$\{a_{m , n}\}$ is a double sequence of real numbers with the following property.
For every $\epsilon$ we will get a natural number $k$ such that $|a_{m , n} - a_{p , q}| < \epsilon$ for all natural numbers $m , n , p , q > k$.
How can I prove that $$\lim_{m \to \infty}\lim_{n \to \infty} a_{m , n} ,\quad \lim_{n \to \infty}\lim_{m \to \infty} a_{m , n} $$ exist, and $$\lim_{m \to \infty}\lim_{n \to \infty} a_{m , n} = \lim_{n \to \infty}\lim_{m \to \infty} a_{m , n}?$$
Can anyone please help me? I think I can show the second part once I am able to prove the first part.
The given condition does not imply the existence of the horizontal and vertical limits. Example: $$ a_{m,n} = \frac{(-1)^{m+n}}{\min(m,n)} \, . $$ For $m , n , p , q > k$ with $k >2/\epsilon$ is $$ |a_{m , n} - a_{p , q}| \le |a_{m , n}| + |a_{p , q}| \le \frac 1k + \frac 1k < \epsilon \, , $$ but $\lim_{m \to \infty} a_{m , n}$ does not exist for any $n$, and $\lim_{n \to \infty} a_{m , n}$ does not exist for any $m$.
$$ \begin{matrix} 1 & -1 & 1 & -1 & 1 & -1 & \cdots \\ -1 & \frac 12 & -\frac 12 & \frac 12 & -\frac 12 & \frac 12 & \cdots \\ 1 & -\frac 12 & \frac 13 & -\frac 13 & \frac 13 & -\frac 13 & \cdots \\ -1& \frac 12 & -\frac 13 & \frac 14 & -\frac 14 & \frac 14 & \cdots \\ 1& -\frac 12 & \frac 13 & -\frac 14 & \frac 15 & -\frac 15 & \cdots \\ -1& \frac 12 & -\frac 13 & \frac 14 & -\frac 15 & \frac 16 & \cdots \\ \vdots &\vdots &\vdots &\vdots &\vdots &\vdots & \ddots \end{matrix} $$
However, if the horizontal and vertical limits $$ b_n = \lim_{m \to \infty} a_{m , n} \\ c_m = \lim_{n \to \infty} a_{m , n} $$ exist for all $m$ respectively $n$, then the iterated limits $\lim_{n \to \infty} b_n$ and $\lim_{m \to \infty} c_m$ exist as well, and they are equal.