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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.

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    $\begingroup$ Do you know if it has a single non-trivial self-homeomorphism? $\endgroup$
    – YCor
    Commented Nov 8, 2017 at 17:33
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    $\begingroup$ @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 :( $\endgroup$
    – Taras Banakh
    Commented Nov 8, 2017 at 19:53
  • $\begingroup$ It's a lovely question! Do you think your former argument could possibly have carried from $1$ to $2$, and so on..? $\endgroup$
    – Dominic van der Zypen
    Commented Nov 9, 2017 at 13:01
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    $\begingroup$ @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. $\endgroup$
    – Taras Banakh
    Commented Nov 9, 2017 at 14:00

1 Answer 1

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[Edit, Dec 6, 2019] I have a pleasure to inform that this problem was finally resolved in affirmative by T.Banakh, D.Spirito and S.Turek who proved the following

Theorem. The Golomb space is topologically rigid.

Edit: now published: The Golomb space is topologically rigid. Comment. Math. Univ. Carolin. 62 (2021), no. 3, 347–360.

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  • $\begingroup$ 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. $\endgroup$
    – YCor
    Commented Nov 17, 2017 at 23:55
  • $\begingroup$ @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. $\endgroup$
    – Taras Banakh
    Commented Nov 18, 2017 at 5:48
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    $\begingroup$ @YCor I wrote down the proof (consisting of 10 lemmas). $\endgroup$
    – Taras Banakh
    Commented Nov 18, 2017 at 7:39
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    $\begingroup$ Could you edit to clarify "the announced proof of the rigidity of the Golomb space contained a gap"? which announced proof? you wrote this a while ago but now the main question seems to refer to another announced proof. $\endgroup$
    – YCor
    Commented Jul 1, 2019 at 10:13
  • $\begingroup$ @YCor Now I wrote "my proof" instead of "announced proof" $\endgroup$
    – Taras Banakh
    Commented Jul 2, 2019 at 11:17

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