A Brown space is a topological space $X$  such that for all non-empty open subsets $U$  and $V$  of $X\mathrm{,}$  we have ${\text{cl}}_X\text{}\left(U\right)\cap {\text{cl}}_X\text{}\left(V\right)\ne \varnothing \mathrm{.}$  It is clear that Brown spaces are connected and not completely Hausdorff. Given $a\mathrm{,}b\in \mathbb{N}\mathrm{,}$  whose greatest common divisor is 1, we consider the arithmetic progression $P_G\mathrm{<mo\; stretchy="false">(</mo>}a\mathrm{,}b\mathrm{<mo\; stretchy="false">)</mo><mo>=</mo><mo>\{</mo>}b+an:n\in \mathbb{N}\cup \mathrm{<mo>\{</mo>0<mo>\}</mo><mo>\}</mo>.}$  The family ${\mathcal{B}}_G$  of all such arithmetic progressions is a base for a topology ${\tau }_G$  on $\mathbb{N}\mathrm{.}$  In this paper we show that for every $d\in \mathbb{N}\mathrm{,}$  the set $P_G\mathrm{<mo\; stretchy="false">(</mo>1,}d\mathrm{<mo\; stretchy="false">)</mo>}$  is a Brown space which is dense in $\mathrm{<mo\; stretchy="false">(</mo>}\mathbb{N}\mathrm{,}{\tau }_G\mathrm{<mo\; stretchy="false">)</mo>.}$  In particular, $\mathrm{<mo\; stretchy="false">(</mo>}\mathbb{N}\mathrm{,}{\tau }_G\mathrm{<mo\; stretchy="false">)</mo>}$  is a Brown space. We also show that for each prime number $p$  and every natural number $c\mathrm{,}$  such that the greatest common divisor between $p$  and $c$  is 1, the set $P_G\mathrm{<mo\; stretchy="false">(</mo>}p\mathrm{,}c\mathrm{<mo\; stretchy="false">)</mo>}$  is totally separated. We write some consequences of such result. For example that the space $\mathrm{<mo\; stretchy="false">(</mo>}\mathbb{N}\mathrm{,}{\tau }_G\mathrm{<mo\; stretchy="false">)</mo>}$  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 $\mathbb{Z}$  and by $\mathbb{N}$  the sets of integers and of natural numbers, respectively, and we let ${\mathbb{N}}_0\mathrm{<mo>=</mo>}\mathbb{N}\cup \mathrm{<mo>\{</mo>0<mo>\}</mo>.}$  We also denote by $\mathbb{P}$  the set of prime numbers and consider that $\mathbb{P}\subset \mathbb{N}\mathrm{.}$  Given $a\mathrm{,}b\in \mathbb{N}\mathrm{,}$  the symbol $\langle a\mathrm{,}b\rangle$  denotes the greatest common divisor of a and b and we consider the infinite arithmetic progressions $P\mathrm{<mo\; stretchy="false">(</mo>}a\mathrm{,}b\mathrm{<mo\; stretchy="false">)</mo><mo>=</mo><mo>\{</mo>}b+an:n\in {\mathbb{N}}_0\mathrm{<mo>\}</mo><mo>=</mo>}b+a{\mathbb{N}}_0$  and

$P_G\mathrm{<mo\; stretchy="false">(</mo>}a\mathrm{,}b\mathrm{<mo\; stretchy="false">)</mo><mo>=</mo><mo>\{</mo>}b+an:n\in {\mathbb{N}}_0\mathrm{<mo>\}</mo><mo>=</mo>}b+a{\mathbb{N}}_0\mathrm{,}providedthat\langle a\mathrm{,}b\rangle \mathrm{<mo>=</mo>1.}$  (1)

For $a\in \mathbb{N}$  and $b\in \mathbb{Z}$  we also consider the infinite arithmetic progressions $P_F\mathrm{<mo\; stretchy="false">(</mo>}a\mathrm{,}b\mathrm{<mo\; stretchy="false">)</mo><mo>=</mo><mo>\{</mo>}b+az:z\in \mathbb{Z}\mathrm{<mo>\}</mo><mo>=</mo>}b+a\mathbb{Z}andM\mathrm{<mo\; stretchy="false">(</mo>}a\mathrm{<mo\; stretchy="false">)</mo><mo>=</mo><mo>\{</mo>}an:n\in \mathbb{N}\mathrm{<mo>\}</mo>.}$

Clearly $P\mathrm{<mo\; stretchy="false">(</mo>}a\mathrm{,}b\mathrm{<mo\; stretchy="false">)</mo><mo>=</mo>}{P}_F\mathrm{<mo\; stretchy="false">(</mo>}a\mathrm{,}b\mathrm{<mo\; stretchy="false">)</mo>}\cap {\mathbb{N}}_0$  and $P\mathrm{<mo\; stretchy="false">(</mo>}a\mathrm{,}b\mathrm{<mo\; stretchy="false">)</mo><mo>=</mo>}{P}_G\mathrm{<mo\; stretchy="false">(</mo>}a\mathrm{,}b\mathrm{<mo\; stretchy="false">)</mo>}$  if and only if $\langle a\mathrm{,}b\rangle \mathrm{<mo>=</mo>1.}$  Note that $M\mathrm{<mo\; stretchy="false">(</mo>}a\mathrm{<mo\; stretchy="false">)</mo><mo>=</mo>}P\mathrm{<mo\; stretchy="false">(</mo>}a\mathrm{,}a\mathrm{<mo\; stretchy="false">)</mo>.}$  In 1955 Furstenberg H³ showed in that the family ${\mathcal{B}}_F\mathrm{<mo>=</mo><mo>\{</mo>}{P}_F\mathrm{<mo\; stretchy="false">(</mo>}a\mathrm{,}b\mathrm{<mo\; stretchy="false">)</mo>}:\mathrm{<mo\; stretchy="false">(</mo>}a\mathrm{,}b\mathrm{<mo\; stretchy="false">)</mo>}\in \mathbb{N}\times \mathbb{Z}\mathrm{<mo>\}</mo>}$  is a base for a topology ${\tau }_F$  on $\mathbb{Z}\mathrm{.}$  The topological space $\mathrm{<mo\; stretchy="false">(</mo>}\mathbb{Z}\mathrm{,}{\tau }_F\mathrm{<mo\; stretchy="false">)</mo>}$  is second countable and $T_4\mathrm{,}$  and hence metrizable. Moreover each basic set $P_F\mathrm{<mo\; stretchy="false">(</mo>}a\mathrm{,}b\mathrm{<mo\; stretchy="false">)</mo>}$  is open and closed in $\mathrm{<mo\; stretchy="false">(</mo>}\mathbb{Z}\mathrm{,}{\tau }_F\mathrm{<mo\; stretchy="false">)</mo>,}$  so this space is zero-dimensional and not connected.

In 1959 and 1962 Golomb SW^4,5 showed inthat the family

${\mathcal{B}}_G\mathrm{<mo>=</mo><mo>\{</mo>}{P}_G\mathrm{<mo\; stretchy="false">(</mo>}a\mathrm{,}b\mathrm{<mo\; stretchy="false">)</mo>}:\mathrm{<mo\; stretchy="false">(</mo>}a\mathrm{,}b\mathrm{<mo\; stretchy="false">)</mo>}\in \mathbb{N}\times \mathbb{N}and\langle a\mathrm{,}b\rangle \mathrm{<mo>=</mo>1<mo>\}</mo>}$ (2)

is a base for a topology ${\tau }_G$  on $\mathbb{N}\mathrm{.}$  Indeed

${\tau }_G\mathrm{<mo>=</mo><mo>\{</mo>}\varnothing \mathrm{<mo>\}</mo>}\cup \mathrm{<mo>\{</mo>}U\subset \mathbb{N}:foreachb\in Uthereisa\in \mathbb{N}suchthat\langle a\mathrm{,}b\rangle \mathrm{<mo>=</mo>1}and{P}_G\mathrm{<mo\; stretchy="false">(</mo>}a\mathrm{,}b\mathrm{<mo\; stretchy="false">)</mo>}\subset U\mathrm{<mo>\}</mo>.}$

