0
$\begingroup$

Suppose $a_n,n=1,2,...$ are real numbers and $A_n=(-\infty,a_n)$. Are both $\varlimsup A_n$ and $\varliminf A_n$ $\mathbb{R}$?

$$\varliminf A_n\equiv \bigcup^\infty_{n=1}\bigcap^\infty_{n=k}A_n=(-\infty,\varliminf a_n)=(-\infty,\sup_{n\geq1}(\inf_{k\geq n}a_k))$$

Since $\inf a_k$ is increasing, the supremum of $\inf a_k$ seems can only be the extended real number $+\infty$. Similarly, $$\varlimsup A_n\equiv\bigcap^\infty_{n=1}\bigcup^\infty_{n=k}A_n=(-\infty,\varlimsup a_n)=(-\infty,\inf_{n\geq1}(\sup_{k\geq n} a_k))$$ Since $\sup a_k=+\infty$ for any $k\geq n,n\in\mathbb{N}$, the inifinimum is also $+\infty$. Therefore, $\varliminf A_n=\varlimsup A_n=\mathbb{R}$.

I am wondering if I am correct? When I first looked at the question I am not sure how to deal with $(-\infty,\sup_{n\geq1}(\inf_{k\geq n}a_k))$ and $(-\infty,\sup_{n\geq1}(\inf_{k\geq n}a_k))$. I naturally think the left end should be some values depending on the sequence of $a_n$. My second intuition is that one of the limits should equal to null set and another one $\mathbb{R}$. But from my reasoning, it turns out the limit exists and is $\mathbb{R}$. I am not very comfortable with what I have gotten from the question. Have I made mistakes somewhere?

Besides, I also have a more general question to ask. I did not formally study college-level mathematics, and I was only taught calculus at the pre-college level. When I studied myself the concept of the "limit", it is usually in the $\epsilon-\delta$ fashion. The limit of the sequence is similar as compared to the limit of a function because if we want to establish a limit of a sequence exists, we only need to show "for any given $\epsilon, \exists M\in N, n>M\Rightarrow|x_n-C|<\epsilon$", where $C$ is simoly the limit of the sequence. Why do we take all the troubles to say something like if $\varliminf a_n=\varlimsup a_n$ then $\lim a_n$ exists?

$\endgroup$
5
  • 1
    $\begingroup$ What exactly are you assuming about the $a_k$? Please state that at the outset. E.g. what about $a_k = 0$ for all $k$? $\endgroup$ – Hans Engler Sep 24 '20 at 16:58
  • $\begingroup$ @HansEngler isn't $a_k$ any subsequence of $a_n$? $\endgroup$ – JoZ Sep 24 '20 at 17:00
  • $\begingroup$ No it's the same sequence with a different variable name for the index. $\endgroup$ – Hans Engler Sep 24 '20 at 17:03
  • $\begingroup$ @HansEngler $a_n$ are real numbers, that's all the information I have got. I assume it is simply an infinite sequence with real numbers... $\endgroup$ – JoZ Sep 24 '20 at 17:05
  • 1
    $\begingroup$ Work out the case where $a_n = 0$ for all $n$. Also work out the cases where $a_n = -n$ and where $a_n = n$ for all $n$. This will shed light on the question. $\endgroup$ – Hans Engler Sep 24 '20 at 17:08
1
$\begingroup$

The following arguments can be made rigorous but I think it is important to understand them in a colloquial way:

  • $\liminf_nA_n$ is the set of all $x$ (in the OP setting $x\in\mathbb{R}$) that belong to all but finitely many $A_n$'s.
  • $\liminf_na_n=\sup_n\inf_{k\geq n}a_k$ is the infimum of all sub sequential limits (as extended real numbers) $A$ of $\{a_n\}$ (If there are no sub sequential limits, for example $a_n=(-1)^nn$ has no convergent subsequence in $\mathbb{R}$, but it does in the extended real numbers $\overline{\mathbb{R}}=\mathbb{R}\cup\{-\infty,\infty\}$; hence $\liminf_na_n=-\infty$). Moreover, $\liminf_na_n\in A$.

Let $A_n-(-\infty,a_n)$. In general $$(-\infty,\liminf_na_n)\subset\liminf_nA_n\subset(-\infty,\liminf_na_n]$$

  • If $x<\liminf_na_n$, then only finitely many (or none) $a_n$'s satisfy $a_n\leq x$ (otherwise there would be a subsequence $a_{n_k}$ converging (in the sense of extended real numbers) to a limit $a_*\leq x$ contradicting the definition of $\liminf_na_n$). Consequently $x\in(-\infty,a_n)$ for all but finitely many $a_n$'s. That is $$(-\infty,\liminf_na_n)\subset\liminf_nA_n$$
  • Conversely, if $x\in\liminf_nA_n$ , then $x<a_n$ for all but finitely many $a_n$'s. Hence $x\leq\liminf_na_n$ and so $$\liminf_nA_n\subset(-\infty,\liminf_na_n]$$
  • The following examples shows that $(-\infty,\liminf_na_n)=\liminf_nA_n$ may not hold: $A_n=(-\infty,\frac{1}{n})$. Clearly $\liminf_nA_n=(-\infty,0]\supsetneq(-\infty,\liminf_na_n)=(-\infty,0)$.

A similar conclusion holds for $\limsup_n$: $$(-\infty,\limsup_na_n)\subset\limsup_nA_n\subset(-\infty,\limsup_n]$$

Here

  • $\limsup_nA_n$ is the set of all $x$ that belong to infinitely many $A_n$'s.
  • $\limsup_na_n$ is the supremum of the set of sub sequential limits (as extended real numbers) $A$ of $\{a_n\}$.

The remaining details are left as exercise.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.