Research Article Volume 1 Issue 6
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
A Brown space is a topological space XX such that for all non-empty open subsets UU and VV of X,X, we have clX(U)∩clX(V)≠∅.clX(U)∩clX(V)≠∅. It is clear that Brown spaces are connected and not completely Hausdorff. Given a,b∈ℕ, whose greatest common divisor is 1, we consider the arithmetic progression PG(a,b)={b+an:n∈ℕ∪{0}}. The family BG of all such arithmetic progressions is a base for a topology τG on ℕ. In this paper we show that for every d∈ℕ, the set PG(1,d) is a Brown space which is dense in (ℕ,τG). In particular, (ℕ,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 (ℕ,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.
We denote by ℤ and by ℕ the sets of integers and of natural numbers, respectively, and we let ℕ0=ℕ∪{0}. We also denote by ℙ the set of prime numbers and consider that ℙ⊂ℕ. Given a,b∈ℕ, 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:n∈ℕ0}=b+aℕ0 and
PG(a,b)={b+an:n∈ℕ0}=b+aℕ0, provided that 〈a,b〉=1. (1)
For a∈ℕ and b∈ℤ we also consider the infinite arithmetic progressions PF(a,b)={b+az:z∈ℤ}=b+aℤ and M(a)={an:n∈ℕ}.
Clearly P(a,b)=PF(a,b)∩ℕ0 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)∈ℕ×ℤ} is a base for a topology τF on ℤ. The topological space (ℤ,τF) is second countable and T4, and hence metrizable. Moreover each basic set PF(a,b) is open and closed in (ℤ,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)∈ℕ×ℕ and 〈a,b〉=1} (2)
is a base for a topology τG on ℕ. Indeed
τG={∅}∪{U⊂ℕ: for each b∈U there is a∈ℕ such that 〈a,b〉=1 and PG(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 (ℕ,τG) is called the Golomb space. It is known that (ℕ,τG) is second countable, T2 and connected Theorems 2 and 3,4 and Theorems 2 and 35 . Using the fact that for every p∈ℙ, the set M(p) is closed in (ℕ,τG), Golomb SW4 proved in both Theorem 14 & Theorem 15 that the set ℙ is infinite. The proof of the connectedness of (ℕ,τG), as presented by Golomb SW,4 uses Number Theory. As it is indicated in4,5 “a proof of the connectedness of (ℕ,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 (ℕ,τG), though he did not publish his work. The abstract of his talk, published in11 is the following one:
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 Y⊂X, 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=U∪V and U∩V=∅. Then clX(U)∩clX(V)=U∩V=∅, 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 (ℕ,τG) is a Brown space (Theorem 3.4). We will also show that, for every d∈ℕ, the subset PG(1,d) is a Brown space in (ℕ,τG) (Theorem 3.3). Since (ℕ,τG) is second countable, it is also Lindelöf. By the space (ℕ,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 (ℕ,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 (ℕ,τG) is not T3 by showing that for the closed set M(2) in (ℕ,τG) that do not contain the point 1, there are no open subsets U y V in (ℕ,τG) such that 1∈U,M(2)⊂V and U∩V=∅.
Since compact T2 spaces as well as locally compact and T2 spaces are T3 , 14 the space (ℕ,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.
In this paper we will use notation and results from both Number Theory and from General Topology. Concerning Number Theory, if c,d∈ℤ and c≠0, then the symbol c|d means that there exists a∈ℤ such that d=ca. > If c,d∈ℤ and m∈ℕ−{1}, then the symbol c≡d (modm) means that m|(c−d). The next result is proved in.15
Theorem 2.1 Let a,b,q,r∈ℤ 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∈ℤ . If p∈ℙ is such that 〈b,p〉=1, then for each n,s∈ℕ0, 〈pn+b,ps〉=1. (3)
Proof. If s=0, then
〈pn+b,ps〉=〈pn+b,p0〉=〈pn+b,1〉=1, for each n∈ℕ0.
