User felipeg - MathOverflowmost recent 30 from http://mathoverflow.net2013-05-19T16:04:58Zhttp://mathoverflow.net/feeds/user/18384http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/128259/is-function-from-topological-group-to-metric-space-borelIs function from topological group to metric space Borel?FelipeG2013-04-21T16:52:51Z2013-04-21T19:05:58Z
<p>Let $G$ be a pseudometrizable compact abelian topological group, $X$ a compact
metric space and $f:X\rightarrow G$ a continuous bijective function.</p>
<p>Suppose there exists $g\in G$ such that if $d_{G}(g_{1},g_{2})\leq\epsilon$ then there exists
$n$ such that
$d_{X}(f^{-1}g^{n}g_{1},f^{-1}g^{n}g_{2})\leq\epsilon.$</p>
<p>If $G$ is not metrizable then in general $f^{-1}$ is not continuous but can we conclude $f^{-1}$ is Borel? </p>
http://mathoverflow.net/questions/128183/existence-of-limit-measureExistence of limit measureFelipeG2013-04-20T17:46:28Z2013-04-20T20:50:45Z
<p>Let $X$ be a separable metric space, $\mu_{n}$ a sequence of Borel probability measures
and $\mathcal{C}$ be a family of sets that is closed under finite unions and
interections, and that contains all the balls. If $\mu_{n}(A)$ converges for
every $A\in\mathcal{C}$, does there exists a Borel measure $\mu_{\infty}$
such that $\mu_{\infty}(E)=\lim\mu_{n}(E)$ for every $E\in\mathcal{C}?$</p>
<p>From Theorem 4.3 in <a href="http://msp.org/pjm/1964/14-3/pjm-v14-n3-p23-p.pdf" rel="nofollow">this paper</a> we can get this result when $X$ is locally compact. Here Sion makes an outer measure and then shows open sets are measurable. His proof definitively uses local compactness. </p>
<p>Does anybody know if the result is true when $X$ is not necessarily locally compact?</p>
http://mathoverflow.net/questions/123252/extension-of-measures-from-the-ball-sigma-algebra-to-the-borel-sigma-algebraExtension of measures from the ball sigma-algebra to the borel sigma-algebraFelipeG2013-02-28T18:32:52Z2013-02-28T19:52:30Z
<p>Let $X$ be a metric space, $\Sigma_{1}$ the borel sigma algebra and
$\Sigma_{2}$ the sigma algebra generated by balls (open and closed). </p>
<p>If $\mu$ is a probability measure on $\Sigma_{2}$ can it be extended to a
measure on $\Sigma_{1}?$</p>
http://mathoverflow.net/questions/121036/do-ergodic-isometries-have-discrete-spectrumDo ergodic isometries have discrete spectrum?FelipeG2013-02-07T03:19:11Z2013-02-07T03:19:11Z
<p>Let $X$ be a metric space, $\mu$ a Borel probability measure, and
$T:X\rightarrow X$ be an ergodic measure preserving isometry. </p>
<p>Is $(X,\mu,T)$ measure theoretically isomorphic to a minimal isometry on a
compact metric space (equivalently it has discrete spectrum)?</p>
<p>I understand Krieger representation theorem states that ergodic MPT are
measure theoretically isomorphic to minimal systems on a compact metric spaces.
I would like to know if structure like isometry can be conserved. </p>
<p>The general question I am interested in is: do ergodic isometries have discrete spectrum?</p>
http://mathoverflow.net/questions/111949/weak-convergence-and-cesaro-convergence-of-mu-n-e-imply-convergence-of-muWeak convergence, and Cesaro convergence (of mu_n (E) ) imply convergence (of mu_n (E))?FelipeG2012-11-09T23:16:57Z2012-11-12T19:55:25Z
<p>Let $X$ be a compact metric Borel space. Suppose $\mu_{n}(A)\rightarrow\mu(A)$
for all $\mu-$continuity sets $A$ (sets with zero boundary measure), where $\mu_{n}$ is a sequence of probability measures.
