$R$ is ring with $a\subset R$ ideal. $M$ as $R-$module and submodule $N\subset M$. What is difference between $a$-adic topology and subspace topology? $R$ is ring with $a\subset R$ ideal. $M$ is an $R-$module and submodule $N\subset M$. 
"We note that, when $N$ is a submodule of $M$, the $a$-adic topology of $N$ may be different from the topology of $N$ as a subspace of $M$."
The theorem following above assertion indicates that if $M$ is a noetherian module, the $a$-adic topology of $N$ coincides with the subspace topology on $N$ when $M$ has the $a$-adic topology. 
$\textbf{Q:}$ What is difference between $a-$adic topology and subspace topology and what is the example s.t. $a-$adic topology differs from subspace topology? $a$-adic topology is defined by $a^nM$ as open neighborhood basis of $0$ as in $p$-adic sense.
To see $\textbf{Q'}$'s claim of equivalence when $N$ is finite module over noetherian ring $R$, I denote $N'$ as $N$ in $M'$ $a-$adic subspace topology and $N$ as the $a-$adic topology. It is clear that $Id_N: N\to N'$ is continuous by $a^nN\subset a^nM\cap N$. It suffices to show $Id_N:N'\to N$ is also continuous. From Artin-Rees, I have $\exists r\geq 0$ s.t. for all $n>r$ $a^nM\cap N=a^{n-r}(a^rM\cap N)$. Note that statement holds for all $n>r$. In particular, $n-r\geq 1$ holds as well. Thus we have $a^nM\cap N\subset a^{n-r}N$.(Note that you can choose $n-r$ to start with.) Therefore, this map is also continuous. Thus $Id_N$ is homeomorphism.[This is basically Nagata's proof in the book.]
$\textbf{Q':}$ What is topological interpretation of Artin-Rees lemma? Subspace topology of $N$ as $M$ with $a$-adic topology is exactly $N$ own $a$-adic topology?
Ref. Nagata, Local rings, Chpt 2, Sec 16. 
 A: You sorta ask many questions in one, so it is difficult for me to know what exactly you are looking for.  However, I hope that the following example would allow you to answer what you need.
Let $M$ be the ring $\Bbb Z[x_1,x_2,\ldots]$ (allowing infinitely many terms in a product, as long as all the powers are finite), and $N=(x_1\cdot x_2\cdot\ldots)M$.
Let ${\frak a}=x_1M+x_2M+x_3M+\ldots$.  Then the $v_{\frak a}$-adic topology on $M$ acts as $$v_{\frak a}(x_1^{a_1}\cdot x_2^{a_2}\cdot\ldots)=a_1+a_2+\ldots.$$  Then $v_{\frak a}$ is a non-trivial valuation on $M$.  However, $v_{\frak a}(y)=\infty$ for all $y\in N$.  Therefore, $v_{\frak a}$ reduces to the trivial topology, when viewed as a subspace topology on $N$.  
On the other hand, we can still have the $v_{\frak a}$-adic topology on $N$, via $$v_{\frak a}(x_1^{a_1}\cdot x_2^{a_2}\cdot\ldots)=(a_1-1)+(a_2-1)+\ldots.$$
This $v_{\frak a}$-dic topology on $N$ is not trivial, therefore not equal to the subspace topology (which is trivial).
The idea is, if you have a Noetherian ring, then it's not "big enough" to make valuations level out on a non-trivial submodule -- and therefore, such esoteric examples are not possible.
