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:

"The ideals pnZp form a basis of neighborhoods of 0; since xpnZp is equivalent to vp(x)n, the topology on Zp is defined by the distance d(x,y)=evp(xy). Since Zp is compact, it is complete. Finally, if x=(xn) is an element of Zp, and if ynZ is such that ynxn (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,BF, then ABF.
2) If AF and ABP(S), then BF.

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 gG, then consider the set of all neighborhoods of x denoted V(g).
V(g):={VP(S) |  Uτ,gUV}
We can easily check the two properties.
1) Let V1,V2V(g), then there exist U1 and U2 such that gUiVi for i=1,2. Then clearly, gU1U2V1V2 with U1U2τ.
2) If V1V2 with V1V(g), then there exists Uτ such that gUV1V2. This immediately shows that V2V(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 gG.

Definition: A filter base B is a subset of P(S) satisfying
1) B is nonempty, and if for any A,BB there exists CB such that CAB.
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 filterBFF
For a filter F, any A,BFABAB with ABF. 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: VV(g) if and only if there exists BB(g) such that BV. We call such B(g) a neighborhood base of g.

Let A,BB(g), then A and B are neighborhoods of g by themselves, as set contains itself. AB is thus a neighborhood, and there exists an CB(g) such that CAB 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 VV(g), then by definition, BV by some BB(g). By the definition of a filter, VF. 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:XY be a map between topological spaces X and Y. Then f is called continuous if for all open set VY, the inverse image f1(V)X is open. However, there exists a neighborhood definition which resembles the εδ definition.

Definition: A map f:XY between topological spaces is said to be continuous at a point xX if for a neighborhood W of f(x), there exists V a neighborhood of x such that f(V)W.

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

Proof)
Let f be continuous. Fix xX and let W be a neighborhood of f(x). Then there exists a open set W containing f(x) with WW. This implies that xf1(W) with f1(W) open by continuity.  f1(W) becomes the neighborhood of x with f(f1(W))WW.

Conversely, let f be continuous at every point in X. Now consider an open set WY. If f1(W) is empty, then there is nothing to prove. If f1(W) is not empty, then for every xf1(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 UxVx. As Vxf1(W), Uxf1(W). Clearly the union of all Ux equals f1(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×GG defined by (g,h)gh and the inverse GG defined by gg1 is continuous when G×G is endowed with the product topology.

The two inclusions maps i1:GG×G and i2:GG×G defined by i1(g)=(g,h) and i2(g)=(e,h) for some hG are continuous as i1j(U1×U2)=Uj for j=1,2 with Uj open.

Furthermore, if f:XX and g:YY, then we can define the f×g:X×YX×Y by (f×g)(x,y)=(f(x),g(y)). Then observing that (f×g)1(U1×U2)=f1(U1)×g1(U2), we see that the (cartesian) product of continuous functions is continuous.

The composition
G×GG×GG
defined by
(g,h)(g,h1)gh1
is continuous.

Conversely, suppose the above composition is continuous in some group G endowed with some topology. Then
GG×GG
defined by g(e,g)g1 is continuous. Also,
G×GG×GG
defined by (g,h)(g,h1)g(h1)1=gh is contiuous. Instead of saying that addition and inverse map are continuous, we could simply prove that (g,h)gh1 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 hG,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 eG
1) For all UV(e), there exists VV(e) such that VVU ,
2) For all UV(e), there exists VV(e) such that V1U.
3) For all UV(e),gG, there exists VV(e) such that gVg1U.

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 UV,eU. Take WV such that WWU. There exists VV such that V1W. We have that V=VWV. Also, V1V1, hence V1W. We conclude by eV(V)1WWU.

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={UG |  gU, UgV such that gUgU}
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 gG. Also, we denote gV={gV | VV.

First we show that V(g)gV. Let VV(g), then there exists Uτ such that gUV. By the construction of τ, there exists UgV such that gUgUV. Then this implies that Ugg1V. Because UgV and V is a filter, g1VVVgV.

We prove the converse. Let VV. We define
U={hgV |  WV such that hWgV}
Clearly, gVgVgU and U is a subset of gV. So it suffices to show that Uτ. Let xU, then by definition, there exists W such that xWgV. Let OV such that OOWxOOxWgV. Because O contains the identity, xOgV. Also, as xoOgV for all oO, we see that xOU 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)xy1. Fix x,yG. Let UV. By 3), there exists WV such that yWy1UWy1y1U. We can choose VV such that VVW, and similar to what we did above, there exists VV such that VV,V1V. To conclude,  VV1VVW. Let O=xV×yV, then it is clearly open as xV,yV are open. Furthermore, f(O)=xVV1y1xWy1xy1U.

In other words, fix a point (x,y), then for any neighborhood of xy1 which will have the form xy1U, there exists a neighborhood of (x,y) which is xV×yV such that f(xV×yV)xy1U, thus continuous.

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