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Mathematical and Theoretical Physics

Research Article Volume 1 Issue 6

Brown spaces and the Golomb topology

Maira Madriz Mendoza,2 Jos Del Carmen Alberto Dom nguez,3 Gerardo Acosta1

1Institute of Mathematics, National Autonomous University of Mexico, Mexico
2Autonomous Technological Institute of Mexico, Mexico
3Academic Division of Basic Sciences, Universidad Juárez Autónoma de Tabasco, Mexico

Correspondence: Acosta G, Institute of Mathematics,National Autonomous University of Mexico, D.F. 04510, Mexico

Received: September 28, 2018 | Published: December 18, 2018

Citation: Acosta G, Madriz-Mendoza M, Domínguez JDCA. Brown spaces and the Golomb topology. Open Acc J Math Theor Phy. 2018;1(6):242-247 DOI: 10.15406/oajmtp.2018.01.00042

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Abstract

A Brown space is a topological space X  such that for all non-empty open subsets U  and V  of X,  we have clX(U)clX(V).  It is clear that Brown spaces are connected and not completely Hausdorff. Given a,b,  whose greatest common divisor is 1, we consider the arithmetic progression PG(a,b)={b+an:n{0}}.  The family BG  of all such arithmetic progressions is a base for a topology τG  on .  In this paper we show that for every d,  the set PG(1,d)  is a Brown space which is dense in (,τG).  In particular, (,τG)  is a Brown space. We also show that for each prime number p  and every natural number c,  such that the greatest common divisor between p  and c  is 1, the set PG(p,c)  is totally separated. We write some consequences of such result. For example that the space (,τG)  is not connected im kleinen at each of its points. This generalizes a result of Kirch AM.1 We also present a simpler proof of a result presented by Szczuka P.2 Some general properties of Brown spaces are also presented in this paper.

Introduction

We denote by  and by  the sets of integers and of natural numbers, respectively, and we let 0={0}.  We also denote by  the set of prime numbers and consider that .  Given a,b,  the symbol a,b  denotes the greatest common divisor of a and b and we consider the infinite arithmetic progressions P(a,b)={b+an:n0}=b+a0  and

PG(a,b)={b+an:n0}=b+a0,providedthata,b=1.  (1)

For a  and b  we also consider the infinite arithmetic progressions PF(a,b)={b+az:z}=b+aandM(a)={an:n}.

Clearly P(a,b)=PF(a,b)0  and P(a,b)=PG(a,b)  if and only if a,b=1.  Note that M(a)=P(a,a).  In 1955 Furstenberg H3 showed in that the family BF={PF(a,b):(a,b)×}  is a base for a topology τF  on .  The topological space (,τF)  is second countable and T4,  and hence metrizable. Moreover each basic set PF(a,b)  is open and closed in (,τF),  so this space is zero-dimensional and not connected.

In 1959 and 1962 Golomb SW4,5 showed inthat the family

BG={PG(a,b):(a,b)×anda,b=1} (2)

is a base for a topology τG  on .  Indeed

τG={}{U:foreachbUthereisasuchthata,b=1andPG(a,b)U}.

In,5 whose first edition was published in 1970, τG  is called the “relatively prime integer topology”, though in this paper, as well as in all the papers by Szczuka P,2,6–10 we call τG  the Golomb topology and the topological space (,τG)  is called the Golomb space. It is known that (,τG)  is second countable, T2  and connected Theorems 2 and 3,4 and Theorems 2 and 35 . Using the fact that for every p,  the set M(p)  is closed in (,τG),  Golomb SW4 proved in both Theorem 14 & Theorem 15 that the set  is infinite. The proof of the connectedness of (,τG),  as presented by Golomb SW,4 uses Number Theory. As it is indicated in4,5 “a proof of the connectedness of (,τG),  without reference to Number Theory, was presented by Brown M11 in the April 1953 meeting of the American Mathematical Society”, held in New York. Brown M11 studied the space (,τG),  though he did not publish his work. The abstract of his talk, published in11 is the following one:

    1. A countable connected Hausdorff space. The points are the positive integers. Neighborhoods are sets of integers {a+bx},  where a and b are relatively prime to each other ( x=1,2,3, ). Let {a+bx}  and {c+dx}  be two neighborhoods. It is shown that bd  is a limit point of both neighborhoods. Thus, the closures of any two neighborhoods have a nonvoid intersection. This is a sufficient condition that a space be connected.

