Positive subsequence of a double sequence

Question

Suppose for the double sequence $\{a_{m,\,n}\} \subset \mathbb{R}$ we have $\lim_{n\rightarrow\infty}\lim_{m\rightarrow \infty} a_{m,\,n} = L$, where $L \in (0,+\infty]$. In particular, there exists some $n_0$ such that $n > n_0$ implies $\lim_{m \rightarrow \infty} a_{m,\,n} > 0$. Is it possible to find a subsequence $\{a_{m_k,\,n}\}$ such that $a_{m_k,\,n} > 0$ for all $n > n_0$?

Answer 1

Well if I get you right, the answer is no. Think of a sequence of the form $$a_{m,n}=1-\frac{b_n}m.$$ Since for each $n\in \mathbb N$ $$\lim_{m\to\infty} a_{m,n}=1>0,$$ and then for $n>n_0=1$ we have $\lim_{m\to\infty} a_{m,n}>0,$ there should be a subsequence $a_{m_k,n}$ such that for each $n>1$ and —I think you omitted this— for each $k\in \mathbb N$) $$a_{m_k,n}>0.$$

But suppose, for instance, that $b_n=n$ (or any other sequence which is unbounded above)... In this case, that condition holds if and only if $$1-\frac nm>0\iff m>n.$$ So no matter how big you choose $m_k$, there will always be infinite numbers $n$ such that $a_{m_k,n}\le 0$.

Note also that this says that the result is not true not only for the chosen $n_0$, but that no matter which $n_0$ such that the first condition holds you choose, the conclusion is always false for this sequence.