequence limits of a sequence of real numbers is equal to the set of nonempty closed subsets of the extended real numbers. That is, $[0,1]$ can be replaced with any nonempty closed subset of the extended real numbers, and no other set can be a set of subsequence limits of a sequence of real numbers. See Given closed $C \subseteq \mathbb{R}$ find a sequence with subsequences convergent to every point in $C$ and nowhere else and its linked questions. – Dave L. Renfro Dec 03 '24 at 18:39
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    @ColinTan Any enumeration of the rational numbers between $0$ and $1$ is sufficient. Why do you think that's not the case? – jjagmath Dec 04 '24 at 00:43
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    @ColinTan: Your first comment is almost correct: you just have to delete the word "not". – TonyK Dec 04 '24 at 01:03
  • @123321lll, could I request perhaps you ask specifically whether the sequence ${|\sin(n) } $ has every real between 0 and 1 as a subsequential limit as a separate question? – Colin Tan Dec 04 '24 at 08:07
  • @ColinTan ,sorry I see the comment now ,I will ask the question. – 123321lll Dec 06 '24 at 10:20
  • @ColinTan the question is there – 123321lll Dec 06 '24 at 10:36
  • thank you, 123321lll for asking. – Colin Tan Dec 06 '24 at 11:58
  • @TonyK, actually, I still don't understand. Why does every enumeration of the rationals between 0 and 1 have every real as a subsequential limit? Could someone share a proof with me? – Colin Tan Dec 07 '24 at 01:35
  • Do you know that each number in $[0,1]$ has a decimal expansion? If so, perhaps you could use that. – GEdgar Dec 07 '24 at 02:14
  • @GEdgar, lets say we are thinking of $\pi - 3 = 0.14159\ldots$. But the numbers $0.1, 0.14, 0.141,\dots$ might not appear in ${a_n} $ in increasing order. – Colin Tan Dec 07 '24 at 04:47
  • @ColinTan: Given $x\in[0,1]$ and an enumeration $(q_i)$, then for every $n$ there esists a rational $q_{i_n}$ within distance $1/n$ of $x$. The sequence $(q_{i_n})$ tends to $x$. (To make this work, we have to make $(i_n)$ a strictly increasing sequence, but this is easy to do.) – TonyK Dec 07 '24 at 11:14
  • ok. Thanks, TonyK. User65023 has written up this solution as an answer. – Colin Tan Dec 07 '24 at 14:31
  • @ColinTan... If the term 0.14 is not later in my enumeration, I could use 0.140 or 0.1400 etc. One of these is later in my enumeration. – GEdgar Dec 07 '24 at 14:42
  • $0.14=0.140$. If by enumeration, you mean one $a_i$ for each rational number, then you cannot have both $0. 14$ and $0.140$ in your sequence. – Colin Tan Dec 08 '24 at 08:06

3 Answers3

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Here's a very simple sequence:

$$\frac{1}{2},\frac{1}{3},\frac{2}{3},\frac{1}{4},\frac{2}{4},\frac{3}{4},\frac{1}{5},\frac{2}{5},\frac{3}{5},\frac{4}{5},\frac{1}{6},\frac{2}{6},\frac{3}{6},\frac{4}{6},\frac{5}{6},\frac{1}{7},\cdots$$

jjagmath
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Actually, take $a_n$ to be any dense sequence in $[0,1]$. Any enumeration of the sequence works, because after any finite number of elements, the remaining tail of the sequence is still dense. If the tail of any dense sequence wasn't dense, then the entire sequence wouldn't be dense. Once you have a dense sequence, then for any real number $a \in [0,1]$, there exists a subsequence $n_k$ such that $\lim_{k\to\infty} a_{n_k} = a$.

The Other Terry
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This solution is based on Jean Abou Samra's comment.

Choose a bijection $f : \mathbb{N} \to ([0,1] \cap \mathbb{Q}) \times \mathbb{N}$. Then the composite $a: \mathbb{N} \xrightarrow{f} ([0,1] \cap \mathbb{Q}) \times \mathbb{N} \xrightarrow{\text{proj}_1}([0,1] \cap \mathbb{Q}) $ defines a sequence $a(n)$, which is just the first coordinate of $f(n)$.

Then, for each given $c \in [0,1]$, choose a sequence $\{q_i\} $ of rationals between 0 and 1 that converges to $c$. Let $n_1$ be the least index such that $f(n_1) = (q_1, k_1)$ for some $k_1$. Then $a(n_1) = q_1$, by definition. For each $i > 1$, let $n_i > n_{i - 1}$ be the least index such that $f(n_i) = (q_i, k_i)$ for some $k_i$. Such $n_i$ exists since $q_i$ occurs as the first coordinate of $a(n) $ for infinitely many indices $n$. Then $a(n_i) = q_i$, by definition again.

Therefore $\lim a(n_i) =\lim q_i = c$.

Colin Tan
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