# Prove that if x > limsup s_n, then x is not the limit of any subsequence

## Homework Statement

Directly from the definition, for a sequence $$(s_n)_{n \in \mathbb{N}} \subseteq \mathbb{R}$$ prove that if $$x > \limsup s_n$$, then $$x$$ is not the limit of any subsequence of $$(s_n)$$. (i.e. Do not use the fact that $$\limsup s_n$$ is the supremum of the set of subsequential limits.)

## Homework Equations

I have been told by my instructor that my proof will fail due to problems with inequalities --- but I fail to see where it would fail; i.e. are there any errors where $$>$$ should be $$\ge$$ or vice-versa?

## The Attempt at a Solution

Let's first show that for any sequence $$(s_n)$$ and any number $$M$$, if $$s_n > M$$ for all $$n \in \mathbb{N}$$, then $$\limsup s_n \geq M$$. Since $$s_n > M$$ for all $$n \in \mathbb{N}$$, then it follows that $$s_n > M$$ for all $$n > N$$ for some $$N \in \mathbb{N}$$. Now, let's denote $$v_N = \sup\{s_n : n > N\}$$. Since $$v_N$$ is the least upper bound for $$s_n$$ when $$n > N$$, then it follows that we have $$v_N \geq s_n > M$$ for $$n > N$$. And note that we have the property $$v_N \geq v_{N+1} \geq \ldots$$ and so $$(v_N)$$ is monotonically decreasing and since $$(s_n)$$ is bounded, by the monotone convergence theorem, $$\lim_{N \to \infty}v_N$$ exists and is real. Moreover, this is precisely the definition of the limit superior and hence, we have that $$\lim_{N \to \infty}v_N = \lim_{N \to \infty}\sup\{s_n : n > N\} = \limsup s_n$$. Thus, we have $$\limsup s_n > M$$ as desired.

For contradiction, suppose that $$x$$ is the limit of some subsequence $$(s_{n_k})$$ of the sequence $$(s_n)$$. Let's first denote $$v = \limsup s_n$$. Note that if $$\lim_{k \to \infty}s_{n_k} = x > v$$, then there exists some $$N$$ such that for all $$k > N$$, we have $$s_{n_k} > v$$. Now, consider $$s_{n_k} > v$$ for all $$k > N$$. Then by the result above, and since removing finite number of terms $$1 \leq k \leq N$$ does not affect convergence results, it implies $$\limsup s_{n_k} > v = \limsup s_n$$ --- contradiction, since $$\{s_{n_k}: k > N\} \subseteq \{s_n: n > N\}[tex], it implies [tex] \sup\{s_{n_k} : k > N\} \leq \sup\{s_n : n > N\}[tex] and thus$$ \lim_{N \to \infty}\sup\{s_{n_k} : k > N\} \leq \lim_{N \to \infty}\sup\{s_n : n > N\}[/tex].