Timeline for Amenability and ultrafilters
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
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| Apr 27, 2012 at 14:51 | comment | added | Clinton Conley | Misha, the use of generic ergodicity was really me being lazy, and isn't essential to the argument. The main point is that following the outlined argument, you can build in any nonmeager BP subset of $2^\mathbb{Z}$ large finite collections of pairwise almost disjoint sets (any pair has finite intersection). If you had a nontrivial Baire measurable measure $f$ on $2^\mathbb{Z}$, there'd be some $r>0$ with $f^{-1}([r,1])$ nonmeager (say $r = 1/2$), and by building enough almost disjoint sets you'd get a contradiction (I suppose all you need is for singletons to be null). | |
| Apr 26, 2012 at 17:30 | vote | accept | Misha | ||
| Apr 26, 2012 at 17:30 | comment | added | Misha | @Clinton: Thank you, I understand most of it though I still have to learn more about generic ergodicity (beyond the definition). | |
| Apr 21, 2012 at 4:24 | comment | added | Clinton Conley | Inductively build finite binary strings $u_n$, $v_n$, $w_n$ in $2^{[-k_n,k_n]}$ for some large $k_n$ such that (a) $u_n$, $v_n$, $w_n$ have disjoint supports, (b) the basic open neighborhood determined by $u_n$ (and $v_n$ and $w_n$) is contained in $U_n$, and (c) $u_{n+1}$ extends $u_n$ (and $v$ and $w$). In the end, you've built strings $u$, $v$, $w$ in $\bigcap_n U_n$ of disjoint support, and thus three disjoint subsets of $\mathbb{Z}$ each with measure $1/2$, a contradiction. | |
| Apr 21, 2012 at 4:19 | comment | added | Clinton Conley | Misha, Here's a sketch of the argument I have in mind. Let $f:2^\mathbb{Z} \to \mathbb{R}$ be our putative Baire measurable measure. By generic ergodicity of the shift of $\mathbb{Z}$ on $2^\mathbb{Z}$, $f$ is constant on a comeager set. Moreover, since complementation (in $\mathbb{Z}$, viewed as an automorphism of $2^\mathbb{Z}$) is a homeomorphism, WLOG we may assume this constant equals $1/2$. Now fix dense open sets $U_n \subseteq 2^\mathbb{Z}$ whose intersection is contained in the $f$-preimage of $1/2$. [cont.] | |
| Apr 20, 2012 at 19:28 | comment | added | Misha | @Clinton: Thank you for the answer. I am probably missing something simple, but how do you show non-existence of a finitely-additive Baire measurable probability measures on $2^{\mathbb Z}$? | |
| Apr 19, 2012 at 19:32 | comment | added | Clinton Conley | In the previous comment I really meant ZFDC + BP to avoid silly pathologies. | |
| Apr 19, 2012 at 17:49 | comment | added | Clinton Conley | Simon's answer is more complete, but an easy way to see that A1 does not imply A2 in ZF is to work in a model of ZF + all sets of reals have the property of Baire. Certainly the integers will have a Folner sequence in this model, but a straightforward Baire category argument shows that no shift-invariant finitely additive probability measure on the integers is Baire measurable (as a function from $2^\mathbb{Z}$ to $\mathbb{R}$). | |
| Apr 19, 2012 at 17:09 | answer | added | Simon Thomas | timeline score: 14 | |
| Apr 19, 2012 at 16:59 | history | asked | Misha | CC BY-SA 3.0 |