# Calculate $\liminf$ and $\limsup$ for odd and even interval $A_n$

$$A_n = (-n^{-1},1+n^{-1})$$ if n is odd, and $$A_n = (1-n^{-1},2+n^{-1})$$ if n is even.

Calculate $$\overline{\lim}_nA_n$$ and $$\underline{\lim}_nA_n$$ (aka $$\limsup$$ and $$\liminf$$ notation wise).

I'm given that $$\overline{\lim}_{n\to\infty}A_n=\bigcap_{n\ge 1}\bigcup_{k\ge n}A_k$$, and $$\underline{\lim}_{n\to\infty}A_n=\bigcup_{n\ge 1}\bigcap_{k\ge n}A_k$$, and I have worked out:

\begin{aligned} \overline{lim}_nA_n=&((-1,2)\cup(-.33,1.33)\cup(-.2,1.2)\cup\cdots\cup(.5,2.5)\cup(.75,2.25)\cup(.83,2.16)\cup\cdots)\\ &\cap((-.33,1.33)\cup(-.2,1.2)\cup\cdots\cup(.75,2.25)\cup(.83,2.16)\cup\cdots))\cap\cdots\\ =&((-1,2)\cup(.5,2.5))\cap((-.33,1.33)\cup(.75,2.25))\cap\cdots\\ =&(-1,2.5)\cap(-.33,2.25)\cap(-.2,2.16)\cap\cdots\\ =&(0,2) \end{aligned} \begin{aligned} \underline{lim}_nA_n=&((-1,2)\cap(-.33,1.33)\cap(-.2,1.2)\cap\cdots\cap(.5,2.5)\cap(.75,2.25)\cap(.83,2.16)\cap\cdots)\\ &\cup((-.33,1.33)\cap(-.2,1.2)\cap\cdots\cap(.75,2.25)\cap(.83,2.16)\cap\cdots)\cup\cdots\\ =&((0,1)\cap(1,2))\cup((0,1)\cap(1,2))\cup\cdots\\ =&\emptyset\cup\emptyset\cup\cdots\\ =&\emptyset \end{aligned}

Could someone let me know whether what I've done is correct?

My logic is that $$\overline{lim}_nA_n=(0,2)$$ is the set of numbers which belong to infinitely many $$A_n$$, and that $$\underline{lim}_nA_n=\emptyset$$ is the set of real numbers which belong to all $$A_n$$ for sufficiently large $$n$$, and since the limits of the odd and even $$A_n$$ are $$(0,1)$$ and $$(1,2)$$, which have $$\emptyset$$ as their intersection, there are no numbers that can be in both all the odd and even $$A_n$$'s.

• $1 \in A_n$ for all $n\in \mathbb N$ and the countable intersection of open sets can be closed (a typical example would be $\cap_{n\in \mathbb N} (-1/n, 1+1/n) = [0,1]$ Oct 11, 2021 at 12:21
• Well certainly the steps that look like "${}={}$ an endless sequence of unreadable decimal sequences that are terminated at the hundredths place" are incorrect because the endpoints of the actual intervals are not multiples of $0.01$. Readability might improve if you were to explicitly find $\bigcup_{k \geq n} A_k$ and $\bigcap_{k \geq n} A_k$ and not replace the intervals with low resolution approximations. Oct 11, 2021 at 12:40
• @postmortes so then $\overline{lim}_nA_n=[0,2]$ and $\underline{lim}_nA_n=[1,1]$? Oct 11, 2021 at 12:56
• @EricTowers then $\bigcup_{k\ge n}A_k=A_{2n-1}\cup A_{2n}$ and $\bigcap_{k\ge n}A_k=[1,1]$? Oct 11, 2021 at 12:57

Start with the intervals themselves to understand what we're looking at. When $$k=2n+1$$ for $$n\in \mathbb N$$ we see that $$A_k$$ is always slightly larger than the interval $$[0,1]$$ Note that this a closed interval, even though $$A_k$$ is always open, because $$0\in A_k$$ (you should make sure you understand this; consider the limit of $$-1/k$$ as $$k\rightarrow \infty$$). So any intersection of the $$A_k$$ when $$k$$ is odd will contain $$[0,1]$$.
Now let $$k=2n$$ for $$n\in \mathbb N$$. By the same reasoning we see that $$[1,2] \in A_k$$ for all even $$k$$ and so any intersection of $$A_{2n}$$ must contain $$[0,2]$$.
Now we're ready to work out the $$\limsup$$ and $$\liminf$$: we have $$\limsup_n A_n = \overline{\lim}_n A_n = \bigcap_{n\geq 1} \bigcup_{k \geq n} A_k$$. The union of any $$A_k$$ must contain $$[0,1] \cup [1,2]$$ because there will always be an even and an odd $$k$$ in the union $$\bigcup_{k \geq n} A_k$$ (if there weren't, and you could prove it, you could tell us if countable-infinity were even or odd!). Thus the intersection of all of these unions will contain $$[0,1]\cup[1,2] = [0,2]$$. Finally we check that no other point can sneak in there: if it did it has the form $$-1/n$$ or $$2+1/n$$ as they are the only points that can "stick out" from $$[0,2]$$. However, any you pick will eventually be excluded when $$k\geq n$$, so we're good: $$\limsup _n A_n = [0,2]$$.
For the $$\liminf_n A_n = \bigcup_{n \geq 1} \bigcap_{k \geq n} A_k$$ we note that $$A_k \cap A_{k+1} = \left(1-(k+1)^{-1}, 1+k^{-1} \right)$$. Since we take a countable intersection over $$k \geq n$$, this is the same as considering $$\lim_{k \rightarrow \infty} \left(1-(k+1)^{-1}, 1+k^{-1} \right) = \{1\}$$ Taking unions of $$\{1\}$$ still only gives us $$\{1\}$$ however. This time it's easier to see that there can be no other points in the $$\liminf$$ because the intersection reduces down to a single point.