# liminf and limsup of sequence of numbers and sets

I am struggling with the following question.

Let $$\{a_n\}_{n=1}^{\infty}$$ be a real sequence and $$A_n = (-\infty, a_n](n\geq 1)$$. Prove the following.

1. $$\exists \lim_{n\to \infty}A_n \Rightarrow \exists \lim_{n\to \infty}a_n \in [-\infty,\infty]$$

2. Converse of 1 does not hold.

(Hint: 1. Prove that $$\liminf a_n < \limsup a_n \Rightarrow \liminf A_n \subsetneq \limsup A_n$$ , 2: Counterexample is $$a_n =(-1)^n/n.$$

My approach to 1. was as follows.

(1)$$\bigcap_{k\geq n}(-\infty, a_k] = (-\infty, \inf_{k\geq n}a_k]$$

if $$x\in \bigcap_{k\geq n}(-\infty, a_k]$$ then $$\forall k\geq n, x\leq a_k.$$ This means that $$x$$ is a lower bound of $$\{a_k\}_{k\geq n}$$ . By definition of inf, $$x\leq \inf_{k\geq n}a_k$$ Therefore $$x\in (-\infty, \inf_{k\geq n}a_k]$$.

if $$x\in (-\infty, \inf_{k\geq n}a_k]$$ then since $$\inf_{k\geq n}a_k \leq a_k (\forall k\geq n)$$,$$x\in (-\infty, a_k](\forall k\geq n)$$. Therefore $$x\in \bigcap_{k\geq n}(-\infty, a_k]$$.

This means that $$\bigcap_{k\geq n}(-\infty, a_n] = (-\infty, \inf_{k\geq n}a_k]$$.

(2)$$\bigcup_{n=1}^{\infty}(-\infty, \inf_{k\geq n}a_k] = (-\infty, \lim_{n\to \infty}\inf_{k\geq n}a_k]$$

As $$\{\inf_{k\geq n}a_k\}_n$$ is a monotone increase sequence and bounded by each element of $$\{a_n\}_n$$, $$\{\inf_{k\geq n}a_k\}_n$$ converges to its least upper bound. Therefore $$\inf_{k\geq n}a_k \leq \lim_{n\to \infty}\inf_{k\geq n}a_k (\forall n \in \mathbb{N}$$).

If $$x\in \bigcup_{n=1}^{\infty}(-\infty, \inf_{k\geq n}a_k]$$ then $$\exists n\in \mathbb{N}, x\in (-\infty, \inf_{k\geq n}a_k]$$. Since $$\inf_{k\geq n}a_k \leq \lim_{n\to \infty}\inf_{k\geq n}a_k (\forall n \in \mathbb{N}$$), $$x\in (-\infty, \lim_{n\to \infty}\inf_{k\geq n}a_k]$$.

If $$x\in (-\infty, \lim_{n\to \infty}\inf_{k\geq n}a_k]$$, as $$\lim_{n\to \infty}\inf_{k\geq n}a_k$$ is the least upper bound, $$\exists m\in \mathbb{N}, x\in (-\infty, \inf_{k\geq m}a_k]$$. therefore $$x\in \bigcup_{n=1}^{\infty}(-\infty, \inf_{k\geq n}a_k]$$.

This means that$$\bigcup_{n=1}^{\infty}(-\infty, \inf_{k\geq n}a_k] = (-\infty, \lim_{n\to \infty}\inf_{k\geq n}a_k]$$.

From (1) and (2), $$\liminf_{n\to \infty}A_n = (-\infty, \liminf_{n\to \infty}a_n]$$. By using the same method, we can prove that $$\limsup_{n\to \infty}A_n = (-\infty, \limsup_{n\to \infty}a_n]$$. If $$\liminf a_n < \limsup a_n$$ then obviously $$\liminf A_n \subsetneq \limsup A_n$$.

I thought this was correct, but if $$\liminf_{n\to \infty}A_n = (-\infty, \liminf_{n\to \infty}a_n]$$ hold then the converse of 1. should hold as $$\exists \lim_{n\to \infty} a_n \Leftrightarrow \liminf a_n = \limsup a_n$$. So I thought the proof above has a flaw, but I couldn't find one. Also, I could not figure out how to use the hint given in 2.

Since $$\lim\limits_n A_n$$ exists, we have that:

$$\liminf A_n=\limsup A_n\Leftrightarrow\bigcup_{n=1}^\infty\bigcap_{k=n}^∞A_k=\bigcap_{n=1}^\infty\bigcup_{k=n}^∞A_k.$$ Note that: $$\bigcap_{k=n}^∞A_k=\bigcap_{k=n}^∞(-\infty,a_k]=\left(-\infty,\inf\limits_{k\geq n}a_k\right],$$ as you said. However, also not that, if $$b_n$$ is an increasing sequence with $$b_n\to b$$ and $$b_n, then it holds that:

$$\bigcup_{n=1}^\infty(-\infty,b_n]=(-\infty,b),$$ since $$b\not\in(-\infty,b_n)$$ for every $$n=1,2,\ldots$$. So, what actually holds is that:

$$\liminf A_n=I(-\infty,\liminf a_n),$$ where by $$I(a,b)$$ we denote any interval with endpoints $$a,b\in\overline{\mathbb{R}}$$, $$a. Similarly,

$$\limsup A_n=I(-\infty,\limsup a_n).$$

From this, both results are almost immediate.

1. If $$\limsup A_n=\liminf A_n$$ then $$I(-\infty,\limsup a_n)=I(-\infty,\liminf a_n),$$ so $$\liminf a_n=\limsup a_n$$ and, hence $$a_n$$ converges.
2. For $$a_n=(-1)^n/n$$, we have $$\limsup A_n=\ldots=(-\infty,0]$$ and $$\liminf A_n=(-\infty,0).$$ (both are left for you to verify, since, knowing the results, it is relatively easy to prove them).