This abstract served Clark PL,12 in 2017, the authors of,12 to coin the following term:

Definition 1.1 A Brown space is a topological space X  such that for all non-empty open subsets U  and V  of X,  we have clX(U)clX(V).  If X  is a topological space and YX,  we say that Y  is a Brown space in X  if Y,  as a subspace of X,  is a Brown space.

The following result appears in [2, Proposition 6].

Theorem 1.2 Each Brown space X  is connected.

Proof. If X  is not connected, then there exist non-empty open and closed subsets U  and V  of X  such that X=UV  and UV=.  Then clX(U)clX(V)=UV=,  a contradiction to the fact that X  is a Brown space.

In this paper we will present some general properties of Brown spaces. We will give an explicit proof of the fact that (,τG)  is a Brown space (Theorem 3.4). We will also show that, for every d,  the subset PG(1,d)  is a Brown space in (,τG)  (Theorem 3.3). Since (,τG)  is second countable, it is also Lindelöf. By the space (,τG)  is not T312. 4 Since every second coutable space is Lindelöf, 14 and every Lindelöff and T3  space is T4 14 and hence T312,  the space (,τG)  is not T3.  Without using all these results from General Topology (that lead to the fact that every non-empty countable connected T3  space is a one-point-set), in both Theorem~44 and Theorem~4,5 Golomb SW5 proved that (,τG)  is not T3  by showing that for the closed set M(2)  in (,τG)  that do not contain the point 1,  there are no open subsets U  y V  in (,τG)  such that 1U,M(2)V  and UV=.

Since compact T2  spaces as well as locally compact and T2  spaces are T3 , 14 the space (,τG)  is neither compact nor locally compact (compare with Theorem 54 and Theorem 55). The paper is divided in three sections. After this Introduction, in Section 2 we write the notation as well as some preliminary results that we will use in the paper. In this section we also write some general properties of Brown spaces. In Section 3 we write properties of the Golomb space as well as of some subsets of it.

Notation and preliminary results

In this paper we will use notation and results from both Number Theory and from General Topology. Concerning Number Theory, if c,d  and c0,  then the symbol c|d  means that there exists a  such that d=ca. > If c,d  and m{1},  then the symbol cd(modm)  means that m|(cd).  The next result is proved in.15

Theorem 2.1 Let a,b,q,r  be such that a=bq+r.  Then a,b=b,r.

The following result was used in both [13, p. 169] and [1, p. 902] without proof.13,2 We will use it in Section 3 so, for completeness, we present a proof here.

Theorem 2.2 Let b . If p  is such that b,p=1,  then for each n,s0, pn+b,ps=1. (3)

Proof. If s=0,  then

pn+b,ps=pn+b,p0=pn+b,1=1,foreachn0.

If s=1  then, since p,b=1,  by Theorem 2.1,

pn+b,ps=pn+b,p=p,b=1,foreachn0.  (4)

Now assume that there exist s0{1}  and n00  such that g=pn0+b,ps0>1.  Let q  be such that q|g.  Then q|(pn0+b)  and q|ps0,  so q=p.  Hence q|(pn0+b)  and q|p,  so q|pn0+b,p.  This implies, using (4) with n=n0,  that q|1,  a contradiction. Thus pn+b,ps=1,  for every s,n0.  

