bio | website | |
---|---|---|

location | ||

age | ||

visits | member for | 4 years, 8 months |

seen | Jan 31 '14 at 6:12 | |

stats | profile views | 4,695 |

Feb 13 |
awarded | Enlightened |

Jan 17 |
awarded | Enlightened |

Jan 17 |
awarded | Nice Answer |

Nov 25 |
awarded | Yearling |

Aug 6 |
awarded | Notable Question |

Nov 25 |
awarded | Yearling |

Sep 9 |
awarded | Good Question |

Jun 25 |
awarded | Excavator |

Jun 25 |
awarded | Pundit |

Mar 29 |
awarded | Good Answer |

Jan 9 |
comment |
reference for “X compact <=> C_b(X) separable” (X metric space)
I don't have a reference, but I'd suggest this argument: if $X$ is not compact, there is an infinite closed discrete subset $D$ of $X$. For every $A \subset D$ choose a continuous function $f_A \colon X \to [0,1]$ such that $f_A|_A = 1$ and $f_A|_{D \setminus A} = 0$ (by Urysohn). This gives an uncountable family of continuous bounded functions such that $\|f_A - f_B\| = 1$ whenever $A \neq B$. Alternatively, embed $\ell^\infty$ using a similar trick. If $X$ is compact then $C(X)$ is separable by Stone-Weierstrass. |

Jan 9 |
comment |
Why are abelian groups amenable?
In the definition of $K$ you should replace $\lVert \Phi\rVert \leq 1$ by $\Phi(\chi_G) = 1$ (which together with $\Phi(F) \geq 0$ implies $\lVert \Phi \rVert = 1$). The way you define it, $0 \in K$ is always a fixed-point, whether $G$ is amenable or not. |

Jan 9 |
revised |
Why are abelian groups amenable?
fixed LaTeX |

Jan 9 |
comment |
Why are abelian groups amenable?
(and the $F$ and $F_n$ must have positive measure, of course). |

Jan 9 |
comment |
Why are abelian groups amenable?
I'm not sure what exactly the deleted comments you responded to said, but it is not true that Følner limits you to discrete groups. A locally compact group $G$ with Haar measure $\mu$ is amenable if and only if for every compact $K$ and every $\varepsilon \gt 0$ there is a compact set $F$ such that $\mu(F \mathbin{\Delta} kF) \leq \varepsilon \mu(F)$ for all $k \in K$. If $G$ happens to be $\sigma$-compact then you can arrange $F$ to be from an exhaustive sequence $F_n$. |

Jan 4 |
comment |
About the definition of Borel and Radon measures
Thanks! Meanwhile, I remembered an old construction due to Oxtoby dx.doi.org/10.1090/S0002-9947-1946-0018188-5 which produces a non-trivial (invariant and inner regular) Borel measure on every separable completely metrizable group. If the group is not locally compact then every open set has infinite measure, but there are always sets of finite measure, thus Borel regularity fails. This construction can be adapted to give a non-Borel regular Borel measure even on $\mathbb{R}$, by working on the set of irrationals and using that they are homeomorphic to $\Bbb{Z^N}$. |

Jan 1 |
comment |
The Practical Impact of Set-Theoretic Axioms on Measure Theory
A standard application of Martin's axiom en.wikipedia.org/wiki/Martin%27s_axiom is the existence of Banach limits satisfying some measurability conditions (medial means in the sense of Mokobodzki). See my answer to math.stackexchange.com/q/54554 for some details and references and part 4 of that answer for a basic sample application that might illustrate their power. |

Jan 1 |
comment |
About the definition of Borel and Radon measures
What would be a good example of a non-Borel regular Borel measure on a second countable metric space? All the standard procedures for constructing measures I'm aware of seem to yield measures satisfying this condition. The examples I was able to produce either live on large (i.e. non-separable) spaces or fail to measure all Borel sets. |

Jan 1 |
revised |
Math Zeitgeist 2012
spelling of Agol's last name |

Dec 30 |
comment |
Continuum Hypothesis
The point of "naturality" is exactly what I meant by emphasizing "algebraic". Thanks a lot for all the additional pointers and all these interesting links in your answer. |