5

Let $(x_d)_{d\in D}$ be a net net of real numbers. Limit superior of a net is defined as $$\limsup x_d = \lim_{d\in D} \sup_{e\ge d} x_e = \inf_{d\in D} \sup_{e\ge d} x_e.$$ See, for example, Limsups of nets.

We can replace $\inf$ by $\lim$ since a monotone net is convergent. (If we allow also the values $\pm\infty$.)

A number $p$ is a cluster point of the net $(x_d)_{d\in D}$ if, for every neighborhood $U$ of $p$ and for any $d_0\in D$ there exists $d\ge d_0$ such that $x_d\in U$. (In the other words, the set $x^{-1}[U]=\{d\in D; x_d\in U\}$ is cofinal in $D$.)

Question: How to show that limit superior of $(x_d)_{d\in D}$ is also a cluster point of $(x_d)_{d\in D}$?

This question came up in comments to another question. Since this topic might crop up from time to time, I consider this useful enough to be posted in a separate topic.

Martin Sleziak
  • 56,060

1 Answers1

4

Let $p=\limsup x_d = \lim\limits_{d\in D} \sup\limits_{e\ge d} x_e$. Let us denote $y_d=\sup\limits_{e\ge d} x_e$.

Let $U=(p-\varepsilon,p+\varepsilon)$ be a neighborhood of $p$. Let $d_0\in D$.

Since $\lim y_d=p$, there is $d_1$ such that $d\ge d_1$ implies $y_d \in U$.

So for any $d\ge\max\{d_0,d_1\}$ we have $$x-\varepsilon< \sup\limits_{e\ge d} x_e < x+\varepsilon,$$ which implies that there is an $e\in D$ such that $e\ge d_0$ and $x_e\in (x-\varepsilon,x+\varepsilon)$.

Basically the same reasoning would work work also for $p=+\infty$ and for a neighborhood of the form $(k,\infty]$.

Martin Sleziak
  • 56,060