Timeline for On the "infinitely often in" relation between subsets of $\mathbb{N}$
Current License: CC BY-SA 4.0
16 events
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Jan 8, 2019 at 17:15 | comment | added | Noah Schweber | @PierrePC Indeed, and that's actually how I (mis)read the question. :P | |
Jan 8, 2019 at 17:15 | history | edited | Noah Schweber | CC BY-SA 4.0 |
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Jan 8, 2019 at 16:34 | comment | added | Pierre PC | Not a fascinating remark, but the proof also holds with a limsup in the definition of $\mu$. | |
Jan 8, 2019 at 16:11 | comment | added | Noah Schweber | @Wojowu You're quite right, I remembered it incorrectly. | |
Jan 8, 2019 at 16:10 | history | edited | Noah Schweber | CC BY-SA 4.0 |
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Jan 8, 2019 at 16:09 | comment | added | Wojowu | I don't think Szemeredi gives us the result - the infinitely many arithmetic progressions might all have different differences and not be translates of one another. | |
Jan 8, 2019 at 15:54 | history | edited | Noah Schweber | CC BY-SA 4.0 |
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Jan 8, 2019 at 15:54 | vote | accept | Dominic van der Zypen | ||
Jan 8, 2019 at 15:54 | comment | added | Dominic van der Zypen | Beautiful @NoahSchweber - and lightning fast! | |
Jan 8, 2019 at 15:53 | comment | added | Noah Schweber | @Wojowu Yeah, I caught that at the same time. | |
Jan 8, 2019 at 15:52 | history | edited | Noah Schweber | CC BY-SA 4.0 |
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Jan 8, 2019 at 15:50 | comment | added | Wojowu | In the third paragraph you first write $\mu(B)=\epsilon$, and immediately after $\mu(B)>\epsilon$. I think you want the first equality to be an inequality as well. | |
Jan 8, 2019 at 15:48 | history | edited | Noah Schweber | CC BY-SA 4.0 |
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Jan 8, 2019 at 15:44 | history | undeleted | Noah Schweber | ||
Jan 8, 2019 at 15:43 | history | deleted | Noah Schweber | via Vote | |
Jan 8, 2019 at 15:42 | history | answered | Noah Schweber | CC BY-SA 4.0 |