Rudin's theorem 3.17 on $\lim \sup x_n$

Question

The theorem is as follows: Given $s = \lim \sup x_n$ and any $y > s$, there exists an integer $N$ such that $n \geq N$ $\implies $ $x_n < y$.

Naively I am thinking the theorem also implies there exists an integer $N$ such that $n \geq N$ $\implies $ $x_n \leq s$. Is is true? Probably not but how to refute this?

Answer 1

Consider the sequence $(x_{n})_{n\geq 1}$ defined by $x_{n}=1+\frac{(-1)^{n}}{n}$. So the sequence goes $$1-1,1+\frac{1}{2},1-\frac{1}{3},1+\frac{1}{4},1-\frac{1}{5},...$$ Then $\limsup{x_{n}}=1$ but for any $N$, there exists $n>N$ with $x_{n}>1$.

Answer 2

Perhaps too obvious:

$x_n$ a strictly decreasing sequence that converges to $s$;

We have

$x_n > s$, for $n \in \mathbb{N}$; and

$ \limsup x_n=\lim x_n =s$.