Prove, by nonstandard reasoning, that the limit superior of a sequence is a cluster point.

Question

I'm working through Goldblatt's Lectures on the Hyperreals, and I've found myself quite stuck on this exercise:

Prove, by nonstandard reasoning, that both the limit superior and the limit inferior are cluster points of the sequence $s$. (Exercise 6.8.1, page 67)

For background, the limit superior is defined as the least upper bound of the set of cluster points of a bounded sequence $s$, and the cluster points as the standard parts ("shadows") of the unlimited terms of $s$.

Answer 1

Here's my answer, which I'm semi-satisfied with. If someone has a better one, please post it and I'll accept that.

Let $L = \limsup s$. Since $L$ is the least upper bound of the cluster points of $s$, it follows that for each real $\epsilon > 0$ there is some unlimited $N$ such that $s_N \ge L - \epsilon$; thus $s_n \ge L - \epsilon$ for infinitely many limited $n$. But this is precisely the (equivalent) standard definition of a cluster point, so we can conclude that $L$ is itself a cluster point of $s$.

Answer 2

Some of the comments concern my original answer, which assumed that $\limsup$ is defined as the limit of suprema. Instead, as the question points out, Goldblatt defines the $\limsup$ of the sequence $s: \mathbb{N} \rightarrow \mathbb{R}$ as $$\limsup_{n \rightarrow \infty} s_n = \sup \left\{ \mathrm{sh}\:\overline{s}_K \:|\: K \in \!\!~^\star\mathbb{N}_\infty \right\} $$ where $\overline{s}:\!\!~^\star\mathbb{N} \rightarrow \!\!~^\star\mathbb{R}$ denotes the usual extension of $s$ to the hypernaturals.

Write $\limsup_{n \rightarrow \infty} s_n = L$. According to Theorem 6.6.1 (page 66), we can show that $L$ is a cluster point of $s$ by proving that $\overline{s}_K \approx L$ for some unlimited $K \in \!\!~^\star\mathbb{N}$.

Assume for a contradiction that for all $K \in \ \!\!~^\star\mathbb{N}_\infty$ there is a real number $\varepsilon > 0$ such that $\left|\overline{s}_K - \!\!~^\star L\right| \geq \!\!~^\star\varepsilon$. Now, take your favorite unlimited hypernatural $\omega \in \!\!~^\star\mathbb{N}$. If some $M \in \!\!~^\star\mathbb{N}$ satisfies $M > \omega$, then clearly $M \in \!\!~^\star\mathbb{N}_\infty$. Using the fact that $\omega^{-1}$ is infinitesimal, we then get that $\left| \overline{s}_M - \!\!~^\star L \right| \geq \!\!~^\star\varepsilon > \omega^{-1}$. Putting these observations together, we have $$\exists \omega \in \!\!~^\star\mathbb{N}.\: \forall M \in \!\!~^\star\mathbb{N}.\: M > \omega \rightarrow \left|\:\!\!~^\star L - \overline{s}_M\right| \geq \omega^{-1}. $$ By Transfer we can conclude that there is in fact a natural $n \in \mathbb{N}$ satisfying $$\forall M \in \!\!~^\star\mathbb{N}.\: M > \!\!~^\star n \rightarrow \left|\:\!\!~^\star L - \overline{s}_M\right| \geq \!\!~^\star n^{-1}. $$ But since $\mathrm{sh}\:\overline{s}_K$ and $\overline{s}_K$ are infinitesimally close, we get that $L - \mathrm{sh}\:\overline{s}_K$ must also be larger than some fixed real number $\delta \in \left(0,n^{-1}\right)$ for any unlimited $K$. This means that the real number $L - \delta < L$ is an upper bound for the set $\left\{ \mathrm{sh}\:\overline{s}_K \:|\: K \in \!\!~^\star\mathbb{N}_\infty \right\} \subseteq \mathbb{R}$, a contradiction.