Limsup’s and Liminf’s (Again)

A while back, we discussed what the limit superior and limit inferior are. (See here) In that article, we defined these two concepts as follows,

Definition (Limit Superior): Let $(a_n)_{n\in \N}$ be a bounded sequence of real numbers and define the sequence $(\alpha_n)_{n\in \N}$ by $\alpha_n := \sup{\{a_k\;:\;k\geq n\}}.$ If the limit exists, define

$\limsup_{n\rightarrow \infty}{(a_n)}:=\lim_{n\rightarrow \infty}{(\alpha_n)}.$

Similarly, define the limit inferior:

Definition (Limit Inferior): Let $(a_n)_{n\in \N}$ be a bounded sequence and define the sequence $(\beta_n)_{n\in \N}$ by $\beta_n := \inf{\{a_k\;:\;k\geq n\}}.$ If the limit exists, define

$\liminf_{n\rightarrow \infty}{(a_n)}:=\lim_{n\rightarrow \infty}{(\beta_n)}.$

***Note we may abbreviate these by $\limsup{(a_n)}$ and $\liminf{(a_n)}.$ ***

In that article, we built some intuition for what these strange-looking limits are. In particular, we saw that $(\alpha_n)_n$ form a monotone decreasing sequence and $(\beta_n)_n$ form a monotone increasing sequence. This means, by the monotone convergence theorem (in the extended reals, that is, the real numbers with $\infty$ and $-\infty$ unioned in) the sequences $(\alpha_n)_n$ and $(\beta_n)_n$ always converge. Better yet, if $(a_n)_{n}$ is a bounded seqeuence then so is $(\alpha_n)_n$ and $(\beta_n)_n$ and therefore $(\alpha_n)_n$ and $(\beta_n)_n$ converge in $\R.$ In particular, $\limsup{(a_n)}$ and $\liminf{(a_n)}$ always exist.

From there, we saw that we can find a subsequence $(a_{n_k})_{k}$ of our original sequence that converges to $\limsup{(a_n)}$ and $\liminf{(a_n)}.$ This means that $\limsup{(a_n)}$ and $\liminf{(a_n)}$ are subsequential limits of $(a_n)_{n}.$ This is the main content of the Bolzano-Weierstrass Theorem (also covered here).

As it turns out, there are many books and sources out there (most famously Baby Rudin) that define $\limsup{(a_n)}$ and $\liminf{(a_n)}$ in terms of possible subsequential limits of $(a_n)_{n}.$ In the article, we would like to connect these two definitions so that you can choose whichever you like best when working out problems.

The Other Definition

Definition 2 (Limsup and Liminf): Let $(a_n)_{n}$ be a sequence of real numbers and let $L$ be the set of subsequential limits of $(a_n)_{n}.$ Then, we define the limit superior and limit inferior as

$\limsup_{n\rightarrow \infty}{(a_n)}:=\sup{(L)}.$

$\liminf_{n\rightarrow \infty}{(a_n)}:=\inf{(L)}.$

Our Goal For Today

We aim to show that the two definitions are the same. In particular, we want to show that $\lim_{n\rightarrow \infty}{(\alpha_n)} = \sup{(L)}$ and $\lim_{n\rightarrow \infty}{(\beta_n)} = \inf{(L)}.$ We will show that $\lim_{n\rightarrow \infty}{(\alpha_n)} = \sup{(L)}$ and leave the other as an excercise. Note, that $\lim_{n\rightarrow \infty}{(\beta_n)} = \inf{(L)}$ follows from $\lim_{n\rightarrow \infty}{(\alpha_n)} = \sup{(L)}$ by considering the negative of a sequence.

