Suppose $A$ is a Noetherian ring, $M$ a finite $A$-module and $N \subset M$ a submodule. Furthermore, $\mathfrak{a} \subset A$ is an ideal.
Consider the $\mathfrak{a}$-adic topology on M, i.e. the filtration $... \mathfrak{a}^3M \subset \mathfrak{a}^2M \subset \mathfrak{a}M \subset M$ and the associated linear topology on $M$. Analogously, $N$ is topologized via the filtration $... \mathfrak{a}^3N \subset \mathfrak{a}^2N \subset \mathfrak{a}N \subset N$.
How can I show that the $\mathfrak{a}$-adic topology on $N$ is equivalent to the subspace topology of $N \subset M$?
I may use Artin-Rees, i.e.
$\exists n_o \in \mathbb{N}:\forall n > n_0: \mathfrak{a}^nM \cap N = \mathfrak{a}^{n-n_0}(\mathfrak{a}^{n_0}M \cap N)$
Thanks for any suggestions.
Edit: $A$ is commutative.