# Limit superior is a cluster point of a net

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 $$U$$.)

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.

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 $$(k,\infty]$$.