To prove that $\lim_{x \to 0} \arcsin x =0$ using the definition is it enough to argue like this: let $0<\epsilon \le 1$ be arbitrary. Setting $\delta_\epsilon =\sin \epsilon$, if $|x|<\delta_\epsilon$ then $-\sin \epsilon <x<\sin \epsilon$ and so, being $\text{arcsin}$ increasing and odd, we have $-\epsilon<\arcsin x <\epsilon$. I assumed $0<\epsilon<1$ because $\arcsin \sin t = t$ if and only if $t \in [-1,1]$; I proved previously that assuming "small" $\epsilon$ doesn't affect the validity
@TedShifrin Uuuh sure, I got confused with sin of arcsin instead of arcsin of sin. Thanks for pointing that out.
I am ashamed because I always use the example: "What is $\arcsin (\sin 2)$?" with my physicists friend to shock them because very often they automatically answer $2$. Karma is real :D
$an+b(n^2+1)=1\implies a=\frac{1-b}{n}-bn$. And if we let $b=nk+1$ for some $k\in \mathbb{Z}$ then $a=\frac{1-(nk+1)}{n}-bn=k-bn$ so $a\in \mathbb{Z}$. Does this look good? I'll also think about no prime dividing both
Something feels strange about this: If $v$ is a Lipschitz continuous function, then $\max(v,0)$ has gradient equal to $0$ almost everywhere on the level set $\{v = 0 \}$. Which means if $v \geq 0$ everywhere and is Lipschitz, then on $\{v = 0 \}$, $\nabla v = 0$ almost everywhere, which seems .. wrong? Am I misunderstanding something here?
wait nvm. That isn't weird at all, the only way $\nabla v = 0$ can fail is on the measure-theoretic boundary of $\{ v = 0\}$, and that always has lebesgue measure zero
I'm reading a statistics paper that includes a proposition that states in part, "Suppose that the law of $X$ has a dense support on $\mathcal X$." In the paper, $(\mathcal X, d_{\mathcal X})$ is a metric space and $(\mathcal X, \Sigma, \mu_X)$ is a probability space, and I believe $X$ would be a random variable from $\mathcal X$ to $\mathbb R$.
My issue is that the quote discusses the support of the law on $\mathcal X$. But the law of a random variable is the image measure, so it's a function on $\mathcal B(\mathbb R)$. So how can a function whose domain is $\mathcal B(\mathbb R)$ have anything to do with $\mathcal X$? Can this be reconciled?
Here's a quote from p. 20 of Kallenberg's probability theory textbook: "For any measure $\mu$ on a topological space $S$, we define the support $\text{supp } \mu$ as the set of points $s \in S$ such that $\mu B > 0$ for every neighborhood $B$ of $s$.
Given an $n$-dimensional smooth manifold $M$, $N(M)$ is defined as the minimal cardinality of an open cover $\{U_i\}$ of $M$ such that each $U_i$ admits a smooth embedding into $\Bbb{R}^n$. What will be $N(\Bbb{R}P^3)$? I figured out that $N(\Bbb{R}P^2)$ will be $3$, but I don't have any idea for the above
Since the Dirichlet function differs from the constant function to zero on a null set, I thought the removal of discontinuities would/should result in the said constant function.
Okay, but in your version of the Dirichlet function, it's discontinuous everywhere, not just as at the rationals. Though there are variants which are continuous at all the irrationals.
I mean... For Dirichlet function, removing discontinuities on the rationals would "cost" less than removing discontinuities on the irrationals. 'Cause the set of rationals is a null set.
Are the discontinuities of $$f(x) = \begin{cases}1,&x\in[0,1]\\0&x\notin[0,1]\end{cases}$$ removable? In the same way, it would "cost less" to set $f$ equal to 0 everywhere than to set it equal to 1!!!
@DannyuNDos That isn't the "usual" classification, AFAIK. In an elementary calc class, I typically talk about removable discontinuities (essentially, we can change the value of the function at a removable discontinuity in order to make the function continuous at that point), jump discontinuities (left and right limits exist, but are not equal), and oscillatory discontinuities (the function is bounded, but the limit doesn't exist).
I don't generally consider "blow ups" to be discontinuities, since we typically don't define functions at such points. It doesn't make sense to talk about functions being discontinuous at points which aren't in their domain.
But, I suppose, you could define $f(x) = 1/x$ for $x \ne 0$ and $f(x) = 1$ for $x=0$, and talk about the "infinite" discontinuity at zero. Seems a little artificial, though.
@DannyuNDos No, they are oscillatory. At the points where the Thomae function is discontinuous, the limits don't exist. This is similar to $\sin(1/x)$ at zero (where you define the function however you like at zero).