Concerning General Topology, given a topological space X  and AX,  we denote by clX(A)  and by intX(A)  the closure and the interior of A  in X,  respectively. In particular, for A,  the symbol clX(A)  denotes the closure of A  in (,τG).  We consider the closed interval [0,1]  with its usual topology. Recall that a topological space X  is said to be

    1. T1  if for each xX,  the set {x}  is closed in X;  
    2. T2  or Hausdorff if for every x,yX  such that xy,  there exist open sets U  and V  in X  so that xU,   yV  and UV=;  
    3. T212  or completely Hausdorff if for every x,yX  with xy,  there exist open sets U  and V  such that xU,   yV  and clX(U)clX(V)=;  
    4. regular if for every xX  and each closed set C  in X  such that xC,  there exist open sets U  and V  in X  so that xU,   CV  and UV=;  
    5. T3  if X  is regular and T1;  
    6. T312  if X  is T1  and for each xX  and every closed set C  in X  so that xC,  there is a continuous function f:X[0,1]  such that f(x)=0  and f(C){1};  
    7. T4  if X  is T1  and for every closed subsets C  and D  of X  with CD=,  there exist open sets U  and V  in X  such that CU,   DV  and UV=.  

It is known that T4T312T3T212T2T1.  It is also known that none of the previous implications is reversible. The following result is easy to prove.

Theorem 2.3 No Brown space X,  with at least two points, is T212.  

By Theorem 2.3 no connected T212,  with at least two points, is a Brown space.

Let X  be a topological space and x  in X.  We say that X  is indiscrete in x  or that x  is an indiscrete point of X  if the only open subset of X  that contains x  is X  itself. We say that X  is indiscrete if its topology is the indiscrete topology. Note that X  is indiscrete if and only if every point of X  is indiscrete. The following result is proved in [2, Proposition 6].

Theorem 2.4 Let X  be a topological space. Then

  1. if X  contains an indiscrete point, then X  is a Brown space;
  2. if X  is a Brown space, then X  is regular if and only if X  is indiscrete.

Note that a T1  space contains no indiscrete points. By Theorem 2.4, no connected T3  space, with at least two points, is a Brown space. We also have that if X  is a connected regular space without indiscrete points, then X  is not a Brown space. Hence the converse of Theorem 1.2 is not true.

We say that a topological property P  is

    1. hereditary if for any space X  that has the property P,  every subspace of X  also has the property P;  
    2. multiplicative if for any family {Xs:sS}  of topological spaces with the property P,  the Cartesian product sSXs,  with the product topology, also has the property P;  
    3. factorizable if for any family {Xs:sS}  of topological spaces if the Cartesian product sSXs,  with the product topology, has the property P  then each factor Xs  also has the property P.  

Theorem 2.5 Let X  and Y  be topological spaces and f:XY  be a continuous and surjective function. If X  is a Brown space, then Y  is a Brown space.

Proof. Let U  and V  be non-empty open subsets of Y.  Since f  is continuous and surjective, f1(U)  and f1(V) are non-empty open subsets of X.  Hence, since X  is a Brown space, clX(f1(U))clX(f1(U)).  By continuity of X  we have clX(f1(U))clX(f1(U))f1(clY(U))f1(clY(V))=f1(clY(U)clY(V)).

Hence clY(U)clY(V),  so Y  is a Brown space.

By Theorem 2.5 being a Brown space is a topological property. Moreover.

Theorem 2.6 Being a Brown space is both a multiplicative and a factorizable property.

Proof. Let {Xs:sS}  be a family of non-empty topological spaces. Let X=sSXs  and assume that X  has the product topology. For each tS  the projection pt:XXt  defined for any x=(xs)sSX  by pt(x)=xt  is continuous and surjective. Hence if X  is a Brown space then, by Theorem 2.5, Xt  is a Brown space too. This shows that being a Brown space is a factorizable property.

