I am aware of the question (Topology on $\mathbf{Z}_p$)

One can use two definitions of the p-adic integers, the one that sees them as infinite formal sums $$\sum_{i=0}^\infty a_i p^i$$ and the one that sees them as inverse limit $$\varprojlim \mathbb Z/p^i\mathbb Z$$. One can show that those two are isomorphic as rings.

Now the infinite formal sums can be endowed with the p-adic norm, which gives rise to a topological structure induced by the norm; and the inverse limit inherits the product topology (we give each group the discrete topology).

I am trying to show that those two topologies are equal.

So given a

The p-adic norm induces a metric in which a neighbourhood base of $$0$$ is given by the open balls $$\{B(0,\frac{1}{p^n})\}$$. Infinite formal sums in $$B(0,\frac{1}{p^n})$$ are divisible by $$p^n$$, and henceforth correspond to sequences in the inverse limit belonging to the subset $$\left(\underbrace{\{0\}\times \{0\}}_{n\text{ times}}\times \prod_{k\geq n+1} \mathbb Z/p^k\mathbb Z\right)\cap \varprojlim \mathbb Z/p^i\mathbb Z$$, obviously an open set.

Is this correct?

Conversely, an open basis set in $$(\mathbb Z_p)_{\text{proj}}$$ is of the form $$\left(\prod_{j\in K\subset \mathbb N\text{ finite}} U_j \times \prod_{j\in \mathbb N\setminus K} \mathbb Z/p^j\mathbb Z\right)\cap \varprojlim \mathbb Z/p^\ell \mathbb Z$$ where $$U_j\subset \mathbb Z/p^j\mathbb Z$$ is open .

How do I show that this corresponds to an open set in the p-adic metric?

• Both topology are $+$ invariant and $p^n \Bbb{Z}_p$ is a basis of neighborhoods of $0$ in both case ($U$ is open in the metric topology iff $\forall a\in U$ it contains $|a-x|< r$ for some $r$, and $U$ is open in the $\varprojlim$ topology iff $\forall a\in U$ it contains $\ker(x\to x -a\bmod p^n)$ for some $n$) Commented Nov 18, 2019 at 10:13

Let $$\Bbb Z_p$$ denote the ring of $$p$$-adic ring as infinite formal sums. For every $$n\in\Bbb N$$ let $$\varepsilon_n:\Bbb Z_p\to\Bbb Z/p^n\Bbb Z$$ denote the reduction modulo $$p^n$$. Each of these is open and continuous, because for every $$a\in\Bbb Z$$ we have $$\varepsilon_n^{-1}\{a+p^n\Bbb Z\}=a+p^n\Bbb Z_p$$ Since the projective limit is endowed with the initial topology, the induced ring homomorphism $$\varepsilon:\Bbb Z_p\xrightarrow\sim\varprojlim\Bbb Z/p^n\Bbb Z$$ is continuous as well. Let $$\lambda_n:\varprojlim\Bbb Z/p^n\Bbb Z\to\Bbb Z/p^n\Bbb Z$$ denote the limit cone. Then \begin{align} \varepsilon\{a+p^n\Bbb Z_p\} &=\varepsilon\varepsilon_n^{-1}\{a+p^n\Bbb Z\}\\ &=\varepsilon^{-1}\varepsilon\lambda_n^{-1}\{a+p^n\Bbb Z\}\\ &=\lambda_n^{-1}\{a+p^n\Bbb Z\} \end{align} thus $$\varepsilon$$ is an open mapping hence an homeomorphism.