Mathematics - 2018-01-21 (page 5 of 6)

So what I was trying to do was prove that if $X^*$ is separable or $X$ is reflexive, you can find a sequence $x_n$ such that $\|x_n\| = 1$ but $x_n \to 0$ weakly

Akiva Weinberger

7:40 PM

Like, all the points "look the same"

Balarka Sen

@Akiva The long line is very noncompact, right?

Leaky Nun

2

A: Long line is connected and compact

You can actually show that the long line is path-connected, which shows that it is connected. Pick any two points $x=\langle\alpha,s\rangle$ and $y=\langle\beta,t\rangle$ on the long line, with $x<y$. If $\alpha=\beta,$ then $s<t$ and the long line interval $[x,y]$ is readily homeomorphic to the ...

Alessandro Codenotti

@AkivaWeinberger So like there's an homeomorphism swapping the "corner" and points in the middle? Weird

Daminark

If $X^*$ is separable, the unit ball in $X$ is metrizable

Akiva Weinberger

No, not if you close it off at the end @BalarkaSen

Balarka Sen

7:41 PM

I mean it's not paracompact, even

@Akiva Hm?

Akiva Weinberger

What's paracompact?

Daminark

So in particular it's first countable, so all you have to do is say aight, we know in general for the dual space that the weak closure of the unit sphere is the unit ball

Balarka Sen

I do not think I know of that construction

Paracompact means every cover has a locally finite refinement

You can choose a subcover so that every point gets hits by finitely many elements of the subcover

Akiva Weinberger

This is $\big(\omega_1\times[0,1)\big)\cup\{\omega_1\}$

Daminark

Reflexive was throwing me off. But all you gotta do is find a closed separable subspace, that'll be separable, so its dual will be separable, then apply it there and everything's good

Akiva Weinberger

7:43 PM

@BalarkaSen The thing is, once you choose an open set containing the point $\omega_1$, it has to contain all but countably many ordinals in it as well

so you can choose a finite subcover of the remainder easily

Alessandro Codenotti

@Daminark Stupid question: what's wrong with just picking a sequence $y_n\rightharpoonup 0$ but $y_n\not\rightarrow 0$ and taking $x_n=y_n/||y_n||$?

Balarka Sen

Paracompact is essential for getting a partition of unity, which as a manifold the long line does not admit, so send that space to the deepest core of the hell to starve and self-mutilate and die and rot and decompose and turn into nothingness

Leaky Nun

oh right, if you compactify the long ray then it is compact

Akiva Weinberger

topology.jdabbs.com/spaces/S000039

Balarka Sen

@AkivaWeinberger Aha.

I do not know this stuff

Akiva Weinberger

7:44 PM

"Extended long line", then

Daminark

@Alessandro will such a sequence always exist?

Leaky Nun

right @AkivaWeinberger

Daminark

It won't actually

In general

Look at $\ell^1$, weak = strong convergence

Alessandro Codenotti

something something Schur property :/

Balarka Sen

@Akiva Wait, so what does the local structure near the two infinities look like?

Akiva Weinberger

7:45 PM

@AlessandroCodenotti You know what, I think the idea is,

you can embed a square in a cube such that any given point in the square lands on any given point in the cube

Balarka Sen

I guess it's like an infinite broom at the endpoints or whatever

Alessandro Codenotti

And $(\ell^1)^\star$ isn't separable so we're fine

Akiva Weinberger

This thing is infinite-dimensional though so it's kinda one dimension smaller than itself

Leaky Nun

@Daminark a bunch of examples

Akiva Weinberger

I don't know if you need it to be a surjective function though

Daminark

7:46 PM

Yeah that webpage makes me want to throw up

Akiva Weinberger

@BalarkaSen Two?

Alessandro Codenotti

it's the third time in 10 minutes someone links it lol

Akiva Weinberger

Are you making this long on both ends?

Daminark

Right?

Balarka Sen

@AkivaWeinberger Yeah, let's just take two-point compactification

Alessandro Codenotti

7:47 PM

Ok, I'm not following your reasoning as to why this works if $X^\star$ is separable

Balarka Sen

I am guessing the end result would not be path connected

Because of that broom thing happening

Akiva Weinberger

> The compactness of the Hilbert cube can also be proved without the Axiom of Choice by constructing a continuous function from the usual Cantor set onto the Hilbert cube.

Hmm

@BalarkaSen I don't understand

It should be path connected

Oh, wait

Balarka Sen

Really?

Daminark

@Alessandro so if you have a first countable space $A$ and some subset $B$

Akiva Weinberger

Nah, never mind

Sorry

Daminark

7:48 PM

$x\in \overline{B}$ iff there's some sequence $x_n$ in $B$ which converges to $x$

Akiva Weinberger

I don't know what you mean by "broom" though

Daminark

So now, if you look at the weak topology on any Banach space, you know the closure of the unit sphere is the closed unit ball

Akiva Weinberger

$L := \omega_1 \times [0,1)$ in the order topology is path connected, but not compact. $L^*$, the one-point compactification of $L$, is however not path connected. $0$ cannot be joined with $\infty$ by a continuous path. It is "too far away" — kahen Oct 4 '13 at 19:32

Alessandro Codenotti

@Daminark wait when did it become first countable?

