Is $\lim\limits_{n \rightarrow \infty} \lim\limits_{m \rightarrow \infty} |a_n-a_m|=0$ equivalent to $(a_n)_{n\in \mathbb{N}}$ being a Cauchy? Are these statements equivalent? 


*

*$ (a_n)_{n\in \mathbb{N}} $ is cauchy

*$\lim\limits_{n \rightarrow \infty} \lim\limits_{m \rightarrow \infty} |a_n-a_m|=0$ 

*$\lim\limits_{m \rightarrow \infty} \lim\limits_{n \rightarrow \infty} |a_n-a_m|=0$ 


I want to prove it using epsilon delta, but I have great trouble in rewriting the above statements using the epsilon delta definition. Furthermore, I do believe all the statements to be equivalent.
 A: Show that $\boldsymbol{(2)}$ or $\boldsymbol{(3)}$ imply $\boldsymbol{(1)}$
Choose an $\epsilon\gt0$. We can find an $N$ so that for all $n\ge N$,
$$
\lim_{m\to\infty}\left|a_n-a_m\right|\le\epsilon/4
$$
That means that for each $n\ge N$, there is an $M_n$ so that for all $m\ge M_n$,
$$
\left|a_n-a_m\right|\le\epsilon/3
$$
So given $n_1,n_2\ge N$, choose $m_1\ge\max(M_{n_1},N)$ and $m_2\ge\max(M_{m_1},M_{n_2})$
$$
\begin{align}
\left|a_{n_1}-a_{n_2}\right|
&\le\left|a_{n_1}-a_{m_1}\right|+\left|a_{m_1}-a_{m_2}\right|+\left|a_{n_2}-a_{m_2}\right|\\
&\le\epsilon/3+\epsilon/3+\epsilon/3\\
&=\epsilon
\end{align}
$$
Thus, $a_n$ is Cauchy.

Show that $\boldsymbol{(1)}$ implies $\boldsymbol{(2)}$ and $\boldsymbol{(3)}$
After some discussion, the OP has clarified that the norms they are considering are real-valued.
The other direction is trickier than it appears at first. First, we must show that for each $n$,
$$
\lim\limits_{m\to\infty}|a_n-a_m|\tag{$\ast$}
$$
exists. So, for each $n$, consider the sequence $|a_n-a_m|$. Choose an $\epsilon\gt0$, then find the $N$ so that for all $m_1,m_2\ge N$, we have $|a_{m_1}-a_{m_2}|\le\epsilon$. Then
$$
\big||a_n-a_{m_1}|-|a_n-a_{m_2}|\big|\le|a_{m_1}-a_{m_2}|\le\epsilon
$$
This means that, for each $n$, the sequence $|a_n-a_m|$ is Cauchy and real, hence convergent. Thus, $(\ast)$ exists.
Next, choose any $\epsilon\gt0$. Then we can find an $N$ so that for all $m,n\ge N$, $|a_n-a_m|\le\epsilon$. This means that
$$
\lim_{m\to\infty}|a_n-a_m|\le\epsilon\tag{$\ast\ast$}
$$
Since we can choose any $\epsilon\gt0$ and get an $N$ so that if $n\ge N$, we have $(\ast\ast)$, it follows that
$$
\lim_{n\to\infty}\lim_{m\to\infty}|a_n-a_m|=0
$$
A: Points (2) and (3) are trivially equivalent (just change the roles of $n$ and $m$).
Now, suppose the sequence takes values in a complete space (e.g., $\mathbb{R}$).
Hint 1: Suppose $(a_{n})_{n}$ is Cauchy so that it has a limit $a$.
Then,
\begin{align*}
\lim_{n\rightarrow\infty}\lim_{m\rightarrow\infty}\left|a_{n}-a_{m}\right| & =\lim_{n\rightarrow\infty}\left|a_{n}-\lim_{m\rightarrow\infty}a_{m}\right| & \text{(by continuity)}\\
 & =\lim_{n\rightarrow\infty}\left|a_{n}-a\right| & \text{(the sequence has a limit)}
\end{align*}
(try to finish the proof).
Hint 2: For the reverse direction, suppose 
$$
\lim_{n\rightarrow\infty}\lim_{m\rightarrow\infty}\left|a_{n}-a_{m}\right|=0.
$$
Let $\epsilon>0$. Then, by the definition of limit, we can find $N$
such that for all $n\geq N$,
$$
\lim_{m\rightarrow\infty}\left|a_{n}-a_{m}\right|<\epsilon/2
$$
(try to finish the proof).
A: Let $\{a_n\}$ be Cauchy. Then for $\forall \varepsilon>0$, $\exists N\in\mathbb{N}$ such that
$$ |a_m-a_n|<\varepsilon, \forall m, n\ge N $$
namely
$$ \lim_{m\to\infty}\lim_{n\to\infty}|a_m-a_n|=0. \tag{1}$$
Conversely, suppose (1) holds. Then for $\forall \varepsilon>0$, $\exists N_1\in\mathbb{N}$ such that
$$ \lim_{n\to\infty}|a_m-a_n|<\varepsilon/2, \forall m\ge N_1. \tag{2}$$
From (2), $\exists N_2\in\mathbb{N}$ such that
$$ |a_m-a_n|<\varepsilon, \forall m\ge N_1, \forall n>N_2. $$
Lent $N=\max\{N_1,N_2\}$. Then for above $\varepsilon>0$, one has
$$ |a_m-a_n|<\varepsilon, \forall m, n\ge N. $$
Namely $\{a_n\}$ is Cauchy.