Now assume that each Xs  is a Brown space. Let U  and V  be two non-empty open subsets of X.  Fix xU  and yV  and assume that B=sSBs  and C=sSCs  are basic subsets of X  such that xBU  and yCV.  For each sS,  the sets Bs  and Cs  are non-empty and open in the Brown space Xs,  so clXs(Bs)clXs(Cs).  Then

clX(B)clX(C)=(sSclXs(Bs))(sSclXs(Cs))=sS(clXs(Bs)clXs(Cs)).

Hence clX(B)clX(C)clX(U)clX(V),  so X  is a Brown space. This shows that being a Brown space is a multiplicative property.

In Theorem 3.5 we will show that being a Brown space is not a hereditary property. If X  and Y  are topological spaces, f:XY  is a quotient mapping and X  is a Brown space then, by Theorem 2.5, Y  is a Brown space too.

A topological space X  is said to be

    1. hereditarily disconnected if no non-empty connected subset of X  contains more that one point;
    2. totally separated if for every x,yX  with xy,  there exist open sets U  and V  in X  such that xU,   yV,   X=UV  and UV=;  
    3. zero-dimensional is X  is T1  and has a base consisting of open and closed sets.

Hereditarily disconnected space are also called totally disconnected. In14 it is shown that zero-dimensional spaces are hereditarily disconnected. In14 it is proved that if X  is hereditarilly disconnected and locally compact, then X  is zero-dimensional. Note that a space X  is hereditarily disconnected if and only if, for each xX,  the component Cx  of X  that contains x  is a one-point-set, namely {x}.  Note also that X  is totally separated if and only if, for every xX,  the quasicomponet Qx  of X  that contains x  is a one-point-set, namely {x}.  Since CxQx  for each xX 14, totally separated spaces are hereditarily disconnected. If X  is compact and T2  then Cx=Qx  for every xX 14 so, in compact and T2  spaces the notions of being hereditarily disconnected and of being totally separated coincide.

Let X  be a topological space and a,bX.  We say that X  is connected between a  and b  if for every open sets U  and V  in X  such that aU,   bV  and UV=,  we have XUV.  It is straight forward to see that if X  if T2  then X  is totally separated if and only if for each x,yX  with xy,  the space X  is not connected between x  and y.

Let X  be a topological space and xX.  We say that

    1. X  is locally connected at x,  if for any open subset V  of X  such that xV,  there exists an open and connected subset U  of X  such that xUV;  
    2. X  is connected im kleinen at x,  if for any open subset V  of X  such that xV,  there exists a connected subset U  of X  such that xintX(U)UV;  
    3. X  is almost connected im kleinen at x,  if for any open subset V  of X  such that xV,  there exists a closed and connected subset U  of X  such that intX(U)  and UV;  
    4. X  is almost connected im kleinen if X  is almost connected im kleinen at each of its points;
    5. X  is locally connected if X  is locally connected at each of its points.

Clearly, if X  is locally connected at x,  then X  is connected im kleinen at x.  The converse of this implication is not true. We construct an example in 2,  with the usual topology. For each i,  let qi=(1i,0)2  and let Li,0  be the straight line segment from qi  to qi+1 . For each (i,n)×,  we consider

pi,n=1i+1(1,1n)2,

as well as the straight line segment Li,n  from pi,n  to qi.  Formally, for each (i,n)×,  let

Li,n={(x,y)2:1i+1x1iandy=1n(1ix)},

where y=1n(1ix)  is the equation of the straight line in 2  that contains the points pi,n  y qi.  For each i,  we define Xi=n{0}Li,n.

Then Xb=cl2(iXi) is a topological space which is compact, connected, connnected im kleinen at (0,0)  and not locally connected at (0,0).  The space Xb,  called the infinite broom in [5, p. 139], appears in16 Example ~27.15, p. 201], though there is no any detailed proof of the fact that Xb,  is connected im kleinen at (0,0).  and not locally connected at (0,0).  In16 there is a detailed proof of this, though it is written in Spanish. In16 it is shown that if a topological space X  is connected im kleinen at each of its points, then X  is locally connected. Let

Y=(nLi,0){qi:i}Xb.

