Thomas E.
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 Mar10 comment A doubt in Probability essentials by Protter @user3503589: There is probably a mistake in the print. What is certain is, that once you leave the absolute values you have an equality, and he is, essentially, arguing that if you drop the absolute values the estimate grows. Which seems to be false. But you can keep the absolute values and the argument goes through. Mar10 answered A doubt in Probability essentials by Protter Mar10 comment A doubt in Probability essentials by Protter What is the definition of $p_{\omega}$? Is $\Omega$ a countable space here? Mar9 comment Measure of $E$ is the limit of the measure of the open set $\mathcal{O}_n$ @dragon: yeah, precisely what I meant. I'm glad you got it. Mar9 revised Measure of $E$ is the limit of the measure of the open set $\mathcal{O}_n$ added 5 characters in body Mar9 comment Measure of $E$ is the limit of the measure of the open set $\mathcal{O}_n$ @dragon: There is no reverse inequality needed to prove. Since $E=\bigcap_{n=1}^{\infty}O_{n}$ and $O_{n}\supseteq O_{n+1}$ for all $n$, then $m(E)=\lim_{n\to\infty}m(O_{n})$ by convergence of measure since $m(O_{1})<\infty$ as $E$ was compact. Are you aware of the result that if $O_{n}\supseteq O_{n+1}$ for all $n$ and $E=\bigcap O_{n}$, then $m(E)=\lim_{n\to\infty}m(O_{n})$ if $m(O_{1})<\infty$? Mar9 revised Measure of $E$ is the limit of the measure of the open set $\mathcal{O}_n$ added 483 characters in body Mar9 answered Measure of $E$ is the limit of the measure of the open set $\mathcal{O}_n$ Jan6 awarded Yearling Dec21 awarded Constituent Dec12 awarded Caucus Oct20 comment The Complex projective space is homeomorphic to the n-sphere What definition of the complex projective space do you use? Oct15 comment Question about the closure of a set @Luis: That sequence does not have a limit. Oct15 answered Question about the closure of a set Oct15 reviewed Approve Question about the closure of a set Oct2 comment The set of rational numbers in the interval $(0,1)$ cannot be expressed as the intersection of a countable collection of open sets @YiorgosS.Smyrlis. Got it. Thanks. Oct2 comment The set of rational numbers in the interval $(0,1)$ cannot be expressed as the intersection of a countable collection of open sets @YiorgosS.Smyrlis: Why do $U_{n}$ have to be dense? The statement only says that rationals in $(0,1)$ can not be expressed as a countable intersection of open sets, not necessarily dense ones. Sep30 awarded Explainer Sep28 answered Prove that $\{x_{n}\} \subseteq A$ a Cauchy sequence $\Rightarrow$ $\{f(x_{n}\}\} \subseteq y$ is a Cauchy sequence. Sep12 comment Why is differential geometry called differential geometry? Usually the Riemannian metric of a smooth manifold captures it's geometric properties such as curvatures and angles. One doesn't need to integrate on the manifold to capture these notions.