In,⁵ whose first edition was published in 1970, ${\tau }_G$  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 ${\tau }_G$  the Golomb topology and the topological space $\mathrm{<mo\; stretchy="false">(</mo>}\mathbb{N}\mathrm{,}{\tau }_G\mathrm{<mo\; stretchy="false">)</mo>}$  is called the Golomb space. It is known that $\mathrm{<mo\; stretchy="false">(</mo>}\mathbb{N}\mathrm{,}{\tau }_G\mathrm{<mo\; stretchy="false">)</mo>}$  is second countable, $T_2$  and connected Theorems 2 and 3,⁴ and Theorems 2 and 3⁵ . Using the fact that for every $p\in \mathbb{P}\mathrm{,}$  the set $M\mathrm{<mo\; stretchy="false">(</mo>}p\mathrm{<mo\; stretchy="false">)</mo>}$  is closed in $\mathrm{<mo\; stretchy="false">(</mo>}\mathbb{N}\mathrm{,}{\tau }_G\mathrm{<mo\; stretchy="false">)</mo>,}$  Golomb SW⁴ proved in both Theorem 1⁴ & Theorem 1⁵ that the set $\mathbb{P}$  is infinite. The proof of the connectedness of $\mathrm{<mo\; stretchy="false">(</mo>}\mathbb{N}\mathrm{,}{\tau }_G\mathrm{<mo\; stretchy="false">)</mo>,}$  as presented by Golomb SW,⁴ uses Number Theory. As it is indicated in^4,5 “a proof of the connectedness of $\mathrm{<mo\; stretchy="false">(</mo>}\mathbb{N}\mathrm{,}{\tau }_G\mathrm{<mo\; stretchy="false">)</mo>,}$  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 $\mathrm{<mo\; stretchy="false">(</mo>}\mathbb{N}\mathrm{,}{\tau }_G\mathrm{<mo\; stretchy="false">)</mo>,}$  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 $\mathrm{<mo>\{</mo>}a+bx\mathrm{<mo>\}</mo>,}$  where a and b are relatively prime to each other ( $x\mathrm{<mo>=</mo>1,2,3,}\dots$ ). Let $\mathrm{<mo>\{</mo>}a+bx\mathrm{<mo>\}</mo>}$  and $\mathrm{<mo>\{</mo>}c+dx\mathrm{<mo>\}</mo>}$  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 $X\mathrm{,}$  we have ${\text{cl}}_X\text{}\left(U\right)\cap {\text{cl}}_X\text{}\left(V\right)\ne \varnothing \mathrm{.}$  If $X$  is a topological space and $Y\subset X\mathrm{,}$  we say that $Y$  is a Brown space in $X$  if $Y\mathrm{,}$  as a subspace of $X\mathrm{,}$  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 $X\mathrm{<mo>=</mo>}U\cup V$  and $U\cap V\mathrm{<mo>=</mo>}\varnothing \mathrm{.}$  Then ${\text{cl}}_X\text{}\left(U\right)\cap {\text{cl}}_X\text{}\left(V\right)\mathrm{<mo>=</mo>}U\cap V\mathrm{<mo>=</mo>}\varnothing \mathrm{,}$  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 $\mathrm{<mo\; stretchy="false">(</mo>}\mathbb{N}\mathrm{,}{\tau }_G\mathrm{<mo\; stretchy="false">)</mo>}$  is a Brown space (Theorem 3.4). We will also show that, for every $d\in \mathbb{N}\mathrm{,}$  the subset $P_G\mathrm{<mo\; stretchy="false">(</mo>1,}d\mathrm{<mo\; stretchy="false">)</mo>}$  is a Brown space in $\mathrm{<mo\; stretchy="false">(</mo>}\mathbb{N}\mathrm{,}{\tau }_G\mathrm{<mo\; stretchy="false">)</mo>}$  (Theorem 3.3). Since $\mathrm{<mo\; stretchy="false">(</mo>}\mathbb{N}\mathrm{,}{\tau }_G\mathrm{<mo\; stretchy="false">)</mo>}$  is second countable, it is also Lindelöf. By the space $\mathrm{<mo\; stretchy="false">(</mo>}\mathbb{N}\mathrm{,}{\tau }_G\mathrm{<mo\; stretchy="false">)</mo>}$  is not $T_{3\frac{1}{2}}\mathrm{.}$ ⁴ Since every second coutable space is Lindelöf, ¹⁴ and every Lindelöff and $T_3$  space is $T_4$ ¹⁴ and hence $T_{3\frac{1}{2}}\mathrm{,}$  the space $\mathrm{<mo\; stretchy="false">(</mo>}\mathbb{N}\mathrm{,}{\tau }_G\mathrm{<mo\; stretchy="false">)</mo>}$  is not $T_3\mathrm{.}$  Without using all these results from General Topology (that lead to the fact that every non-empty countable connected $T_3$  space is a one-point-set), in both Theorem~4⁴ and Theorem~4,⁵ Golomb SW⁵ proved that $\mathrm{<mo\; stretchy="false">(</mo>}\mathbb{N}\mathrm{,}{\tau }_G\mathrm{<mo\; stretchy="false">)</mo>}$  is not $T_3$  by showing that for the closed set $M\mathrm{<mo\; stretchy="false">(</mo>2<mo\; stretchy="false">)</mo>}$  in $\mathrm{<mo\; stretchy="false">(</mo>}\mathbb{N}\mathrm{,}{\tau }_G\mathrm{<mo\; stretchy="false">)</mo>}$  that do not contain the point $\mathrm{1,}$  there are no open subsets $U$  y $V$  in $\mathrm{<mo\; stretchy="false">(</mo>}\mathbb{N}\mathrm{,}{\tau }_G\mathrm{<mo\; stretchy="false">)</mo>}$  such that $1\in U\mathrm{,}M\mathrm{<mo\; stretchy="false">(</mo>2<mo\; stretchy="false">)</mo>}\subset V$  and $U\cap V\mathrm{<mo>=</mo>}\varnothing \mathrm{.}$

Since compact $T_2$  spaces as well as locally compact and $T_2$  spaces are $T_3$ , ¹⁴ the space $\mathrm{<mo\; stretchy="false">(</mo>}\mathbb{N}\mathrm{,}{\tau }_G\mathrm{<mo\; stretchy="false">)</mo>}$  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 $c\mathrm{,}d\in \mathbb{Z}$  and $c\ne \mathrm{0,}$  then the symbol $c|d$  means that there exists $a\in \mathbb{Z}$  such that $d\mathrm{<mo>=</mo>}ca\mathrm{.}$ > If $c\mathrm{,}d\in \mathbb{Z}$  and $m\in \mathbb{N}-\mathrm{<mo>\{</mo>1<mo>\}</mo>,}$  then the symbol $c\equiv d\mathrm{<mo\; stretchy="false">(</mo>}modm\mathrm{<mo\; stretchy="false">)</mo>}$  means that $m|\mathrm{<mo\; stretchy="false">(</mo>}c-d\mathrm{<mo\; stretchy="false">)</mo>.}$  The next result is proved in.¹⁵

Theorem 2.1 Let $a\mathrm{,}b\mathrm{,}q\mathrm{,}r\in \mathbb{Z}$  be such that $a\mathrm{<mo>=</mo>}bq+r\mathrm{.}$  Then $\langle a\mathrm{,}b\rangle \mathrm{<mo>=</mo>}\langle b\mathrm{,}r\rangle \mathrm{.}$

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 $b\in \mathbb{Z}$ . If $p\in \mathbb{P}$  is such that $\langle b\mathrm{,}p\rangle \mathrm{<mo>=</mo>1,}$  then for each $n\mathrm{,}s\in {\mathbb{N}}_0\mathrm{,}$ $\langle pn+b\mathrm{,}{p}^s\rangle \mathrm{<mo>=</mo>1.}$ (3)

Proof. If $s\mathrm{<mo>=</mo>0,}$  then

$\langle pn+b\mathrm{,}{p}^s\rangle \mathrm{<mo>=</mo>}\langle pn+b\mathrm{,}{p}^0\rangle \mathrm{<mo>=</mo>}\langle pn+b\mathrm{,1}\rangle \mathrm{<mo>=</mo>1,}foreachn\in {\mathbb{N}}_0\mathrm{.}$

If $s\mathrm{<mo>=</mo>1}$  then, since $\langle p\mathrm{,}b\rangle \mathrm{<mo>=</mo>1,}$  by Theorem 2.1,

$\langle pn+b\mathrm{,}{p}^s\rangle \mathrm{<mo>=</mo>}\langle pn+b\mathrm{,}p\rangle \mathrm{<mo>=</mo>}\langle p\mathrm{,}b\rangle \mathrm{<mo>=</mo>1,}foreachn\in {\mathbb{N}}_0\mathrm{.}$  (4)

