For $n \in \mathbb N^*$, $A_n = (\sin (n) -1,\sin (n) + 1)$.
How to show that $\limsup A_n = [-2,2], \liminf A_n = \{0\}$?
What if $A_n = (\sin (n -1),\sin (n + 1))$ instead?
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Your taking the limsup of a sequence of sets? – zhw. Mar 16 '16 at 20:10
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Perhaps you mean for two different sequences? For $sin n-1$ and $sin n+1$? – MathematicianByMistake Mar 16 '16 at 20:11
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1Ah sorry, it's a typo. I will edit my question. – c.robert Mar 16 '16 at 20:12
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Is it $\sin(n)\pm 1$ or $\sin(n\pm1)$? – sinbadh Mar 16 '16 at 20:13
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It is the former. – c.robert Mar 16 '16 at 20:14
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The question is not accurate. if we talk $\sup$ and $\inf$ in two dimesnion, what is the order of set here? – runaround Mar 16 '16 at 20:18
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@runaround. The question is well defined. Exist the concepts of $\lim\sup$ and $\lim\inf$ for sequences of sets – sinbadh Mar 16 '16 at 20:20
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Infimum and supremum are defined on sets, while $A_n$ here are two-element sets, so $\sup A_n = \sin n + 1$ and $\inf A_n = \sin n - 1$, neither is even converging. So it is either partial limits meant or union of the sets $A$ up to $n$. – Andrew Mar 16 '16 at 20:23
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2@AndrewMiloradovsky $A_n$ is the interval $(\sin(n)-1,\sin(n)+1)$. Then, by definition, $\lim\sup A_n=\cap_{n=1}^\infty\cup_{m=n}^\infty A_n$ and $\lim\inf A_n=\cup_{n=1}^\infty\cap_{m=n}^\infty A_n$ – sinbadh Mar 16 '16 at 20:26
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@sinbadh Oops. Didn't mention that the brackets aren't curved... – Andrew Mar 16 '16 at 20:28
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@sinbadh, that is what I am wondering from question. THX. – runaround Mar 16 '16 at 20:28
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@SiXUlm: your extra question is interesting. – c.robert Mar 16 '16 at 20:32
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@c.robert sorry, I don't mean to edit directly your post. Just don't know how to undo it. – SiXUlm Mar 16 '16 at 20:33
2 Answers
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Given a sequence of sets $A_1, A_2, \ldots$, the lim sup of the $A_n$ is the set of points that are seen in infinitely many of the $A_n$, while $\liminf A_n$ is the set of points that are seen in all but a finite number of the $A_n$.
Your sets $A_n$ are intervals of radius $1$ centered on $\sin(n)$. As $n$ varies, the centers of these intervals weave back and forth between $-1$ and $1$ (visiting points in between). I agree that every point between $-2$ and $2$ (exclusive) must be visited infinitely often by your $A_n$, since $\{\sin(n):n=1,2,\ldots\}$ is dense in the interval $[-1,1]$, which implies that for every small $\epsilon>0$ there exists infinitely many $n$ for which $\sin(n)>1-\epsilon$; for those $n$ we have $2-\epsilon<\sin(n)+1$, which means $2-\epsilon\in A_n$. Similarly there are infinitely many $n$ for which $-2+\epsilon\in A_n$. But $\sin(n)$ is never $-1$ or $1$ when $n$ is an integer (since $\pi$ is irrational), so $-2$ and $2$ do not belong to any of the $A_n$. Conclude $\limsup A_n=(-2,2)$.
As for the lim inf, only the point $0$ is found in every $A_n$, since $-1<\sin(n)<1$ for every integer $n$. Any nonzero number won't belong to $\liminf A_n$, because if $\epsilon >0$ then (by the same claim as cited earlier) there are infinitely many $n$ where $\sin(n)>1-\epsilon$. For those $n$ we have $-\epsilon<\sin(n)-1$, which means $-\epsilon\not\in A_n$. Similiarly if $\epsilon <0$ then there are infinitely many $n$ for which $\epsilon\not\in A_n$.
grand_chat
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Note you only need to know $\limsup \sin n $ and $\liminf \sin n$ to solve this.
The $\limsup$ part has been answered here, and here, the $\inf$ part should be similar.
YoTengoUnLCD
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Thanks. However I don't really understand why $\liminf$ reduces to one point? – c.robert Mar 16 '16 at 20:25