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If $\lim\sup$ of $\{s_n\}$ equals $M$, prove that the $\lim\sup$ of any subsequence $\{s_{n_k}\}$ is $\le M$

PROOF: Given that $\{s_n\}$ is a sequence of real numbers such that $\displaystyle \limsup_{n\to\infty}s_n = M$, then by definition we know that $\{s_n\}$ is bounded above and $\displaystyle \lim_{n\to\infty} M_n = M$, where $M_n = \sup\{s_n, s_{n+1}, s_{n+2}, \ldots\}$, which implies that any subsequence $\{s_{n_k}\}$ is also bounded above.

So, $$\limsup_{n\to\infty}s_n = \lim_{n\to\infty}M_n = M$$

We know that $\{M_n\}$ is a nonincreasing sequence (i.e., $M_1 \ge M_2 \ge M_3 \ge \cdots$), and we also know that $n_k \ge k$ for all $k\in \mathbb N$. So $M_n \ge M_{n_k}$ for all $k\in\mathbb N$. Hence, observe that $$M = \lim_{n\to\infty}M_n = \limsup_{n\to\infty}s_n \underset{(*)}{\ge} \lim_{k\to\infty}M_{n_k} = \limsup_{k\to\infty}s_{n_k}.$$

So the $\limsup$ of any subsequence of $\{s_n\}$ is $\le M$.

Is this proof valid? I feel like I make too large of a jump at $(*)$. I should also note that trivially if $\limsup_{k\to\infty}s_{n_k} = -\infty$, then $\limsup_{k\to\infty}s_{n_k} = -\infty < M$.

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You do skip something in step ($*$); I would modify it as follows.

Define $M_n$ as you have and define $N_{n_j}=\sup\{s_{n_j},s_{n_{j+1}},\ldots\}$. Then, $M_{n_j}\geq N_{n_j}$ because the former entails taking supremum over more elements. Moreover, $M_n\to\limsup s_n$ implies $M_{n_j}\to\limsup s_n$. It follows that $$ M\geq\limsup s_n=\lim M_{n_j}\geq\lim N_{n_j}=\limsup s_{n_j}. $$

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Your $M_{n_k}$ is not correctly defined (it shouldn't be a subsequence of $M_n$ if the limit should be $\limsup_k s_{n_k}$). A possible path: For $j\geq k\geq 1$ $$ s_{n_j} \leq M_{n_k} $$ so for every $k$ $$ s^*=\limsup_j s_{n_j} \leq M_{n_k}$$ and then $$ s^* \leq \lim_k M_{n_k} =\lim_n M_n=s$$

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