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)
\[
(U_1 \cap H) \times (U_2 \cap H) = (U_1 \times U_2) \cap (H \times H)
\]
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)
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