Now assume that there exist $s_0\in \mathbb{N}-\mathrm{<mo>\{</mo>1<mo>\}</mo>}$  and $n_0\in {\mathbb{N}}_0$  such that $g\mathrm{<mo>=</mo>}\langle p{n}_0+b\mathrm{,}{p}^{s_0}\rangle \mathrm{<mo>></mo>1.}$  Let $q\in \mathbb{P}$  be such that $q|g\mathrm{.}$  Then $q|\mathrm{<mo\; stretchy="false">(</mo>}p{n}_0+b\mathrm{<mo\; stretchy="false">)</mo>}$  and $q|p^{s_0}\mathrm{,}$  so $q\mathrm{<mo>=</mo>}p\mathrm{.}$  Hence $q|\mathrm{<mo\; stretchy="false">(</mo>}p{n}_0+b\mathrm{<mo\; stretchy="false">)</mo>}$  and $q|p\mathrm{,}$  so $q|\langle p{n}_0+b\mathrm{,}p\rangle \mathrm{.}$  This implies, using (4) with $n\mathrm{<mo>=</mo>}{n}_0\mathrm{,}$  that $q|\mathrm{1,}$  a contradiction. Thus $\langle pn+b\mathrm{,}{p}^s\rangle \mathrm{<mo>=</mo>1,}$  for every $s\mathrm{,}n\in {\mathbb{N}}_0\mathrm{.}$  

Concerning General Topology, given a topological space $X$  and $A\subset X\mathrm{,}$  we denote by ${\text{cl}}_X\text{}\left(A\right)$  and by ${\text{int}}_X\text{}\left(A\right)$  the closure and the interior of $A$  in $X\mathrm{,}$  respectively. In particular, for $A\subset \mathbb{N}\mathrm{,}$  the symbol ${\text{cl}}_X\text{}\left(A\right)$  denotes the closure of $A$  in $\mathrm{<mo\; stretchy="false">(</mo>}\mathbb{N}\mathrm{,}{\tau }_G\mathrm{<mo\; stretchy="false">)</mo>.}$  We consider the closed interval $\mathrm{<mo\; stretchy="false">[</mo>0,1<mo\; stretchy="false">]</mo>}$  with its usual topology. Recall that a topological space $X$  is said to be

1. $T_1$  if for each $x\in X\mathrm{,}$  the set $\mathrm{<mo>\{</mo>}x\mathrm{<mo>\}</mo>}$  is closed in $X\mathrm{<mo>;</mo>}$  
2. $T_2$  or Hausdorff if for every $x\mathrm{,}y\in X$  such that $x\ne y\mathrm{,}$  there exist open sets $U$  and $V$  in $X$  so that $x\in U\mathrm{,}$   $y\in V$  and $U\cap V\mathrm{<mo>=</mo>}\varnothing \mathrm{<mo>;</mo>}$  
3. $T_{2\frac{1}{2}}$  or completely Hausdorff if for every $x\mathrm{,}y\in X$  with $x\ne y\mathrm{,}$  there exist open sets $U$  and $V$  such that $x\in U\mathrm{,}$   $y\in V$  and ${\text{cl}}_X\text{}\left(U\right)\cap {\text{cl}}_X\text{}\left(V\right)\mathrm{<mo>=</mo>}\varnothing \mathrm{<mo>;</mo>}$  
4. regular if for every $x\in X$  and each closed set $C$  in $X$  such that $x\notin C\mathrm{,}$  there exist open sets $U$  and $V$  in $X$  so that $x\in U\mathrm{,}$   $C\subset V$  and $U\cap V\mathrm{<mo>=</mo>}\varnothing \mathrm{<mo>;</mo>}$  
5. $T_3$  if $X$  is regular and $T_1\mathrm{<mo>;</mo>}$  
6. $T_{3\frac{1}{2}}$  if $X$  is $T_1$  and for each $x\in X$  and every closed set $C$  in $X$  so that $x\notin C\mathrm{,}$  there is a continuous function $f:X\to \mathrm{<mo\; stretchy="false">[</mo>0,1<mo\; stretchy="false">]</mo>}$  such that $f\mathrm{<mo\; stretchy="false">(</mo>}x\mathrm{<mo\; stretchy="false">)</mo><mo>=</mo>0}$  and $f\mathrm{<mo\; stretchy="false">(</mo>}C\mathrm{<mo\; stretchy="false">)</mo>}\subset \mathrm{<mo>\{</mo>1<mo>\}</mo><mo>;</mo>}$  
7. $T_4$  if $X$  is $T_1$  and for every closed subsets $C$  and $D$  of $X$  with $C\cap D\mathrm{<mo>=</mo>}\varnothing \mathrm{,}$  there exist open sets $U$  and $V$  in $X$  such that $C\subset U\mathrm{,}$   $D\subset V$  and $U\cap V\mathrm{<mo>=</mo>}\varnothing \mathrm{.}$  

It is known that $T_4\Rightarrow T_{3\frac{1}{2}}\Rightarrow T_3\Rightarrow T_{2\frac{1}{2}}\Rightarrow T_2\Rightarrow T_1\mathrm{.}$  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 $X\mathrm{,}$  with at least two points, is $T_{2\frac{1}{2}}\mathrm{.}$  

By Theorem 2.3 no connected $T_{2\frac{1}{2}}\mathrm{,}$  with at least two points, is a Brown space.

Let $X$  be a topological space and $x$  in $X\mathrm{.}$  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 $T_1$  space contains no indiscrete points. By Theorem 2.4, no connected $T_3$  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 $P\mathrm{,}$  every subspace of $X$  also has the property $P\mathrm{<mo>;</mo>}$  
2. multiplicative if for any family $\mathrm{<mo>\{</mo>}{X}_s:s\in S\mathrm{<mo>\}</mo>}$  of topological spaces with the property $P\mathrm{,}$  the Cartesian product ${\displaystyle {\prod }_{s\in S}}{X}_s\mathrm{,}$  with the product topology, also has the property $P\mathrm{<mo>;</mo>}$  
3. factorizable if for any family $\mathrm{<mo>\{</mo>}{X}_s:s\in S\mathrm{<mo>\}</mo>}$  of topological spaces if the Cartesian product ${\displaystyle {\prod }_{s\in S}}{X}_s\mathrm{,}$  with the product topology, has the property $P$  then each factor $X_s$  also has the property $P\mathrm{.}$  

Theorem 2.5 Let $X$  and $Y$  be topological spaces and $f:X\to Y$  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 $Y\mathrm{.}$  Since $f$  is continuous and surjective, $f^{-1}\mathrm{<mo\; stretchy="false">(</mo>}U\mathrm{<mo\; stretchy="false">)</mo>}$  and $f^{-1}\mathrm{<mo\; stretchy="false">(</mo>}V\mathrm{<mo\; stretchy="false">)</mo>}$ are non-empty open subsets of $X\mathrm{.}$  Hence, since $X$  is a Brown space, ${\text{cl}}_X\text{}\left(f^{-1}\mathrm{<mo\; stretchy="false">(</mo>}U\mathrm{<mo\; stretchy="false">)</mo>}\right)\cap {\text{cl}}_X\text{}\left(f^{-1}\mathrm{<mo\; stretchy="false">(</mo>}U\mathrm{<mo\; stretchy="false">)</mo>}\right)\ne \varnothing \mathrm{.}$  By continuity of $X$  we have $\varnothing \ne {\text{cl}}_X\text{}\left(f^{-1}\mathrm{<mo\; stretchy="false">(</mo>}U\mathrm{<mo\; stretchy="false">)</mo>}\right)\cap {\text{cl}}_X\text{}\left(f^{-1}\mathrm{<mo\; stretchy="false">(</mo>}U\mathrm{<mo\; stretchy="false">)</mo>}\right)\subset f^{-1}\mathrm{<mo\; stretchy="false">(</mo>}{\text{cl}}_Y\text{}\left(U\right)\mathrm{<mo\; stretchy="false">)</mo>}\cap f^{-1}\mathrm{<mo\; stretchy="false">(</mo>}{\text{cl}}_Y\text{}\left(V\right)\mathrm{<mo\; stretchy="false">)</mo><mo>=</mo>}{f}^{-1}\mathrm{<mo\; stretchy="false">(</mo>}{\text{cl}}_Y\text{}\left(U\right)\cap {\text{cl}}_Y\text{}\left(V\right)\mathrm{<mo\; stretchy="false">)</mo>.}$

Hence ${\text{cl}}_Y\text{}\left(U\right)\cap {\text{cl}}_Y\text{}\left(V\right)\ne \varnothing \mathrm{,}$  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 $\mathrm{<mo>\{</mo>}{X}_s:s\in S\mathrm{<mo>\}</mo>}$  be a family of non-empty topological spaces. Let $X\mathrm{<mo>=</mo>}{\displaystyle {\prod }_{s\in S}}{X}_s$  and assume that $X$  has the product topology. For each $t\in S$  the projection $p_t:X\to X_t$  defined for any $x\mathrm{<mo>=</mo><mo\; stretchy="false">(</mo>}{x}_s{\mathrm{<mo\; stretchy="false">)</mo>}}_{s\in S}\in X$  by $p_t\mathrm{<mo\; stretchy="false">(</mo>}x\mathrm{<mo\; stretchy="false">)</mo><mo>=</mo>}{x}_t$  is continuous and surjective. Hence if $X$  is a Brown space then, by Theorem 2.5, $X_t$  is a Brown space too. This shows that being a Brown space is a factorizable property.

