Timeline for A natural $\mathbb Q\times \mathbb P$ subset of $\mathbb R$?
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
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| Jan 25, 2021 at 19:21 | comment | added | HenrikRüping | Sure multiplication with an irrational number should do the trick. | |
| Jan 22, 2021 at 11:41 | comment | added | Ivan Meir | Thank you Henrik you are definitely correct but I believe we can just replace $\mathbb{Q}$ by say the quadratic irrationals or even simpler $\sqrt{2} \mathbb{Q}$ and this should work. Let me know what you think - I have updated my answer to reflect this. | |
| Jan 22, 2021 at 11:38 | history | edited | Ivan Meir | CC BY-SA 4.0 |
Added additional step to map to an irrational set homeomorphic to the rationals first to avoid Henrik's objection.
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| Jan 22, 2021 at 8:32 | comment | added | HenrikRüping | Isn't there still a problem with a sequence of the form $a_n=(1+(-1/2)^n,\pi)$. What should H(1) be? The limit of $H(a_{2n})$ forces us to take the finite representation and the limit of $H(a_{2n+1})$ forces us to take the infinite representation. I do not see how to fix it. | |
| Jan 21, 2021 at 14:45 | history | edited | Ivan Meir | CC BY-SA 4.0 |
Clarified that one must choose the infinite binary representation for a rational as observed by Henrik in the commens
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| Jan 21, 2021 at 14:37 | comment | added | Ivan Meir | @HenrikRüping (I think you mean $M(q,0)=T_e$) The problem with setting $p=0$ is that $p\notin \mathbb{P}$, the set of irrational numbers. However I certainly do appreciate your point which is relevant for fractions which have a finite binary representation and when you use this finite form in the construction. I believe you simply need to ensure you always use the infinite binary representation when you construct the mapping and the dense subset. I have updated my answer with this clarification. Thank you for your interesting observation. | |
| Jan 21, 2021 at 9:35 | comment | added | HenrikRüping | I still don't get it. If $M(p,q)$ is continuous, then $M(0,q)=T_e(q)$ would also be continuous. | |
| Jan 19, 2021 at 0:32 | comment | added | Ivan Meir | @HenrikRüping We require that $M(p,q)=T_e(q)+T_o(p)$ is continuous not $T_e$ or $T_o$ individually. | |
| Jan 18, 2021 at 13:14 | comment | added | HenrikRüping | why is e.g. $T_e$ continuous? For example $a_n=1-2^n=0,1...1$ converges to one, but $T_e(a_n)$ converges to $0.0101010101...\neq T_e(1)=1$? | |
| Jun 15, 2020 at 10:26 | history | edited | Ivan Meir | CC BY-SA 4.0 |
Modified section describing generalised dense subset.
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| Jun 15, 2020 at 9:48 | history | answered | Ivan Meir | CC BY-SA 4.0 |