I need help with this problem:
Does exist a sequence $ \{x_n\} $ such that $ \lim \inf \{x_n\} = 0 $ and $ \lim \sup \{x_n\} = 1 $ , and for every real $\alpha \in [0,1] $ exists a subsequence of $ \{x_n\} $ that converges to $\alpha$ ?
I saw a similar question already posted but it didn't include the lim inf and lim sup restriction.
Thanks in advance
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4Go over all the rationals between zero and one – Ofir Jan 23 '17 at 17:36
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would $(1 + \sin n)/2$ suffice? it covers enough of the interval $[0,1]$ that it would seem its terms get arbitrarily close to any chosen real – Rob Jan 23 '17 at 17:36
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1@Ofir surprisingly simple solution, chapeau! – Peter Jan 23 '17 at 17:38
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@Ofir you mean like an enumeration of $\Bbb Q $ ? – mate89 Jan 23 '17 at 17:43
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1@isccha89 Yes, any enumeration of $\mathbb{Q}\cap[0,1]$ turns out to have this property. If this isn't clear to you, try looking at my answer where I give two examples of enumerations that more obviously have this property and try to generalize. – Stella Biderman Jan 23 '17 at 17:45
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Related stronger results: Any closed subset of $\mathbb C$ is the set of limit points of some sequence AND Given closed $C \subseteq \mathbb{R}$ find a sequence with subsequences convergent to every point in $C$ and nowhere else – Dave L. Renfro Jan 23 '17 at 18:07
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Note that any sequence such that $x_{n+1}-x_n\to 0$ fits the bill. – Gabriel Romon Jan 23 '17 at 20:51
4 Answers
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Sure this is possible! There are a number of ways to do this, one of them being to use the Farey sequence. We can recursively build our sequence by starting with $x_0=0$ and $x_1=1$ and then go to the next new rational number in the order that they are in the Farey sequence, breaking ties ($1/4$ and $3/4$ get introduced at the same step) by putting the smaller one first. This sequence slowly "refines" rational estimates, and so since $\mathbb{Q}\cap[0,1]$ is dense in $\mathbb{R}\cap[0,1]$, has a subsequence that converges to every real number. This particular example is very important in Continued Fraction theory.
Analogously, you can do something similar refining by decimal digit. The first few elements of that sequence are $$0,1,0.1,0.2,0.3,\ldots,0.9,0.01,0.02,\ldots, 0.11,0.12,\ldots0.98,0.99,0.001,\ldots$$
It turns out to be the case that any mapping of $\mathbb{N}$ to $\mathbb{Q}\cap[0,1]$ will do as your sequence. Can you figure out why?
Stella Biderman
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@isccha89 I also gave some of the terms of the other sequence I mentioned. See if you can fill in the gaps! – Stella Biderman Jan 23 '17 at 17:48
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Density of $\sin n$ is not easy to prove, thus an overkill. Instead I'd suggest you consider the following lemma:
Let $(u_n)$ be a bounded sequence such that $u_{n+1}-u_n\to 0$. Then the set of subsequential limits of $(u_n)$ is a closed interval $[\alpha, \beta]$.
Consider $u_n=\frac{1}{2}(\sin(\ln n) + 1)$
The mean value theorem says $\displaystyle \frac{\sin(\ln (n+1))-\sin(\ln n)}{n+1 -n} = \frac{\cos(\ln(\xi_n))}{\xi_n}\to 0$
The set of subsequential limits of $(u_n)$ is therefore an interval and surely a subset of $[0,1]$. It suffices to prove that $0$ and $1$ are indeed subsequential limits of $(u_n)$.
To this effect, consider the subsequences given by these indices $\alpha_n=\lfloor \exp(\frac \pi 2+2n\pi) \rfloor$ and $\beta_n=\lfloor \exp(-\frac \pi 2+2n\pi) \rfloor$
Gabriel Romon
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Amazing. Can you recommend me a book for advanced sequences analysis? I am at the level of Bartle's Introduction of Real analysis – mate89 Jan 23 '17 at 23:06
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I would recommend the sequence $$\frac{1}{2}(\sin(n) + 1)$$ because $\{\sin(n) : n\in \mathbb{N}\}$ is dense in $[-1,1]$.
Edit. Here is a nice an constructive proof of the density property of the sine function.
TheGeekGreek
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2I agree with the above comment, and am moderately shocked that this was the first thing you (and another person! See the above comments) came up with. – Stella Biderman Jan 23 '17 at 17:42
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1@TheGeekGreek Becuase this is an example from the third class of solutions that would come to mind for me (I gave the first and copper.hat gave the second). This is more complex than the other two and has a harder proof of correctness. – Stella Biderman Jan 23 '17 at 17:52
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@StellaBiderman That ${e^{in}}$ is dense in the unit circle is basically the same as copper.hat's example. The density of $\sin n$ in $[-1,1]$ follows immediately from this. – zhw. Jan 23 '17 at 18:52
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Take any irrational $\beta$ and let $x_n = n \beta \mod 1$.
copper.hat
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@SRobertJames There are already lots of proofs, for example https://math.stackexchange.com/questions/450493/positive-integer-multiples-of-an-irrational-mod-1-are-dense – copper.hat Oct 23 '22 at 15:35