TITLE: Showing that two definitions of $\limsup$ are equivalent
QUESTION [24 upvotes]: In Rudin, $\limsup$ is defined as follows:

Let $S$ be the set of subsequential limits of $\{s_n\}$. Then
  $$\limsup s_n = \sup S. \tag{1}$$

However, our real analysis instructor defined $\limsup$ in a different manner:

$$\limsup s_n = \lim_{n \to \infty} \sup_{m \ge n} s_m. \tag{2}$$

I am having trouble understanding how these two definitions are equivalent. It would be very helpful to me if somebody could provide a proof with some explanation.
My thoughts on the problem:
I have noticed that the usual trend with these sort of proofs is to prove the upper bound $(1) \le (2)$ and then the lower bound $(1) \ge (2)$ to get the desired conclusion. However, I am unsure how to even begin.

REPLY [5 votes]: Let $s_n \in R$ and $S \subset R \cup \{+\infty,-\infty\}$. Define $s_n^* = \sup\{s_k \mid k\ge n\}$, $s^* = \lim_{n\to\infty}s_n^*$, and $\{s_k\}_{k\ge n} = s_n,s_{n+1},\dots$. If $s_1^* = +\infty$, then there exists a subsequence of $\{s_n\}$ that diverges to $+\infty$, so $\sup S = s^* = +\infty$. $s_1^* = -\infty$ is not possible because $s_n \in R$. Suppose $s_1^* < +\infty$. $\sup S \le s_n^*$ $(n\ge1)$ because $S$ is the set of subsequential limits of $\{s_k\}_{k\ge n}$ $(n\ge1)$. Therefore, $\sup S \le s^*$. Let $k_0 = 1$. For each $\{s_k\}_{k\ge n}$, there exists $k_n \ge n$ such that $k_n \ge k_{n-1}$ and $|s_n^* - s_{k_n}| < 1/2^n$ by the definition of the supremum. Construct a sequence $\{t_n\}$ by removing repeating elements from $s_{k_1},s_{k_2},\dots$. Then, $\{t_n\}$ is a subsequence of $\{s_n\}$ that converges to $s^*$. Therefore, $s^* \in S$ and $s^* \le \sup S$. It completes the proof.