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

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 $s$.

• Nonstandard analysis...I never got it... Oct 4 '14 at 19:50
• It's the first time I hear of this. Interesting! Oct 4 '14 at 19:58

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 $s$.

• 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 Oct 9 '14 at 7:58
• The first half is non-standard (since it talks about unlimited numbers) but the second half uses the standard definition. Hence my semi-satisfaction. Oct 22 '14 at 1:54

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.

• How does your first equality follow from Goldblatt's definition of limit superior? Mar 29 '20 at 5:46
• 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. Mar 29 '20 at 5:50
• @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. Mar 29 '20 at 11:56
• This is nice, but you're right, it's missing the elegance and simplicity of the usual Goldblatt in-text exercise. Mar 29 '20 at 14:50