The topology of the p-adic numbers 2
Our goal for this post is to review topology and prove that Zp 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:
Definition: A filter F on a set S is a subset of P(S), the power group of S satisfying the following properties,
1) If A,B∈F, then A∩B∈F.
2) If A∈F and A⊂B∈P(S), then B∈F.
If the filter contains ∅, 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 τ be a topology on G. Let g∈G, then consider the set of all neighborhoods of x denoted V(g).
V(g):={V∈P(S) | ∃ U∈τ,g∈U⊂V}
We can easily check the two properties.
1) Let V1,V2∈V(g), then there exist U1 and U2 such that g∈Ui⊂Vi for i=1,2. Then clearly, g∈U1∩U2⊂V1∩V2 with U1∩U2∈τ.
2) If V1⊂V2 with V1∈V(g), then there exists U∈τ such that g∈U⊂V1⊂V2. This immediately shows that V2∈V(g).
Because ∅ does not contain an element, it cannot be a neighborhood of g. This shows that V(g) is a proper filter for all g∈G.
Definition: A filter base B is a subset of P(S) satisfying
1) B is nonempty, and if for any A,B∈B there exists C∈B such that C⊂A∩B.
2) ∅∉B
We define the filter F generated by a filter base B to be the smallest filter containing B. In other word,
F=⋂F′ filterB⊂F′F′
For a filter F, any A,B∈F⇒A∩B⊂A∩B with A∩B∈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 B(g)⊂V(g) with the following property: V∈V(g) if and only if there exists B∈B(g) such that B⊂V. We call such B(g) a neighborhood base of g.
Let A,B∈B(g), then A and B are neighborhoods of g by themselves, as set contains itself. A∩B is thus a neighborhood, and there exists an C∈B(g) such that C⊂A∩B by definition. Also, as ∅ is not a neighborhood of any element, ∅∉B(g). This shows that the neighborhood base is a filter base.
Let filter F contain B(g). Let V∈V(g), then by definition, B⊂V by some B∈B(g). By the definition of a filter, V∈F. It is clear now that 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→Y be a map between topological spaces X and Y. Then f is called continuous if for all open set V⊂Y, the inverse image f−1(V)⊂X is open. However, there exists a neighborhood definition which resembles the ε−δ definition.
Definition: A map f:X→Y between topological spaces is said to be continuous at a point x∈X if for a neighborhood W of f(x), there exists V a neighborhood of x such that f(V)⊂W.
Proposition: A map f:X→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∈X and let W be a neighborhood of f(x). Then there exists a open set W′ containing f(x) with W′⊂W. This implies that x∈f−1(W′) with f−1(W′) open by continuity. f−1(W) becomes the neighborhood of x with f(f−1(W′))⊂W′⊂W.
Conversely, let f be continuous at every point in X. Now consider an open set W⊂Y. If f−1(W) is empty, then there is nothing to prove. If f−1(W) is not empty, then for every x∈f−1(W), f(x)∈W⇒ there exist Vx neighborhood of x such that f(Vx)⊂W. For each Vx, there exists Ux open set containing x such that Ux⊂Vx. As Vx⊂f−1(W), Ux⊂f−1(W). Clearly the union of all Ux 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 τ such that the multiplication G×G→G defined by (g,h)↦gh and the inverse G→G defined by g↦g−1 is continuous when G×G is endowed with the product topology.
The two inclusions maps i1:G→G×G and i2:G→G×G defined by i1(g)=(g,h) and i2(g)=(e,h) for some h∈G are continuous as i−1j(U1×U2)=Uj for j=1,2 with Uj open.
Furthermore, if f:X→X′ and g:Y→Y′, then we can define the f×g:X×Y→X′×Y′ by (f×g)(x,y)=(f(x),g(y)). Then observing that (f×g)−1(U1×U2)=f−1(U1)×g−1(U2), we see that the (cartesian) product of continuous functions is continuous.
The composition
G×G→G×G→G
defined by
(g,h)↦(g,h−1)↦gh−1
is continuous.
Conversely, suppose the above composition is continuous in some group G endowed with some topology. Then
G→G×G→G
defined by g↦(e,g)↦g−1 is continuous. Also,
G×G→G×G→G
defined by (g,h)↦(g,h−1)↦g(h−1)−1=gh is contiuous. Instead of saying that addition and inverse map are continuous, we could simply prove that (g,h)↦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∈G,g↦(g,h)↦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∈G
1) For all U∈V(e), there exists V∈V(e) such that VV⊂U ,
2) For all U∈V(e), there exists V∈V(e) such that V−1⊂U.
3) For all U∈V(e),g∈G, there exists V∈V(e) such that gVg−1⊂U.
In fact, the above property characterizes the neighborhood filter. In other words, let V be a filter on G satisfying 1), 2), 3).
