# Is the Golomb countable connected space topologically rigid?

The Golomb space $\mathbb G$ is the set of positive integers endowed with the topology generated by the base consisting of the arithmetic progressions $a+b\mathbb N_0$ with relatively prime $a,b$ and $\mathbb N_0=\{0\}\cup\mathbb N$. It is know that the space $\mathbb G$ is connected and Hausdorff.

It is also easy to check that the multiplication map $\cdot:\mathbb G\times \mathbb G\to\mathbb G$, $\cdot:(x,y)\mapsto xy$, is continuous, so $\mathbb G$ is a commutative topolological semigroup.

Problem. Is the Golomb space $\mathbb G$ topologically homogeneous? Or maybe rigid?

We recall that a topological space $X$ is rigid if its homeomorphism group is trivial.

This problem was motivated by this question, which discusses the relation of the Golomb space to another countable connected Hausdorff space, called the rational projective space $\mathbb QP^\infty$. This space is easily seen to be topologically homogeneous.

• Do you know if it has a single non-trivial self-homeomorphism? – YCor Nov 8 '17 at 17:33
• @YCor No I do not know such a homeomorphism. What is very pityis that I have thought on the problem of homogeneity of the Golomb space about 10 years ago and remember that I proved that 1 is a fixed point of any homeomorphism of $\mathbb G$, which implies that $\mathbb G$ is not topologically homogeneous. But now I cannot find any notes with that proof and also do not remember the idea of the proof :( – Taras Banakh Nov 8 '17 at 19:53
• It's a lovely question! Do you think your former argument could possibly have carried from $1$ to $2$, and so on..? – Dominic van der Zypen Nov 9 '17 at 13:01
• @DominicvanderZypen No if I remember truly, 1 is a special number and the argument does not work for other numbers. Maybe prime number can go to other prime numbers, but I am not sure. – Taras Banakh Nov 9 '17 at 14:00

Unfortunately, the announced proof of the rigidity of the Golomb space contained a gap. So, only the first part of the problem can be answered (at the moment).

Theorem. The Golomb space $\mathbb G$ is not topologicall homogeneous as $h(1)=1$ and $h(\Pi)=\Pi$ for any homeomorphism $h$ of $\mathbb G$.

Here $\Pi$ stands for the set of prime numbers. The proof of this theorem will be divided into a sequence of 9 relatively simple and straightforward lemmas (whose detail proofs can be found here), but the last lemma exploits the powerful Dirichlet Theorem on primes in arithmetic progressions.

First let us fix notation. For a point $x\in\mathbb N$ by $\tau_x=\{U\in\tau:x\in U\}$ we denote the family of open neighborhoods of $x$ in the Golomb topology $\tau$ on $\mathbb N$.

For a number $x\in\mathbb N$ let $\Pi_x$ be the set of all prime divisors of $x$. Two numbers $x,y$ are coprime if $\Pi_x\cap\Pi_y=\emptyset$.

For a number $x\in\mathbb N$ and a prime number $p$ let $l_p(x)$ be the largest integer number such that $p^{l_p(x)}$ divides $x$. The function $l_p(x)$ plays the role of logarithm with base $p$.

A number $x$ is square-free if $l_p(x)\le 1$ for any prime number $p$.

Lemma 1. For a basic open set $a+b\mathbb N_0$ in $\mathbb G$ its closure $$\overline{a+b\mathbb N_0}=\bigcap_{p\in\Pi_b}p\mathbb N\cup (a+p^{l_p(b)}\mathbb N).$$

Lemma 1 implies that the family $$\mathcal F_0=\{F\subset\mathbb N:\exists U_1,\dots,U_n\in\tau\setminus\{\emptyset\}\mbox{ such that }\bigcap_{i=1}^n\bar U_i\subset F\}$$is a filter on $\mathbb N$.

The definition of $\mathcal F_0$ implies that this filter is preserved by any homeomorphism $h$ of $\mathbb G$ (which means that the filter $h[\mathcal F_0]:=\{h(F):F\in\mathcal F_0\}$ coincides with $\mathcal F_0$).

Lemma 2. The filter $\mathcal F_0$ is generated by the base consisting of the sets $q\mathbb N$ with $q$ square-free.

Lemma 3. For every square-free number $q$ and any number $y\ne 1$ coprime with $q$ there are open sets $U_1\in\tau_1$ and $U_y\in\tau_y$ such that $\bar U_1\cap\bar U_y\subset q\mathbb N$.