Now assume that each $X_s$  is a Brown space. Let $U$  and $V$  be two non-empty open subsets of $X\mathrm{.}$  Fix $x\in U$  and $y\in V$  and assume that $B\mathrm{<mo>=</mo>}{\displaystyle {\prod }_{s\in S}}{B}_s$  and $C\mathrm{<mo>=</mo>}{\displaystyle {\prod }_{s\in S}}{C}_s$  are basic subsets of $X$  such that $x\in B\subset U$  and $y\in C\subset V\mathrm{.}$  For each $s\in S\mathrm{,}$  the sets $B_s$  and $C_s$  are non-empty and open in the Brown space $X_s\mathrm{,}$  so ${\text{cl}}_{X_s}\text{}\left(B_s\right)\cap {\text{cl}}_{X_s}\text{}\left(C_s\right)\ne \varnothing \mathrm{.}$  Then

${\text{cl}}_X\text{}\left(B\right)\cap {\text{cl}}_X\text{}\left(C\right)\mathrm{<mo>=</mo>}\left({\displaystyle \prod _{s\in S}}{\text{cl}}_{X_s}\text{}\left(B_s\right)\right)\cap \left({\displaystyle \prod _{s\in S}}{\text{cl}}_{X_s}\text{}\left(C_s\right)\right)\mathrm{<mo>=</mo>}{\displaystyle \prod _{s\in S}}\left({\text{cl}}_{X_s}\text{}\left(B_s\right)\cap {\text{cl}}_{X_s}\text{}\left(C_s\right)\right)\ne \varnothing \mathrm{.}$

Hence $\varnothing \ne {\text{cl}}_X\text{}\left(B\right)\cap {\text{cl}}_X\text{}\left(C\right)\subset {\text{cl}}_X\text{}\left(U\right)\cap {\text{cl}}_X\text{}\left(V\right)\mathrm{,}$  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, $f:X\to Y$  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

1. hereditarily disconnected if no non-empty connected subset of $X$  contains more that one point;
2. totally separated if for every $x\mathrm{,}y\in X$  with $x\ne y\mathrm{,}$  there exist open sets $U$  and $V$  in $X$  such that $x\in U\mathrm{,}$   $y\in V\mathrm{,}$   $X\mathrm{<mo>=</mo>}U\cup V$  and $U\cap V\mathrm{<mo>=</mo>}\varnothing \mathrm{<mo>;</mo>}$  
3. zero-dimensional is $X$  is $T_1$  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 $x\in X\mathrm{,}$  the component $C_x$  of $X$  that contains $x$  is a one-point-set, namely $\mathrm{<mo>\{</mo>}x\mathrm{<mo>\}</mo>.}$  Note also that $X$  is totally separated if and only if, for every $x\in X\mathrm{,}$  the quasicomponet $Q_x$  of $X$  that contains $x$  is a one-point-set, namely $\mathrm{<mo>\{</mo>}x\mathrm{<mo>\}</mo>.}$  Since $C_x\subset Q_x$  for each $x\in X$ ¹⁴, totally separated spaces are hereditarily disconnected. If $X$  is compact and $T_2$  then $C_x\mathrm{<mo>=</mo>}{Q}_x$  for every $x\in X$ ¹⁴ so, in compact and $T_2$  spaces the notions of being hereditarily disconnected and of being totally separated coincide.

Let $X$  be a topological space and $a\mathrm{,}b\in X\mathrm{.}$  We say that $X$  is connected between $a$  and $b$  if for every open sets $U$  and $V$  in $X$  such that $a\in U\mathrm{,}$   $b\in V$  and $U\cap V\mathrm{<mo>=</mo>}\varnothing \mathrm{,}$  we have $X\ne U\cup V\mathrm{.}$  It is straight forward to see that if $X$  if $T_2$  then $X$  is totally separated if and only if for each $x\mathrm{,}y\in X$  with $x\ne y\mathrm{,}$  the space $X$  is not connected between $x$  and $y\mathrm{.}$

Let $X$  be a topological space and $x\in X\mathrm{.}$  We say that

1. $X$  is locally connected at $x\mathrm{,}$  if for any open subset $V$  of $X$  such that $x\in V\mathrm{,}$  there exists an open and connected subset $U$  of $X$  such that $x\in U\subset V\mathrm{<mo>;</mo>}$  
2. $X$  is connected im kleinen at $x\mathrm{,}$  if for any open subset $V$  of $X$  such that $x\in V\mathrm{,}$  there exists a connected subset $U$  of $X$  such that $x\in {\text{int}}_X\text{}\left(U\right)\subset U\subset V\mathrm{<mo>;</mo>}$  
3. $X$  is almost connected im kleinen at $x\mathrm{,}$  if for any open subset $V$  of $X$  such that $x\in V\mathrm{,}$  there exists a closed and connected subset $U$  of $X$  such that ${\text{int}}_X\text{}\left(U\right)\ne \varnothing$  and $U\subset V\mathrm{<mo>;</mo>}$  
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 $x\mathrm{,}$  then $X$  is connected im kleinen at $x\mathrm{.}$  The converse of this implication is not true. We construct an example in ${ℝ}^2\mathrm{,}$  with the usual topology. For each $i\in \mathbb{N}\mathrm{,}$  let $q_i\mathrm{<mo>=</mo>}\left(\frac{1}{i}\mathrm{,0}\right)\in {ℝ}^2$  and let $L_{i\mathrm{,0}}$  be the straight line segment from $q_i$  to $q_{i+1}$ . For each $\mathrm{<mo\; stretchy="false">(</mo>}i\mathrm{,}n\mathrm{<mo\; stretchy="false">)</mo>}\in \mathbb{N}\times \mathbb{N}\mathrm{,}$  we consider

$p_{i\mathrm{,}n}\mathrm{<mo>=</mo>}\frac{1}{i+1}\mathrm{<mo\; stretchy="false">(</mo>1,}\frac{1}{n}\mathrm{<mo\; stretchy="false">)</mo>}\in {ℝ}^2\mathrm{,}$

as well as the straight line segment $L_{i\mathrm{,}n}$  from $p_{i\mathrm{,}n}$  to $q_i\mathrm{.}$  Formally, for each $\mathrm{<mo\; stretchy="false">(</mo>}i\mathrm{,}n\mathrm{<mo\; stretchy="false">)</mo>}\in \mathbb{N}\times \mathbb{N}\mathrm{,}$  let

$L_{i\mathrm{,}n}\mathrm{<mo>=</mo><mo>\{</mo><mo\; stretchy="false">(</mo>}x\mathrm{,}y\mathrm{<mo\; stretchy="false">)</mo>}\in {ℝ}^2:\frac{1}{i+1}\le x\le \frac{1}{i}\text{and}y\mathrm{<mo>=</mo>}\frac{1}{n}\mathrm{<mo\; stretchy="false">(</mo>1}-ix\mathrm{<mo\; stretchy="false">)</mo><mo>\}</mo>,}$

where $y\mathrm{<mo>=</mo>}\frac{1}{n}\mathrm{<mo\; stretchy="false">(</mo>1}-ix\mathrm{<mo\; stretchy="false">)</mo>}$  is the equation of the straight line in ${ℝ}^2$  that contains the points $p_{i\mathrm{,}n}$  y $q_i\mathrm{.}$  For each $i\in \mathbb{N}\mathrm{,}$  we define $X_i\mathrm{<mo>=</mo>}{\displaystyle \underset{n\in \mathbb{N}\cup \mathrm{<mo>\{</mo>0<mo>\}</mo>}}{\cup }}{L}_{i\mathrm{,}n}\mathrm{.}$

