The topology of the p-adic numbers 2

Our goal for this post is to review topology and prove that $\mathbb{Z}_p$ is a topological group/ring. In order to do so, we prove properties of topological groups, inverse limits, and neighborhoods.

This is my attempt to understand the following lines from Serre's A Course in Arithmetic:

"The ideals $p^n\mathbb{Z}_p$ form a basis of neighborhoods of $0$; since $x \in p^n \mathbb{Z}_p$ is equivalent to $v_p(x) \geq n$, the topology on $\mathbb{Z}_p$ is defined by the distance $d(x,y) = e^{-v_p(x-y)}$. Since $\mathbb{Z}_p$ is compact, it is complete. Finally, if $x = (x_n)$ is an element of $\mathbb{Z}_p$, and if $y_n \in \mathbb{Z}$ is such that $y_n \equiv x_n ~(\bmod{~p^n})$, then $\lim y_n = x$, which proves that $\mathbb{Z}$ is dense in $\mathbb{Z}_p$."  

Definition: A filter $\mathcal{F}$ on a set $S$ is a subset of $\mathcal{P}(S)$, the power group of $S$ satisfying the following properties,
1) If $A, B \in \mathcal{F}$, then $A \cap B \in \mathcal{F}$.
2) If $A \in \mathcal{F}$ and $A \subset B \in \mathcal{P}(S)$, then $B \in \mathcal{F}$.

If the filter contains $\varnothing$, then all subsets of $S$ is inside the filter. Hence, we will assume that the filter does not contain the empty set. Such filter is called a proper filter.

We immediately give an example of a filter. Let $\tau$ be a topology on $G$. Let $g \in G$, then consider the set of all neighborhoods of $x$ denoted $\mathcal{V}(g)$.
$$\mathcal{V}(g) := \{ V \in \mathcal{P}(S) ~|~ \exists~U \in \tau, g \in U \subset V \}$$
We can easily check the two properties.
1) Let $V_1, V_2 \in \mathcal{V}(g)$, then there exist $U_1$ and $U_2$ such that $g \in U_i \subset V_i$ for $i=1,2$. Then clearly, $g \in U_1 \cap U_2 \subset V_1 \cap V_2$ with $U_1 \cap U_2 \in \tau$.
2) If $V_1 \subset V_2$ with $V_1 \in \mathcal{V}(g)$, then there exists $U \in \tau$ such that $g \in U \subset V_1 \subset V_2$. This immediately shows that $V_2 \in \mathcal{V}(g)$.

Because $\varnothing$ does not contain an element, it cannot be a neighborhood of $g$. This shows that $\mathcal{V}(g)$ is a proper filter for all $g \in G$.

Definition: A filter base $\mathcal{B}$ is a subset of $\mathcal{P}(S)$ satisfying
1) $\mathcal{B}$ is nonempty, and if for any $A,B \in \mathcal{B}$ there exists $C \in \mathcal{B}$ such that $C \subset A \cap B$.
2) $\varnothing \notin \mathcal{B}$

We define the filter $\mathcal{F}$ generated by a filter base $\mathcal{B}$ to be the smallest filter containing $\mathcal{B}$. In other word,
$$\mathcal{F} = \bigcap_{\substack{\mathcal{F}' ~\text{filter} \\ \mathcal{B} \subset \mathcal{F}'}} \mathcal{F}'$$
For a filter $\mathcal{F}$, any $A, B \in \mathcal{F} \Rightarrow A \cap B \subset A \cap B$ with $A \cap B \in \mathcal{F}$. Hence, every proper filter is a filter base. This implies that a filter can be viewed as some sort of completion of a filter base.

Suppose we have $\mathcal{B}(g) \subset \mathcal{V}(g)$ with the following property: $V \in \mathcal{V}(g)$ if and only if there exists $B \in \mathcal{B}(g)$ such that $B \subset V$. We call such $\mathcal{B}(g)$ a neighborhood base of $g$.

Let $A, B \in \mathcal{B}(g)$, then $A$ and $B$ are neighborhoods of $g$ by themselves, as set contains itself. $A \cap B$ is thus a neighborhood, and there exists an $C \in \mathcal{B}(g)$ such that $C \subset A \cap B$ by definition. Also, as $\varnothing$ is not a neighborhood of any element, $\varnothing \notin \mathcal{B}(g)$. This shows that the neighborhood base is a filter base.

