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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.

Thank you for reading.

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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 $b_n<b$, 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 $a,b\in\overline{\mathbb{R}}$, $a<b$. 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).
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