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We define an embedding of the set of prim numbers into the Cantor set as follows:

First we recall that the cantor set $\mathcal{C}$ is homeomorphic to $(\mathbb{Z}/10\mathbb{Z})^\omega $ since the latter is a compact metrizable space without any isolated point. So according to topological characterization of the Cantor set the classical Cantor set is homeomorphic to $(\mathbb{Z}/10\mathbb{Z})^\omega $.

The space of prime numbers is denoted by $\mathcal{Prime}$.

We define the embedding $\mathcal{E}:\mathcal{P}\to (\mathbb{Z}/10\mathbb{Z})^\omega $ as follows:

$$\mathcal{E}(p)=(a_1,a_2,\ldots,a_n,\ldots)$$

where the decimal expansion of $\sqrt{p}=b_nb_{n-1}\ldots b_{1}b_0/a_1a_2\ldots a_n\ldots$

So in this way we may consider the space of prime numbers $\mathcal{Prime}$ as a subspace of the Cantor set $\mathcal{C}$.

Is $\mathcal{Prime}$ a compact set? Is it an open subset of $\mathcal{C}$?

What would be a number theoretical interpretations for these topological questions?

Ali Taghavi
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    It may not be that easy to get much out of this doing this in base 10. I have a tiny suspicion that if there is anything more interesting here that base 3 would be the one to use. – JoshuaZ Nov 04 '21 at 21:12
  • Number-theretical conclusions will not follow from base 10 expansions. – Gerald Edgar Nov 04 '21 at 21:34

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Your set is a countable dense subset of $(\mathbb{Z}/10\mathbb{Z})^\omega$; cf. Lucia's response here. Hence it is neither open, nor compact.

GH from MO
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    Countable sets can sometimes be compact. But I suspect this set is dense, and therefore not compact. – Gerald Edgar Nov 04 '21 at 21:33
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    @GeraldEdgar This one is not closed though (since it's a proper dense subset) – Wojowu Nov 04 '21 at 21:34
  • @GeraldEdgar I think he means it can not be an open set since it is countable. That is true so when I was giving the question i did not pay attention to this fact – Ali Taghavi Nov 04 '21 at 22:13
  • Thank you and +1 for your answer – Ali Taghavi Nov 04 '21 at 22:14
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    @GeraldEdgar: I said in my post that it was dense, and I gave as a reference Lucia's answer to another MO post. So it cannot be closed (equivalently it cannot be compact). – GH from MO Nov 04 '21 at 22:33
  • What can be said about number theoretical interpretations for such topological consideration? – Ali Taghavi Nov 05 '21 at 08:17
  • Any way itseems that there is a delayed telepathy between my post and the post of Marty you linked. – Ali Taghavi Nov 05 '21 at 09:15
  • @AliTaghavi: If you like my answer, please accept it officially (so that it turns green). Thanks in advance! – GH from MO Nov 05 '21 at 15:16
  • I realy like your answer. thaks again for your answer. But what about my 2 previous comment(and the final part of my question). For example can one say any things about closedness of all twin pairs in the product space $\mathcal{C}\times \mathcal{C}$? – Ali Taghavi Nov 05 '21 at 19:18
  • @AliTaghavi: A question like "what would be a number-theoretical interpretation etc." is really open ended and subjective. Hence it does not fit this website. About twin pairs in $\mathcal{C}\times\mathcal{C}$: I bet that they form a dense subset, hence it is not closed. But for the time being, we don't even know if it is an infinite set. – GH from MO Nov 06 '21 at 00:31
  • The standard function from $[0,1]to \mathcal{C}$ is not a continuous map. But the density of a set implies thats its image is a dense set. So this is motivation to consider the following: Is there a non continuous function from R to R which is surjective and send every dense set to a dense set. As another question, as a consequence of density you pointed out to: Let A and B countable dense subsets of the Cantor set: Does there exist a homeomorphism of the Cantor set which carriy A to B? – Ali Taghavi Nov 07 '21 at 21:22
  • @AliTaghavi: Thanks for accepting my answer. I suggest that you ask your new questions in new posts (not in comments to an old post). – GH from MO Nov 07 '21 at 21:28
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    I thank you for your very helpful answer. – Ali Taghavi Nov 10 '21 at 02:34