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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 $s$.

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  • $\begingroup$ Nonstandard analysis...I never got it... $\endgroup$
    – Troy Woo
    Oct 4 '14 at 19:50
  • $\begingroup$ It's the first time I hear of this. Interesting! $\endgroup$ Oct 4 '14 at 19:58
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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 $s$.

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  • $\begingroup$ Just ask for curisoty, I know nothing about non-standard analysis. Is this a "non-standard" proof? Because it seems to be the same as what we would do with the standard analysis $\endgroup$ Oct 9 '14 at 7:58
  • $\begingroup$ The first half is non-standard (since it talks about unlimited numbers) but the second half uses the standard definition. Hence my semi-satisfaction. $\endgroup$ Oct 22 '14 at 1:54
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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.

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  • $\begingroup$ How does your first equality follow from Goldblatt's definition of limit superior? $\endgroup$
    – BallBoy
    Mar 29 '20 at 5:46
  • $\begingroup$ In fact Goldblatt uses the result from Exercise 6.8.1 to prove this equality (Theorem 6.8.5, p.69), so I don't think this can be the argument he had in mind. $\endgroup$
    – BallBoy
    Mar 29 '20 at 5:50
  • $\begingroup$ @BallBoy I replaced my answer with an argument that works for Goldblatt's definition. It does not stray far from the semi-standard argument, and is not nearly as elegant as the one I gave for the more usual "limits of suprema" definition. I'll see if I can come up with something better. $\endgroup$
    – Z. A. K.
    Mar 29 '20 at 11:56
  • $\begingroup$ This is nice, but you're right, it's missing the elegance and simplicity of the usual Goldblatt in-text exercise. $\endgroup$
    – BallBoy
    Mar 29 '20 at 14:50

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