Question about limsup's and subsequences

Question

I'm a little confused about the definition of $\limsup$ as it pertains to a particular problem. The statement I am trying to prove is as follows:

"Define a set $S$ by $S=\{s \in \mathbb{R} \mid \exists \text{ some subsequence } \{a_{n_k}\}_{k \in \mathbb{N}} \text{ such that } \lim_{k \to \infty}\{a_{n_k}\}=s\}$.
Show that $\limsup_{n \to \infty}a_n=\sup{S}=\max{S}$.

I know that one possible definition of $\limsup$ is $\limsup_{n \to \infty}a_n = \inf_{n \in \mathbb{N}}\sup_{k \geq n}a_k$, but I'm unsure if I should be using a different definition for this problem.

Any help is greatly appreciated.

Answer 1

To show that $\limsup _{n \to \infty} a_n = \sup S$ it would be better to use the definition of $\limsup$ as the supremum of limit points in the real sequence $(a_n)$. A basic result in Real Analysis tells us that if $(a_n)$ is a sequence of real numbers, a point $\alpha \in \mathbb R$ is a limit point of $(a_n)$ if and only if there exists a subsequence $(a_{n_k})$ of $(a_n)$ such that $\lim_{k \to \infty}a_{n_k} = \alpha$. Now if you define $$S' = \{\alpha \in \mathbb R : \alpha \text{ is a limit point of } (a_n)\},$$ then by the above we have that $S' = S$, so $$\limsup_{n \to \infty}a_n= \sup S' = \sup S.$$

The fact that $\sup S = \max S$ follows from noting that $\sup S \in S$, but you need to assume that $S \neq \varnothing$, otherwise $\max S$ is not well-defined; here for example it suffices to impose that $(a_n)$ is bounded, and by Bolzano-Weierstrass it follows that $S \neq \varnothing$. Assuming this last condition, note first that $S = S'$ is by definition a closed set, as it is the set of limit points of the sequence $(a_n)$; to show that $\sup S \in S $ it suffices to show that for any $\epsilon >0$ the $\epsilon$-ball around $\sup S$ contains infinitely many points from $(a_n)$, showing thus that $\sup S$ is a limit point of $(a_n)$, as required.

By definition of $\sup S$, for all $\epsilon >0$ there is $b \in S$ such that $\sup S - \epsilon < b \leq \sup S$; since $\sup S \leq \sup S + \epsilon$, $b \in S$ is in the $\epsilon$-ball around $\sup S$, and since $b$ is a limit point of $(a_n)$, there are infinitely many points of $(a_n)$ around any open neighbourhood of $b$; in particular, $(\sup S - \epsilon, \sup S + \epsilon)$ contains infinitely many points of $(a_n)$ for any $\epsilon > 0$, and we're done.

Answer 2

You don't mention it in your statement but it seems that you have a specific sequence $\{a_n\}$ that $S$ depends on. You can definitely get it with this definition but a better one for this is

$$limsup_{n\rightarrow\infty}a_n = lim_{n\rightarrow\infty}(sup_{m\geq n}x_m)$$

In this case the right hand side naturally gives you a subsequence that approaches what you want.