Theorem: Let $(a_n)_{n}$ be a sequence of real numbers, $(\alpha_n)_{n\in \N}$ be defined as in Defininition (Limit Superior), and let $L$ be defined as in Definition 2 (Limsup and Liminf). Then, the two definitions of $\limsup{(a_n)}$ are equivalent. I.e.,

$\limsup{(a_n)}=\lim_{n\rightarrow \infty}{(\alpha_n)} = \sup{(L)}.$

Scratch Work: Let’s get some notation out of the way. Let $\lim_{n\rightarrow \infty}{(\alpha_n)} = A$ and $\sup{(L)} =B.$ Thus, our goal is to show that $A = B.$ However, using Theorem (LimSup Subsequence and LimInf Subsequence) from our original article (I’m just going to keep referencing this article until you go and look at it!) we know that there is a subsequence $(a_{n_k})_{k}$ that converges to $A.$ And, since $B$ is the supremum of the subsequential limits we already have $A\leq B.$ This is great, since a majority of the work for the proof of this theorem is found in that proof. Thus, all we have to do is demonstrate that $A\geq B.$

The key idea is to try to show that any arbitrary subsequential limit must be less than $A.$ This would then imply that the supremum of all such sequential limits, $B$ , is less than or equal to $A.$ The way we do it is to use the fact that $\alpha_n\rightarrow A$ in order to bound the $\alpha_n$ and $\alpha_n := \sup{\{a_k\;:\;k\geq n\}}$ to bound the $a_k$ for $k\geq N.$ Have some fun and see if you can work out just a little bit of the full proof. All the details are provided below when you’re ready!

Proof: (Click in the Discovery)

We focus on the case where $(a_n)_{n}$ is bounded. This is because if $(a_n)_{n}$ is unbounded, then the statement is “clear” (after some mental effort).¹ In fact, this might be a fun challenge problem for you to try out!

GOAL: $B\leq A.$

To this end, let $\varepsilon>0.$ Recall that we noted that $(\alpha_n)_{n\in \N}$ is a monotone decreasing sequence. In particular, since $\alpha_n\rightarrow A,$ there is some $N\in \N$ such that

\alpha_N\leq A+\varepsilon.

Note, since $\alpha_n := \sup{\{a_k\;:\;k\geq n\}},$ it follows that $a_k\leq A+\varepsilon$ for all $k\geq N.$

Now consider some convergent subsequence $(a_{k_{\ell}})_{{\ell}}$ of $(a_k)_k.$ It follows that $a_{k_{\ell}} \leq A+\varepsilon$ for all ${\ell}\geq N.$ (This is because $k_{\ell}\geq \ell$ for all strictly increasing sequences of integers $k_{\ell}.$ ) Hence, the subsequential limit $a_{k_{\ell}} \rightarrow a,$ is such that $a\leq A+\varepsilon.$ Since $(a_{k_{\ell}})_{{\ell}}$ is an arbitrary convergent subsequence of $(a_k)_{k},$ it follows that $B\leq A + \varepsilon.$ Moreover, since $\varepsilon>0$ was arbitrary, it follows $B\leq A.$ Thus, concluding our proof.

$\square$

Great working with you—until next time!

Today’s article was short and sweet. This is simply because in order to fully understand this article, you must have also read this article. That article is somewhat long, so I tried to keep this one shorter to compensate. I hope that you found this helpful. I know that limsups can be pretty challenging when you first come across them, so the more ways to think about them, the better! With that being said, remember that math(s) can be challenging, so don’t let that make you loose the joy and fun that math(s) can be. The challenge is part of the fun (hopefully, but not always!), so in the words of Meet the Robinsons, Keep Moving Forward.

See you next time!

Be Kind. Be Curious. Be Compassionate. Be Creative.

And Have Fun!

Footnote(s):

The words clear and obvious are perhaps the most dangerous words in all of mathematics. They have unraveled proofs, caused needless stress for students learning new material by making them think they are not good enough, and caused unnecessary aggravation that could have been mitigated if only the author(s) had put a little more effort into explaining what they meant by “clear” or “obvious”. With that said, I hope I mitigated at least some stress by calling out that what I said was clear, though it may not be, in fact, clear! I chose to leave out the full description in this case so I could give you some things to think about and work on, since I am not assigning homework! ↩︎

A Kick in the Discovery