The $I$-adic topology of $N$ coincides with the topology induced by the $I$-adic of $M$ on the subspace $N$. I'm having trouble understanding the proof for the following proposition.

Let $A$ be a Noetherian ring, $M$ a finite $A$-module, $N\subset M$ a
  submodule, $I$ an ideal of $A$. Then the $I$-adic topology of $N$
  coincides with the topology induced by the $I$-adic of $M$ on the
  subspace $N$.

So I get that the I-adic topology on $N$ is $\mathcal{B}_1=\{I^nN\}_{n=1,2,...}$ and the topology induced by the $I$-adic topology of $M$ on $N$ is $\mathcal{B}_2=\{I^nM\cap N\}_{n=1,2,...}$.
So I guess I need to show, that taking $B_1\in \mathcal{B}_1$ and $b_1\in B_1$ we have to find $B_2\in \mathcal{B}_2$ such that $b_1\in B_2$ and $B_2\subset B_1$ (and vice versa)?
Now the proof for the proposition states, that it follows from the Artin-Rees lemma since $I^nN\subset I^nM\cap N\subset I^{n-c}N$. Now I see this would imply that $B_2\subset B_1$ (and vice versa) if we use the notation above, but doesn't we also need to show that there exist elements $b_1$ which is in $B_2$ (and vice versa)?
Meaning we need som element say $r_1\in I^nM\cap N$ which is also in $I^nN$, and some element $r_2\in I^{n-c}N$ which is also in $I^nM\cap N$?
Also, I'm not sure of the inclusions $I^nN\subset I^nM\cap N\subset I^{n-c}N$. My guess for the last inclusions follows from the fact that $I^nM\cap N=I^{n-c}(I^cM\cap N)\subset I^{n-c}N$ since $I^cM\cap N\subset N$, is that correct? For the first inclusion I'd say that it follows from the fact that  $I^nN\subset AN=N$ and $I^nN\subset I^nM$?
 A: (In this post $A\subset B$ and $B\supset A$ mean that $A$ is a subset of $B$.)
Let $R$ be a commutative ring with one, $I$ an ideal, $M$ a module, $N$ a submodule, $\mathcal N$ the set of neighborhoods of $0$ for the $I$-adic topology of $N$, and $\mathcal N'$ the set of neighborhoods of $0$ for the topology on $N$ induced by the $I$-adic topology of $M$. 
The inclusion $\mathcal N'\subset\mathcal N$ is easily verified.
Consider the conditions
$(1)\ \exists a\in\mathbb N,\ \forall b\in\mathbb N\ \Big(I^b(N\cap I^aM)=N\cap I^{a+b}M\Big),$
$(2)\ \forall b\in\mathbb N,\ \exists a\in\mathbb N\ \Big(I^bN\supset N\cap I^{a+b}M\Big),$
$(3)\ \mathcal N\subset\mathcal N'$.
Clearly $(1)$ implies $(2)$. 
Claim: $(2)$ implies $(3)$.
Proof: Let $X$ be in $\mathcal N$. Then $X\supset I^bN$ for some $b$, and we get 
$$
X\supset I^bN\supset N\cap I^{a+b}M\
$$ 
for some $a$, and thus $X\in \mathcal N'$.
A: See Corollary 10.10 in Atiyah Macdonald. This follows from the Artin--Rees lemma:

Let $I$ be an ideal in a Noetherian ring $R$; let $M$ be a finitely generated $R$-module and let $N$ a submodule of $M$. Then there exists an integer $k \geqslant 1$ so that $ I^nM\cap N=I^{n-k}((I^kM)\cap N)$ for each $n\geqslant k$.

One can find this statement and a proof in, say, Wikipedia.  
