# Math Help - An exercise about limsup/liminf of a subset sequence

1. ## An exercise about limsup/liminf of a subset sequence

Let $E_{n} = \{x \in \[0,2\pi] : (sin(nx))/n > 0\}$ with n natural number.
Calculate: E', E'' where E' = liminf En, E'' = limsup En for n that goes to $\infty$

Looking at the goniomethrical discus i should say that liminf is the empty set and that limsup is $[0,2\pi]$ \ $\{0,\pi, 2\pi\}$, but how could i formalize it (if is correct) ?

2. ## Re: An exercise about limsup/liminf of a subset sequence

What is the infimum of $E_1$? Of $E_2$? Of $E_n$? Given any $n \in \mathbb{N}$, choose any $x \in \left(0,\dfrac{\pi}{2n}\right)$. Then $0<\dfrac{\sin(nx)}{n} < \dfrac{\sin\left(n\dfrac{\pi}{2n}\right)}{n} = \dfrac{\sin\left(\dfrac{\pi}{2}\right)}{n} = \dfrac{1}{n}$. So, the infimum of $E_n$ is $0$ for all $n$. Hence, $E' = 0$.

The suprema are a little different. How do you find the supremum of $E_n$? We need only look at the interval $\left[ \dfrac{n-1}{n}2\pi,2\pi \right]$. We know $\sin(nx)> 0$ on the interval $\left(\dfrac{n-1}{n}2\pi, \dfrac{2n-1}{n}\pi\right)$ and $\sin(nx)<0$ on $\left(\dfrac{2n-1}{n}\pi,2\pi\right)$. So, the supremum of $E_n$ would be $\dfrac{2n-1}{n}\pi$. So, what is $\lim_{n\to \infty} \dfrac{2n-1}{n}\pi$?

3. ## Re: An exercise about limsup/liminf of a subset sequence

the first answer should be $(0,\dfrac{\pi}{n}\right )$ and the other subsets depending on the periodicity so $(\dfrac{2k\pi}{n}\right,\dfrac{(2k+1)\pi}{n}\right )$ with the limitation that 2k+1/n < 2, 2k/n > 0 with k positive integer.

the second answer is looking at the sign of the prime derivative of sin(nx)/n.

the third answer should be $2\pi$. So the limsup should be { $2\pi$}

P.S. When you say 0 for E' you mean the subset {0} i suppose, anyhow thanks a lot for your help. I was struggling about this one.

4. ## Re: An exercise about limsup/liminf of a subset sequence

You seem a little confused about what an infimum or supremum is. It is a greatest lower bound or least upper bound respectively. You had an example before where you were looking at a set of sets. A set of sets can be viewed as a poset where the partial order is containment. $A \le B$ if $A\subseteq B$. In that context, an infimum or supremum will be a set.

In the context of the current problem, you do not have a set of sets. You have a set of points of the real line. In this context, a greatest lower bound or a least upper bound will be a point, not a set. So, the first answer is not the set $\left(0,\dfrac{\pi}{n}\right)$. How do we know that $0$ is the infimum of the set? We know it is a lower bound since $0 \le x$ for all $x \in \left(0,\dfrac{\pi}{n}\right)\subseteq E_n \subseteq [0,2\pi]$. Given any $y > 0$, $0<\dfrac{y}{2} which shows that $y$ is not a lower bound. Hence, 0 is the greatest lower bound, and therefore the infimum for each $E_n$. So, $\inf E_1, \inf E_2, \ldots = 0,0,\ldots$ is an infinite sequence of zeros. The limit of an infinite sequence of zeros is zero. That is what is meant by the limit inferior ( $\liminf$). So, when I say 0 for E', I mean 0, not {0}.

Similarly for the supremum.

5. ## Re: An exercise about limsup/liminf of a subset sequence

Looking at my book of Mathematical Analysis III i found this definition: $E' = \liminf_{k\rightarrow +\infty}\; E_k = \bigcup_{k=1}^{+\infty} \bigcap_{n=k}^{+\infty} E_n$ so is a set, not a value.
I thought you was talking about $\bigcap_{n=k}^{+\infty} E_n$ when you said infimum in this case.

6. ## Re: An exercise about limsup/liminf of a subset sequence

Anyhow if i expressed myself bad what the exercise asks is to calculate E', E'' defined as $E' = \liminf_{k\rightarrow +\infty}\; E_k = \bigcup_{k=1}^{+\infty} \bigcap_{n=k}^{+\infty} E_n, E'' = \limsup_{k\rightarrow +\infty}\; E_k = \bigcap_{k=1}^{+\infty} \bigcup_{n=k}^{+\infty} E_n$ for $\{E_{n}\}_{n\in\Bbb N}$

P.S. i'm sorry for the double post but doesn't let me edit the previous one.

7. ## Re: An exercise about limsup/liminf of a subset sequence

Using that definition, you are probably correct about $E' = \emptyset, E'' = [0,2\pi] \setminus \{0,\pi,2\pi\}$.

Here is one possible approach to a formal proof:

Obviously, $\forall n\in\mathbb{N},\, \{0,\pi, 2\pi\} \cap E_n = \emptyset$, so those points cannot be in $E'$ or $E''$.

If $x \in \bigcap_{n\ge k}E_n$, then $x \in E_n$ for all $n\ge k$. Given $x \in (0,2\pi)$, suppose that for every $k\ge 1$, there exists $n\ge k$ with $x \notin E_n$. Then, obviously $x \notin \bigcap_{n\ge k}E_n$ for any $k \ge 1$. That obviously implies $x \notin \bigcup_{k\ge 1}\bigcap_{n\ge k} E_n$. So, if you can show that given any $x \in (0,2\pi)$ and any $k\ge 1$, there exists $n\ge k$ with $x \notin E_n$, then you have proven $E' \cap (0,2\pi) = \emptyset$.

For $E''$, if for all $k\ge 1$, there exists $n\ge k$ with $x \in E_n$, then $x \in E''$. (This condition is dual to $x \notin E'$ in a similar way to how intersections are dual to unions).

If you need any more help setting up the argument, I have some ideas.

8. ## Re: An exercise about limsup/liminf of a subset sequence

thanks a lot SlipEternal