$\Bbb {Z}_p \cong lim_{←n} \Bbb{Z}_p/p^n \Bbb{Z}_p$,$a→(・・・,amodp^2,amodp$),(set theoretic bijection) implies $ \Bbb{Z}_p$ is complete as metric space ?
Completeness of ring is often defined as completion as ring $lim_{←n} \Bbb{Z}_p/p^n \Bbb{Z}_p$ is isom(as ring)to $ \Bbb{Z}_p$ itself. I want to know relation between the definition of completeness using cushy sequence. Titled question is what I want to know. 'Yes or No' answer and just hint is also appreciated (ofcourse selecontained answer is really welcomed). Thank you for your help.
This is more or less true, but set theoretical bijection isn't the key. There is an algebraic completion, which makes the projective limit $\lim \mathbb Z/p^n\mathbb Z$ a completion. It coincides with the metric completion in this case.
To be a little bit more precise, consider a (unital commutative) ring $R$ with an ideal $I$. We can define a (pseudo)metric on $R$ by $d(a,b) = r^{\max \{n | a-b\in I^n\}}$. (When $\cap_n I^n=\{0\}$, this is a metric, otherwise we get a metric on the quotient ring.) It can be shown that the uniform structure induced by the metric doesn't depend on the choice of $0<r<1$ (in the $p$-adic case, $r=\frac{1}{p}$ to make the product formula hold). Therefore the completion doesn't depend on the choice of $r$ either.
Now we can complete $R$ with respect to the metric $d$. This turns out to be isomorphic to the algebraic completion $\lim R/I^n$, and the proof is more or less straightforward.
More generally, we could have valuations on a ring which are not necessarily rank 1 (hence cannot be defined by a metric), but we still can make completions along them. This is quite useful in the theory of adic and perfectoid spaces. See e.g. Brian Conrad's notes.