Moore–Osgood theorem

Remarks

Proof idea: uniform convergence gives an $M$ such that $d(a_{m,n}, a_{m',n}) < 2\varepsilon$ for all $n$ when $m, m' \ge M$. Letting $n \to \infty$ gives $d(b_m, b_{m'}) \le 2\varepsilon$, so $(b_m)$ is a Cauchy sequence and has a limit $L$ by completeness. The triangle inequality $d(c_n, L) \le d(c_n, a_{M,n}) + d(a_{M,n}, b_M) + d(b_M, L)$ then shows $c_n \to L$, and the same splitting gives the double limit. Example: for $a_{m,n} = \frac{1}{m} + \frac{1}{n}$ the convergence in $m$ is uniform, and all limits equal $0$. Without uniformity the conclusion fails: $a_{m,n} = \frac{m}{m+n}$ gives $\lim_m \lim_n a_{m,n} = 0$ but $\lim_n \lim_m a_{m,n} = 1$. Without completeness it also fails: in $\mathbb{Q}$, take rationals $x_k \to \sqrt{2}$ and $a_{m,n} = x_{\min(m,n)}$; both hypotheses hold, but $b_m = x_m$ has no limit in $\mathbb{Q}$.

Going deeper

For a student

This is the abstract principle behind many interchange results in analysis. The theorem that a uniform limit of continuous functions is continuous, and the theorem that uniform limits commute with integration, follow the same pattern. For double series, it gives conditions under which summing by rows, by columns, or over squares gives the same value.