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

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?

• What exactly are you assuming about the $a_k$? Please state that at the outset. E.g. what about $a_k = 0$ for all $k$? – Hans Engler Sep 24 '20 at 16:58
• @HansEngler isn't $a_k$ any subsequence of $a_n$? – JoZ Sep 24 '20 at 17:00
• No it's the same sequence with a different variable name for the index. – Hans Engler Sep 24 '20 at 17:03
• @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... – JoZ Sep 24 '20 at 17:05
• 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. – 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 $$x for all but finitely many $$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.