Let filter $\mathcal{F}$ contain $\mathcal{B}(g)$. Let $V \in \mathcal{V}(g)$, then by definition, $B \subset V$ by some $B \in \mathcal{B}(g)$. By the definition of a filter, $V \in \mathcal{F}$. It is clear now that $\mathcal{B}(g)$ generates the neighborhood filter of $g$. There are many literature which calls a neighborhood base as a fundamental system of neighborhoods.

Next topological property that we will review is continuity. Many (including myself) is familiar with the following definition. Let $f : X \rightarrow Y$ be a map between topological spaces $X$ and $Y$. Then $f$ is called continuous if for all open set $V  \subset Y$, the inverse image $f^{-1}(V) \subset X$ is open. However, there exists a neighborhood definition which resembles the $\varepsilon-\delta$ definition.

Definition: A map $f: X \rightarrow Y$ between topological spaces is said to be continuous at a point $x \in X$ if for a neighborhood $W$ of $f(x)$, there exists $V$ a neighborhood of $x$ such that $f(V) \subset W$.

Proposition: A map $f: X \rightarrow Y$ between topological spaces is continuous if and only if $f$ is continuous at every point $x$ in $X$.

Proof)
Let $f$ be continuous. Fix $x \in X$ and let $W$ be a neighborhood of $f(x)$. Then there exists a open set $W'$ containing $f(x)$ with $W' \subset W$. This implies that $x \in f^{-1}(W')$ with $f^{-1}(W')$ open by continuity.  $f^{-1}(W)$ becomes the neighborhood of $x$ with $f(f^{-1}(W')) \subset W' \subset W$.

Conversely, let $f$ be continuous at every point in $X$. Now consider an open set $W \subset Y$. If $f^{-1}(W)$ is empty, then there is nothing to prove. If $f^{-1}(W)$ is not empty, then for every $x \in f^{-1}(W)$, $f(x) \in W \Rightarrow$ there exist $V_x$ neighborhood of $x$ such that $f(V_x) \subset W$. For each $V_x$, there exists $U_x$ open set containing $x$ such that $U_x \subset V_x$. As $V_x \subset f^{-1}(W)$, $U_x \subset f^{-1}(W)$. Clearly the union of all $U_x$ equals $f^{-1}(W)$, thus open.

Now we are ready to learn about topological groups.

Definition: A topological group $G$ is a group endowed with a topology $\tau$ such that the multiplication $G \times G \rightarrow G$ defined by $(g,h) \mapsto gh$ and the inverse $G \rightarrow G$ defined by $g \mapsto g^{-1}$ is continuous when $G \times G$ is endowed with the product topology.

The two inclusions maps $i_1 : G \rightarrow G \times G$ and $i_2: G \rightarrow G \times G$ defined by $i_1(g) = (g,h)$ and $i_2(g) = (e, h)$ for some $h \in G$ are continuous as $i_j^{-1}(U_1 \times U_2) = U_j$ for $j=1,2$ with $U_j$ open.

Furthermore, if $f: X \rightarrow X'$ and $g: Y \rightarrow Y'$, then we can define the $f \times g : X \times Y \rightarrow X' \times Y'$ by $(f \times g)(x,y) = (f(x), g(y))$. Then observing that $(f \times g)^{-1}(U_1 \times U_2) = f^{-1}(U_1) \times g^{-1}(U_2)$, we see that the (cartesian) product of continuous functions is continuous.

The composition
$$G \times G \rightarrow G \times G \rightarrow G$$
defined by
$$(g,h) \mapsto (g, h^{-1}) \mapsto gh^{-1}$$
is continuous.

Conversely, suppose the above composition is continuous in some group $G$ endowed with some topology. Then
$$G \rightarrow G \times G \rightarrow G$$
defined by $g \mapsto (e,g) \mapsto g^{-1}$ is continuous. Also,
$$G \times G \rightarrow G \times G \rightarrow G$$
defined by $(g,h) \mapsto (g, h^{-1}) \mapsto g(h^{-1})^{-1} = gh$ is contiuous. Instead of saying that addition and inverse map are continuous, we could simply prove that $(g,h) \mapsto gh^{-1}$ is continuous to show that a group is a topological group.

Then continuity of the injections implies that the left translation and right translation are continuous because for a fixed $h \in G, g \mapsto (g,h) \mapsto gh$ is continuous. Composition of both left and right translation gives an internal automorphism which is continuous.