Then $X_b\mathrm{<mo>=</mo>}{\text{cl}}_{{}^2}\text{}\left({\displaystyle \underset{i\in \mathbb{N}}{\cup }}{X}_i\right)$ is a topological space which is compact, connected, connnected im kleinen at $\mathrm{<mo\; stretchy="false">(</mo>0,0<mo\; stretchy="false">)</mo>}$  and not locally connected at $\mathrm{<mo\; stretchy="false">(</mo>0,0<mo\; stretchy="false">)</mo>.}$  The space $X_b\mathrm{,}$  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 $X_b\mathrm{,}$  is connected im kleinen at $\mathrm{<mo\; stretchy="false">(</mo>0,0<mo\; stretchy="false">)</mo>.}$  and not locally connected at $\mathrm{<mo\; stretchy="false">(</mo>0,0<mo\; stretchy="false">)</mo>.}$  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

$Y\mathrm{<mo>=</mo>}\left({\displaystyle \underset{n\in \mathbb{N}}{\cup }}{L}_{i\mathrm{,0}}\right)-\left\{q_i:i\in \mathbb{N}\right\}\subset X_b\mathrm{.}$

Note that $X_b$  is almost connected im kleinen at any point $p\in Y$  and not connected im kleinen at such point $p\mathrm{.}$  Note also that if a $T_3$  space is connected im kleinen at $x\in X\mathrm{,}$  then it is almost connected im kleinen at $x\mathrm{.}$

In the space $\mathrm{<mo\; stretchy="false">(</mo>}\mathbb{N}\mathrm{,}{\tau }_G\mathrm{<mo\; stretchy="false">)</mo>}$  a non-empty subset $U$  of $\mathbb{N}$  is open if and only if, for every $b\in U\mathrm{,}$  there exists $a\in \mathbb{N}$  such that $\langle a\mathrm{,}b\rangle \mathrm{<mo>=</mo>1}$  and $P_G\mathrm{<mo\; stretchy="false">(</mo>}a\mathrm{,}b\mathrm{<mo\; stretchy="false">)</mo>}\subset U$  (compare with).⁷ Hence $U$  is infinite. In particular any subset of $\mathbb{N}$  with non-empty interior in $\mathrm{<mo\; stretchy="false">(</mo>}\mathbb{N}\mathrm{,}{\tau }_G\mathrm{<mo\; stretchy="false">)</mo>}$  is infinite. Let $a\mathrm{,}x\in \mathbb{N}\mathrm{.}$  In Szczuka P⁸ presented some results that involve the set $\text{cl}\left(P\mathrm{<mo\; stretchy="false">(</mo>}a\mathrm{,}x\mathrm{<mo\; stretchy="false">)</mo>}\right)\mathrm{.}$  In,⁷ she showed that if $x_1\equiv x\mathrm{<mo\; stretchy="false">(</mo>}moda\mathrm{<mo\; stretchy="false">)</mo>,}$   $x_1\le a$  and $\langle a\mathrm{,}x\rangle \mathrm{<mo>=</mo>1,}$  then $P\mathrm{<mo\; stretchy="false">(</mo>}a\mathrm{,}{x}_1\mathrm{<mo\; stretchy="false">)</mo>}\subset \text{cl}\left(P_G\mathrm{<mo\; stretchy="false">(</mo>}a\mathrm{,}x\mathrm{<mo\; stretchy="false">)</mo>}\right)\mathrm{.}$  In⁸ she showed that if $a$  and $x$  are odd and $\langle a\mathrm{,}x\rangle \mathrm{<mo>=</mo>1,}$  then $\text{cl}\left(P_G\mathrm{<mo\; stretchy="false">(</mo>}a\mathrm{,}x\mathrm{<mo\; stretchy="false">)</mo>}\right)\mathrm{<mo>=</mo>}\text{cl}\left(P_G\mathrm{<mo\; stretchy="false">(</mo>2}a\mathrm{,}x\mathrm{<mo\; stretchy="false">)</mo>}\right)\mathrm{.}$  We show the following result.

Theorem 3.1 Let $a\mathrm{,}x\in \mathbb{N}$  be so that $\langle a\mathrm{,}x\rangle \mathrm{<mo>=</mo>1.}$  Then $M\mathrm{<mo\; stretchy="false">(</mo>}a\mathrm{<mo\; stretchy="false">)</mo>}\subset \text{cl}\left(P_G\mathrm{<mo\; stretchy="false">(</mo>}a\mathrm{,}x\mathrm{<mo\; stretchy="false">)</mo>}\right)\mathrm{.}$

Proof. Let $ab\in M\mathrm{<mo\; stretchy="false">(</mo>}a\mathrm{<mo\; stretchy="false">)</mo>}$  and let $W$  be an open subset of $\mathrm{<mo\; stretchy="false">(</mo>}\mathbb{N}\mathrm{,}{\tau }_G\mathrm{<mo\; stretchy="false">)</mo>}$  such that $ab\in W\mathrm{.}$  There exists $d\in \mathbb{N}$  such that $\langle d\mathrm{,}ab\rangle \mathrm{<mo>=</mo>1}$  and $P_G\mathrm{<mo\; stretchy="false">(</mo>}d\mathrm{,}ab\mathrm{<mo\; stretchy="false">)</mo>}\subset W\mathrm{.}$  Assume first that $a\mathrm{<mo>=</mo>1.}$  Then $P_G\mathrm{<mo\; stretchy="false">(</mo>}a\mathrm{,}x\mathrm{<mo\; stretchy="false">)</mo><mo>=</mo><mo>\{</mo>}x+n:n\in {\mathbb{N}}_0\mathrm{<mo>\}</mo>}$  so if $x\le ab\mathrm{,}$  then $ab\in P_G\mathrm{<mo\; stretchy="false">(</mo>}a\mathrm{,}x\mathrm{<mo\; stretchy="false">)</mo>}\subset \text{cl}\text{}\left(P_G\mathrm{<mo\; stretchy="false">(</mo>}a\mathrm{,}x\mathrm{<mo\; stretchy="false">)</mo>}\right)\mathrm{.}$  If $ab\mathrm{<mo><</mo>}x$  then, since $P_G\mathrm{<mo\; stretchy="false">(</mo>}d\mathrm{,}ab\mathrm{<mo\; stretchy="false">)</mo>}$  is infinite, there is $n\in \mathbb{N}$  such that $dn+ab\ge x\mathrm{,}$  so $dn+ab\in P_G\mathrm{<mo\; stretchy="false">(</mo>}d\mathrm{,}ab\mathrm{<mo\; stretchy="false">)</mo>}\cap P_G\mathrm{<mo\; stretchy="false">(</mo>}a\mathrm{,}x\mathrm{<mo\; stretchy="false">)</mo>}\subset W\cap P_G\mathrm{<mo\; stretchy="false">(</mo>}a\mathrm{,}x\mathrm{<mo\; stretchy="false">)</mo>.}$  This shows that $ab\in \text{cl}\text{}\left(P_G\mathrm{<mo\; stretchy="false">(</mo>}a\mathrm{,}x\mathrm{<mo\; stretchy="false">)</mo>}\right)\mathrm{.}$

Now assume that $a\ge 2$  By showing that $P_G\mathrm{<mo\; stretchy="false">(</mo>}d\mathrm{,}ab\mathrm{<mo\; stretchy="false">)</mo>}\cap P_G\mathrm{<mo\; stretchy="false">(</mo>}a\mathrm{,}x\mathrm{<mo\; stretchy="false">)</mo>}\ne \varnothing$  we will obtain that $W\cap P_G\mathrm{<mo\; stretchy="false">(</mo>}a\mathrm{,}x\mathrm{<mo\; stretchy="false">)</mo>}\ne \varnothing \mathrm{.}$  To prove that $P_G\mathrm{<mo\; stretchy="false">(</mo>}d\mathrm{,}ab\mathrm{<mo\; stretchy="false">)</mo>}\cap P_G\mathrm{<mo\; stretchy="false">(</mo>}a\mathrm{,}x\mathrm{<mo\; stretchy="false">)</mo>}\ne \varnothing \mathrm{,}$  assume first that $d\mathrm{<mo>=</mo>1.}$  Then $ab+x\in P_G\mathrm{<mo\; stretchy="false">(</mo>}a\mathrm{,}x\mathrm{<mo\; stretchy="false">)</mo>}$  and $ab+x\mathrm{<mo>=</mo>1}\cdot x+ab\in P_G\mathrm{<mo\; stretchy="false">(</mo>}d\mathrm{,}ab\mathrm{<mo\; stretchy="false">)</mo>.}$  Now assume that $d\ge 2.$  Since $d\ge 2.$  and $a\ge \mathrm{2,}$  an element $z\in P_G\mathrm{<mo\; stretchy="false">(</mo>}d\mathrm{,}ab\mathrm{<mo\; stretchy="false">)</mo>}\cap P_G\mathrm{<mo\; stretchy="false">(</mo>}a\mathrm{,}x\mathrm{<mo\; stretchy="false">)</mo>}$  satisfies the system of congruencies

