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.
