# 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 known 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 topological 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