Prove $\,\displaystyle\limsup_{n\to\infty} s_n=m\,$ if $\,m=\sup\left\{s_n \mid h\ge 1\right\}\lt \infty\,$ and sup not attained

Question

I'm trying to prove the following for my analysis course:

Let $\;m=\sup\left\{s_n \mid n\ge 1\right\}\lt \infty\;$ and suppose that the supremum is not attained. Prove that $\,\limsup\limits _{n\to\infty} s_n=m$.

To me it seems as if I have to prove that the lim sup is equal to the sup, but I don't have a clue where to start.

Can anybody guide me in the right direction, please?

Answer 1

Let $S\subseteq\mathbb{R}$ be a set that is bounded above.

Then evidently for any $s\in\mathbb{R}$ the set $S-\left\{ s\right\} $ is bounded above with $\sup\left(S-\left\{ s\right\} \right)\leq\sup S$.

If $s<\sup S$ then some $t\in S$ exists with $s<t$ so that $\sup\left(S-\left\{ s\right\} \right)$ serves as upperbound for $S$ and consequently we also have $\sup\left(S-\left\{ s\right\} \right)\geq\sup S$.

So acually: $$s<\sup S\implies\sup\left(S-\left\{ s\right\} \right)=\sup S$$

This is the ground of $a_n=a_{n+1}$ below.

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Define $a_{n}:=\sup\left\{ s_{k}\mid k\geq n\right\} $.

Then by definition $a_{1}=\sup\left\{ s_{k}\mid k\geq1\right\} $ and $\lim_{n\to\infty}a_{n}=\limsup\ s_{n}$.

Also it is evident that $a_{n}\geq a_{n+1}$ for each $n$.

However the fact that the supremum is not attained takes care of $a_{n}=a_{n+1}$ for each $n$.

Then $a_1=\lim_{n\to\infty}a_n$.

Answer 2

Consider the intervals $ [m - \epsilon, m) $, the claim is that no matter how small $ \epsilon $ is, there are always infinite number of points of $ s_n $ in it.

Now suppose that is not the case, for some $ \epsilon $ there exist only a finite number of points there, now it is obvious that within this finite number of points, there exist one that is closest to $ m $, now that distance must not be 0 because the problem tell us the supremum is not attained, but then isn't true that we just found the maximum? $ m $ is not the supremum anymore.

Now I think you can take it from here.

Answer 3

Let $\alpha_n$ be the supremum of the set $\{s_k : k \ge n\}$. By definition we have $\limsup s_n = \lim \alpha_n$. Thus, we want to proof that $\lim \alpha_n = \alpha_1$. Note that $\alpha_n$ is decreasing and, consequently, $\lim \alpha_n \le \alpha_k \le \alpha_1$ for every $k$.

If we had $\lim \alpha_n = \alpha_1$, then we would have $\alpha_1 = \lim \alpha_n \le \alpha_k \le \alpha_1$ and, thus, $\alpha_n = \alpha_1$ for every $n \in \mathbb{N}$. Consequently, we should try to prove that the sequence $\alpha_n$ is constantly $\alpha_1$ because this fact is equivalent to the statement.

In order to prove this new assertion you must use the characterization of the supremum.