$z\equiv ab\mathrm{<mo\; stretchy="false">(</mo>}modd\mathrm{<mo\; stretchy="false">)</mo>}andz\equiv x\mathrm{<mo\; stretchy="false">(</mo>}moda\mathrm{<mo\; stretchy="false">)</mo>.}$  (5)

Hence, an element in $P_G\mathrm{<mo\; stretchy="false">(</mo>}d\mathrm{,}ab\mathrm{<mo\; stretchy="false">)</mo>}\cap P_G\mathrm{<mo\; stretchy="false">(</mo>}a\mathrm{,}x\mathrm{<mo\; stretchy="false">)</mo>}$  is a solution of the system (5). Conversely, every solution of the system (5), is an element of $P_G\mathrm{<mo\; stretchy="false">(</mo>}d\mathrm{,}ab\mathrm{<mo\; stretchy="false">)</mo>}\cap P_G\mathrm{<mo\; stretchy="false">(</mo>}a\mathrm{,}x\mathrm{<mo\; stretchy="false">)</mo>.}$  Let us show, then, that the system (5) has a solution. If $\langle d\mathrm{,}a\rangle \ne \mathrm{1,}$  then there exists $p\in \mathbb{P}$  such that $p|d$  and $p|a\mathrm{.}$  Then $p|d$  and $p|ab\mathrm{,}$  so $\langle d\mathrm{,}ab\rangle \ne \mathrm{1,}$  a contradiction. Hence $\langle d\mathrm{,}a\rangle \mathrm{<mo>=</mo>1}$  and by the Chinese Reminder Theorem, the system (5) has a solution. This shows that $P_G\mathrm{<mo\; stretchy="false">(</mo>}d\mathrm{,}ab\mathrm{<mo\; stretchy="false">)</mo>}\cap P_G\mathrm{<mo\; stretchy="false">(</mo>}a\mathrm{,}x\mathrm{<mo\; stretchy="false">)</mo>}\ne \varnothing \mathrm{.}$  Hence $W\cap P_G\mathrm{<mo\; stretchy="false">(</mo>}a\mathrm{,}x\mathrm{<mo\; stretchy="false">)</mo>}\ne \varnothing$  and then $ab\in \text{cl}\left(P_G\mathrm{<mo\; stretchy="false">(</mo>}a\mathrm{,}x\mathrm{<mo\; stretchy="false">)</mo>}\right)\mathrm{.}$

Corollary 3.2 For every $d\in \mathbb{N}\mathrm{,}$  the set $P_G\mathrm{<mo\; stretchy="false">(</mo>1,}d\mathrm{<mo\; stretchy="false">)</mo>}$  is dense in $\mathrm{<mo\; stretchy="false">(</mo>}\mathbb{N}\mathrm{,}{\tau }_G\mathrm{<mo\; stretchy="false">)</mo>.}$  

Proof. Let $d\in \mathbb{N}\mathrm{.}$  Put $a\mathrm{<mo>=</mo>1.}$  Then $\langle a\mathrm{,}d\rangle \mathrm{<mo>=</mo>1,}$   $M\mathrm{<mo\; stretchy="false">(</mo>}a\mathrm{<mo\; stretchy="false">)</mo><mo>=</mo>}\mathbb{N}$  and, by Theorem 3.1, $\mathbb{N}\mathrm{<mo>=</mo>}M\mathrm{<mo\; stretchy="false">(</mo>}a\mathrm{<mo\; stretchy="false">)</mo>}\subset \text{cl}\left(P_G\mathrm{<mo\; stretchy="false">(</mo>}a\mathrm{,}d\mathrm{<mo\; stretchy="false">)</mo>}\right)\mathrm{,}$  so $P_G\mathrm{<mo\; stretchy="false">(</mo>}a\mathrm{,}d\mathrm{<mo\; stretchy="false">)</mo><mo>=</mo>}{P}_G\mathrm{<mo\; stretchy="false">(</mo>1,}d\mathrm{<mo\; stretchy="false">)</mo>}$  is dense in $\mathrm{<mo\; stretchy="false">(</mo>}\mathbb{N}\mathrm{,}{\tau }_G\mathrm{<mo\; stretchy="false">)</mo>.}$  

Theorem 3.3 For every $d\in \mathbb{N}\mathrm{,}$  the set $P_G\mathrm{<mo\; stretchy="false">(</mo>1,}d\mathrm{<mo\; stretchy="false">)</mo>}$  is a Brown space in $\mathrm{<mo\; stretchy="false">(</mo>}\mathbb{N}\mathrm{,}{\tau }_G\mathrm{<mo\; stretchy="false">)</mo>.}$  In particular, $P_G\mathrm{<mo\; stretchy="false">(</mo>1,}d\mathrm{<mo\; stretchy="false">)</mo>}$  is connected.

Proof. Let $U$  and $V$  be two non-empty open subsets of $P_G\mathrm{<mo\; stretchy="false">(</mo>1,}d\mathrm{<mo\; stretchy="false">)</mo>.}$  Since $P_G\mathrm{<mo\; stretchy="false">(</mo>1,}d\mathrm{<mo\; stretchy="false">)</mo>}$  is open in $\mathrm{<mo\; stretchy="false">(</mo>}\mathbb{N}\mathrm{,}{\tau }_G\mathrm{<mo\; stretchy="false">)</mo>,}$  the sets $U$  and $V$  are open in $\mathrm{<mo\; stretchy="false">(</mo>}\mathbb{N}\mathrm{,}{\tau }_G\mathrm{<mo\; stretchy="false">)</mo>,}$  Let $x\in U$  and $y\in V\mathrm{.}$  Then there exist $a\mathrm{,}b\in \mathbb{N}$  such that $\langle a\mathrm{,}x\rangle \mathrm{<mo>=</mo>}\langle b\mathrm{,}y\rangle \mathrm{<mo>=</mo>1}$  and $P_G\mathrm{<mo\; stretchy="false">(</mo>}a\mathrm{,}x\mathrm{<mo\; stretchy="false">)</mo>}\subset U$  and $P_G\mathrm{<mo\; stretchy="false">(</mo>}b\mathrm{,}y\mathrm{<mo\; stretchy="false">)</mo>}\subset V\mathrm{.}$  If $a\mathrm{<mo>=</mo>1,}$  then $P_G\mathrm{<mo\; stretchy="false">(</mo>}a\mathrm{,}x\mathrm{<mo\; stretchy="false">)</mo><mo>=</mo><mo>\{</mo>}x+n:n\in {\mathbb{N}}_0\mathrm{<mo>\}</mo>.}$  Since $P_G\mathrm{<mo\; stretchy="false">(</mo>}b\mathrm{,}y\mathrm{<mo\; stretchy="false">)</mo>}$  is infinite, there is $n_0\in {\mathbb{N}}_0$  such that $y+b{n}_0\ge x$  and then

$y+b{n}_0\in P_G\mathrm{<mo\; stretchy="false">(</mo>}a\mathrm{,}x\mathrm{<mo\; stretchy="false">)</mo>}\cap P_G\mathrm{<mo\; stretchy="false">(</mo>}b\mathrm{,}y\mathrm{<mo\; stretchy="false">)</mo>}\subset U\cap V\cap P_G\mathrm{<mo\; stretchy="false">(</mo>1,}d\mathrm{<mo\; stretchy="false">)</mo>}\subset \text{cl}\left(U\right)\cap \text{cl}\left(V\right)\cap P_G\mathrm{<mo\; stretchy="false">(</mo>1,}d\mathrm{<mo\; stretchy="false">)</mo><mo>=</mo>}$ ${\text{cl}}_{P_G\mathrm{<mo\; stretchy="false">(</mo>1,}d\mathrm{<mo\; stretchy="false">)</mo>}}\text{}\left(U\right)\cap {\text{cl}}_{P_G\mathrm{<mo\; stretchy="false">(</mo>1,}d\mathrm{<mo\; stretchy="false">)</mo>}}\text{}\left(V\right)\mathrm{.}$

