Consider the set $S=\{2,3,\ldots\}$ equipped with the operation $n\cdot m=n^m$.
Question: Do there exist a mean on $S$ which is left and right invariant with respect to $\cdot$?
Thanks in advance,
Valerio
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asked Jun 10, 2011 at 21:02
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1$\begingroup$ What do you require from a mean? I guess a mean with good properties is almost impossible. If your mean is really left-invariant, it should satisfy $M\left(S\right)=M\left(p^S\right)$ for each prime $p$ (where $M$ denotes your mean). But the sets $p^S$ are disjoint for varying $p$, and thus $M\left(S\right)$ must be zero lest it be greater or equal to an infinite sum of itself. $\endgroup$– darij grinbergCommented Jun 10, 2011 at 21:10
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$\begingroup$ darij: when considering means in the sense of amenability, one usually only requires finite additivity not countable additivity. (Your definition/argument would tells us there are no means on the usual additive group of the integers) $\endgroup$– Yemon ChoiCommented Jun 10, 2011 at 21:12
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2$\begingroup$ Well, $M\left(S\right)$ would be still $=M\left(2^S\right)+M\left(3^S\right)+M\left(S\setminus 2^S\setminus 3^S\right)\geq M\left(2^S\right)+M\left(3^S\right) = 2M\left(S\right)$, leading to $M\left(S\right)=0$. Unless $M\left(S\right)$ is allowed to be negative... $\endgroup$– darij grinbergCommented Jun 10, 2011 at 21:14
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$\begingroup$ yes, it was very easy indeed! sorry for the stupid question :( $\endgroup$– Valerio CapraroCommented Jun 10, 2011 at 21:20
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1 Answer
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Posting it as an answer, since OP confirmed I did understand the question right.
Such a mean cannot exist. In fact, if it would, we could denote it by $M$ and compute:
$M\left(S\right)=M\left(2^S\right)+M\left(3^S\right)+M\left(S\setminus 2^S\setminus 3^S\right)$ (since the sets $2^S$, $3^S$ and $S\setminus 2^S\setminus 3^S$ are pairwise disjoint and cover $S$)
$\geq M\left(2^S\right)+M\left(3^S\right)$ (since the mean is positive)
$=M\left(S\right)+M\left(S\right)$ (since the mean is left-invariant, so that $M\left(p^S\right)=M\left(S\right)$ for every $p\in\mathbb S$)
and thus $0\geq M\left(S\right)$, contradiction.
We only used the left-invariance of $M$.
answered Jun 10, 2011 at 21:48
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1$\begingroup$ Actually, the original poster seems to have confirmed that he understood your answer, which is not quite the same as understanding the question. I like it as an answer though, given my limited understanding of both question and answer. (I hope (possibly trivial) means make sense for arbitrary binary operations.) Gerhard "Understands Understanding Is Oft Misunderstood" Paseman, 2011.06.10 $\endgroup$ Commented Jun 10, 2011 at 22:00