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I looked over lots of questions of limit superior but I couldn't find one that addressed my question exactly. The closest is this and compares the definitions of 3 limit superiors. I think the first one they explained (must) be equivalent to one of mine but it wasn't clear to me.

I want to show that if $E = \{ x \in \mathbb R \cup \pm \infty \mid \exists \ ( s_{n_k} ), n_1<n_2< ... , s_{n_k} \to x \}$ i.e. $E$ is the set of all sub sequential limits of $(s_n)$ then the following is true:

$$ \sup E = \lim_{n \to \infty} \left[ \sup_{m \geq n } s_m \right] = s^*$$

I think the idea and intuition is clear to me however, making it 100% formal has been a challenge for me.

The intuition seems simple:

The LHS is just the largest limit amongst all the possible limits of a sequence. The RHS says, ok for every tail end, choose the largest element. Let that be a new sequence. Of course that must be in E. Since we are always choosing the "largest element" of the sequence (or at least an upper bound that can get arbitrarily close to the largest element of the sequence) then it must be that by choosing the sequence of largest elements we must approach the largest limit. If this wasn't true then there would be a sequence approaching a larger limit but neighbourhoods around that must be elements of the form $ \sup_{ m \geq n } s_m $ so that must be false.

Can someone provide me some help to formalize this?

The cases where we go to infinity or minus infinity seemed easy since the sup would just choose "infinity" immediately or if minus infinity is the largest limit then it doesn't matter with diverging decreasing sequence we choose, they all go down forever.

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  • $\begingroup$ Define $\ell = \lim_{n\to\infty} [ \sup_{m\ge n} s_m ]$. To prove that $\sup E = \ell$, it is enough to prove both $\sup E \le \ell$ and $\sup E \ge \ell$. Splitting the problem up into two subproblems in this way is a common device in analysis that makes it easier for you to untangle the formalism required. $\endgroup$ – Greg Martin May 14 '18 at 3:44
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In analysis, showing that two quantities equal often involves showing that one bounds the other. We write $\mathsf{L} = \lim_{n\to\infty} \sup_{m\geq n} s_m$ for brevity.

1. $\sup E \leq \mathsf{L}$.

Let $\alpha$ be any real number satisfying $\alpha < \sup E$. Then there exist $\ell \in E$ and a subsequence $(s_{n_k})$ such that $\alpha < \ell = \lim_{k\to\infty} s_{n_k}$. Since $s_{n_k} \leq \sup_{m \geq n_k} s_m$ for each $k$, taking $k \to \infty$ gives

$$\alpha < \ell \leq \mathsf{L}$$

Since this is true for all $\alpha < \sup E$, taking $\alpha \uparrow \sup E$ proves the desired inequality.

2. $\sup E \geq \mathsf{L}$.

Fix $\alpha_k \uparrow \mathsf{L}$ and invoke the following algorithm:

  • Set $n_0 = 0$ for convenience.

  • If $k \geq 1$ and $n_{k-1}$ is defined, then there exists $N$ such that $\sup_{m\geq n} s_m > \alpha_k$ for all $n \geq N$. In particular, there exists $n_k \geq \max\{N, n_{k-1}+1\}$ such that $s_{n_k} > \alpha_k$.

Then $\alpha_k < s_{n_k} \leq \sup_{m\geq n_k} s_m$ holds and by the squeezing theorem, $s_{n_k} \to \mathsf{L}$ as $k\to\infty$, hence we have $\mathsf{L} \in E$ and $\mathsf{L} \leq \sup E$.


So, what is going on in both directions?

  1. This shows that any subsequential limit is bounded by $\mathsf{L}$, which is quite obvious by thinking that $\mathsf{L}$ is kind of 'supremum over tail'.

  2. This realizes $\mathsf{L}$ is a subsequential limit, by algorithmically choosing elements that are close to the supremum.

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