Hence ${\text{cl}}_{P_G\mathrm{<mo\; stretchy="false">(</mo>1,}d\mathrm{<mo\; stretchy="false">)</mo>}}\text{}\left(U\right)\cap {\text{cl}}_{P_G\mathrm{<mo\; stretchy="false">(</mo>1,}d\mathrm{<mo\; stretchy="false">)</mo>}}\text{}\left(V\right)\ne \varnothing \mathrm{.}$  Similarly, if $b\mathrm{<mo>=</mo>1}$  we have ${\text{cl}}_{P_G\mathrm{<mo\; stretchy="false">(</mo>1,}d\mathrm{<mo\; stretchy="false">)</mo>}}\text{}\left(U\right)\cap {\text{cl}}_{P_G\mathrm{<mo\; stretchy="false">(</mo>1,}d\mathrm{<mo\; stretchy="false">)</mo>}}\text{}\left(V\right)\ne \varnothing \mathrm{.}$  Now assume that $a\ge 2$  and $b\ge 2.$  By Theorem 3.1, $M\mathrm{<mo\; stretchy="false">(</mo>}a\mathrm{<mo\; stretchy="false">)</mo>}\subset \text{cl}\text{}\left(P_G\mathrm{<mo\; stretchy="false">(</mo>}a\mathrm{,}x\mathrm{<mo\; stretchy="false">)</mo>}\right)$  and $M\mathrm{<mo\; stretchy="false">(</mo>}b\mathrm{<mo\; stretchy="false">)</mo>}\subset \text{cl}\left(P_G\mathrm{<mo\; stretchy="false">(</mo>}b\mathrm{,}y\mathrm{<mo\; stretchy="false">)</mo>}\right)$  so, in particular, $abd\in \text{cl}\left(P_G\mathrm{<mo\; stretchy="false">(</mo>}a\mathrm{,}x\mathrm{<mo\; stretchy="false">)</mo>}\right)\cap \text{cl}\left(P_G\mathrm{<mo\; stretchy="false">(</mo>}b\mathrm{,}y\mathrm{<mo\; stretchy="false">)</mo>}\right)\mathrm{.}$  Since $a\ge 2$  and $b\ge \mathrm{2,}$  we have $d\mathrm{<mo><</mo>}abd\mathrm{,}$  so $abd\in P_G\mathrm{<mo\; stretchy="false">(</mo>1,}d\mathrm{<mo\; stretchy="false">)</mo>.}$  Then

$abd\in \text{cl}\text{}\left(P_G\mathrm{<mo\; stretchy="false">(</mo>}a\mathrm{,}x\mathrm{<mo\; stretchy="false">)</mo>}\right)\cap \text{cl}\left(P_G\mathrm{<mo\; stretchy="false">(</mo>}b\mathrm{,}y\mathrm{<mo\; stretchy="false">)</mo>}\right)\cap P_G\mathrm{<mo\; stretchy="false">(</mo>1,}d\mathrm{<mo\; stretchy="false">)</mo>}\subset \text{cl}\left(U\right)\cap \text{cl}\left(V\right)\cap P_G\mathrm{<mo\; stretchy="false">(</mo>1,}d\mathrm{<mo\; stretchy="false">)</mo><mo>=</mo>}$ ${\text{cl}}_{P_G\mathrm{<mo\; stretchy="false">(</mo>1,}d\mathrm{<mo\; stretchy="false">)</mo>}}\text{}\left(U\right)\cap {\text{cl}}_{P_G\mathrm{<mo\; stretchy="false">(</mo>1,}d\mathrm{<mo\; stretchy="false">)</mo>}}\text{}\left(V\right)\mathrm{.}$

Hence ${\text{cl}}_{P_G\mathrm{<mo\; stretchy="false">(</mo>1,}d\mathrm{<mo\; stretchy="false">)</mo>}}\text{}\left(U\right)\cap {\text{cl}}_{P_G\mathrm{<mo\; stretchy="false">(</mo>1,}d\mathrm{<mo\; stretchy="false">)</mo>}}\text{}\left(V\right)\ne \varnothing \mathrm{,}$ so $P_G\mathrm{<mo\; stretchy="false">(</mo>1,}d\mathrm{<mo\; stretchy="false">)</mo>}$  is a Brown space in $\mathrm{<mo\; stretchy="false">(</mo>}\mathbb{N}\mathrm{,}{\tau }_G\mathrm{<mo\; stretchy="false">)</mo>.}$  Since Brown spaces are connected, $P_G\mathrm{<mo\; stretchy="false">(</mo>1,}d\mathrm{<mo\; stretchy="false">)</mo>}$  is connected.

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

Theorem 3.4 $\mathrm{<mo\; stretchy="false">(</mo>}\mathbb{N}\mathrm{,}{\tau }_G\mathrm{<mo\; stretchy="false">)</mo>}$  is a Brown space. In particular, $\mathrm{<mo\; stretchy="false">(</mo>}\mathbb{N}\mathrm{,}{\tau }_G\mathrm{<mo\; stretchy="false">)</mo>}$  is connected and it is not $T_{2\frac{1}{2}}\mathrm{.}$
Proof. Since $P_G\mathrm{<mo\; stretchy="false">(</mo>1,1<mo\; stretchy="false">)</mo><mo>=</mo>}\mathbb{N}$  the result follows from Theorem 3.3 and the fact that Brown spaces are not $T_{2\frac{1}{2}}\mathrm{.}$

Now we prove the following result.

Theorem 3.5 Let $c\mathrm{,}p\in \mathbb{N}$  be such that $p\in \mathbb{P}$  and $\langle p\mathrm{,}c\rangle \mathrm{<mo>=</mo>1.}$  Take $a\mathrm{,}b\in P_G\mathrm{<mo\; stretchy="false">(</mo>}p\mathrm{,}c\mathrm{<mo\; stretchy="false">)</mo>}$  such that $a\mathrm{<mo><</mo>}b$  and $n\mathrm{,}m\in {\mathbb{N}}_0$  so that $a\mathrm{<mo>=</mo>}pm+c\mathrm{,}b\mathrm{<mo>=</mo>}pn+c$  and $0\le m\mathrm{<mo><</mo>}n\mathrm{.}$  Then

$U\mathrm{<mo>=</mo>}{\displaystyle \underset{i\mathrm{<mo>=</mo>0}}{\overset{m}{\cup }}}{P}_G\mathrm{<mo\; stretchy="false">(</mo>}{p}^{n+1}\mathrm{,}pi+c\mathrm{<mo\; stretchy="false">)</mo>}andV\mathrm{<mo>=</mo>}{\displaystyle \underset{j\mathrm{<mo>=</mo>}m+}{\overset{p^n-1}{\cup }}}{P}_G\mathrm{<mo\; stretchy="false">(</mo>}{p}^{n+1}\mathrm{,}pj+c\mathrm{<mo\; stretchy="false">)</mo>}$  (6)

are open subset of $\mathrm{<mo\; stretchy="false">(</mo>}\mathbb{N}\mathrm{,}{\tau }_G\mathrm{<mo\; stretchy="false">)</mo>}$  such that $a\in U\mathrm{,}$ $b\in V\mathrm{,}$ $U\cap V\mathrm{<mo>=</mo>}\varnothing$  and $P_G\mathrm{<mo\; stretchy="false">(</mo>}p\mathrm{,}c\mathrm{<mo\; stretchy="false">)</mo><mo>=</mo>}U\cup V\mathrm{.}$  In particular, $P_G\mathrm{<mo\; stretchy="false">(</mo>}p\mathrm{,}c\mathrm{<mo\; stretchy="false">)</mo>}$  is not connected.

