Topology of numbers

A topology of numbers 

Natural*

Introduction

 Topology is one of the most modern branches of mathematics and  its relationship with other areas of this science is fundamental.  

The study of  topology focuses on qualitative properties and not on quantitative ones  of mathematical objects, its extensive development has given rise to different areas, such as topology: algebraic, differential, sets, geometric.  

The present work belongs to the topology of sets.  In the bibliography the reader will be able to find sources for each of these topics.

 The topology of a set

 Any set can be given a structure of topological space.

 To do this, it is enough to distinguish a family τ of subsets of X, whose

 elements are given the name of open subsets of X. Such a family

 must satisfy the following three properties:

 a) the total space X and the empty set ∅ are in τ ;

 b) the arbitrary union of elements in τ is also an element of τ ;

 c) the finite intersection of elements in τ is also an element of τ .

The pair (X, τ ) is called a topological space.

 The usual topology of the reals

 As an example of a topological space, let us consider the set of real numbers R. To endow this set with a topology, consider

the family τ = {U ⊂ R|  if x ∈ U there exist a, b ∈ R, such that x ∈ (a, b) ⊂ U}.

It is easy to prove that (R, τ ) is a topological space.  To this topology of

R is known as the usual topology of R. This is because it can be give many other topologies, but this one in particular is the most important of them.

 Examples and counterexamples

 In the investigation of the properties of topological spaces it is;

It is important to construct spaces that fulfill certain properties interesting.  

Let's think about the question:

 Is there a topological space X such that of all bijective functions  f: X → X, only the identity is continuous?

 If there were a topological space with this property, it would be an example pathological with respect to what our intuition expects.

 If there were no space with that property, then we would have a theo- rema that could guarantee the existence of bijective functions, continuousand distinct from the identity in any topological space.

 In counterexamples the opposite is true: one can suspect that a certain proposition is true, but suddenly a person arrives and shows us a space that satisfies the hypothesis, but not the conclusion of our statement maci'on;  that is, it shows us a counterexample to our reasoning.

No doubt, at that time we will look for another direction in our work.

 Thus, we can observe a kind of duality in the examples and counterexamples, whose main purpose is to be a compass in the mathematical task.

 Properties of topological spaces

 Due to the great generality of the definition of a topological space, it is

 It is necessary to add more properties to aspire to unravel its structure.

 We have seen that a topological space is a pair (X, τ ) where τ is a family of subsets of X which are called subsets X open.

Definition 1: A subset A ⊂ X of X is said to be a set closed of X if its complement is open, that is, if X − A ∈ τ .

 Definition 2: Let A be a subset of X, we define the closure of A,denoted A¯, as the smallest closed set of X that contains A.

 Definition 3: Let (X, τ ) be a topological space.  A subset β of τ it says that it is a basis for the topology τ of the space X, if it satisfies the two following conditions: 

i) for all x ∈ X there exists U ∈ τ such that x ∈ U;

 ii) given U, V ∈ τ and x ∈ U ∩ V , there exists W ∈ τ such that x ∈ W ⊂ U ∩ V .

 Definition 4: A topological space is Hausdorff if for every pair ofdistinct points x, y there exist open sets U, V , containing x and a  respectively, such that U ∩ V = ∅.

 Definition 5: A space X is connected if X is not the union of two subcon- non-empty disjoint open together of X.

 Definition 6: A topological space X is locally connected if given any

 For any open U of X and an element x ∈ U, there exists an open V such that

 x ∈ V ⊂ U and V cannot be written as the union of two open spaces of X.

 For example, the reals with the usual topology is a Hausdorff space,

 connected, locally connected.

 Our space

 We are going to consider the set of natural numbers and we are going to endow it with of a topology and thus obtain a topological space which we will denote by x.

 The topology that we are going to give to the naturals will be the one induced by the following family of subsets of N.

 Given a, b ∈ N, with (a, b) = 1, define Ub

 (a) = {a + bk ∈ N/k ∈ Z}.

 Be

 β = {Ub (a)/a, b ∈ N}.

 We are going to prove that β is a basis for some topology of N.

 Let a, b ∈ N such that (a, b) = 1. Then a ∈ Ub(to);  that is, given any point of N, there exists an element of β that contains it.

 To conclude that β is a basis, we will use the following lemma:

 Motto 1: If Ux (y), ub (a) ∈ β, then Ux (y) ∩Ub(a) 6= ∅ if and only if

 (x, b)|y − a.

Demonstration.  

