Simple exercise about Limsup/Liminf of sets

elementary-set-theorylimitslimsup-and-liminfreal-analysissolution-verification

I tried the following exercise regarding the limsup/liminf of sets. In the following let $(A_n)_{n \in \mathbb{N}}$ be a sequence of events. Calculate the set-theoretic $\limsup_{n \to \infty}$ and $\liminf_{n \to \infty}$ of the following sequences of sets

  1. Let $A_n = \{1\}$ if $n$ is even and $A_n = \{0\}$ if $n$ is odd.
  2. $$ A_n =
    \begin{cases}
    (0,3+1/n) \quad & \text{if } n \text{ is odd} \\
    (-1-1/n, 2] \quad & \text{if } n \text{ is even} \\
    \end{cases} $$
  3. $A_n = [\sin(n) – 1, \sin(n) + 1]$

Here is what I have tried.

  1. $\limsup_{n \to \infty} A_n = \{0,1\}$, since for every $n \in \mathbb{N}$ one has $\cup_{k > n} A_n = \{0,1\}$, moreover we have $\liminf_{n \to \infty} A_n = \emptyset$, since $A_{n} \cap A_{n+1} = \emptyset$.
  2. $\limsup_{n \to \infty} A_n = (-1,3]$ since $A_n \cup A_{n+1} = (-1-1/n, 3+1/n] =(0,3+1/n) \cup (-1-1/n,2]$, moreover $\liminf_{n \to \infty} A_n = (0,2]$, since this intervalls is the only one that remains under intersection up to finitely many $n$.
  3. Here I am not 100% sure. One can prove the following fact $\{\sin(n) : n \in \mathbb N\}$ is dense in $[-1,1]$. Hence for every $y \in [-1,1]$ one can find a sequence of rational points $(n_k)_{k \in \mathbf N}$ s.t.
    $$
    \sin(n_k) = y
    $$
    hence one can find a sequence $(n_k)$ and $(n^{\prime}_{k})$ such that $y \in \{-1,1\}$ is approximated, hence
    $$
    \limsup_{n \to \infty} A_n = [-2,2]
    $$
    since all the the limit of both sequences is contained in $\cup_{k>n} a_k$ for all $n \in \mathbb N$.
    For the second part I believe $\liminf_{n \to \infty} A_n = [-1,1]$, since for all other points one can find a natural number such that the intervall in not contain in $\textbf{all}$ next intervalls.

Sorry for the messy writeup, is this correctly solved, especially last part?

Thanks in advance.

Best Answer

Let us do the second case: \begin{align*} \limsup_{n\to\infty}A_{n} = \bigcap_{n\geq 1}\bigcup_{k\geq n}A_{k} & = \bigcap_{n\geq 1}\bigcup_{k\geq n}(A_{2k}\cup A_{2k-1})\\\\ & = \bigcap_{n\geq 1}\bigcup_{k\geq n}\left[\left(-1 - \frac{1}{2k},2\right]\cup\left(0, 3 + \frac{1}{2k-1}\right)\right]\\\\ & = \bigcap_{n\geq 1}\bigcup_{k\geq n}\left(-1 - \frac{1}{2k}, 3 + \frac{1}{2k-1}\right)\\\\ & = \bigcap_{n\geq 1}\left(-1 - \frac{1}{2n},3 + \frac{1}{2n - 1}\right)\\\\ & = [-1,3] \end{align*}

Hopefully this helps!

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