Double Limit implies Successive Limits

Question

I know it seems very stupid question, but is it right that:

Suppose $X$ being a complete metric space. Then: $$\lim_{(m,n)}x_{(m,n)}\quad\text{exists} \quad\Rightarrow\quad \lim_n\lim_m x_{(m,n)}\quad\text{exists}$$ ...obviously, the limits agre if so and the converse fails in general.

I'm missing the proof that for large enough fixed $n_0$: $$x_{(m,n)}\quad(m,n)\text{-cauchy} \quad\Rightarrow\quad x_{(m,n_0)}\quad (m)\text{-cauchy}$$

Thanks in advance! Cheers, Alex.

Answer 1

Define $f(m,n)=(-1)^{m}/n+(-1)^n/m$.
Here $\lim_{(m,n)}f(m,n)$ exists, but neither $\lim_{m}f(m,n)$ nor $\lim_{n}f(m,n)$ exist.

Define $g(m,n)=m/(m+n)-n/(m+n)$.
Here $\lim_{(m,n)}g(m,n)$ not exists, but both $\lim_{m}g(m,n)$ and $\lim_{n}g(m,n)$ exist.

Answer 2

Oh just got a counterexample: $$x_{(m,n)}:=\frac{1}{n}\sin(m)$$ Here, the double limit exists while the successive limits don't.

Answer 3

Well, if $x_{(m,n)}$ is $(m,n)\text{-cauchy}$, then it's also $(m,n)\text{-convergent}$ ( because $X$ complete), and so: $$\exists \lim_{m,n}x_{m,n} \Rightarrow \exists\lim_n\lim_mx_{m,n}\Rightarrow \\ \Rightarrow\exists n_0\in \mathbb N \mid\forall p,q\ge n_0, |x_{p,n_0}-x_{q,n_0}|<\varepsilon, \forall\varepsilon>0.$$ Is that it? I hope so

Edit: this only holds if the successive limits exist, maybe someone knows a condition for that?