Timeline for Regarding scaling in sumsets
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| May 27 at 20:08 | comment | added | D.W. | It's not obvious to me why you write $(K^2)^{3+2}$ instead of $K^{3+2}$. I think it would help to show the reasoning in a bit more detail in the answer. | |
| May 27 at 20:06 | comment | added | D.W. | Might you be open to editing your answer to incorporate that information into the proof? On Stack Exchange, instead of putting clarifications and additional information in the comments, we'd prefer that you revise the answer so it reads well for someone who encounters it for the first time, and so that people don't need to read the comments to follow. | |
| May 27 at 19:20 | comment | added | RFZ | @D.W., Let $K\geq 1$. An additive set $H$ is said to be a $K$-approximate group if it is symmetric (so $H=-H$), contains the origin, and $H+H$ can be covered by at most $K$ translates of $H$. A few remarks: 1) When we say that $A$ is an additive set in an abelian group $G$, it means that $A$ is a finite non-empty subset of $G$. 2) $H+H$ can be covered by at most $K$ translates of $H$ means that $H+H \subseteq H+B$, where $|B| \leq K$. | |
| May 27 at 19:13 | comment | added | RFZ | @D.W., The answer to your first question: Generally, if the set $A$ has a small doubling or difference constant, then the size of the set $|nA-mA|$ is relatively small compared to the size of $|A|$. More precisely, if $|A+A| \leq C|A|$ or $|A-A| \leq C|A|$, then for any non-negative integers $n$ and $m$, we have: $|nA-mA| \leq C^{n+m}|A|$. This inequality is known as the Plünnecke-Ruzsa inequality. So in our case $n=3$ and $m=2$. | |
| May 27 at 17:35 | comment | added | D.W. | Can you explain what is the definition of "$K_1$ approximate group"? How does the cardinality of $X$ imply that $H$ is a $K_1$ approximate group? Why does being a $K_1$ approximate group imply an upper bound on $|nH|$? | |
| May 27 at 17:33 | comment | added | D.W. | Can you explain why we can conclude $|3A-2A| \le (K^2)^{3+2} |A|$? Why does that step follow? | |
| May 27 at 17:30 | history | edited | D.W. | CC BY-SA 4.0 |
Improve English
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| May 27 at 17:09 | history | edited | RFZ | CC BY-SA 4.0 |
added 931 characters in body
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| May 26 at 0:51 | history | answered | RFZ | CC BY-SA 4.0 |