in this step testing whether $f: \mathbb R^n \to \mathbb R$ is measurable, the guy considers whether the set $f^{-1}((-\infty,\alpha])$ is measurable for all $\alpha$. Are sets of that form all you need to check in general?
So @Balarka we consider a Riemann surface $X$ and some finite $G\leq\operatorname{Aut}(X)$ acting on $X$. So we followed the proof that $X/G$ is a Riemann surface, which explicitely constructs the charts, but we're lacking intuition
More specifically if we take as $X$ the riemann sphere and as $G$ the identity and the map sending each point to it's antipodal, the quotient should be a riemann sphere again, right?
(It's like quotienting $\Bbb R\Bbb P^1=S^1$ identifying antipodal points, just with $\Bbb C\Bbb P^1$)
@AlessandroCodenotti Hm, unclear what kind of intuition you're looking for. A (compact) Riemann surface is a topological 2-manifold $X$ with a holomorphic structure on the charts. If you quotient by a finite group on $X$ acting freely, $X/G$ is again a topological 2-manifold with the projection $X \to X/G$ a covering space. When $G$ is a subgroup of the group of biholomorphic automorphisms of $X$, $G$ preserves the complex structure, whereby $X/G$ also gets a complex structure
So, eg, take the surface $\Sigma_{g}$ of genus $g$. Consider an axis passing through it which intersects the surface at $4g$ points. Rotate the whole thing by an angle of $\pi$ counterclockwise about the axis. This is an action of $\Bbb Z/2$ on the surface of genus $g$
The quotient is $S^2$.
If you give everything the right holomorphic structure, it turns out this is actually a Riemann surface quotient.
So there is a holomorphic branched cover $\Sigma_g \to \Bbb P^1$ from any compact Riemann surface to $\Bbb P^1$.
Actually, better description of the last cover. Take any compact Riemann surface $X$ and let $f$ be a meromorphic function on $X$ (there always exists one). Then $f$ extends to a map $\tilde{f} : X \to \Bbb P^1$ by sending the poles to infinity. This is the branched cover.
@AlessandroCodenotti An orientation-reversing one, isn't it?
@konoa note that you have to use some nontrivial theorem for this
the statement about f.g. generated Artinian k-algebras being finite-dimensional directly implies Zariski's lemma as a special case. And Zariski's lemma pretty easily implies the Nullstellensatz
@MatheinBoulomenos Right, the point is, the antipodal map sends the upper hemisphere orientation-reversed to the lower hemisphere. When considering $\Bbb C$, the upper hemisphere is $|z| > 1$, and lower is the disk $|z| < 1$. $f(z) = 1/z$ switches upper to lower, but leaves the "position of the points" invariant. This is reflection about the equatorial plane. $f(z) = 1/\overline{z}$, on the other hand, not only flips upper/lower hemispheres, but also flips the "position of points"
We're basically going to turn into the Trento university group chat then. Still better than a Heidelberg university group chat, where scary algebraists like @Mathein resides
Set and measure theorists over algebraists any day
@MatheinBoulomenos Cheeers, there's a step in the proof of FLT for reg. primes that says that $(\alpha, \beta)(1 - \zeta_p)$ is the greatest common divisor of the ideals $(\alpha + \zeta_p^i \beta)$ for all $i \in \lbrace 0, \dots, p-1\rbrace$
and I'm not sure why!
I know that $(1-\zeta_p) \mid (\alpha + \zeta_p^i \beta)$ for all $i$ and $(1 - \zeta_p)^2 \mid (\alpha + \beta)$ only
woops
and in fact I can see that $(\alpha, \beta) \supset (\alpha + \zeta_p^i \beta)$ for all of the $i$, so is it just because $(1 - \zeta_p)$ and $(1 - \zeta_p)^2$ are coprime that makes $(\alpha, \beta)(1 - \zeta_p)$ the gcd of those ideals?
is this case 1 of FLT? iirc, you just want to say that if $z^p=x^p+y^p$ and $x,y,z,p$ are coprime then the ideals $(x+\zeta_p^iy)$ for $i\in \{0,1, \dots, p-1\}$ are coprime
but you need to put in more assumptions than you stated above. I assume that you have $\prod_{i=0}^{p-1}(\alpha+\zeta_p^i\beta)=\alpha^p+\beta^p=\gamma^p$, where $p \mid \gamma$? Then the statement about the gcd seems more plausible
if $I$ contains $\alpha+\beta$ and $\alpha+\zeta_p\beta$, it also contains $\alpha+\beta-(\alpha+\zeta_p\beta)=(1-\zeta_p)\beta$ and $\alpha+\zeta_p\beta-\zeta_p(\alpha+\beta)=(1-\zeta_p)\alpha$
since we can write $\operatorname{gcd}_{\Bbb Z}(\alpha,\beta)=x\alpha+y\beta$, $I$ also contains $\operatorname{gcd}_{\Bbb Z}(\alpha,\beta)(1-\zeta_p)$
here $x,y\in \Bbb Z$ (I assume that $\alpha, \beta \in \Bbb Z$, also)
it contains $(1-\zeta_p)\alpha$ and $(1-\zeta_p)\beta$, so it contains the ideal generated by that which is $((1-\zeta_p)\alpha,(1-\zeta_p)\beta)=(1-\zeta_p)(\alpha,\beta)$
@LeakyNun, why does this hold: Let $A:\Bbb R^2\to \Bbb R^2$ be a linear transformation with eigenvalues $\dfrac23$ and $\dfrac95$, then there exists a nonzero vector $ v\in \Bbb R^2$ such that $\lVert Av\rVert=\lVert v\rVert$?
