**Proof.**
It follows from Lemma 15.79.10 that (1) implies (2). Assume (2). Then $(A_ n)$ is ML by Lemma 15.79.11. For $m \geq n$ let $A_{n, m} = \mathop{\mathrm{Im}}(A_ m \to A_ n)$ so that $A_ n = A_{n, n} \supset A_{n, n + 1} \supset \ldots $. Note that $(A_ n)$ is zero as a pro-object if and only if for every $n$ there is an $m \geq n$ such that $A_{n, m} = 0$. Note that $(A_ n)$ is ML if and only if for every $n$ there is an $m_ n \geq n$ such that $A_{n, m} = A_{n, m + 1} = \ldots $. In the ML case it is clear that $\mathop{\mathrm{lim}}\nolimits A_ n = 0$ implies that $A_{n, m_ n} = 0$ because the maps $A_{n + 1, m_{n + 1}} \to A_{n, m}$ are surjective. This finishes the proof.
$\square$

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