If s=1 then, since 〈p,b〉=1, by Theorem 2.1,
〈pn+b,ps〉=〈pn+b,p〉=〈p,b〉=1, for each n∈ℕ0. (4)
Now assume that there exist s0∈ℕ−{1} and n0∈ℕ0 such that g=〈pn0+b,ps0〉>1. Let q∈ℙ 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,n∈ℕ0.
Concerning General Topology, given a topological space X and A⊂X, we denote by clX(A) and by intX(A) the closure and the interior of A in X, respectively. In particular, for A⊂ℕ, the symbol clX(A) denotes the closure of A in (ℕ,τG). We consider the closed interval [0,1] with its usual topology. Recall that a topological space X is said to be
It is known that T4⇒T312⇒T3⇒T212⇒T2⇒T1. 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
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
Theorem 2.5 Let
X
and
Y
be topological spaces and
f:X→Y
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,
f−1(U)
and
f−1(V)
are non-empty open subsets of
X.
Hence, since
X
is a Brown space,
clX(f−1(U))∩clX(f−1(U))≠∅.
By continuity of
X
we have
∅≠clX(f−1(U))∩clX(f−1(U))⊂f−1(clY(U))∩f−1(clY(V))=f−1(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:s∈S} be a family of non-empty topological spaces. Let X=∏s∈SXs and assume that X has the product topology. For each t∈S the projection pt:X→Xt defined for any x=(xs)s∈S∈X 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 x∈U and y∈V and assume that B=∏s∈SBs and C=∏s∈SCs are basic subsets of X such that x∈B⊂U and y∈C⊂V. For each s∈S, 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)=(∏s∈SclXs(Bs))∩(∏s∈SclXs(Cs))=∏s∈S(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:X→Y 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
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 x∈X, 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 x∈X, the quasicomponet Qx of X that contains x is a one-point-set, namely {x}. Since Cx⊂Qx for each x∈X 14, totally separated spaces are hereditarily disconnected. If X is compact and T2 then Cx=Qx for every x∈X 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,b∈X. We say that X is connected between a and b if for every open sets U and V in X such that a∈U, b∈V and U∩V=∅, we have X≠U∪V. It is straight forward to see that if X if T2 then X is totally separated if and only if for each x,y∈X with x≠y, the space X is not connected between x and y.
Let X be a topological space and x∈X. We say that
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 ℝ2, with the usual topology. For each i∈ℕ, let qi=(1i,0)∈ℝ2 and let Li,0 be the straight line segment from qi to qi+1 . For each (i,n)∈ℕ×ℕ, we consider
pi,n=1i+1(1,1n)∈ℝ2,
as well as the straight line segment Li,n from pi,n to qi. Formally, for each (i,n)∈ℕ×ℕ, let
Li,n={(x,y)∈ℝ2:1i+1≤x≤1i and y=1n(1−ix)},
where y=1n(1−ix) is the equation of the straight line in ℝ2 that contains the points pi,n y qi. For each i∈ℕ, we define Xi=∪n∈ℕ∪{0}Li,n.
Then Xb=cl2(∪i∈ℕXi) 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=(∪n∈ℕLi,0)−{qi:i∈ℕ}⊂Xb.
Note that Xb is almost connected im kleinen at any point p∈Y and not connected im kleinen at such point p. Note also that if a T3 space is connected im kleinen at x∈X, then it is almost connected im kleinen at x.
In the space (ℕ,τG) a non-empty subset U of ℕ is open if and only if, for every b∈U, there exists a∈ℕ such that 〈a,b〉=1 and PG(a,b)⊂U (compare with).7 Hence U is infinite. In particular any subset of ℕ with non-empty interior in (ℕ,τG) is infinite. Let a,x∈ℕ. In Szczuka P8 presented some results that involve the set cl(P(a,x)). In,7 she showed that if x1≡x (moda), x1≤a 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∈ℕ be so that 〈a,x〉=1. Then M(a)⊂cl(PG(a,x)).
