How to read $\bigcap\limits_{r=1}^{\infty}\bigcup\limits_{k=1}^{\infty}\bigcap\limits_{n,m\ge k}^{\infty}\{x:|f_n-f_m|<\frac{1}{r}\}$ 
Can somebody explain me step by step why does this $\bigcap\limits_{r=1}^{\infty}\bigcup\limits_{k=1}^{\infty}\bigcap\limits_{n,m\ge k}^{\infty}\{x:|f_n-f_m|<\frac{1}{r}\}$ represents the set of points, where $f_n$ converges to $f$ (Assume $f_n\to f$ as $(n\to \infty)$)

Since $f_n$ converges the sequence is Cauchy and therefore there exists a $k$, such that for $n,m\ge k$ their difference is smaller than $r$ and then why do we take the union of all $k$'s and then intersect with all $r$'s in which order do we have to increase the indices ? 
 A: We are given some space $X$ (not mentioned) and a sequence $(f_n)_{n\geq0}$ of functions $f_n: \>X\to{\mathbb R}$. The set $A$ defined in the title of the question is defined using three nested "quantors", and has to be "read" from the inside out. 
(a) For given $r$,  $m$,  $n$ the set
$$A_{rmn}:=\biggl\{x\in X\>\biggm| \>|f_n(x)-f_m(x)|<{1\over r}\biggr\}$$
consists of all $x\in X$ for which the individual $f_n$ and $f_m$ have values differing by less than ${1\over r}$.
(b) For given $r$ and $k$ the set
$$A_{rk}:=\bigcap_{m,\ n\geq k}A_{rmn}=\bigcap_{m,\, n\geq k}\biggl\{x\in X\>\biggm| \>|f_n(x)-f_m(x)|<{1\over r}\biggr\}$$
consists of the $x\in X$ for which all $f_n$ and $f_m$ with numbers $m$, $n\geq k$ differ by less than ${1\over r}$.
(c) In part (b) the $k$ was given in advance, and a small given $k$ is a very strong condition on an $x\in X$ to be accepted in $A_{rk}$. By forming the set
$$A_r:=\bigcup_{k\geq 1} A_{rk}=\bigcup_{k\geq 1}\ \bigcap_{m,\, n\geq k}\biggl\{x\in X\>\biggm| \>|f_n(x)-f_m(x)|<{1\over r}\biggr\}$$
we are generous: Any $x$ which ultimately, i.e., for sufficiently large $k$, qualifies for acceptance in $A_{rk}\>$, is accepted in $A_r\>$.
(d) But in the end we shall be stricter in the "tolerance", which is  measured by ${1\over r}$. Only points $x\in X$ that pass the acceptance test of $A_r$ for all $r\geq 1$  shall be in the final set $A$:
$$A:=\bigcap_{r\geq 1}A_r=\bigcup_{k\geq 1} A_{rk}=\bigcap_{r\geq 1}\ \bigcup_{k\geq 1}\ \bigcap_{m,\, n\geq k}\biggl\{x\in X\>\biggm| \>|f_n(x)-f_m(x)|<{1\over r}\biggr\}\ .$$
Reading (c) and (d) we see that $A$ collects all points $x\in X$ for which the sequence $n\mapsto f_n(x)$ is a Cauchy sequence. Therefore $A\subset X$ is the set of points $x\in X$ for which we have a limiting value $f(x):=\lim_{n\to\infty} f_n(x)$.
A: Since $f$ isn't mentioned, you can say it is the set of points $x$ where $\{f_n(x)\}$ converges. Restate with quantifiers: A point $x$ belongs to the set if and only if: for every $r \in \mathbb N$ there exists $k \in \mathbb N$ so that for all $n,m \in \mathbb N$ with $n,m \ge k$ you have $|f_n(x) - f_m(x)| < \dfrac 1r$. This is precisely the statement that $\{f_n(x)\}$ is Cauchy. If your underlying space is complete, this is exactly what you needed.
A: Let us call your set $A$. 
The $x \in A$ iff
$$\forall r \in \mathbb{N} \quad \exists k \in \mathbb{N} \quad \textrm{such that} \quad [n,m \geq k \Longrightarrow |f_n(x) - f_m(x)| < \frac{1}{r} ].  \quad (1)$$
This shows that the sequence $(f_n(x))$ is Cauchy for all $x \in \mathbb{R}$. So we can define $f(x):= \lim_{n \rightarrow \infty} f_n(x).$
You can then check that statement $(1)$ implies that
$$\forall r \in \mathbb{N} \quad \exists k \in \mathbb{N} \quad \textrm{such that} \quad [n \geq k \Longrightarrow |f_n(x) - f(x)| < \frac{1}{r} ] \quad; $$
in other words, $f_n \rightarrow f$ in the pointwise sense. 