Note that Xb  is almost connected im kleinen at any point pY  and not connected im kleinen at such point p.  Note also that if a T3  space is connected im kleinen at xX,  then it is almost connected im kleinen at x.

Properties of the golomb space

In the space (,τG)  a non-empty subset U  of  is open if and only if, for every bU,  there exists a  such that a,b=1  and PG(a,b)U  (compare with).7 Hence U  is infinite. In particular any subset of  with non-empty interior in (,τG)  is infinite. Let a,x.  In Szczuka P8 presented some results that involve the set cl(P(a,x)).  In,7 she showed that if x1x(moda),   x1a  and a,x=1,  then P(a,x1)cl(PG(a,x)).  In8 she showed that if a  and x  are odd and a,x=1,  then cl(PG(a,x))=cl(PG(2a,x)).  We show the following result.

Theorem 3.1 Let a,x  be so that a,x=1.  Then M(a)cl(PG(a,x)).

Proof. Let abM(a)  and let W  be an open subset of (,τG)  such that abW.  There exists d  such that d,ab=1  and PG(d,ab)W.  Assume first that a=1.  Then PG(a,x)={x+n:n0}  so if xab,  then abPG(a,x)cl(PG(a,x)).  If ab<x  then, since PG(d,ab)  is infinite, there is n  such that dn+abx,  so dn+abPG(d,ab)PG(a,x)WPG(a,x).  This shows that abcl(PG(a,x)).

Now assume that a2  By showing that PG(d,ab)PG(a,x)  we will obtain that WPG(a,x).  To prove that PG(d,ab)PG(a,x),  assume first that d=1.  Then ab+xPG(a,x)  and ab+x=1x+abPG(d,ab).  Now assume that d2.  Since d2.  and a2,  an element zPG(d,ab)PG(a,x)  satisfies the system of congruencies

zab(modd)andzx(moda).  (5)

Hence, an element in PG(d,ab)PG(a,x)  is a solution of the system (5). Conversely, every solution of the system (5), is an element of PG(d,ab)PG(a,x).  Let us show, then, that the system (5) has a solution. If d,a1,  then there exists p  such that p|d  and p|a.  Then p|d  and p|ab,  so d,ab1,  a contradiction. Hence d,a=1  and by the Chinese Reminder Theorem, the system (5) has a solution. This shows that PG(d,ab)PG(a,x).  Hence WPG(a,x)  and then abcl(PG(a,x)).

Corollary 3.2 For every d,  the set PG(1,d)  is dense in (,τG).  

Proof. Let d.  Put a=1.  Then a,d=1,   M(a)=  and, by Theorem 3.1, =M(a)cl(PG(a,d)),  so PG(a,d)=PG(1,d)  is dense in (,τG).  

Theorem 3.3 For every d,  the set PG(1,d)  is a Brown space in (,τG).  In particular, PG(1,d)  is connected.

Proof. Let U  and V  be two non-empty open subsets of PG(1,d).  Since PG(1,d)  is open in (,τG),  the sets U  and V  are open in (,τG),  Let xU  and yV.  Then there exist a,b  such that a,x=b,y=1  and PG(a,x)U  and PG(b,y)V.  If a=1,  then PG(a,x)={x+n:n0}.  Since PG(b,y)  is infinite, there is n00  such that y+bn0x  and then

y+bn0PG(a,x)PG(b,y)UVPG(1,d)cl(U)cl(V)PG(1,d)= clPG(1,d)(U)clPG(1,d)(V).

Hence clPG(1,d)(U)clPG(1,d)(V).  Similarly, if b=1  we have clPG(1,d)(U)clPG(1,d)(V).  Now assume that a2  and b2.  By Theorem 3.1, M(a)cl(PG(a,x))  and M(b)cl(PG(b,y))  so, in particular, abdcl(PG(a,x))cl(PG(b,y)).  Since a2  and b2,  we have d<abd,  so abdPG(1,d).  Then

abdcl(PG(a,x))cl(PG(b,y))PG(1,d)cl(U)cl(V)PG(1,d)= clPG(1,d)(U)clPG(1,d)(V).

