# $\limsup x_n = \sup{z_k}$ where ${z_k}$ is a set of limit points for the sequence ${x_n}$

There is a theorem saying that:

$\limsup x_n = \sup{z_k}$ where ${z_k}$ is a set of limit points for the sequence ${x_n}$.

All sets and sequences are real.

The definition of limit superior is: $$\limsup x_n = \lim_{n \to \infty} \sup_{m \ge n}\{x_m\}=\inf_{n} \sup_{m \ge n}\{x_m\}$$

Limit point for a sequence is a point such that in any neighborhood of it there is a point from the sequence other than the original point itself.

If $\lim \sup x_n$ is finite, then clear enough it is a limit point itself, and there might be no other limit point which exceed $\lim \sup x_n$.

But how to proceed with the proof if $\lim \sup x_n= \pm \infty$?

• How are you defining $\limsup$? In some texts, the theorem you state is the definition. – Hayden Mar 14 '15 at 21:56
• @Hayden $\lim \sup x_n = \lim_{n \to \infty} \sup_{m \ge n}\{x_m\}=\inf_{n} \sup_{m \ge n}\{x_m\}$. – Sergey Zykov Mar 14 '15 at 21:59
• problem is defining the neighborhood of $\pm \infty$. If I am not wrong, take $(a ,\infty)$ neighborhood for $\infty$ and likewise for $-\infty$ – Harish Mar 14 '15 at 22:01

Suppose $\limsup x_n=\infty$, by the $\inf \sup$ definition it means $\sup_{m\ge n} x_m=\infty$ for all $n$, in particular for $n=1$ it gives $\sup x_n=\infty$.
And, if $\limsup x_n=-\infty$, we need to prove that $x_n$ has no other limit point, i.e. $x_n\to -\infty$.
For this, take any $A\in\Bbb R$ and as $\lim_n(\sup_{m\ge n}x_m )= -\infty$, there is a limit index $N_0$ such that $n\ge N_0\implies \sup_{m\ge n}x_m<A$, and it follows that for all $m\ge N_0$, we have $x_m<A$. $\$ -QED-
We proved a slightly stronger theorem: $$\limsup x_n=\max \{\text{limit points}\}\,.$$
• I am afraid I am not getting your logic. In the first paragraph you show that $\limsup x_n=\infty \implies \sup x_n=\infty$, i.e. the sequence is unbounded from above, but does it really mean that $\sup z_k = \infty$ i.e. the set of limit points is unbounded from above, i.e. does not matter how far you go up along $\mathbb R$ line you will be finding new limit points? – Sergey Zykov Mar 15 '15 at 12:05
• In fact a sequence $1,1,2,\frac{1}{2},...,n,\frac{1}{n},...$ has $\limsup x_n=\infty$ but only one limit point $0$, thus $\sup{z_k}=0$. So $\limsup x_n \ne \sup{z_k}$, so the theorem is wrong? – Sergey Zykov Mar 15 '15 at 12:08
• With the part $\limsup x_n=-\infty \implies$ there is no limit points, thus $\sup z_k = -\infty$ I tend to agree, indeed, if there were a limit point $z_K$, then it would contain an element (different from itself) of the sequence in its any neighborhood, so it would contain an infinite amount of elements in any its neighborhood, but sooner a later you will find $N$ such that $\sup_{m\ge N}x_m<z_K$ means all $x_n < z_k$ with $n \ge N$, thus $z_k$ has a neighborhood which has only a finite number of sequence elements and consequently has a neighborhood with no elements at all. – Sergey Zykov Mar 15 '15 at 12:22