Balarka Sen

@AkivaWeinberger I was thinking that near the infinities it looks like the infinite broom, sorta

Daminark

7:50 PM

If $X^*$ is separable, the weak topology on the closed unit ball in $X$ is metrizable

Balarka Sen

This shit

Alessandro Codenotti

@Daminark uhm, ok

Balarka Sen

I do want to know what a neighborhood of the infinity looks like, though, rigorously

Does it have a well-known homeomorphism type?

Akiva Weinberger

@BalarkaSen This is an ordered set

Daminark

This is the proof that if $X$ is separable, the closed unit ball in $X^*$ with the weak* topology is metrizable

Alessandro Codenotti

7:51 PM

@Daminark yeah, that's the result I was thinking about

Balarka Sen

Yeah I know. The topology comes from the lexicographic ordering on $\omega_1 \times [0, 1)$

Leaky Nun

why $(\frac12,1]$ lol @BalarkaSen

seems so arbitrary

Akiva Weinberger

So how could it be a broom

Daminark

There's a similar argument to show that if $X^*$ is separable, the unit ball in $X$ is weak metrizable

Alessandro Codenotti

oh, ok

Daminark

7:52 PM

Like, Brezis doesn't even bother going through it, they just say yeah the proof is identical with roles swapped

Balarka Sen

@Akiva I wasn't claiming it to be, that's just what came to mind.

I don't actually know what it looks like

@LeakyNun You could take any bit of the x-axis and it'd be true

Any bit disjoint from the origin

Alessandro Codenotti

Yeah, I was just reading it, we didn't do it in the functional analysis course so I just skipped it when I went through Brezis's book

Balarka Sen

Integer broom topology

In general topology, a branch of mathematics, the integer broom topology, is an example of a topology on the so-called integer broom space X. == Definition of the integer broom space == The integer broom space X is a subset of the plane R2. Assume that the plane is parametrised by polar coordinates. The integer broom contains the origin and the points (n,θ) ∈ R2 such that n is a non-negative integer, and θ ∈ {1/k : k ∈ N and k ≥ 1}. The image on the right gives an illustration for 0 ≤ n ≤ 5 and 1/15 ≤ θ ≤ 1. Geometrically, the space consists of a series of convergent sequences. For a fixed n,...

Hmm

Akiva Weinberger

I mean, any complete order topology looks like a line to me

36

Q: Why is the Hilbert Cube homogeneous?

The Hilbert Cube $H$ is defined to be $[0,1]^{\mathbb{N}}$, i.e., a countable product of unit intervals, topologized with the product topology. Now, I've read that the Hilbert Cube is homogeneous. That is, given two points $p, q\in H$, there is a homeomorphism $f:H\rightarrow H$ with $f(p)=q$. ...

general-topology

@AlessandroCodenotti

Balarka Sen

@AkivaWeinberger It shouldn't be a manifold with boundary at the endpoints!

Looking like a line to me means homeomorphic to [0, 1]. Sorry.

I won't accept order bullshit

Akiva Weinberger

7:56 PM

@BalarkaSen Well, yeah, 'cause a neighborhood of the endpoints look like another copy of the (compact) long line!

Balarka Sen

:P

Ah, really?

I suppose you're right. Weird!

Alessandro Codenotti

@AkivaWeinberger looks like an hard result, thanks

Akiva Weinberger

Look at the edit on the original question @Ale

Balarka Sen

skidaddle skidoodle your dick is now a noodle

►

This is the new meme everyone

This is what we were waiting for

Leaky Nun

what the hell

1988: maybe 30 years later people will have flying cars

2018: skidaddle skidoodle your dick is now a noodle

usukidoll

8:04 PM

?

Leaky Nun

a meme within a meme @usukidoll

usukidoll

Ohhhhhhh

Tuki

guys if i have function $f(x,y)=x^3+3y^4$ i want to define a function for tangent plane on point $f(2,1)$ $$ \nabla f(x,y)=\begin{bmatrix} 3x^2 \\ 12y^3 \end{bmatrix} $$ and in point $$\nabla f(2,1)=\begin{bmatrix} 12 \\ 12 \end{bmatrix}$$ i can call tangent plane function $w(x,y)$, $$ w(x,y)=f(2,1)+f_x(2,1)(x-2)+f_y(2,1)(y-1) $$ $$ w(x,y)=12x+12y-25 $$ is this correct ?

Balarka Sen

stackedit.io is really useful

Daminark

8:22 PM

@BalarkaSen welp

Tuki

your right

Meow Mix

my right is your left

Tuki

@Semiclassical how would i define tangent plane ?

Semiclassical

well, it would be right if there was an actual line defined there

as it stands, $w(x,y)$ is just a function

Balarka Sen

9:00 PM

Hey @Ted

Ted Shifrin

Hi, Balarka.

