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Suppose we have a sequence $(x_n)_n\subset [a,b]$, we know by compactness there must be at least some limit point, i.e. there exists a subsequence $n_k$ and $x\in[a,b]$ such that

$$x_{n_k}\xrightarrow{k\to\infty}{} x$$

I'm wondering whether we can say anything about the integer sequence $n_k$, can we always find an $x$ limit point such that $n_k$ goes to infinity slowly? Say, at most exponential? I have the feeling that probably you can guarantee at most exponential growth, meaning that there's an $x$ and $n_k$ such that $\sqrt[k]{n_k}$ converges and $x_{n_k}\to x$. I've been trying to find a pigeonhole argument using binary intervals, but haven't been able to materialize it (which maybe means that you just can't guarantee any kind of slow growth, but that feels geometrically counterintuitive to me)

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  • $\begingroup$ @SassatelliGiulio I care abou the sequence $n_k$, not $x_{n_k}$, if $x_n$ is convergent, then we can take $n_k=k$, so in particular it's linear! (And so subexponential) $\endgroup$ Commented yesterday
  • $\begingroup$ Try and see for $x_n=q/2^k$ where $0<q<2^k$ odd and ordered after $k$ so $1/2, 1/4, 3/4, 1/8, 3/8, 5/8, 7/8$ etc; my feeling is that will give a maximal element for the separation you seek $\endgroup$ Commented 22 hours ago
  • $\begingroup$ @Conrad: That does show (I think) that you can't necessarily do better than exponential, but leaves the question of whether exponential is always possible. $\endgroup$ Commented 21 hours ago
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    $\begingroup$ Maybe not indeed by constructing very long blocks of equal elements separated as above (so at any step instead of one $q/2^k$ pick $k!$ or whatever of them $\endgroup$ Commented 19 hours ago

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There's no slowest rate, at least in the sense of "lower rate".

Let $r : \mathbb{N} \to \mathbb{N}$ be any rapidly increasing function; suppose without loss of generality that $r(k) > k$ for all $k$. I claim there is a bounded sequence $a_n$ such that for every convergent subsequence $a_{n_k}$, we have $n_k \ge r(k)$ infinitely often. In particular, if we take something like $r(k) = k^k$, we see that $\sqrt[k]{n_k}$ can be unbounded.

Define the strictly increasing sequence $s_i$ recursively by $s_0 = 0$ and $s_{i+1} = r(s_i)$. Now define the sequence $a_n$ as:

$$a_n = \begin{cases} 0, & s_i \le n < s_{i+1} \text{ for some even $i$ } \\ 1, & \text{otherwise}. \end{cases} $$

So terms $s_0, \dots, s_{1}-1$ are zeros; terms $s_1, \dots, s_2-1$ are ones; terms $s_2, \dots, s_{3}-1$ are zeros; and so on.

Let $a_{n_k}$ be a convergent subsequence. Suppose first that $a_{n_k} \to 1$, so that $a_{n_k} = 1$ for all large enough $k$. In particular, we will have $a_{n_k} = 1$ for all $k$ of the form $k = s_i$ where $i$ is even and sufficiently large. For such $k$, since $n_k \ge k = s_i$ and terms $s_i, \dots, s_{i+1}-1$ are all zeros, we must have $n_k \ge s_{i+1} = r(s_i) = r(k)$.

If instead $a_{n_k} \to 0$, the argument is similar, replacing "even" by "odd".

(Thanks to Conrad whose comment was a good hint.)

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  • $\begingroup$ Thanks! That's so counterintuitive, to me, even if your construction is still quite 'natural' $\endgroup$ Commented 12 hours ago

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