@MikeMiller Sorry for the delayed reply ... For $x→0$ , the limit is 1 ... Again , I wrote ∣sinx∣ as ±sinx and found the limit to be ±1 ... It isn't okay , is it ? O can't figure it out ...
Write $(\alpha+\beta\zeta_p^j)=(1-\zeta_p)^k w$ with $w$ coprime to $1-\zeta_p$ (this is an equation of elements), so as ideals we have $(\alpha+\beta\zeta_p^j)=(1-\zeta_p)^k\cdot (w)$. Also write $(\alpha+\beta\zeta_p^j)=(\alpha,\beta)J$ with an ideal $J$. By assumption $(\alpha,\beta)$ is coprime to $(1-\zeta_p)$, so $(\alpha,\beta)$ divides $(w)$, thus $w$ is actually contained in $(\alpha,\beta)$. This shows that $(\alpha+\zeta_p^j\beta) \in (1-\zeta_p)(\alpha,\beta)$
@Silent So A is diagonalizable, i.e. two eigenvectors form a basis. Let $Av_1 = \dfrac23 v_1$ and $Av_2 = \dfrac95 v_2$. Then, any vector $v$ can be expressed as $av_1 + bv_2$. Then, $\|Av\| = \|v\| \iff \left(\dfrac49-1\right)a^2\|v_1\|^2 + \left(\dfrac{81}{25}-1\right)b^2\|v_2\|^2 + \dfrac{36}{25}ab(v_1 \cdot v_2) = 0$
one could ask what the discriminant of that thing is
@LeakyNun @Silent consider the following map $\Bbb R^2 \setminus \{0\} \to \Bbb R$, $v \mapsto \frac{\|Av\|}{\|v\|}$ this is continuous. If we plug in an eigenvector for $\frac{2}{3}$, we get $\frac{2}{3}$, if we plug in an eigenvector for $\frac{9}{5}$, we get $\frac{9}{5}$. Since $\Bbb R^2 \setminus \{0\}$ is connected, the image has to be connected, so it also needs to contain $1$, as $\frac{2}{3} \leq 1 \leq \frac{9}{5}$
my proofs shows that if $A: \Bbb R^n \to \Bbb R^n$ is linear where $n \geq 2$, then for any positive real number $\lambda$ that is between the absolute values of two eigenvalues, there is some $v \in \Bbb R^n$ such that $\|Av\|=|\lambda|\|v\|$
ah I see what you mean, you restrict the map the the subspace generated by two eigenvectors with distinct eigenvalues than you use the argument for two dimensions
So I've just learned about the Fréchet derivative on general Banach spaces. If I have two real Banach spaces $E,F$ why isn't the derivative defined as $\lim_{x\rightarrow x_0}\frac{f(x)-f(x_0)}{\|x-x_0\|}$? This isn't equivalent to the Fréchet derivative right?
I'm sorry, of course it's equivalent to the Fréchet derivative.
@JannikPitt: Of course it is not. That formula doesn't remotely work even in $\Bbb R^1$ or $\Bbb R^2$. What kind of object do you think you're defining?
Okay I graphed, I can see it looks like a christmas tree
The max value $f$ takes is $\frac{1}{2}$
I'm thinking I can then split $[0, 1]$ into $[0, \frac{1}{2}]$ and $[\frac{1}{2}, 1]$ which would correspond to a first really crude partition of $P_1 = \{0, \frac{1}{2}, 1\}$
Then $P_2 = \{0, \frac{1}{3}, \frac{1}{2}, \frac{2}{3}, 1\}$
If $(X,d)$ is a metric space that isn't totally bounded, how can I show there exists an $\epsilon > 0$ and collection $\{x_n\}$ of points such that $\{B(x_n,\epsilon)\}_{n=1}^\infty$ is a disjoint collection of open sets?
x = (x,y,z), a = (a,b,c), t = (l,m,n) of the vectors x, a, t. Write a set of equations which determine the coordinates of the point where this line intersects the plane in matrix form Ax = d, where px+qy+rz=s
