# Does the interchangeability of double limits imply that the “concurrent limit” exists?

Let the sequence $$\{a_{n,m}\}$$ be a doubly indexed real sequence. I was wondering if the assumption that $$\lim_{n\to\infty} \lim_{m\to\infty} a_{n,m} = \lim_{m\to\infty} \lim_{n\to\infty} a_{n,m} = L,$$ where $$L$$ is real, would imply that the following "concurrent limit" would also exist and equal $$L$$. $$\lim_{n,m\to\infty} a_{n,m} \stackrel{?}= L.$$

I am defining $$\lim_{n,m\to\infty} a_{n,m} = L$$ to mean that for each $$\epsilon > 0$$ there exists $$K > 0$$ such that $$n,m>K \Rightarrow |a_{n,m}-L|<\epsilon.$$

My intuition tells me that this conjecture is false, since the rapidity of convergence of $$\lim_{m\to\infty} a_{n,m}$$ to $$b_n$$ may depend on $$n$$, so $$m$$ and $$n$$ may not be able to approach $$\infty$$ in a sufficiently similar manner. However, I could not construct any counterexample.

I also wonder whether the weaker statement that

$$\lim_{n\to\infty} a_{n,n} \stackrel{?}= L$$

must hold.

I would appreciate any helpful insights.

• Consider a_{n,m}=\frac{1}{n-m} – R.Jackson Jul 24 '19 at 22:25
• That is a simple and clever counterexample. I did not think of making the concurrent limit undefined. Thank you for your insight! – EnjoyingMath Jul 24 '19 at 22:59

Counterexample to both conjectures. Let $$a_{n,m}=\delta_{nm}$$ (Kronecker delta). Then for all $$n$$, $$\lim_{m\to\infty}a_{n,m}=0$$ and similarly for all $$m$$ $$\lim_{n\to\infty} a_{n,m}=0.$$ But, for any $$1>\epsilon>0$$ and $$K>0$$, we have $$\lvert a_{K+1,K+1}-0 \rvert=1>\epsilon.$$ And indeed $$\lim_{n\to\infty}a_{n,n}=1.$$