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?

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    $\begingroup$ 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
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    $\begingroup$ 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

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 $\limsup_n$: $$(-\infty,\limsup_na_n)\subset\limsup_nA_n\subset(-\infty,\limsup_n]$$


  • $\limsup_nA_n$ 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.


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