Proposition 7

Let A be a filtered ring, E a filtered A-module and $(E_n)$ filtration of E; suppose that $E_0 = E$, that the $E_n$ are submodules of E, that the A-modules $E/E_n$ are Artinian and finally that the topological group E is Hausdorff and complete. Then the intersection of a decreasing sequence of non-empty closed affine subsets of E is non-empty.

We have seen in no. 6 that, since E is Hausdorff and complete, it is identified with $\tilde{E} = \lim_{\leftarrow} E/E_n$. Let $(W_p)$ be a decreasing sequence of non-empty closed affine subsets of E and, for all $n \geq 0$, let $W_{p,n}$ be the canonical image of $W_p$ in $E/E_n$; we are going to construct a sequence $x = (x_n) \in \tilde{E}$ such that $x_n \in W_{p,n}$ for all $p$ and all $n$; hence $x \in W_p + E_n$ for all $p$ and all $n$ and, as the $W_p$ are closed, $x \in W_p$ for all $p$ (no. 5), which will prove the proposition.

As $E/E_0 = 0$, we shall take $x_0 = 0$. Suppose that the $x_i$ are defined for $0 \leq i \leq n - 1$ and let $W'_{p,n}$ be the set of elements of $W_{p,n}$ whose canonical image in $E/E_{n-1}$ is $x_{n-1}$; as $x_{n-1} \in W_{p,n-1}$ and $W_{p,n-1}$ is the canonical image of $W_{p,n}$, $W'_{p,n}$ is non-empty and is obviously an affine subset of $E/E_n$; moreover the sequence $(W'_{p,n})_{p \in \mathbf{N}}$ is decreasing. As $E/E_n$ is Artinian, this sequence is stationary (otherwise the sequence of submodules of $E/E_n$ which are the directions of the $W'_{p,n}$ would be strictly decreasing, which is absurd). It is sufficient then to take $x_n$ to be an element of $\bigcap_{p \in \mathbf{N}} W'_{p,n}$ and the construction of $(x_n)$ can then be performed by induction.