Sequencial chracterisation of $\limsup f(x)$ I am trying to understand the definition of $\limsup_{x\to a}f(x)$. There is a Wikipedia article that gives quite a few definitions. One of them is:

I think I understand it more or less. However, I came across the following post definition of limsup of a function. There it is said that $\limsup f(x)$ can be characterised in terms of sequences in the following way:

I tried to prove it, but I got stuck. I also tried to find a book that explains it, but I could't find a single book that covers $\limsup_{x\to a}f(x)$.Does anybody have any books that talk about it?
Many thanks in advance!
 A: DEFINITION
Suppose $X$ is a metric space, $E\subset X$, $f$ maps $E$ into $\mathbb{R}^1$,
and $p$ is a limit point of $E$.
Define a set $L$ of extended real numbers as follows: If $\{p_n\}$ is a sequence
in $E$ converging to $p$ with $p_n\not=p$, and if $f(p_{n_k})\rightarrow l$
for some subsequence $\{p_{n_k}\}$, then $l\in L$.
Put
$$\limsup_{x\rightarrow p}f(x)=\sup L$$
and
$$\liminf_{x\rightarrow p}f(x)=\inf L\mbox{.}$$
In other words, $\limsup\limits_{x\rightarrow p}f(x)$ is defined to be the supremum of the set consisting of $\limsup\limits_{n\rightarrow\infty}f(p_n)$, where $p_n\rightarrow p$, $p_n\not=p$. (Note that $\sup\bigcup\limits_{\alpha\in A}E_\alpha=\sup\{\sup E_\alpha:\alpha\in A\}$.)
THEOREM 1
(a) If $y>\limsup_{x\rightarrow p}f(x)$, then there exists $\delta>0$ such that
$f(x)<y$ whenever $0<d(x,p)<\delta$, $x\in E$. 
(b) If $y<\lim\sup_{x\rightarrow p}f(x)$, then for every $\delta>0$, there exists
$x\in E$ with $0<d(x,p)<\delta$, such that $f(x)>y\mbox{.}$
A similar result is valid for $\liminf$.
PROOF
(a) Suppose to every $\delta>0$ corresponds $x\in E$ such that $0<d(x,p)<\delta$
and $f(x)\geq y$. Take $\delta=\frac{1}{n}$, $n=1,2,3...$. We then obtain a sequence
$x_n\rightarrow p$, $x_n\not=p$ such that
$$f(x_n)\geq y\mbox{.}$$
Letting $n\rightarrow\infty$, we get
$$\limsup_{x\rightarrow p}f(x)\geq \limsup_{n\rightarrow\infty}f(x_n)\geq y\mbox{.}$$
This proves (a).
(b) Since $y<\sup L$, $l>y$ for some $l\in L$, which means that there exists a sequence
$x_n\rightarrow p$, $x_n\not=p$ such that $f(x_n)\rightarrow l$. Since $l>y$, $f(x_n)>y$
for all $n$ from some definite index $N$ onwards. For any $\delta>0$, there exists $n>N$
such that $0<d(p,x_n)<\delta$, since $x_n\rightarrow p$. Evidently, $f(x_n)>y$. $\square$
THEOREM 2
If $\liminf_{x\rightarrow p}f(x)=\limsup_{x\rightarrow p}f(x)=l$, then $f(x)\rightarrow l$
as $x\rightarrow p$.
PROOF
This follows immediately from Theorem 1(a).
THEOREM 3
If $g$ is another mapping of $E$ into $\mathbb{R}^1$, and if there exists $\delta>0$ such
that $0<d(p,x)<\delta$, $x\in E$ imply that
$$\tag{1}f(x)\leq g(x)\mbox{,}$$
then
$$\tag{2}\liminf f(x)\leq\liminf g(x)\mbox{,}$$
$$\tag{3}\limsup f(x)\leq\limsup g(x)\mbox{,}$$
as $x\rightarrow p$.
PROOF
We show (3). If $\limsup f(x)>y>\limsup g(x)$, then, by Theorem 1, there exist $\delta>\delta'>0$
such that
$$\tag{4}g(x)<y\quad(x\in E,0<d(x,p)<\delta')\mbox{;}$$
also, there exists $\xi\in E$ with $0<d(\xi,p)<\delta'$ such that
$$\tag{5}f(\xi)>y\mbox{.}$$
(4) and (5) contradict (1).
THEOREM 4
If $\lim\limits_{x\rightarrow p}f(x)=l$, then
$$\tag{6}\limsup\limits_{x\rightarrow p}f(x)=\limsup\limits_{x\rightarrow p}f(x)=l\mbox{.}$$
PROOF
Let $p_n\rightarrow p$ with $p_n\not=p$. Then $f(p_n)\rightarrow l$, so that
$$\limsup\limits_{n\rightarrow\infty}f(p_n)=\liminf\limits_{n\rightarrow\infty}f(p_n)=l\mbox{.}$$
Since $\{p_n\}$ is arbitrary, (6) follows.
