Timeline for Is every positive-density set of positive integers almost rational?
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
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| Oct 10, 2019 at 7:10 | comment | added | YCor | @MCS no, even with rational density it's not "almost rational". Indeed if it has density $t$ then it will meet every arithmetic progression $D$ with density $$t$ inside $D$. In particular, it will not contain, up to density zero, an arithmetic progression at all. | |
| Oct 9, 2019 at 18:53 | comment | added | MCS | Interesting. That makes sense. But, I wonder: if the density is rational, is the set then necessarily almost rational? | |
| Oct 9, 2019 at 4:45 | comment | added | YCor | Indeed, write (mod $\pi$) instead of (mod 2) in my previous comment and this yields irrational density. | |
| Oct 9, 2019 at 3:31 | comment | added | LeechLattice | Any set with irrational natural density works. | |
| Oct 8, 2019 at 22:51 | comment | added | MCS | @YCor: do you have a proof that your set is not almost-rational? | |
| Oct 8, 2019 at 21:16 | comment | added | YCor | $\{n:\sqrt{2}n$ mod $2\;\in [0,1]\}$ | |
| Oct 8, 2019 at 21:15 | history | edited | YCor | CC BY-SA 4.0 |
removed capitals from title
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| Oct 8, 2019 at 20:25 | comment | added | Gerhard Paseman | OK then. How about Thue Morse? The set of all positive numbers with an even number of bits in their binary expansion? Gerhard "Is Grabbing At Bits Now" Paseman, 2019.10.08. | |
| Oct 8, 2019 at 20:23 | comment | added | Wojowu | @GerhardPaseman I imagine this strategy should work, but the details might be a little cumbersome | |
| Oct 8, 2019 at 20:21 | comment | added | Gerhard Paseman | @Wojowu, of course I realized that after commenting, not before. What if we make the removed set thicker? Gerhard "Now Thinking Of Thicker Sets" Paseman, 2019.10.08. | |
| Oct 8, 2019 at 20:19 | comment | added | Wojowu | @GerhardPaseman Complement of powers of two has density $1$. | |
| Oct 8, 2019 at 20:16 | comment | added | Gerhard Paseman | How about the complement of powers of two? Or diagonalize against an enumeration of arithmetic progressions with the next element removed at least twice as large as the previous removed element. Gerhard "Can Think Of Thinner Sets" Paseman, 2019.10.08. | |
| Oct 8, 2019 at 19:58 | history | asked | MCS | CC BY-SA 4.0 |