Proof. Since $p\in \mathbb{P}$  and $\langle p\mathrm{,}c\rangle \mathrm{<mo>=</mo>1,}$  by Theorem 2.2,

$\langle p^{n+1}\mathrm{,}pi+c\rangle \mathrm{<mo>=</mo>1,}foreachi\in {\mathbb{N}}_0\mathrm{.}$

Then we can consider the sets $U$ and $V$  indicated in (6), and they are open in $\mathrm{<mo\; stretchy="false">(</mo>}\mathbb{N}\mathrm{,}{\tau }_G\mathrm{<mo\; stretchy="false">)</mo>.}$  Since $0\le m\mathrm{<mo><</mo>}n{\mathrm{<mo><</mo>2}}^n\le p^n\mathrm{,}$  we have $m+1\le n\le p^n-\mathrm{1,}$  so

$a\mathrm{<mo>=</mo>}pm+c\mathrm{<mo>=</mo>}{p}^{n+1}\mathrm{<mo\; stretchy="false">(</mo>0<mo\; stretchy="false">)</mo>}+pm+c\in P_G\mathrm{<mo\; stretchy="false">(</mo>}{p}^{n+1}\mathrm{,}pm+c\mathrm{<mo\; stretchy="false">)</mo>}\subset {\displaystyle \underset{i\mathrm{<mo>=</mo>0}}{\overset{m}{\cup }}}{P}_G\mathrm{<mo\; stretchy="false">(</mo>}{p}^{n+1}\mathrm{,}pi+c\mathrm{<mo\; stretchy="false">)</mo><mo>=</mo>}U$

and

$b\mathrm{<mo>=</mo>}pn+c\mathrm{<mo>=</mo>}{p}^{n+1}\mathrm{<mo\; stretchy="false">(</mo>0<mo\; stretchy="false">)</mo>}+pn+c\in P_G\mathrm{<mo\; stretchy="false">(</mo>}{p}^{n+1}\mathrm{,}pn+c\mathrm{<mo\; stretchy="false">)</mo>}\subset {\displaystyle \underset{j\mathrm{<mo>=</mo>}m+1}{\overset{p^n-1}{\cup }}}{P}_G\mathrm{<mo\; stretchy="false">(</mo>}{p}^{n+1}\mathrm{,}pj+c\mathrm{<mo\; stretchy="false">)</mo><mo>=</mo>}V\mathrm{.}$

Hence $a\in U$  and $b\in V$  so $U$  and $V$  are infinite. Now assume that there is a point $z\in U\cap V\mathrm{.}$  Let $i\in \mathrm{<mo>\{</mo>0,1,}\dots \mathrm{,}m\mathrm{<mo>\}</mo>}$  and $j\in \mathrm{<mo>\{</mo>}m+\mathrm{1,}m+\mathrm{2,}\dots \mathrm{,}{p}^n-\mathrm{1<mo>\}</mo>}$  be such that $z\in P_G\mathrm{<mo\; stretchy="false">(</mo>}{p}^{n+1}\mathrm{,}pi+c\mathrm{<mo\; stretchy="false">)</mo>}$  and $z\in P_G\mathrm{<mo\; stretchy="false">(</mo>}{p}^{n+1}\mathrm{,}pj+c\mathrm{<mo\; stretchy="false">)</mo>.}$  Since $p^{n+1}\mathrm{<mo>></mo>1,}$  it follows that

$z\equiv pi+c\mathrm{<mo\; stretchy="false">(</mo>}mod{p}^{n+1}\mathrm{<mo\; stretchy="false">)</mo>}andz\equiv pj+c\mathrm{<mo\; stretchy="false">(</mo>}mod{p}^{n+1}\mathrm{<mo\; stretchy="false">)</mo>.}$

Hence $pi+c\equiv pj+c\mathrm{<mo\; stretchy="false">(</mo>}mod{p}^{n+1}\mathrm{<mo\; stretchy="false">)</mo>,}$  so $pi\equiv pj\mathrm{<mo\; stretchy="false">(</mo>}mod{p}^{n+1}\mathrm{<mo\; stretchy="false">)</mo>.}$  This implies that $i\equiv j\mathrm{<mo\; stretchy="false">(</mo>}mod{p}^n\mathrm{<mo\; stretchy="false">)</mo>.}$  However, since the set $\mathrm{<mo>\{</mo>0,1,}\dots \mathrm{,}m\mathrm{,}m+\mathrm{1,}\dots \mathrm{,}{p}^n-\mathrm{1<mo>\}</mo>}$  is a complete system of reminders modulus $p^n$  and $i\le m\mathrm{<mo><</mo>}m+1\le j\mathrm{,}$  we have $i\cancel{\equiv }j\mathrm{<mo\; stretchy="false">(</mo>}mod{p}^n\mathrm{<mo\; stretchy="false">)</mo>.}$  From this contradiction we infer that $U\cap V\mathrm{<mo>=</mo>}\varnothing \mathrm{.}$

Now we show that $U\cup V\mathrm{<mo>=</mo>}{P}_G\mathrm{<mo\; stretchy="false">(</mo>}p\mathrm{,}c\mathrm{<mo\; stretchy="false">)</mo>.}$  If $z\in U\mathrm{,}$  then there exists $i\in \mathrm{<mo>\{</mo>0,1,}\dots \mathrm{,}m\mathrm{<mo>\}</mo>}$  such that $z\in P_G\mathrm{<mo\; stretchy="false">(</mo>}{p}^{n+1}\mathrm{,}pi+c\mathrm{<mo\; stretchy="false">)</mo>.}$  Let $z_0\in {\mathbb{N}}_0$  be such that $z\mathrm{<mo>=</mo>}{p}^{n+1}{z}_0+pi+c\mathrm{.}$  Then $z\mathrm{<mo>=</mo>}p\mathrm{<mo\; stretchy="false">(</mo>}{p}^n{z}_0+i\mathrm{<mo\; stretchy="false">)</mo>}+c\in P_G\mathrm{<mo\; stretchy="false">(</mo>}p\mathrm{,}c\mathrm{<mo\; stretchy="false">)</mo>.}$  Similarly, if $z\in V\mathrm{,}$  there is $j\in \mathrm{<mo>\{</mo>}m+\mathrm{1,}m+\mathrm{2,}\dots \mathrm{,}{p}^n-\mathrm{1<mo>\}</mo>}$  such that $z\in P_G\mathrm{<mo\; stretchy="false">(</mo>}{p}^{n+1}\mathrm{,}pj+c\mathrm{<mo\; stretchy="false">)</mo>.}$  Let $z_1\in {\mathbb{N}}_0$  be such that $z\mathrm{<mo>=</mo>}{p}^{n+1}{z}_1+pj+c\mathrm{.}$  Then $z\mathrm{<mo>=</mo>}p\mathrm{<mo\; stretchy="false">(</mo>}{p}^n{z}_1+j\mathrm{<mo\; stretchy="false">)</mo>}+c\in P_G\mathrm{<mo\; stretchy="false">(</mo>}p\mathrm{,}c\mathrm{<mo\; stretchy="false">)</mo>.}$  This shows that $U\cup V\subset P_G\mathrm{<mo\; stretchy="false">(</mo>}p\mathrm{,}c\mathrm{<mo\; stretchy="false">)</mo>.}$  To prove the other inclusion, let $z\in P_G\mathrm{<mo\; stretchy="false">(</mo>}p\mathrm{,}c\mathrm{<mo\; stretchy="false">)</mo>.}$  Then there exists $k\in {\mathbb{N}}_0$  such that $z\mathrm{<mo>=</mo>}pk+c\mathrm{.}$  Using the Division Algorithm we obtain $s\mathrm{,}t\in {\mathbb{N}}_0$  such that $k\mathrm{<mo>=</mo>}{p}^{n}s+t$  and $0\le t\mathrm{<mo><</mo>}{p}^n\mathrm{.}$  Hence

$z\mathrm{<mo>=</mo>}pk+c\mathrm{<mo>=</mo>}p\mathrm{<mo\; stretchy="false">(</mo>}{p}^{n}s+t\mathrm{<mo\; stretchy="false">)</mo>}+c\mathrm{<mo>=</mo>}{p}^{n+1}s+pt+c\mathrm{.}$ (7)

Since $0\le t\mathrm{<mo><</mo>}{p}^n$  and $0\le m\mathrm{<mo><</mo>}{p}^n\mathrm{,}$  we obtain

$t\in \mathrm{<mo>\{</mo>0,1,}\dots \mathrm{,}{p}^n-\mathrm{1<mo>\}</mo><mo>=</mo><mo>\{</mo>0,1,}\dots \mathrm{,}m\mathrm{<mo>\}</mo>}\cup \mathrm{<mo>\{</mo>}m+\mathrm{1,}m+\mathrm{2,}\dots \mathrm{,}{p}^n-\mathrm{1<mo>\}</mo>.}$

Hence, by (7),

$z\in \left({\displaystyle \underset{i\mathrm{<mo>=</mo>0}}{\overset{m}{\cup }}}{P}_G\mathrm{<mo\; stretchy="false">(</mo>}{p}^{n+1}\mathrm{,}pi+c\mathrm{<mo\; stretchy="false">)</mo>}\right)\cup \left({\displaystyle \underset{j\mathrm{<mo>=</mo>}m+1}{\overset{p^n-1}{\cup }}}{P}_G\mathrm{<mo\; stretchy="false">(</mo>}{p}^{n+1}\mathrm{,}pj+c\mathrm{<mo\; stretchy="false">)</mo>}\right)\mathrm{<mo>=</mo>}U\cup V\mathrm{.}$