Let $R$ be a topological ring (commutative with $1$) and let $R((t))$ the ring of Laurent power series. So, it is the ring containing the formal power series:
$$\sum_{i\ge m} a_i t^i,\quad m\in\mathbb Z,\;a\in R$$
It is easy to see that $R((t))$ can be obtained as an ind-pro objects in the following way:
$$R((t))=\varinjlim_{m\in\mathbb Z}\left (\varprojlim_{n\in\mathbb N} \frac{t^m R[[t]]}{t^{m+n} R[[t]]}\right)\quad (\ast)$$
If we topologize $\frac{t^m R[[t]]}{t^{m+n} R[[t]]}\cong \bigoplus^n_{i=1} R$ with the product topology, we then obtain a topology on $R((t))$ by equation $(\ast)$.
One can show that: $$\cdots\supset t^m R[[t]]\supset t^{m+1}R[[t]]\supset\cdots\quad (\ast\ast)$$
Is a local basis (of open sets) at the point $0$. Here my question:
By the above construction it seems that we obtain always the same topology on $R((t))$, independently from the starting topology on $R$! This because the local basis at $0$ is always $(\ast\ast)$. How is it possible? There should be some mistake in my reasoning.
It is simply not true that $(**)$ is a local basis at $0$ in general. Indeed, these sets are not even open if $R$ is not discrete.
For instance, let us consider $tR[[t]]$. If it is open in $R((t))$, then it must be open in $R[[t]]$ with the inverse limit topology. But the inverse limit topology on $R[[t]]$ is just the product topology, identifying $R[[t]]$ with $R^\mathbb{N}$ in the obvious way (indeed, you can prove the inverse limit topology has the same universal property as the product topology, since we are using the product topology on each quotient $R[[t]]/t^nR[[t]]$). In this identification, $tR[[t]]$ is the set of elements of $R^\mathbb{N}$ whose $0$th coordinate is $0$. This won't be open in the product topology unless $\{0\}$ is open in $R$, i.e. unless $R$ is discrete.