We first show that for all U∈V,e∈U. Take W∈V such that WW⊂U. There exists V∈V such that V−1⊂W. We have that V′=V∩W∈V. Also, V′−1⊂V−1, hence V′−1⊂W. We conclude by e∈V′(V′)−1⊂WW⊂U.
Next we will define a topology such that V is the collection of all neighborhoods of e, and that the such topology makes G into a topolgocial group. Consider the following,
τ={U⊂G | ∀ g∈U,∃ Ug∈V such that gUg⊂U}
The fact that ∪gUi=g(∪Ui) and ∩g(Ui)=g(∩Ui) implies that τ is a topology. In this topology, we let V(g) be the collection of all neighborhoods of g∈G. Also, we denote gV={gV | V∈V.
First we show that V(g)⊂gV. Let V∈V(g), then there exists U∈τ such that g∈U⊂V. By the construction of τ, there exists Ug∈V such that gUg⊂U⊂V. Then this implies that Ug⊂g−1V. Because Ug∈V and V is a filter, g−1V∈V⇒V∈gV.
We prove the converse. Let V∈V. We define
U={h∈gV | ∃ W∈V such that hW⊂gV}
Clearly, gV⊂gV⇒g∈U and U is a subset of gV. So it suffices to show that U∈τ. Let x∈U, then by definition, there exists W such that xW⊂gV. Let O∈V such that OO⊂W⇒xOO⊂xW⊂gV. Because O contains the identity, xO⊂gV. Also, as xoO⊂gV for all o∈O, we see that xO⊂U by the definition of U. This O is exactly the Ug needed for U to be open.
Last thing we need to show is the continuity of the map f defined by (x,y)↦xy−1. Fix x,y∈G. Let U∈V. By 3), there exists W∈V such that yWy−1⊂U⇒Wy−1⊂y−1U. We can choose V′∈V such that V′V′⊂W, and similar to what we did above, there exists V∈V such that V⊂V′,V−1⊂V′. To conclude, VV−1⊂V′V′⊂W. Let O=xV×yV, then it is clearly open as xV,yV are open. Furthermore, f(O)=xVV−1y−1⊂xWy−1⊂xy−1U.
In other words, fix a point (x,y), then for any neighborhood of xy−1 which will have the form xy−1U, there exists a neighborhood of (x,y) which is xV×yV such that f(xV×yV)⊂xy−1U, thus continuous.
This is my attempt to understand the following lines from Serre's A Course in Arithmetic:
"The ideals pnZp form a basis of neighborhoods of 0; since x∈pnZp is equivalent to vp(x)≥n, the topology on Zp is defined by the distance d(x,y)=e−vp(x−y). Since Zp is compact, it is complete. Finally, if x=(xn) is an element of Zp, and if yn∈Z is such that yn≡xn (mod pn), then limyn=x, which proves that Z is dense in Zp."
Definition: A filter F on a set S is a subset of P(S), the power group of S satisfying the following properties,
1) If A,B∈F, then A∩B∈F.
2) If A∈F and A⊂B∈P(S), then B∈F.
If the filter contains ∅, 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 τ be a topology on G. Let g∈G, then consider the set of all neighborhoods of x denoted V(g).
V(g):={V∈P(S) | ∃ U∈τ,g∈U⊂V}
We can easily check the two properties.
1) Let V1,V2∈V(g), then there exist U1 and U2 such that g∈Ui⊂Vi for i=1,2. Then clearly, g∈U1∩U2⊂V1∩V2 with U1∩U2∈τ.
2) If V1⊂V2 with V1∈V(g), then there exists U∈τ such that g∈U⊂V1⊂V2. This immediately shows that V2∈V(g).
Because ∅ does not contain an element, it cannot be a neighborhood of g. This shows that V(g) is a proper filter for all g∈G.
Definition: A filter base B is a subset of P(S) satisfying
1) B is nonempty, and if for any A,B∈B there exists C∈B such that C⊂A∩B.
2) ∅∉B
We define the filter F generated by a filter base B to be the smallest filter containing B. In other word,
F=⋂F′ filterB⊂F′F′
For a filter F, any A,B∈F⇒A∩B⊂A∩B with A∩B∈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 B(g)⊂V(g) with the following property: V∈V(g) if and only if there exists B∈B(g) such that B⊂V. We call such B(g) a neighborhood base of g.
Let A,B∈B(g), then A and B are neighborhoods of g by themselves, as set contains itself. A∩B is thus a neighborhood, and there exists an C∈B(g) such that C⊂A∩B by definition. Also, as ∅ is not a neighborhood of any element, ∅∉B(g). This shows that the neighborhood base is a filter base.
Let filter F contain B(g). Let V∈V(g), then by definition, B⊂V by some B∈B(g). By the definition of a filter, V∈F. It is clear now that 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→Y be a map between topological spaces X and Y. Then f is called continuous if for all open set V⊂Y, the inverse image f−1(V)⊂X is open. However, there exists a neighborhood definition which resembles the ε−δ definition.
