The topology of the p-adic numbers 1

Let $A_n = \mathbb{Z}/p^n \mathbb{Z}$, then $A_n$ with the canonical epimorphism $\varphi_n: A_n \rightarrow A_{n-1}$ defines a projective system. The projective limit $\mathbb{Z}_p$ is the subring of
$$\prod_{n \geq 1} A_n$$
containing $(x_n)$ such that $\varphi_n(x_n) = x_{n-1}$. Suppose we give $A_n$ the discrete topology and $\prod A_n$ the product topology. By Tychonoff theorem, we have a compact space $\prod A_n$.  We call $\mathbb{Z}_p$ the $p$-adic integers.

Theorem 1 The $p$-adic integers $\mathbb{Z}_p$ is closed subset of $\prod A_n$.
Proof) We prove that the complement $\mathbb{Z}_p^c$ is open. Let $(x_n) \in \mathbb{Z}_p^c$, then we have some positive number $m$ such that $\varphi(x_m) \neq x_{m-1}$. Consider the following set
$$U = \{ x_1 \} \times \cdots \times \{ x_m \} \times \prod_{n > m} A_n$$
which is open because there are only finitely many proper subsets of $A_n$. Clearly $x_n \in U$, and for any other $(y_n) \in U$, we have that $y_m = x_m, y_{m-1} = x_{m-1} \Rightarrow \varphi_m(y_m) \neq y_{m-1}$. This means that $(y_m) \notin \mathbb{Z}_p$. This shows that $U \subset \mathbb{Z}_p^c$ and hence $\mathbb{Z}_p$ closed.

This turns $\mathbb{Z}_p$ into a compact space when given the subspace topology.

We will, however, approach the topology of the $p$-adic integers by using a more number-theoretic method. By the fundamental theorem of arithmetic, for any integer $m$, there exists a unique $n$ such that $p^n | m$ yet $p^{n+1} \not| m$. We define the $p$-adic valuation $v_p: \mathbb{Z} \rightarrow \mathbb{N}$ by $v_p(m) = n$. We let $v_p(0) = \infty$. The $p$-adic valuation satisfies the following properties,

Properties
     1) $v_p(xy) = v_p(x) + v_p(y)$
     2) $v_p(x+y) \geq \min(v_p(x), v_p(y))$

We have that $x = p^ex', y= p^fy'$ where $e= v_p(x)$ and $f = v_p(y)$ with $x', y'$ not divisible by $p$. Then $xy = p^{e+f}x'y'$ with $x'y'$ coprime to $p$. This show the first property. For the second property, we let $e$ be the minimum with out loss of generosity. Then $x+y = p^e(x'+p^{f-e}y')$ but because $p \not| x' \Rightarrow p \not| x'+ p^{f-e}y'$ this implies that $v_p(x+y) \geq e = \min(v_p(x), v_p(y))$.

We extend $v_p$ to $\mathbb{Q}$ as follows. For all $q \in \mathbb{Q}, q = x/y$ for some integers $x,y$. $v_p(q) = v_p(x) - v_p(y)$. In the case when $x = 0$, we let $\infty - n = 0$ for any integer $n \in \mathbb{Z}$. It is easy to check well-definedness. Let $x_1/y_1 = x_2/y_2 \Rightarrow x_1y_2 = y_1x_2$.
$$\begin{array}{llcl} & v_p(x_1y_2) & = & v_p(y_1x_2) \\ \Rightarrow & v_p(x_1) + v_p(y_2) & = & v_p(y_1) + v_p(x_2) \\ \Rightarrow & v_p(x_1) - v_p(x_2) & = & v_p(y_1) - v_p(y_2) \\ \Rightarrow & v_p(x_1/x_2) & = & v_p(y_1/y_2) \end{array}$$
We define another map $|x|_p = p^{-v_p(x)}$, then by the above properties, we have

Properties
     1) $|x|_p \geq 0$ and equality if and only if $x = 0$
     2) $|xy|_p = |x|_p|y|_p$
     3) $|x+y|_p \leq \max(|x|_p, |y|_p) \leq |x|_p + |y|_p$

$p^{-v_p(xy)} = p^{-v_p(x)-v_p(y)} = p^{-v_p(x)}p^{-v_p(y)}$
$v_p(x+y) \geq \min(v_p(x), v_p(y)) \Rightarrow -v_p(x+y) \leq \max(-v_p(x), -v_p(x))$

with the have that $p^x \geq 0$ for all real number $x$ give us the desired results. This makes $\mu(x,y) = |x-y|_p$ into a metric, hence $\mathbb{Q}$ becomes a metric space respect to $\mu$. We may define Cauchy sequence via this absolute value and prove that the set of all Cauchy sequences is, in fact, a ring. The subset of all null sequences forms a maximal ideal, and the quotient ring becomes a field. This field is called a completion of $\mathbb{Q}$ via the valuation $v_p$, and we denote it by $\mathbb{Q}_p$. We can prove that $\mathbb{Q}_p$ can also be a metric space by extending $v_p$ further to $\mathbb{Q}_p$ and contains $\mathbb{Q}$ as a dense subspace.

The set
$$\mathbb{Z}_p := \{ x \in \mathbb{Q}_p ~|~ v_p(x) \geq 0 \}$$
is a subring of $\mathbb{Z}_p$. The notation would not cause any confusion with our original $p$-adic integers $\mathbb{Z}_p$ because we have that $\mathbb{Z}_p$ defined using valuation and completion and $\mathbb{Z}_p$ defined using projective limit are isomorphic (both algebraically and topologically).