An intuitive question - say $f$ is a real function which is differentiable n times at a real point $x_0$. Now suppose $Q$ is a polynomial s.t for each $m \leq n$, the $m$'th derivative of the function function $R(x) = Q(x) - f(x)$ at $x_0$ equals zero. Is it the same as saying that $Q$ is $f$'s Taylor polynomial around $x_0$ of at least order $n$?
@Thorgott: I want to show that a certain map induces a weak equivalence from S^1 to $\Omega CP^\infty$.
I know that $\pi_k (\Omega CP^\infty)=0$ for k not equal to 2 and is 0 otherwise.
the map that does this is: induced by the inclusion $S^2\to CP^\infty$.
in the following way: the groups $[S^2, CP^\infty]_*\simeq [S^1, \Omega CP^\infty]_*$
Here is what I did (please let me know in case you find my proof incomplete): I denote the inclusion $S^2\to CP^\infty$ by $i$. Since S^2 is 3 skeleton, it is clear that the pair (CP^oo, CP^1) is 3 connected, and therefore i induces isomorphisms between $\pi_k S^2$ and $\pi_k (CP^\infty)$ for k=1,2.
Let's consider only the non trivial case k=2.
i is a generator of the group $\pi_2(S^2)$ so its isomorphic image say $i'$ (under the isomorphism $[S^2, CP^\infty]_*= [S^1, \Omega CP^\infty]_*$) is also a generator of $\pi_1(\Omega CP^\infty)$, which is isomorphic to $\pi_2(CP^\infty)$.
So the induced map gives isomorphism of homotopy groups.
@Thorgott (pinging again to indicate-my message is complete now.)
@冥王Hades Let me know when you're about to bring a total eclipse. I want to watch too. But for your own enjoyment, wouldn't it be more fun and pleasing to watch God does it than yourself? Sometimes I would rather listen to a world-class performance than hearing myself play a piece that I like.
@冥王Hades The next "God-scheduled" total eclipse will pass your hometown (!) on April 8, 2024. If you plan a sideshow of total eclipse to upstage this one, please let me know when/where since being in other part of Canada it's too far for me to watch. @XanderHenderson
Timeanddate.com has a nice very graphics and detailed map showing where a Total Solar Eclipse can be seen between 1900 to 2199. Looks like my best time is not until Aug 22, 2044 and I may have to travel to Calgary for that. For you, @冥王Hades, DFW is close enough to smack center next year!
It seems AE don't, but if they don't then in this theorem they must state that $\varphi(1_k) \neq 0$, right? For otherwise, if it maps to 0 then this claim needn't be true?
@TedShifrin Can you elaborate? That would seem to fail to be a ring homomorphism no matter what, right? That is, $\varphi(n^2) = 2n^2 \neq \varphi(n)\varphi(n) = 4n^2$?
My book defines a ring as the barest of definitions (without a multiplicative identity or commutativity)
So their ring homomorphism definition is only that it respect addition and multiplication between the rings
Hmmm ... You're right, EE18. .... What am I meaning to say? Oh, you never want to build in commutativity. Many interesting rings (like matrices) are non-commutative.
A mathematical physics text of mine did build in commutativity (Szekeres!) I guess matrices were "viewed" as an algebra over a field so maybe that was the reason, but it was weird...
A question now that I've done the proof...I didn't understand AE's proof a priori. In particular, they suggested tacitly that we could restrict and corestrict $\varphi$ such that it was between $K^\times$ and $K^{'^{\times}}$. It was the corestriction which was not obvious to me. That is, how could we be sure that $\varphi(a) \neq 0$ for all $a \in K^\times$? Prof. Shifrin, I guess your earlier argument also leads us to $\varphi(a) \neq 0$ as required?
Well, the kernel of a ring homomorphism is an ideal and the only ideals in a field are $(0)$ and the whole field. So only $0$ can map to $0$. We need to prove this bare-hands in this case, I imagine.
Suppose $a\ne 0$ and $\varphi(a)=0$. Then for any $x\in F$, $\varphi(x)=\varphi(aa^{-1}x)=\varphi(a)\varphi(a^{-1}x)=0$.
Is there some bounding relation between the number of prime ideals and the number of maximal ideals of a ring? of course the former is bigger than the later
Oh. I had missed that the original comment asserted repeated differentiability at $x_0$ and just saw your comment about differentiability on an interval. I don't think that's needed.
@TedShifrin If we don't care about the Taylor series converging to the function on a neighborhood, then we don't need differentiability in a neighborhood.
@robjohn It's just important not to fall into the trap into which I've fallen recently that if $P$ is a polynomial of degree $\ge k$ that agrees with $f$ to order $k$ at $0$, then $f$ must be $k$ times diff at $0$.
