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If $\{a_n\}$ is a sequence of real numbers, prove that $\limsup_{n \to \infty}{a_n} = \lim_{N \to \infty} \sup{\{a_n: n \ge N\}}$.

Please, verify this proof:

Suppose $\alpha = \limsup_{n \to \infty}{a_n} $. It means that $\alpha = \sup{\mathcal{L}(\{a_n\})}$, where ${\mathcal{L}(\{a_n\})}$ is a set of subsequential limits of $a_n$.

Let A = ${\mathcal{L}(\{a_n\})}$. Let $B = {\mathcal{L}(\{a_n: n \ge N\})}$.

$\gamma = \lim{a_{n_k}}$ means: $\forall \epsilon > 0$ $ \exists N \in \mathbb{N}$, s. t. if $n_k \ge M$, then $|a_{n_k} - \gamma| < \epsilon$.

Consider $\{a_{n_k}: n_k > L\}$. It is clear that, by the definition of a limit, if $\gamma = \lim{a_{n_k}}$, then $\gamma = \lim\{{a_{n_k}}: n_k > L\}$.

Thus we proved that $A \subset B$. Similarly, using the definition of the limit, we prove that $B \subset A$. Therefore $A = B$ and ${\mathcal{L}(\{a_n\})} = {\mathcal{L}(\{a_n: n \ge N\})}$. It follows that $\sup\{{\mathcal{L}(\{a_n\})}\} = \sup{\{\mathcal{L}(\{a_n: n \ge N\})\}}$.

By the definition of limsup the last equation is equivalent to $\limsup\limits_{n \to \infty}{a_n} = \limsup\limits_{N \to \infty}{\{a_n: n \ge N\}}$, as desired.

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    $\begingroup$ Do you mean $\sup\{a_n: n \ge N\}$? $\endgroup$ – GFauxPas Nov 20 '16 at 2:11
  • $\begingroup$ Yes, exactly. I have just edited the post. $\endgroup$ – User 1234 Nov 20 '16 at 2:15
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It's not clear (proven) that $\limsup_{N\to\infty} \{a_n : n \geq N\}$ in your conclusion is the same as $\lim_{N \to\infty} \sup\{a_n : n \geq N\}$, which is what you seek.

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    $\begingroup$ Do you mean that I need to prove $\limsup\limits_{n \to \infty}\{a_n: n \ge N\} = \lim_{n \to \infty} \sup{\{a_n: n \ge N\}}$ as an additional step or the above approach to the proof is wrong (because in this case proof seems to grow too big and complicated)? $\endgroup$ – User 1234 Nov 20 '16 at 3:21

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