If $a_{n,m}$ is a double sequence in a metric space such that $a_{n,m} \rightarrow_m a_n$ uniformly on $n$ and $a_{n,m} \rightarrow_n a$ for all $m$, then

$$a_{n} \rightarrow a.$$

Indeed for any $\epsilon > 0$ let $m_0$ be such that for all $n$ and $m > m_0$ we have $$|a_{n,m}- a_n| < \epsilon/2$$ and let $n(m_0)$ such that for all $n>n(m_0)$ it is

$$|a_{n,m_0+1}- a| < \epsilon/2$$

Then for any $n>n(m_0)$ we also have:

$$|a_n- a| < |a_{n,m_0+1}- a| + |a_{n} - a_{n,m_0+1}| < \epsilon$$

If we drop the uniformity assumption of the convergence on $m$ to be uniform on $n$ then $a_n \rightarrow a$ should not be true. I am looking for a counterexample.

Or is such assumption not even required? Thanks.


1 Answer 1


Let $a_{n,m} = \begin{cases} 1 & \text{if}\ n < m\\ 0 & \text{if}\ n \geq m \end{cases}$. Then $a_{n,m} \to_{m} a_n = 1$ for every $n$, and $a_{n,m} \to_{n} a = 0$ for every $m$. In this case, $a_n \not\to a$.

  • $\begingroup$ very easy thanks. I worked out n/(n+m) in the meanwhile. $\endgroup$
    – Mr_3_7
    Jan 8, 2014 at 13:05

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