First let us fix the terminology.

The space (1) is known in General Topology as the Golomb space. More precisely, the *Golomb space* $\mathbb G$ is the set $\mathbb N$ of positive integers, endowed with the topology generated by the base consisting of arithmetic progressions $a+b\mathbb N_0$ where $a,b$ are relatively prime natural numbers and $\mathbb N_0=\{0\}\cup\mathbb N$.

Let us call the space (2) the *rational projective space* and denote it by
$\mathbb QP^\infty$.

Both spaces $\mathbb G$ and $\mathbb QP^\infty$ are countable, connected and Hausdorff but they are not
homeomorphic. A topological property distinguishing these spaces will
be called the oo-regularity.

**Definition.** A topological space $X$ is called *oo-regular*
if for any non-empty disjoint open sets $U,V\subset X$ the subspace $X\setminus(\bar U\cap\bar V)$ of $X$ is regular.

**Theorem.**

The rational projective space $\mathbb QP^\infty$ is oo-regular.

The Golomb space $\mathbb G$ is not oo-regular.

*Proof.* The statement 1 is relatively easy, so is left to the interested reader.

*The proof of 2.* In the Golomb space $\mathbb G$ consider two basic open sets $U=1+5\mathbb N_0$ and $V=2+5\mathbb N_0$.
It can be shown that $\bar U=U\cup 5\mathbb N$ and $\bar V=V\cup 5\mathbb N$, so $\bar U\cap\bar V=5\mathbb N$.

We claim that the subspace
$X=\mathbb N\setminus (\bar U\cap\bar V)=\mathbb N\setminus 5\mathbb N$ of the Golomb space
is not regular.

Consider the point $x=1$ and its neighborhood $O_x=(1+4\mathbb N)\cap X$ in $X$.
Assuming that $X$ is regular, we can find a neighborhood $U_x$ of $x$ in $X$ such that
$\bar U_x\cap X\subset O_x$.

We can assume that $U_x$ is of basic form
$U_x=1+2^i5^jb\mathbb N_0$ for some $i\ge 2$, $j\ge 1$ and $b\in\mathbb N\setminus(2\mathbb N_0\cup 5\mathbb N_0)$.

Since the numbers $4$, $5^j$, and $b$ are relatively prime, by the Chinese remainder Theorem, the intersection $(1+5^j\mathbb N_0)\cap (2+4\mathbb N_0)\cap b\mathbb N_0$
contains some point $y$. It is clear that $y\in X\setminus O_x$.

We claim that $y$ belongs to the closure of $U_x$ in $X$. We need to check that each basic neighborhood $O_y:=y+c\mathbb N_0$ of $y$ intersects the set $U_x$. Replacing $c$ by $5^jc$, we can assume that $c$ is divisible by $5^j$ and hence $c=5^jc'$ for some $c'\in\mathbb N_0$.

Observe that $O_y\cap U_x=(y+c\mathbb N_0)\cap(1+4^i5^jb\mathbb N_0)\ne\emptyset$ if and only if $y-1\in 4^i5^jb\mathbb N_0-5^jc'\mathbb N_0=5^j(4^ib\mathbb N_0-c'\mathbb N_0)$. The choice of $y\in 1+5^j\mathbb N_0$ guarantees that $y-1=5^jy'$. Since $y\in 2\mathbb N_0\cap b\mathbb N_0$ and $c$ is relatively prime with $y$, the number $c'=c/5^j$ is relatively prime with $4^ib$. So, by the Euclidean Algorithm, there are numbers $u,v\in\mathbb N_0$ such that $y'=4^ibu-c'v$. Then $y-1=5^jy'=5^j(4^ibu-c'v)$ and hence $1+4^i5^ju=y+5^jc'v\in (1+4^i5^jb\mathbb N_0)\cap(y+c\mathbb N_0)=U_x\cap U_y\ne\emptyset$. So, $y\in\bar U_x\setminus O_x$, which contradicts the choice of $U_x$.

**Remark.** Another well-known example of a countable connected space is the Bing space $\mathbb B$. This is the rational half-plane $\mathbb B=\{(x,y)\in\mathbb Q\times \mathbb Q:y\ge 0\}$ endowed with the topology generated by the base consisting of sets $$U_{\varepsilon}(a,b)=
\{(a,b)\}\cup\{(x,0)\in\mathbb B:|x-(a-\sqrt{2}b)|<\varepsilon\}\cup
\{(x,0)\in\mathbb B:|x-(a+\sqrt{2}b)|<\varepsilon\}$$ where $(a,b)\in\mathbb B$ and $\varepsilon>0$.

It is easy to see that the Bing space $\mathbb B$ is not oo-regular, so it is not homeomorphic to the rational projective space $\mathbb QP^\infty$.

**Problem 1.** Is the Bing space homeomorphic to the Golomb space?

**Remark.** It is clear that the Bing space has many homeomorphisms, distinct from the identity.

So, the answer to Problem 1 would be negative if the answer to the following problem is affirmative.

**Problem 2.** Is the Golomb space $\mathbb G$ topologically rigid?

**Problem 3.** Is the Bing space topologically homogeneous?

Since the last two problems are quite interesting I will ask them as separate questions on MathOverFlow.

**Added in an edit.** Problem 1 has negative solution. The Golomb space and the Bing space are not homeomorphic since

1) For any non-empty open sets $U_1,\dots,U_n$ in the Golomb space (or in the rational projective space) the intersection $\bigcap_{i=1}^n\bar U_i$ is not empty.

2) The Bing space contain three non-empty open sets $U_1,U_2,U_3$ such that $\bigcap_{i=1}^3\bar U_i$ is empty.

**Added in a next edit.** Problem 2 has the affirmative answer: the Golomb space $\mathbb G$ is topologically rigid. This implies that $\mathbb G$ is not homeomorphic to the Bing space or the rational projective space (which are topologically homogeneous).

Problem 3 has an affirmative solution: the Bing space is topologically homogeneous.

**Added in Edit made 14.03.2020.** The rational projective space $\mathbb Q P^\infty$ admits a nice topological characterization:

**Theorem.** A topological space $X$ is homeomorphic to $\mathbb Q P^\infty$
if and only if $X$ is countable, first countable, and admits a decreasing sequence of nonempty closed sets $(X_n)_{n\in\omega}$ such that $X_0=X$, $\bigcap_{n\in\omega}X_n=\emptyset$, and for every $n\in\omega$,
(i) the complement $X\setminus X_n$ is a regular topological space, and
(ii) for every nonempty open set $U\subseteq X_n$ the closure $\overline{U}$ contains some set $X_m$.