@TedShifrin if you localize $k[x_i:i\in I]$ at the maximal ideal $(x_i:i\in I)$, you should get a local ring with at least $|\mathcal{P}(I)|$ many prime ideals
@TedShifrin the differentiability at $x_0$ was already given, but the conditions in a neighborhood were meant to insure agreement. However, they were just asking about the Taylor series at that point, and for that, I think what they said is sufficient.
@robjohn Do you have a slick answer to this? I want to just observe that $\partial c/\partial\bar t$ satisfies a linear differential equation along the curve?
Recall that you can color the quadrants of the plane, the octants of space, and in general the orthants of n-space with two colors so that adjacent orthants have different colors. For example you can color a point black if an even number of coordinates is positive and white otherwise.
Draw a sphere surrounding, but not centered on, the origin. We are interested in how much of the sphere is in each color.
Conjecture: The amount of the sphere in each color is equal, regardless of the location of the sphere's center, in even dimensions greater than 2. It does not hold in general for odd dimensions or two dimensions.
This is a weird high-dimensional version of the pizza theorem.
@Thorgot: I have proven that [i] is a generator of $[S^2, CP^\infty]_*$. So by adjunction group isomorphism, [i] goes to a generator [i'] of $[S^1,\Omega CP^\infty]_*$. How do I show that $i': S^1\to \Omega CP^\infty$ induces an isomorphism on $\pi_1$?
@Koro Do you really want to show that, or are you just trying to use that towards showing $[S^2,\Bbb{CP}^\infty]\cong \Bbb Z$? I have the feeling that your question there is harder than the original question
Just to be clear, are you saying your end goal is to know $S^1\to \Omega CP^\infty$ is a weak equivalence, or you are using that towards your actual end goal which is to show that $[S^2,\Bbb{CP}^\infty]\cong \Bbb Z$?
(follows by noting that (CP^oo, S^2) is 3 connected.)
if I take a generator of S^2--> CP^oo and map this via the adjunction isomorphism say $\psi$, then I get a generator $\psi(i)$ of [S^1, \Omega CP^oo]. I'm stuck at showing that $\pi_1(\psi(i))$ is an isomorphism.
Ok, so you know $\pi_2(\Bbb CP^\infty)\cong \pi_1(\Omega\Bbb{CP}^\infty)\cong \Bbb Z$. So generator gets sent to generator, and maybe I'm being silly but doesn't the result follow?
if I take a generator of S^2--> CP^oo and map this via the adjunction isomorphism say $\psi$, then I get a generator $\psi(i)$ of [S^1, \Omega CP^oo]. I'm stuck at showing that $\pi_1(\psi(i))$ is an isomorphism.
i have a question that i've google for some time and i didn't find anything
so we know that given any topological space $X$, the category $\mathrm{Sh}(X)$ is an abelian category
thus we have kernels, images, cokernels, etc. and thus we do also have quotient objects
that said, we can even define the (co)homology of chain complexes of objects in that category, and thus we can define (co)homology
so why do we actually care about the global section functor and derived functors, injective resolutions blablabla to define sheaf cohomology when we can simply quotient out chains of sheaves?
@LucasHenrique Global sections functor is left exact, so it takes short exact sequences $0\to A\to B\to C\to 0$ to $0\to \Gamma A \to \Gamma B\to \Gamma C$ and we can extend this to the right to get a long exact sequence $0\to \Gamma A\to \Gamma B\to \Gamma C \to R^1\Gamma A \to R^1\Gamma B\to \dots$, maybe this helps you understand the map $\Gamma B\to \Gamma C$ better? I guess it matters what you are trying to do
@LucasHenrique You could ask why people care about derived hom or derived tensor too, and there are things people are trying to compute/understand via those
@Thorgott I suppose the point s/he is objecting to, is that the adjunction only gives a bijection of sets, a priori, since this isn't an additive category, so you have to say something about the multiplicative structures
@Thorgott if it were, then all the abstract nonsense effort would be pointless :p. what is sheaf cohomology is pretty much what i'm trying to understand
$\psi(i)$ is a generator of $[S^1, \Omega CP^\infty]_* =Z$ so it could happen that $\pi_1(\psi(i)): Z\to Z$ is $2\times $ or $3\times $ or even a zero map.
@LucasHenrique Left exact covariant functors have right derived functors - take an injective resolution (any will do equally well) and then compute cohomology in some degree. Right exact covariant functors have left derived functors - take a projective resolution (any will do equally well) and then compute cohomology in some degree. That's all there is to it. (The resolutions define isomorphic objects in the derived category to the objects you are resolving )