Hence clPG(1,d)(U)clPG(1,d)(V), so PG(1,d)  is a Brown space in (,τG).  Since Brown spaces are connected, PG(1,d)  is connected.

Now we present an explicit proof of what Brown M11 claimed in assertion (B) of Section 1.

Theorem 3.4 (,τG)  is a Brown space. In particular, (,τG)  is connected and it is not T212.
Proof. Since PG(1,1)=  the result follows from Theorem 3.3 and the fact that Brown spaces are not T212.

Now we prove the following result.

Theorem 3.5 Let c,p  be such that p  and p,c=1.  Take a,bPG(p,c)  such that a<b  and n,m0  so that a=pm+c,b=pn+c  and 0m<n.  Then

U=mi=0PG(pn+1,pi+c)andV=pn1j=m+PG(pn+1,pj+c)  (6)

are open subset of (,τG)  such that aU, bV, UV=  and PG(p,c)=UV.  In particular, PG(p,c)  is not connected.

Proof. Since p  and p,c=1,  by Theorem 2.2,

pn+1,pi+c=1,foreachi0.

Then we can consider the sets U and V  indicated in (6), and they are open in (,τG).  Since 0m<n<2npn,  we have m+1npn1,  so

a=pm+c=pn+1(0)+pm+cPG(pn+1,pm+c)mi=0PG(pn+1,pi+c)=U

and

b=pn+c=pn+1(0)+pn+cPG(pn+1,pn+c)pn1j=m+1PG(pn+1,pj+c)=V.

Hence aU  and bV  so U  and V  are infinite. Now assume that there is a point zUV.  Let i{0,1,,m}  and j{m+1,m+2,,pn1}  be such that zPG(pn+1,pi+c)  and zPG(pn+1,pj+c).  Since pn+1>1,  it follows that

zpi+c(modpn+1)andzpj+c(modpn+1).

Hence pi+cpj+c(modpn+1),  so pipj(modpn+1).  This implies that ij(modpn).  However, since the set {0,1,,m,m+1,,pn1}  is a complete system of reminders modulus pn  and im<m+1j,  we have ij(modpn).  From this contradiction we infer that UV=.

Now we show that UV=PG(p,c).  If zU,  then there exists i{0,1,,m}  such that zPG(pn+1,pi+c).  Let z00  be such that z=pn+1z0+pi+c.  Then z=p(pnz0+i)+cPG(p,c).  Similarly, if zV,  there is j{m+1,m+2,,pn1}  such that zPG(pn+1,pj+c).  Let z10  be such that z=pn+1z1+pj+c.  Then z=p(pnz1+j)+cPG(p,c).  This shows that UVPG(p,c).  To prove the other inclusion, let zPG(p,c).  Then there exists k0  such that z=pk+c.  Using the Division Algorithm we obtain s,t0  such that k=pns+t  and 0t<pn.  Hence

z=pk+c=p(pns+t)+c=pn+1s+pt+c. (7)

Since 0t<pn  and 0m<pn,  we obtain

t{0,1,,pn1}={0,1,,m}{m+1,m+2,,pn1}.

Hence, by (7),

z(mi=0PG(pn+1,pi+c))(pn1j=m+1PG(pn+1,pj+c))=UV.

This shows that PG(p,c)UV.  Hence, UV=PG(p,c).  This completes the first part of the proof. Since PG(p,c)  has been written as the union of two non-empty open subsets of (,τG),  which are disjoint, the set PG(p,c)  is not connected.

Corollary 3.6 Let c,p  be such that p  and p,c=1.  Take a,bPG(p,c)  such that ab.  Then there exist open sets U  and V  in (,τG)  such that aU, bV, UV=  and PG(p,c)=UV.

