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Nov 10, 2021 at 2:34 comment added Ali Taghavi I thank you for your very helpful answer.
Nov 7, 2021 at 21:28 comment added GH from MO @AliTaghavi: Thanks for accepting my answer. I suggest that you ask your new questions in new posts (not in comments to an old post).
Nov 7, 2021 at 21:22 comment added Ali Taghavi 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?
Nov 7, 2021 at 21:17 vote accept Ali Taghavi
Nov 6, 2021 at 0:31 comment added GH from MO @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.
Nov 5, 2021 at 19:18 comment added Ali Taghavi 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 $\mathcal{C}\times \mathcal{C}$?
Nov 5, 2021 at 15:16 comment added GH from MO @AliTaghavi: If you like my answer, please accept it officially (so that it turns green). Thanks in advance!
Nov 5, 2021 at 9:15 comment added Ali Taghavi Any way itseems that there is a delayed telepathy between my post and the post of Marty you linked.
Nov 5, 2021 at 8:17 comment added Ali Taghavi What can be said about number theoretical interpretations for such topological consideration?
Nov 4, 2021 at 22:33 comment added GH from MO @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).
Nov 4, 2021 at 22:14 comment added Ali Taghavi Thank you and +1 for your answer
Nov 4, 2021 at 22:13 comment added Ali Taghavi @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
Nov 4, 2021 at 21:34 comment added Wojowu @GeraldEdgar This one is not closed though (since it's a proper dense subset)
Nov 4, 2021 at 21:33 comment added Gerald Edgar Countable sets can sometimes be compact. But I suspect this set is dense, and therefore not compact.
Nov 4, 2021 at 21:23 history answered GH from MO CC BY-SA 4.0