The continuity of addition, inverse, and the internal automorphism can be translated into the following properties, let $e \in G$
1) For all $U \in \mathcal{V}(e)$, there exists $V \in \mathcal{V}(e)$ such that $VV \subset U$ ,
2) For all $U \in \mathcal{V}(e)$, there exists $V \in \mathcal{V}(e)$ such that $V^{-1} \subset U$.
3) For all $U \in \mathcal{V}(e), g \in G$, there exists $V \in \mathcal{V}(e)$ such that $gVg^{-1} \subset U$.

In fact, the above property characterizes the neighborhood filter. In other words, let $\mathcal{V}$ be a filter on $G$ satisfying 1), 2), 3).

We first show that for all $U \in \mathcal{V}, e \in U$. Take $W \in \mathcal{V}$ such that $WW \subset U$. There exists $V \in \mathcal{V}$ such that $V^{-1} \subset W$. We have that $V' = V \cap W \in \mathcal{V}$. Also, $V'^{-1} \subset V^{-1}$, hence $V'^{-1} \subset W$. We conclude by $e \in V'(V')^{-1} \subset WW \subset U$.

Next we will define a topology such that $\mathcal{V}$ is the collection of all neighborhoods of $e$, and that the such topology makes $G$ into a topolgocial group. Consider the following,
$$\tau = \{ U \subset G ~|~ \forall~g \in U, \exists~U_g \in \mathcal{V} \text{ such that } gU_g \subset U \}$$
The fact that $\cup gU_i = g ( \cup U_i )$ and $\cap g( U_i) = g (\cap U_i )$ implies that $\tau$ is a topology. In this topology, we let $\mathcal{V}(g)$ be the collection of all neighborhoods of $g \in G$. Also, we denote $g\mathcal{V} = \{ gV ~|~ V \in \mathcal{V}$.

First we show that $\mathcal{V}(g) \subset g\mathcal{V}$. Let $V \in \mathcal{V}(g)$, then there exists $U \in  \tau$ such that $g \in U \subset V$. By the construction of $\tau$, there exists $U_g  \in \mathcal{V}$ such that $gU_g \subset U \subset V$. Then this implies that $U_g \subset g^{-1}V$. Because $U_g \in \mathcal{V}$ and $\mathcal{V}$ is a filter, $g^{-1}V \in \mathcal{V} \Rightarrow V \in g\mathcal{V}$.

We prove the converse. Let $V \in \mathcal{V}$. We define
$$U = \{ h \in gV ~|~ \exists~W \in \mathcal{V} \text{ such that } hW \subset gV \}$$
Clearly, $gV \subset gV \Rightarrow g \in U$ and $U$ is a subset of $gV$. So it suffices to show that $U \in \tau$. Let $x \in U$, then by definition, there exists $W$ such that $xW \subset gV$. Let $O \in \mathcal{V}$ such that $OO \subset W \Rightarrow xOO \subset xW \subset gV$. Because $O$ contains the identity, $xO \subset gV$. Also, as $xoO \subset gV$ for all $o \in O$, we see that $xO \subset U$ by the definition of $U$. This $O$ is exactly the $U_g$ needed for $U$ to be open.

Last thing we need to show is the continuity of the map $f$ defined by $(x,y) \mapsto xy^{-1}$. Fix $x,y \in G$. Let $U \in \mathcal{V}$. By 3), there exists $W \in \mathcal{V}$ such that $yWy^{-1} \subset U  \Rightarrow Wy^{-1} \subset y^{-1}U$. We can choose $V' \in \mathcal{V}$ such that $V' V' \subset W$, and similar to what we did above, there exists $V \in \mathcal{V}$ such that $V \subset V', V^{-1} \subset V'$. To conclude,  $VV^{-1} \subset V'V' \subset W$. Let $O = xV \times yV$, then it is clearly open as $xV, yV$ are open. Furthermore, $f(O)  = xVV^{-1}y^{-1} \subset xWy^{-1} \subset xy^{-1}U$.

In other words, fix a point $(x,y)$, then for any neighborhood of $xy^{-1}$ which will have the form $xy^{-1}U$, there exists a neighborhood of $(x,y)$ which is $xV \times yV$ such that $f(xV \times yV) \subset xy^{-1}U$, thus continuous.