Note that (,τG)  is a Brown space and, for each c,p  such that p  and p,c=1,   PG(p,c)  is an open subset of (,τG)  which is not a Brown space, by Theorem 3.5. Hence being a Brown space is not a hereditary property.

Using Corollary 3.6 we obtain the following result.

Theorem 3.7 Let c,p  be such that p  and p,c=1.  Then PG(p,c)  is totally separated.

Since totally separated spaces are hereditarily disconnected, for every c,p  such that p  and p,c=1,  the set PG(p,c)  is hereditarily disconnected.

Now we present several consequences of Theorem 3.7. In 1969 Kirch AM1 showed in1 Theorem 1, that the space (,τG)  is not locally connected. Formally he proved that (,τG)  is not locally conntected at 1. In the following theorem we generalize this result.

Theorem 3.8 For every d  the space PG(1,d)  is neither connected im kleinen nor almost connected im kleinen at each of its points.

Proof. Assume, on the contrary, that there exists d  such that PG(1,d)  is either connected im kleinen or almost connected im kleinen at some point cPG(1,d).  We define a set PG(p,c)  as follows: if c  is odd, we take p=2.  If c  is even, we take p  such that c<p.  In any situation, it follows that p,   p,c=1  and PG(p,c)  is an open subset of PG(1,d)  such that cPG(p,c).  Since PG(1,d)  is either connected im kleinen or almost connected im kleinen at c,  there exists a connected subset C  of PG(1,d)  such that intPG(1,d)(C)  and CPG(p,c).  Since PG(1,d)  is open in (,τG),  we have int(C).  Since all the non-empty open subsets of (,τG) are infinite, the set C is infinite. This contradicts the fact that, by Theorem 3.7, PG(p,c)  is hereditarilly disconnected. Hence PG(1,d)  is neither connected im kleinen nor almost connected im kleinen at c.

Corollary 3.9 The space (,τG)  is neither connected im kleinen nor almost connected im kleinen at each of its points.

As a consequence of Theorem 3.8 for any a,  the space (,τG)  is not locally connected at a.

For a  we define Θ(a)={p:p|a}.  Note that Θ(1)=.

Theorem 3.10 Let a,c.  If P(a,c)  is connected in (,τG)  then Θ(a)Θ(c).  In particular, if a,c=1  then PG(a,c)  is connected if and only if a=1.

Proof. Assume first that P(a,c)  is connected. If Θ(a)Θ(c),  then a>1  and there exists pΘ(a)Θ(c).  Hence p,   p|a  and pc,  so p,c=1.  Since p|a,  we have P(a,c)PG(p,c).  By Theorem 3.7, PG(p,c)  is hereditarily disconnected, so P(a,c)  is a one-point-set, a contradiction. This shows that Θ(a)Θ(c).

Now assume that a,c=1  and that PG(a,c)=P(a,c)  is connected. Then, by the first part of the theorem, Θ(a)Θ(c).  If a2,  then there is p  such that p|a.  Then p|c,  contradicting that a,c=1.  Hence a=1.  In Theorem 3.3 we proved that PG(1,c)  is connected.

In Theorem 3.3 Szczuka P2 showed the following result:

Theorem 3.11 Let a,c.  The arithmetic progression P(a,c)  is connected in (,τG)  if and only if Θ(a)Θ(c).  In particular,

    1. the progression {an:n}  is connected in (,τG)
    2. if a,c=1,  then PG(a,c)  is connected in (,τG)  if and only if a=1.

The proof of the “only if” part of Theorem 3.11 is much simpler is we know that the progressions PG(p,c)  are hereditarily disconnected, as we presented in the proof of Theorem 3.10. In2 it is claimed that the fact that each set PG(1,c)  is connected is obvious.

Acknowledgements

None.

Conflict of interest

Authors declare that there is no conflict of interest.

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