Timeline for When is non amenablity witnessed by a single non measurable set?
Current License: CC BY-SA 3.0
23 events
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| S Aug 7, 2013 at 20:56 | history | suggested | Michael Albanese | CC BY-SA 3.0 |
Replaced \* by *, and fixed "A\sub</i> seteq G"
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| Aug 7, 2013 at 20:41 | review | Suggested edits | |||
| S Aug 7, 2013 at 20:56 | |||||
| May 25, 2011 at 12:11 | history | edited | Justin Moore | CC BY-SA 3.0 |
Corrected error in paragraph 3.
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| Apr 5, 2011 at 15:36 | vote | accept | Justin Moore | ||
| Apr 5, 2011 at 14:10 | history | edited | Justin Moore | CC BY-SA 2.5 |
added details to the paragraph "To connect"
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| Apr 5, 2011 at 2:25 | comment | added | Justin Moore | @Kate: If $E$ is in level $n+1$, $E \cap B_n$ is in $T$. Otherwise there would be a measure in $P(B_n)$ which witnesses this and this measure would also witness $E$ isn't in $B_{n+1}$. | |
| Apr 4, 2011 at 20:09 | comment | added | Kate Juschenko | I assume that you mean that each element of $T$ has a predecessor. That is, if $E$ belongs to $T$ at the level $n+1$ then $E\cap B_n$ also belongs to $T$, right? Could you please explain, why it is so? I am probably missing something. | |
| Apr 4, 2011 at 20:01 | comment | added | Justin Moore | @Kate: Just to be clear, by "tree" I mean a poset in which the predecessors of each element is well ordered by the poset's order. In this case the set of predecessors is finite. Konig's Lemma states that any finitely branching tree which is infinite has an infinite path. "dead ends" are allowed. Start at the bottom and at each stage look for a successor with infinitely many things above it. This builds an infinite chain. | |
| Apr 4, 2011 at 17:29 | comment | added | Kate Juschenko | @Justin, could you please give more details on why exactly T is a tree, ie. why it is connected (it could be just a set of points without an infinite ends, why it is not the case?) | |
| Apr 3, 2011 at 13:56 | history | edited | Justin Moore | CC BY-SA 2.5 |
Fixed o in Folner
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| Apr 3, 2011 at 13:35 | comment | added | Andreas Thom | Justin, thanks a lot. Now I understand. This is a very nice argument! I guess, I have to read it once more, but for the moment it looks convincing. | |
| Apr 3, 2011 at 13:18 | comment | added | Justin Moore | @Andreas,Kate: I have now edited the proof accordingly. The changes are twofold: argue that the version of "B is epsilon-Ramsey for A" where f is just a characteristic function is as strong as the statement for arbitrary maps into the interval. Then prove that if G is not Ramsey as witnessed by A, epsilon, then there is a E for which we can't invariantly measure all translates by elements of A. This is done by a standard compactness argument. Please let me know if you have more questions. If the proofs is now to your satisfaction, please either post a comment or email me. | |
| Apr 3, 2011 at 13:13 | history | edited | Justin Moore | CC BY-SA 2.5 |
Revised substantially to address coments of Andreas and Kate. See my comment.
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| Apr 3, 2011 at 10:06 | comment | added | Kate Juschenko | @Justin: I am also puzzled with the proof. As I understand the question that you are asking is $G$ is amenable iff for every $E\subseteq G$ there exist a proba $\mu$ such that $\mu(gE)=\mu(E)$ for every $g\in G$. Of course one way is obvious. Could you please write more precisely how do you get an invariant mean on $G$. it would more understandable for me if it is written more directly, just a play with sets/measures. I would be very interested to see the proof, since it looks like a strong weakening of the amenability. | |
| Apr 3, 2011 at 2:05 | comment | added | Justin Moore | @Andreas: You are right. I'll try to fix this tomorrow. | |
| Apr 3, 2011 at 1:08 | comment | added | Andreas Thom | Sorry, I still do not understand. Where do you show that the non-existence of such $E$ implies that $G$ is Ramsey? Don't you show (or claim) the converse in the section that starts with "Observe that if $E$...". There you assume some such $E$ exists and conclude that exists no set which is $\varepsilon$-Ramsey with respect to some finite set $A$. Now, how do you conclude that if no such $E$ exists, $G$ is Ramsey? | |
| Apr 2, 2011 at 18:19 | comment | added | Justin Moore | @Andreas: if $G$ is non amenable, then for every $\mu$ there is an $E$ such that two translates of $E$ are measured differently by $\mu$. A priori $E$ depends on $\mu$. I have proved that if $G$ is not amenable then there is an $E$ such that for every $\mu$, $E$ is the set which is not measured invariantly. (Actually, I have assumed that no such $E$ exists, observed that this implies $G$ is Ramsey, and then proved that $G$ is amenable.) | |
| Apr 2, 2011 at 16:23 | comment | added | Andreas Thom | Justin, from your comment I understand that existence of such an $E$ implies that $G$ is not Ramsey. Your argument above (in the the answer) then implies that $G$ is not amenable. But wasn't this clear already? (Being just the contra-positive of saying: If $G$ is amenable, then no such set can exist.) I thought the question is about the converse. Given $G$ non-amenable, does there exists and how can one construct the set $E$? I think, I must miss something very basic here. | |
| Apr 2, 2011 at 14:55 | history | edited | Justin Moore | CC BY-SA 2.5 |
Edited to clarify the connection to the original question.
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| Apr 2, 2011 at 14:50 | comment | added | Justin Moore | @Andreas: If $E \subseteq G$ and there is no $\mu$ which measures all translates of $E$ the same, then there is an $\epsilon$ and a finite $A \subseteq G$ such that there is no $\mu$ which measures all translates of $E$ by elements of $A$ to be within $\epsilon$ of the same number. This $E$ witnesses that there are no $\epsilon$-Ramsey sets with respect to $A$. | |
| Apr 2, 2011 at 13:34 | comment | added | Andreas Thom | Justin, how does this address you original question? How do you construct the set $E$, you were asking for? | |
| Apr 1, 2011 at 20:50 | comment | added | Yemon Choi | I had suspected (though not for any well-defined reason) that your condition might be equivalent to non-amenability. I wonder how your Ramsey-condition relates to other known characterizations of non-amenability. | |
| Apr 1, 2011 at 20:41 | history | answered | Justin Moore | CC BY-SA 2.5 |