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6h
comment Irrational number problem
It's so-called Dirichlet's approximation theorem.
7h
comment A sebset of $\Bbb C^2$
Well, you can try $m(f^{-1}(I)\cap F)$ is small if $m(I)$ is small enough, for any compact set $F$, then $m(f^{-1}(K)\cap F)$ should be zero. I forgot to restrict the domain on a finite measure space.
14h
comment Show that a proper continuous map from $X$ to locally compact $Y$ is closed
In fact, locally compact spaces are compactly generated, and a continuous from a topological space to a compactly generated Hausdorff space is proper if and only if it's closed and preimages of singletons are compact.
15h
comment The sum of reciprocal squares: estimating the remainder
For the asymptotic of the remainder term, try Euler-Maclaurin formula.
16h
accepted Outer measure and Caratheodory's criterion
1d
reviewed Approve suggested edit on Let $\,E = C([0, 1])$. Set $\,X =(E,||.||_\infty)$ and $\,Y =(E,||.||_1)$. …
1d
comment Moving the branch cut of the complex logarithm
It works if we rewrite $f(z)=\log(z-\sqrt[3]2)+\log(z-\sqrt[3]2\omega)+\log(z-\sqrt[3]2\omega^2)$, but I want to obtain good insight on how Riemann surfaces work. For example, what's the exact meaning of $\log(fg)=\log f+\log g$ w.r.t. Riemann surfaces when $f,g$ are multi-valued functions.
1d
comment Moving the branch cut of the complex logarithm
Taking a single-valued branch is always annoying to me. +1
1d
revised Moving the branch cut of the complex logarithm
added 1 character in body
1d
awarded  Announcer
1d
comment A sebset of $\Bbb C^2$
Suppose $f\colon\mathbb C^2\to\mathbb R,(z,w)\mapsto\lvert zw\rvert$. Note that if $I$ is an interval such that $m(I)$ is small, then $f^{-1}(I)$ is measurable and $m(f^{-1}(I))$ is also small.
1d
comment How to prove that $\max\{f,g\}$ is Riemann integrable?
Note that $\max(f,g)=(f+g+\lvert f-g\rvert)/2$.
1d
comment Outer measure and Caratheodory's criterion
@GiuseppeNegro Generally, if $\mathcal A$ is a collection of subsets closed under countable intersection, then there's $A_0\in\mathcal A$ such that $m^*(A_0)=\alpha=\inf_{A\in\mathcal A} m(A)$. We again choose $A_n$ such that $m^*(A_n)\ge\alpha+1/n$, then let $A_0=\bigcap_n A_n\in\mathcal A$ and by def $m^*(A_0)\ge\alpha$, but $m^*(A_0)\le m^*(A_n)\le\alpha+1/n$.
1d
comment Outer measure and Caratheodory's criterion
@GiuseppeNegro $\inf$ in my expression is really $\min$, i.e. if my expression of $m^*(A)$ is right, then it should be a regular measure, since suppose $m(E_n)\le m^*(A)+1/n$ and $E_n\supseteq A$, then consider $E=\bigcap_n E_n$. Since $E\supseteq A$, we have $m(E)\ge m^*(A)$, but $m(E)\le m(E_n)\le m^*(A)+1/n$, thus $m(E)=m^*(A)$.
2d
reviewed Approve suggested edit on How do I count ordered tuples in this inequality?
2d
revised Book suggestions on projective geometry
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2d
comment Proof about Number Fields
Note (elementarily) that if $0\neq f\in\mathbb Z[X]$ and $m/n\in\mathbb Q$ is irreducible, then $m$ divides $f(0)$ and $n$ divides the leading coefficient of $f$.
2d
revised Book suggestions on projective geometry
added 2 characters in body
2d
revised Book suggestions on projective geometry
added 4 characters in body
2d
asked Book suggestions on projective geometry