Lemma 4. For every points $y\ne x\ne 1$, open sets $U_x\in\tau_x$, $U_y\in\tau_y$, and a prime number $p\in\Pi_x$, the set $\bar U_x\cap\bar U_y$ is not contained in $p\mathbb N$.

Lemmas 2-4 imply

Lemma 5. The number $1$ is a fixed point of any homeomorphism $h$ of the space $\mathbb G$.

For any point $x\in\mathbb N$ consider the filter $$\mathcal F_x=\{F\subset\mathbb N:\exists U_x\in\tau_x,\;\exists U_1\in\tau_1\mbox{ such that }\bar U_x\cap\bar U_1\subset F\}.$$

Lemma 6. For two numbers $x,y\in\mathbb N\setminus\{1\}$ the following conditions are equivalent:

$\bullet$ $\mathcal F_x\subset\mathcal F_y$;

$\bullet$ $\Pi_y\subset\Pi_x$.

Lemma 6 implies:

Lemma 7. For every homeomorphism $h$ of the Golomb space $\mathbb G$, and any numbers $x,y\in\mathbb N$ with $\Pi_x\subset\Pi_y$ we get $\Pi_{h(x)}\subset\Pi_{h(y)}$.

Lemma 8. For every homeomorphism $h:\mathbb G\to\mathbb G$ there exists a bijective map $\sigma:\Pi\to\Pi$ of the set $\Pi$ of all prime numbers such that $\Pi_{h(x)}=\sigma(\Pi_x)$ for any $x\in \mathbb N$.

Our final lemma completes the proof of the Theorem.

Lemma 9. $h(\Pi)=\Pi$ for any homeomorphism $h$ of $\mathbb G$.

Proof. It suffices to prove that $h(\Pi)\subset\Pi$. Let $\sigma:\Pi\to\Pi$ be the permutation from Lemma 8. Assuming that $h(\Pi)\not\subset\Pi$, we can find a prime number $p$ such that $h(p)\notin\Pi$. By Lemma 8, $h(p)=q^n$ for some $n>1$ and the prime number $q=\sigma(p)$. Taking into account that $\Pi_{h^{-1}(q)}=\{\sigma^{-1}(q)\}=\{p\}$ and $h^{-1}(q)\ne p=h^{-1}(q^n)$, we conclude that $h^{-1}(q)=p^k$ for some $k>1$. Now consider the continuous map $\gamma:\mathbb G\to\mathbb G$, $\gamma:x\mapsto x^n$, and the continuous map $f=h^{-1}\circ\gamma\circ h:\mathbb G\to\mathbb G$. Observe that $f(p^k)=h^{-1}\circ\gamma\circ h(p^k)=h^{-1}\circ\gamma(q)=h^{-1}(q^n)=p$. For the neighborhood $p+(p^k-1)\mathbb N_0$ of $p$, find a neighborhood $p^k+(p^k-1)b\mathbb N_0$ of $p^k$ such that $f(p^k+(p^k-1)b\mathbb N_0)\subset p+(p^k-1)\mathbb N_0$.

By the classical Dirichlet Theorem, the arithmetic progression $p^k+(p^k-1)b\mathbb N_0$ contains a prime number $r$. Lemma 8 implies that $f(r)=r^l$ for some $l\ge 1$. Taking into account that $r\in p^k+(p^k-1)b\mathbb N_0\subset 1+(p^k-1)\mathbb N_0$, we conclude that $f(r)=r^l=(p+(p^k-1)\mathbb N_0)\cap(1+(p^k-1)\mathbb N_0)^l\subset(p+(p^k-1)\mathbb N_0)\cap(1+(p^k-1)\mathbb N_0)=\emptyset$ and this is a desired contradiction.

Remark. An example of a rigid countable connected Hausdorff space was constructed by Joseph Martin.

• The initial answer contained a sketch of proof that 1 is fixed by any self-homeomorphism, and contained much more mathematical information than the current one. – YCor Nov 17 '17 at 23:55
• @YCor Now I have a proof that each point of the Golomb space is fixed by any homeomorphism, which is much stronger than just fixing 1. I have posted this proof to arXiv and am waiting (48 hours) till it will appear in order to add the link. The proof is 5 pages long and consists of 12 relatively short lemmas. In principle I can write down these 12 lemmas in my answer to make a sketch of the proof here. – Taras Banakh Nov 18 '17 at 5:48
• @YCor I wrote down the proof (consisting of 10 lemmas). – Taras Banakh Nov 18 '17 at 7:39