Proof. Let ab∈M(a) and let W be an open subset of (ℕ,τG) such that ab∈W. There exists d∈ℕ such that 〈d,ab〉=1 and PG(d,ab)⊂W. Assume first that a=1. Then PG(a,x)={x+n:n∈ℕ0} so if x≤ab, then ab∈PG(a,x)⊂cl(PG(a,x)). If ab<x then, since PG(d,ab) is infinite, there is n∈ℕ such that dn+ab≥x, so dn+ab∈PG(d,ab)∩PG(a,x)⊂W∩PG(a,x). This shows that ab∈cl(PG(a,x)).
Now assume that a≥2 By showing that PG(d,ab)∩PG(a,x)≠∅ we will obtain that W∩PG(a,x)≠∅. To prove that PG(d,ab)∩PG(a,x)≠∅, assume first that d=1. Then ab+x∈PG(a,x) and ab+x=1⋅x+ab∈PG(d,ab). Now assume that d≥2. Since d≥2. and a≥2, an element z∈PG(d,ab)∩PG(a,x) satisfies the system of congruencies
z≡ab (modd) and z≡x (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,a〉≠1, then there exists p∈ℙ such that p|d and p|a. Then p|d and p|ab, so 〈d,ab〉≠1, 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 W∩PG(a,x)≠∅ and then ab∈cl(PG(a,x)).
Corollary 3.2 For every d∈ℕ, the set PG(1,d) is dense in (ℕ,τG).
Proof. Let d∈ℕ. Put a=1. Then 〈a,d〉=1, M(a)=ℕ and, by Theorem 3.1, ℕ=M(a)⊂cl(PG(a,d)), so PG(a,d)=PG(1,d) is dense in (ℕ,τG).
Theorem 3.3 For every d∈ℕ, the set PG(1,d) is a Brown space in (ℕ,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 (ℕ,τG), the sets U and V are open in (ℕ,τG), Let x∈U and y∈V. Then there exist a,b∈ℕ 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:n∈ℕ0}. Since PG(b,y) is infinite, there is n0∈ℕ0 such that y+bn0≥x and then
y+bn0∈PG(a,x)∩PG(b,y)⊂U∩V∩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)≠∅. Similarly, if b=1 we have clPG(1,d)(U)∩clPG(1,d)(V)≠∅. Now assume that a≥2 and b≥2. By Theorem 3.1, M(a)⊂cl(PG(a,x)) and M(b)⊂cl(PG(b,y)) so, in particular, abd∈cl(PG(a,x))∩cl(PG(b,y)). Since a≥2 and b≥2, we have d<abd, so abd∈PG(1,d). Then
abd∈cl(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 (ℕ,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
(ℕ,τG)
is a Brown space. In particular,
(ℕ,τG)
is connected and it is not
T212.
Proof. Since
PG(1,1)=ℕ
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∈ℕ be such that p∈ℙ and 〈p,c〉=1. Take a,b∈PG(p,c) such that a<b and n,m∈ℕ0 so that a=pm+c,b=pn+c and 0≤m<n. Then
U=m∪i=0PG(pn+1,pi+c) and V=pn−1∪j=m+PG(pn+1,pj+c) (6)
are open subset of (ℕ,τG) such that a∈U, b∈V, U∩V=∅ and PG(p,c)=U∪V. In particular, PG(p,c) is not connected.
Proof. Since p∈ℙ and 〈p,c〉=1, by Theorem 2.2,
〈pn+1,pi+c〉=1, for each i∈ℕ0.
Then we can consider the sets U and V indicated in (6), and they are open in (ℕ,τG). Since 0≤m<n<2n≤pn, we have m+1≤n≤pn−1, so
a=pm+c=pn+1(0)+pm+c∈PG(pn+1,pm+c)⊂m∪i=0PG(pn+1,pi+c)=U
and
b=pn+c=pn+1(0)+pn+c∈PG(pn+1,pn+c)⊂pn−1∪j=m+1PG(pn+1,pj+c)=V.
