The [Golomb space](https://dml.cz/bitstream/handle/10338.dmlcz/700933/Toposym_01-1961-1_41.pdf) $\mathbb G$ 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. 



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> **Problem.** Is the Golomb space $\mathbb G$ topologically homogeneous? Or maybe rigid? 
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We recall that a topological space $X$ is *rigid* if its homeomorphism group is trivial.


This problem was motivated by [this question](https://mathoverflow.net/questions/46986/countable-connected-hausdorff-space), 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. 



Is the Golomb countable connected space topologically rigid?
[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](https://arxiv.org/abs/1912.01994) who proved the following



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


