An exercise about limsup/liminf of a subset sequence

Let $\displaystyle E_{n} = $\{$x $\in$ $\[0,2$\pi$]$ : (sin(nx))/n > 0$\}$$ with n natural number.

Calculate: E', E'' where E' = liminf E_{n}, E'' = limsup E_{n} for n that goes to $\displaystyle $\infty$$

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

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

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

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

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

the first answer should be $\displaystyle (0,\dfrac{\pi}{n}\right $)$$ and the other subsets depending on the periodicity so $\displaystyle (\dfrac{2k\pi}{n}\right,\dfrac{(2k+1)\pi}{n}\right $)$$ 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 $\displaystyle 2$\pi$$. So the limsup should be {$\displaystyle 2$\pi$$}

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.

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. $\displaystyle A \le B$ if $\displaystyle A\subseteq B$. 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 $\displaystyle \left(0,\dfrac{\pi}{n}\right)$. How do we know that $\displaystyle 0$ is the infimum of the set? We know it is a lower bound since $\displaystyle 0 \le x$ for all $\displaystyle x \in \left(0,\dfrac{\pi}{n}\right)\subseteq E_n \subseteq [0,2\pi]$. Given any $\displaystyle y > 0$, $\displaystyle 0<\dfrac{y}{2}<y$ which shows that $\displaystyle y$ is not a lower bound. Hence, 0 is the greatest lower bound, and therefore the infimum for each $\displaystyle E_n$. So, $\displaystyle \inf E_1, \inf E_2, \ldots = 0,0,\ldots$ 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 ($\displaystyle \liminf$). So, when I say 0 for E', I mean 0, not {0}.

Similarly for the supremum.

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

Looking at my book of Mathematical Analysis III i found this definition: $\displaystyle E' = \liminf_{k\rightarrow +\infty}\; E_k = \bigcup_{k=1}^{+\infty} \bigcap_{n=k}^{+\infty} E_n$ so is a set, not a value.

I thought you was talking about $\displaystyle \bigcap_{n=k}^{+\infty} E_n$ when you said infimum in this case.

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 $\displaystyle E' = \liminf_{k\rightarrow +\infty}\; E_k = \bigcup_{k=1}^{+\infty} \bigcap_{n=k}^{+\infty} E_n, E'' = \limsup_{k\rightarrow +\infty}\; E_k = \bigcap_{k=1}^{+\infty} \bigcup_{n=k}^{+\infty} E_n$ for $\displaystyle $\{E_{n}\}_{n\in\Bbb N}$$

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

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

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

Here is one possible approach to a formal proof:

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

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

For $\displaystyle E''$, if for all $\displaystyle k\ge 1$, there exists $\displaystyle n\ge k$ with $\displaystyle x \in E_n$, then $\displaystyle x \in E''$. (This condition is dual to $\displaystyle x \notin E'$ 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.

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