Definition: A map f:X→Y between topological spaces is said to be continuous at a point x∈X if for a neighborhood W of f(x), there exists V a neighborhood of x such that f(V)⊂W.
Proposition: A map f:X→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∈X and let W be a neighborhood of f(x). Then there exists a open set W′ containing f(x) with W′⊂W. This implies that x∈f−1(W′) with f−1(W′) open by continuity. f−1(W) becomes the neighborhood of x with f(f−1(W′))⊂W′⊂W.
Conversely, let f be continuous at every point in X. Now consider an open set W⊂Y. If f−1(W) is empty, then there is nothing to prove. If f−1(W) is not empty, then for every x∈f−1(W), f(x)∈W⇒ there exist Vx neighborhood of x such that f(Vx)⊂W. For each Vx, there exists Ux open set containing x such that Ux⊂Vx. As Vx⊂f−1(W), Ux⊂f−1(W). Clearly the union of all Ux 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 τ such that the multiplication G×G→G defined by (g,h)↦gh and the inverse G→G defined by g↦g−1 is continuous when G×G is endowed with the product topology.
The two inclusions maps i1:G→G×G and i2:G→G×G defined by i1(g)=(g,h) and i2(g)=(e,h) for some h∈G are continuous as i−1j(U1×U2)=Uj for j=1,2 with Uj open.
Furthermore, if f:X→X′ and g:Y→Y′, then we can define the f×g:X×Y→X′×Y′ by (f×g)(x,y)=(f(x),g(y)). Then observing that (f×g)−1(U1×U2)=f−1(U1)×g−1(U2), we see that the (cartesian) product of continuous functions is continuous.
The composition
G×G→G×G→G
defined by
(g,h)↦(g,h−1)↦gh−1
is continuous.
Conversely, suppose the above composition is continuous in some group G endowed with some topology. Then
G→G×G→G
defined by g↦(e,g)↦g−1 is continuous. Also,
G×G→G×G→G
defined by (g,h)↦(g,h−1)↦g(h−1)−1=gh is contiuous. Instead of saying that addition and inverse map are continuous, we could simply prove that (g,h)↦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∈G,g↦(g,h)↦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∈G
1) For all U∈V(e), there exists V∈V(e) such that VV⊂U ,
2) For all U∈V(e), there exists V∈V(e) such that V−1⊂U.
3) For all U∈V(e),g∈G, there exists V∈V(e) such that gVg−1⊂U.
In fact, the above property characterizes the neighborhood filter. In other words, let V be a filter on G satisfying 1), 2), 3).
We first show that for all U∈V,e∈U. Take W∈V such that WW⊂U. There exists V∈V such that V−1⊂W. We have that V′=V∩W∈V. Also, V′−1⊂V−1, hence V′−1⊂W. We conclude by e∈V′(V′)−1⊂WW⊂U.
Next we will define a topology such that V is the collection of all neighborhoods of e, and that the such topology makes G into a topolgocial group. Consider the following,
τ={U⊂G | ∀ g∈U,∃ Ug∈V such that gUg⊂U}
The fact that ∪gUi=g(∪Ui) and ∩g(Ui)=g(∩Ui) implies that τ is a topology. In this topology, we let V(g) be the collection of all neighborhoods of g∈G. Also, we denote gV={gV | V∈V.
First we show that V(g)⊂gV. Let V∈V(g), then there exists U∈τ such that g∈U⊂V. By the construction of τ, there exists Ug∈V such that gUg⊂U⊂V. Then this implies that Ug⊂g−1V. Because Ug∈V and V is a filter, g−1V∈V⇒V∈gV.
We prove the converse. Let V∈V. We define
U={h∈gV | ∃ W∈V such that hW⊂gV}
Clearly, gV⊂gV⇒g∈U and U is a subset of gV. So it suffices to show that U∈τ. Let x∈U, then by definition, there exists W such that xW⊂gV. Let O∈V such that OO⊂W⇒xOO⊂xW⊂gV. Because O contains the identity, xO⊂gV. Also, as xoO⊂gV for all o∈O, we see that xO⊂U by the definition of U. This O is exactly the Ug needed for U to be open.
Last thing we need to show is the continuity of the map f defined by (x,y)↦xy−1. Fix x,y∈G. Let U∈V. By 3), there exists W∈V such that yWy−1⊂U⇒Wy−1⊂y−1U. We can choose V′∈V such that V′V′⊂W, and similar to what we did above, there exists V∈V such that V⊂V′,V−1⊂V′. To conclude, VV−1⊂V′V′⊂W. Let O=xV×yV, then it is clearly open as xV,yV are open. Furthermore, f(O)=xVV−1y−1⊂xWy−1⊂xy−1U.
In other words, fix a point (x,y), then for any neighborhood of xy−1 which will have the form xy−1U, there exists a neighborhood of (x,y) which is xV×yV such that f(xV×yV)⊂xy−1U, thus continuous.
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