Hence a∈U and b∈V so U and V are infinite. Now assume that there is a point z∈U∩V. Let i∈{0,1,…,m} and j∈{m+1,m+2,…,pn−1} be such that z∈PG(pn+1,pi+c) and z∈PG(pn+1,pj+c). Since pn+1>1, it follows that
z≡pi+c (modpn+1) and z≡pj+c (modpn+1).
Hence pi+c≡pj+c (modpn+1), so pi≡pj (modpn+1). This implies that i≡j (modpn). However, since the set {0,1,…,m,m+1,…,pn−1} is a complete system of reminders modulus pn and i≤m<m+1≤j, we have i≡j (modpn). From this contradiction we infer that U∩V=∅.
Now we show that U∪V=PG(p,c). If z∈U, then there exists i∈{0,1,…,m} such that z∈PG(pn+1,pi+c). Let z0∈ℕ0 be such that z=pn+1z0+pi+c. Then z=p(pnz0+i)+c∈PG(p,c). Similarly, if z∈V, there is j∈{m+1,m+2,…,pn−1} such that z∈PG(pn+1,pj+c). Let z1∈ℕ0 be such that z=pn+1z1+pj+c. Then z=p(pnz1+j)+c∈PG(p,c). This shows that U∪V⊂PG(p,c). To prove the other inclusion, let z∈PG(p,c). Then there exists k∈ℕ0 such that z=pk+c. Using the Division Algorithm we obtain s,t∈ℕ0 such that k=pns+t and 0≤t<pn. Hence
z=pk+c=p(pns+t)+c=pn+1s+pt+c. (7)
Since 0≤t<pn and 0≤m<pn, we obtain
t∈{0,1,…,pn−1}={0,1,…,m}∪{m+1,m+2,…,pn−1}.
Hence, by (7),
z∈(m∪i=0PG(pn+1,pi+c))∪(pn−1∪j=m+1PG(pn+1,pj+c))=U∪V.
This shows that PG(p,c)⊂U∪V. Hence, U∪V=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 (ℕ,τG), which are disjoint, the set PG(p,c) is not connected.
Corollary 3.6 Let c,p∈ℕ be such that p∈ℙ and 〈p,c〉=1. Take a,b∈PG(p,c) such that a≠b. Then there exist open sets U and V in (ℕ,τG) such that a∈U, b∈V, U∩V=∅ and PG(p,c)=U∪V.
Note that (ℕ,τG) is a Brown space and, for each c,p∈ℕ such that p∈ℙ and 〈p,c〉=1, PG(p,c) is an open subset of (ℕ,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∈ℕ be such that p∈ℙ and 〈p,c〉=1. Then PG(p,c) is totally separated.
Since totally separated spaces are hereditarily disconnected, for every c,p∈ℕ such that p∈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 (ℕ,τG) is not locally connected. Formally he proved that (ℕ,τG) is not locally conntected at 1. In the following theorem we generalize this result.
Theorem 3.8 For every d∈ℕ 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∈ℕ such that PG(1,d) is either connected im kleinen or almost connected im kleinen at some point c∈PG(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∈ℙ such that c<p. In any situation, it follows that p∈ℙ, 〈p,c〉=1 and PG(p,c) is an open subset of PG(1,d) such that c∈PG(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 C⊂PG(p,c). Since PG(1,d) is open in (ℕ,τG), we have int (C)≠∅. Since all the non-empty open subsets of (ℕ,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 (ℕ,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∈ℕ, the space (ℕ,τG) is not locally connected at a.
For a∈ℕ we define Θ(a)={p∈ℙ:p|a}. Note that Θ(1)=∅.
Theorem 3.10 Let a,c∈ℕ. If P(a,c) is connected in (ℕ,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 p∈Θ(a)−Θ(c). Hence p∈ℙ, p|a and p∤c, 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 a≥2, then there is p∈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∈ℕ.
The arithmetic progression
P(a,c)
is connected in
(ℕ,τG)
if and only if
Θ(a)⊂Θ(c).
In particular,
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.
None.
Authors declare that there is no conflict of interest.
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