We observe that z ∈ Ux (y)∩Ub (a) if and only if z = y+k1x = to + k2 b for some k1 , k2 ∈ Z, but this is equivalent to the equation

y−a = k2

 b−k1x has a solution k1 , k2 ∈ Z. But this solution exists if and only if (x, b)|y−a.  If the last equation has an integer solution, it can happen that

 z = y +k1x = a+k2

 b is negative and then Ux (y), ub

 (a) would intersect in Z, but we want them to intersect in N. To do this, just add to each member of the equation y + k1x = a + k2 b the term bmx with m a natural large enough that y + (k1 + mb)x = a + (k2 + mx)b be a natural number.

 Let's finish proving that β is a basis.  sean ux(y) ∈ β and U (a) ∈ β not disjoint.  Let z ∈ Ux (y)∩Ub (a), then z = y+k1x = a+k2 b with k1 , k2 ∈ Z. Observe that z ∈ U[x,b] (z).  We affirm that U[x,b] (z) ⊂ Ux (y) ∩ Ub (to).

 Let z + k3 [x,b] ∈ U[x,b] (z). 

 Then, z+k3 [x, b] = 

y + k1x + k3 [x, b] 

= y + µ k1+ k3 [x,b] x ¶ x ∈ Ux (and),

 = a + k2 b+k3 [x, b]

 = a + µ k2 + k3 [x,b] b ¶ b ∈ Ub (to).

 To conclude that U[x,b]

 (z) is an open in β that is contained in the intersection Ux (y) ∩ Ub (a) It is necessary to establish that (z, [x, b]) = 1.

 Lemma 2: If z ∈ Ux (y) ∩ Ub (a), then (z,[x,b]) = 1.

 Demonstration.

Given the equality z = y + k1x, the common divisors of z and x are also of y.  Since (x, y) = 1, we conclude that (z, x) = 1. From likewise (z, b) = 1. Hence (z, xb) = 1 and since [x, b]|xb, we have that (z, [x, b]) = 1. In this way we have proved that U[x,b] (z) ∈ β.  For the so β is a base.

 1 ) X is Hausdorff. 

 Let a, c ∈ N, a 6= c.  So (ac + 1, ac + 1) not divide c − a.  By lemma 1, Uac+1(a) ∩ Uac+1(c) = ∅.

 

2 ) X is connected. 

Here he highlights the reason for the condition (a, b) = 1 in the definition of Ub (a), since otherwise X would not be connected, since it could be separate in the following way:

 X=U2 (1) ∪ U2 (2).

 sean ux (y), ub (a) ∈ β.  

Let's see that ux (y) ∩ Ub (a) 6= ∅.  We affirm that dk ∈ Ud (c) for any k ∈ N and Ud (c) ∈ ​​β.  be up (q) ∈ β such that dk ∈ Up (q), that is, dk = q +pk0 . 

Then (d, p) = 1 since otherwise

(p, q) 6= 1. Therefore, (d, p)|c − q.  By lemma 1,

 you (c) ∩Up (q) 6= ∅.  Of

 here it follows that dk ∈ Ud (c).  So, ux (y)∩Ub (a) 6= ∅

 since bx is in both sets.

 To see that X is connected, assuming the opposite, write X =H ∪ K with H and K open and closed in X such that H 6= ∅ 6= K and H ∩ K = ∅.  Since β is a basis and H and K are open in X, then there exist ux basics

 (y) ⊂ H and Ub(a) ⊂ K. Also, since H, K are closed in X, so ow

 (y) ⊂ H,Ub (a) ⊂ K, and Ux (y) ∩ Ub (a) ⊂ H ∩ K = ∅.

  So ux (y) ∩ Ub(a) = ∅.  This contradiction proves that X is connected.

 3 ) X is not locally connected.

 To prove this we are going to display a point and an open of X that contains, such that there is no other open content in it, which contains at given point and that is connected.

 Let's consider U2

 (1) = {1, 3, 5, .  .  .}, which is a neighborhood of 1.

 If X were locally connected, there would exist an open connected ∅ 6= H ⊂ X such

 that 1 ∈ H ⊂ U2

 (1).  Note that no open of the form U2b

 (1 is connected, since U2b (1) = U4b (1) ∪ U4b (2b+1), u4b (1) ∩ U4b (2b + 1) = ∅, (4b, 1) = 1 = (4b, 2b + 1).

 In particular if b = 1, U2

 (1) = U4 (1)∪U4 (3).  

Since H is connected and 1 ∈ H,

 then H ⊂ U4 (1), but U4(1) = U8 (1) ∪ U8 (5), therefore H ⊂ U8 (1).

 In this way we obtain that H is contained in each term of the

 succession:

 u2 (1) ⊃ U4 (1) ⊃ U8 (1) ⊃ U16(1) ⊃ · · · .

 Therefore H ⊂ ∩n∈NU2 n (1) = {1}, which contradicts that H 6= ∅ and H Is open.  Let's not forget that {1} is closed since X is Hausdorff and not can be open since X is connected.