Let me start by reminding two constructions of topological spaces with such exotic combination of properties:
1) The elements are non-zero integers; base of topology are (infinite) arithmetic progressions with coprime first term and difference.
2) Take $\mathbb{R}^{\infty}\setminus \{0\}$ with product-topology and factorize by the relation $x\sim y \Leftrightarrow x=ty$ for some $t>0$ (infinite-dimensional sphere). Then consider only points with rational coordinates, all but finitely of them vanishing.
The first question is whether are these two examples homeomorphic or somehow related.
The second is an historical one. I've heard that the first example of such space belongs to P. S. Urysohn. What was his example?
Best Answer
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: