The topology of the p-adic numbers 3

Let $G$ be a topological space and let $H$ be a subspace of $G$ endowed with the subspace topology. Then $H \times H$ endowed with the product topology is same as the subspace topology of $H \times H \subset G \times G$ with $G \times G$ endowed with the product topology due to:
$$(U_1 \cap H) \times (U_2 \cap H) = (U_1 \times U_2) \cap (H \times H)$$
Let $G$ be a topological group with a subgroup $H$. We would like to show that $H$ endowed with the subspace topology is also a topological group. Let $p_G$ denote the product map. in $G$ and $p_H$ denote the product map in $H$. We want to show that $p_H$ is continuous. Let $U$ be an open set in $H$, then $U = U_G \cap H$ for some open set $U_G$ in $G$.
$$\begin{array}{lcl} p_H^{-1}(U) & = & \{ (h_1, h_2) \in H \times H ~|~ h_1h_2 \in U \} \\ & = & \{ (g_1, g_2) \in G \times G ~|~ g_1g_2 \in U_G \} \cap H \times H \\ & = & p_G^{-1}(U_G) \cap H \times H \end{array}$$
If $v_G$  and $v_H$ denote the inverse map of $G$ and $H$ respectively.  Suppose $v_G$ is continuous. And suppose $U \subset H$ be an open set in $H$. Then $U = U_G \cap H$ with $U_G$ open in $G$.
$$\begin{array}{lcl} v_H^{-1}(U) & = & \{ h \in H ~|~ h^{-1} \in U \} \\ & = & \{ g \in G ~|~ g^{-1} \in U_G \} \cap H \\ & =& v_G^{-1}(U_G) \cap H \end{array}$$
This two fact shows that subgroup with subspace topology will induce a topological group. Next step is to show that (Cartesian) product of topological groups will induce a topological group.

Proposition: Let $\{ G_i \}$ be a collection of topological groups. Then $G= \prod G_i$ is a topological group with the product topology.

Proof)
Let $\mathcal{V}_G$ be the collection of neighborhood of $e_G$. By the previous post, we have seen that all other neighborhood filters are determined by $\mathcal{V}_G$ simply by translation. Thus, it suffices to show that $\mathcal{V}_G$ satisfies the three properties (which implies the continuity of multiplication, inverse, and the inner automorphism.) However, this is simple once we see that
$$\prod A_i \cdot \prod B_i = \prod A_i  B_i \hspace{3em} (\prod A_i)^{-1} = \prod A_i^{-1} \hspace{3em} g (\prod A_i) g^{-1} = \prod (gA_ig^{-1})$$
Because each component of $\prod U_i \in \mathcal{V}_G$ is a neighborhood in $G_i$, there exists $V_i$ satisfying the desired properties. Then $\prod V_i$ where almost of $V_i = G_i$ shows that $\prod V_i \in \mathcal{V}_G$ and $\prod V_i \prod V_i \subset \prod U_i$, and similarly for all other properties.

All these results can be extended to topological rings.

Given $\mathbb{Z}/p^n \mathbb{Z}$ with the discrete topology, the modular rings are all topological ring. Thus its infinite product is also a topological ring. $\mathbb{Z}_p$ is a subgroup of the infinite product with induced topology, which we have proved to be a topological ring.

Now I can say in confidence that

Proposition: $\mathbb{Z}_p$ is a topological ring.

What is nice about topological groups is that if $\mathcal{V}(e)$ is the collection of neighborhoods of $e$, the identity element, then for all $g \in G$, the collection of neighborhoods of $g$ is $\mathcal{V}(g) = g\mathcal{V}(e)$. Let $V \in \mathcal{V}(g)$, then there exists an open set $U$ containing $g$ and that is contained in $V$. Clearly, $e \in g^{-1}U \subset g^{-1}V$. Because the translation map is continuous in a topological group, $g^{-1}U$ is open, thus $g^{-1}V \in \mathcal{V}(e)$. In other words, $V \in g\mathcal{V}(e)$. Conversely, if $e \in U \subset V$, then $g \in gU \subset gV$ implies that $g\mathcal{V}(e) \subset \mathcal{V}(g)$. We conclude that collection of neighborhoods of $g$ for all $g \in G$ is determined by $\mathcal{V}(e)$. If we go further, this means that the basis of neighborhoods of $e$ is sufficient to determine the topology of a topological group.

In a metric space, it is not hard to see that the open disc around a point is a basis of neighborhoods of a given point. First, we observe that $p^n \mathbb{Z}_p$ form a basis of neighborhoods of $0$. Any open set $U$ in $\mathbb{Z}_p$ is of the form
$$U = \left ( U_1 \times \cdots \times U_m \times \prod_{n > m} \mathbb{Z}/p^n \mathbb{Z} \right )$$
with $0 \in U_i$. Consider $x \in p^m \mathbb{Z}_p$, then for $x = (x_n), x_n = 0$ for $n \leq m$. Then clearly, $x \in U \Rightarrow p^m \mathbb{Z}_p  \subset U$. If $\mathcal{B}(e) = \{ p^n \mathbb{Z}_p$, then for all $V \in \mathcal{V}(e)$ ,there exists an open set $U \subset V \Rightarrow p^m \mathbb{Z}_p \subset U \subset V$ for some $m$ shows that $\mathcal{B}(e)$ is a basis.

For those who aren't familiar with the term basis, it is the filter base of the neighborhood filter. By the following equivalence
$$x \in p^m \mathbb{Z}_p \Leftrightarrow v_p(x) \geq n \Leftrightarrow v_p(x) \leq e^{-m}$$
We see that $p^m \mathbb{Z}_p  = B_{\varepsilon}(e)$ where $B_\varepsilon(e)$ denotes the open disc around $e$ with radius $\varepsilon$. The discreteness of $v_p$ shows that $p^m \mathbb{Z}_p$ is the only possible open discs around $e$, thus coincides with the metric basis. We have thus shown that topology induced by the inverse limit is equivalent to the metric topology induced by the metric $\mu(x,y) = e^{-v_p(x-y)}$.

By basic topology, a compact metric space is complete.

Finally, consider $x \in \mathbb{Z}_p$, then there always exists $y_n \in \mathbb{Z}$ such that $y_n \equiv x_n ~(\bmod{~p^n})$ Then because $y_n - x_n$ is divisible by $p^n$, $p^n | (y_n - x_n) \Rightarrow \mu (y_n - x_n) \leq e^{-n}$. Then as $n \rightarrow \infty$, you see that the limit goes to $0$. This implies that $x$ is a limit point of $\mathbb{Z}$ which shows denseness.

I have thus understood what Serre's four lines that I mentioned in my second post (of the topology of the $p$-adic numbers) means. (Sort of)