(some people call it weak other weak* convergence)</p>
<p>If $E$ is a measurable set such that
$\mu(E)>0$ and the Cesaro average of $\mu_{n}(E)$ converges; can we conclude
that $\mu_{n}(E)$ converges? </p>
<p>Can we conclude this with extra hypothesis?</p>
<p>I am particularly interested in the case when $T:X\rightarrow X$ is a
continuous transformation, $\mu_{n}=T^{n}\mu_{1},$ and $E=\cap T^{-i}A_{i}$
where $A_{i}$ is a sequence of $\mu-$continuity sets. </p>
http://mathoverflow.net/questions/48771/proofs-that-require-fundamentally-new-ways-of-thinking/94167#94167Answer by FelipeG for Proofs that require fundamentally new ways of thinkingFelipeG2012-04-16T01:00:28Z2012-04-16T01:00:28Z<p>The first formal proofs using limits. (the oldest ones I know are in Newton's Principia)</p>
http://mathoverflow.net/questions/87952/a-system-of-equations-for-integersA system of equations for integersFelipeG2012-02-09T02:20:25Z2012-02-10T23:43:52Z
<p>Working with cellular automata I came across a system of equations for unknown integers $R_{k}$ and $C_{k}$ that looks like this.</p>
<p>$\binom{m}{k}=R_{k}+C_{k}+\sum\limits_{j=1}^{k-1}R_{j}C_{k-j}.$</p>
<p>Where 0< k$\leq$ 2m </p>
<p>(for k>m we take $\binom{m}{k}=R_{k}=C_{k}=0$)</p>
<p>Given $R_{1}$, the system has a unique solution. </p>
<p>Has anyone seen something similar?
Do you know if it still possible to solve it, if instead of $\binom{m}{k}$ in the left you put something else? </p>
<p>I just want to know if it's related to something else, and if it's possible to solve more general systems. </p>
http://mathoverflow.net/questions/79531/bijective-function-on-a-dense-setBijective function on a dense setFelipeG2011-10-30T18:28:23Z2011-11-03T14:54:41Z
<p>Suppose X is a complete metric space, and $f:X↦X$ a continuous surjective function. Let D be a dense set. Suppose $f:D↦D$ is injective and $f^{-1}(D)=D$. </p>
<p>Is $f$ injective ?</p>
<p>Is there a family of metric spaces where you can conclude $f$ is injective?</p>
http://mathoverflow.net/questions/128259/is-function-from-topological-group-to-metric-space-borel/128277#128277Comment by FelipeGFelipeG2013-04-22T18:27:40Z2013-04-22T18:27:40ZThanks for your comment. There is something I don't understand.
Are you assuming the closure of the identity is not a single point? Is this is necessarily true?http://mathoverflow.net/questions/128259/is-function-from-topological-group-to-metric-space-borelComment by FelipeGFelipeG2013-04-22T18:13:55Z2013-04-22T18:13:55ZSure. The multiplication of g is a dynamical system, a rotation in a compact abelian group. I want to construct a dynamical isomorphism from a dynamical system in X to G.http://mathoverflow.net/questions/128183/existence-of-limit-measureComment by FelipeGFelipeG2013-04-21T17:18:41Z2013-04-21T17:18:41ZThanks for your comments. Theorem 4.3 talks about two kinds of limits. The other limit is the outer measure constructed in section 3. Theorem 3.3 mentions that the outer measure is Radon (hence every borel set is measurable) if every open set is the countable union of compact sets. http://mathoverflow.net/questions/121036/do-ergodic-isometries-have-discrete-spectrumComment by FelipeGFelipeG2013-02-07T08:41:57Z2013-02-07T08:41:57ZSure, a theorem by Halmos and Von Neumann state that T is transitive (one dense orbit) and isometric, iff it is topologically isomoprhic to a minimal rotation on a compact abelian group iff T is minimal and has discrete topological spectrum. (topological because in this case the operator associated to the dynamical system is acting on the space of continuous functions. )
A good reference for this is An introduction to ergodic theory by Peter Walters. http://mathoverflow.net/questions/119133/a-similar-result-to-lusins-theorem-to-characterize-borel-measuresComment by FelipeGFelipeG2013-01-17T18:10:45Z2013-01-17T18:10:45ZThank you Bill.
Andreas: I am mostly curious about metric spaces but I don't think that assumption is necessary. http://mathoverflow.net/questions/111949/weak-convergence-and-cesaro-convergence-of-mu-n-e-imply-convergence-of-mu/111960#111960Comment by FelipeGFelipeG2012-11-12T20:05:31Z2012-11-12T20:05:31ZThanks, I was not looking for the case when the cesaro means do not converge. I wanted that as a part of the hypothesis. But I think your counter example works, if you take $x_{n}$=0 only sporadically, then you will have cesaro mean convergence but not convergence. http://mathoverflow.net/questions/101593/discrete-rational-spectrum-and-odometersComment by FelipeGFelipeG2012-07-11T22:02:39Z2012-07-11T22:02:39ZI think that works, thanks. http://mathoverflow.net/questions/87952/a-system-of-equations-for-integers/87995#87995Comment by FelipeGFelipeG2012-02-10T18:10:31Z2012-02-10T18:10:31ZYes , that exactly what I meant. by "Given R1, the system has a unique solution." thanks for clearing this out. http://mathoverflow.net/questions/87952/a-system-of-equations-for-integersComment by FelipeGFelipeG2012-02-10T17:42:10Z2012-02-10T17:42:10ZBecause I want things like $R_{m-1}C_{2}$ to be zero.