Balarka Sen

I'm actually computing something in moving frames and think I am succeeding

Ted Shifrin

@Semiclassic @Tuki: I interpret that Tuki means that the tangent plane is the graph of that function. IT's correct (but I haven't checked arithmetic).

Whoa, @Balarka. Do you have a fever?

Balarka Sen

Hahah.

I'll disclose what I am computing if I succeed. I'm pretty excited; I think I am about to find a different proof of Darboux's theorem

Completely geometrically

Meow Mix

hi Ted

wanna see something im working on

Ted Shifrin

9:03 PM

I wondered if you were pursuing that, based on what I caught a while ago. It's really a differential equations proof, and I'm dubious.

Tuki

@TedShifrin yes this is what i meant

Ted Shifrin

hi Meow

Eric Silva

Sup chat

Daminark

@BalarkaSen alright I'm outta here have a good day

Meow Mix

ill email it to you

Ted Shifrin

9:04 PM

Just had my AoPS precalculus kids work out what the fifth roots of unity are explicitly.

Daminark

Hibye Ted, Eric, Meow

:P

Balarka Sen

@TedShifrin Yes, but Moser's proof sheds 0 light on the fact that it's an integrability condition. But I'll let you know what I have after I work it out a little

I'll send you a TeXed up thing

MatheinBoulomenos

@TedShifrin did they work geometrically or algebraically?

Balarka Sen

@Daminark lmao

Eric Silva

@BalarkaSen hmu too my dude

Balarka Sen

9:05 PM

I will!

Ted Shifrin

Well, ultimately, it can be proved as an application of Cartan-Kähler, but I don't think you can do just easy Frobenius, Balarka.

Leaky Nun

hi @TedShifrin @MatheinBoulomenos

Balarka Sen

I am using Frobenius actually

MatheinBoulomenos

Oh, hi @Ted @Balarka @EricSilva @Leaky

Balarka Sen

Hey @Mathein!

Leaky Nun

9:06 PM

22 hours ago, by Leaky Nun

@MatheinBoulomenos what's the point of category theory, seriously?

MatheinBoulomenos

what's the point of your question?

Eric Silva

Sup @MatheinBoulomenos

Leaky Nun

to ask you what category theory is actually about

limsup @EricSilva

Daminark

Hey Mathein!

MatheinBoulomenos

Hi @Daminark

Meow Mix

9:07 PM

@Ted sent

not that you ever said explicitly that you wanted to see it

ÍgjøgnumMeg

@MatheinBoulomenos yo

MatheinBoulomenos

Hey @ÍgjøgnumMeg

Ted Shifrin

Interesting how you have to work in the rank condition, @Balarka.

Leaky Nun

lol, posting others' email-address

Meow Mix

im going to delete it

besides, it's available online anyways

Ted Shifrin

9:09 PM

It's in my profile. No biggie.

Meow Mix

oh i sent to math.uga.edu

should it just be uga.edu?

Ted Shifrin

Either works.

I just got it.

Meow Mix

the rectangles are upside down accidentaly

MatheinBoulomenos

Category theory gives us a framework and a language for thinking about a lot of different things (but not everything can be done categorically). It allows us to see that quite different things can be considered as examples of one unifying concept, for example gcds, a supremum in a poset and the product of topological spaces.

Ted Shifrin

Lots of people have written such packages, Meow. But it's cool to have graphical stuff.

Upside-down rectangles sorta ruin it.

Leaky Nun

9:11 PM

@MatheinBoulomenos nice

Ted Shifrin

Can you see/prove that using trapezoids makes the error noticeably better?

MatheinBoulomenos

Category theory also allows generalizations that can be very fruitful. E.g. abelian categories which basically formalize what is nice about the category of modules over a rings, but it would be very difficult to work with abelian sheaves if we wouldn't have the formalism of abelian categories

Meow Mix

yes but how does that work with the current definition

Balarka Sen

The first nontrivial use of category theory that I ever did was to use Yoneda lemma to prove Hopf degree theorem in TOP

MatheinBoulomenos

Also category theory allows us to do some of the boring stuff that's always the same, such as product are commutative and associative up to isomorphism etc., once and for all so that when we work in a specific category, we can focus on the interesting stuff

Meow Mix

9:12 PM

i dont doubt you, it's a much better approximation

Ted Shifrin

It's an average of two Riemann sums, effectively, right? It's using the average of the left and right endpoints.

Meow Mix

aoh yes

Ted Shifrin

But the error is second order. Surprisingly, when you use parabolic approximations (Simpson's rule) the error become fourth order!!

Eric Silva

@BalarkaSen spooked

Ted Shifrin

You will find proofs in problems in Spivak, Meow. (I wrote these problems.)

Balarka Sen

9:13 PM

@EricSilva It's a clean proof, actually. I sent Daminark the .pdf I wrote it up on once

Ted Shifrin

hi Eric.

Eric Silva

The thing I'm spooked by is TOP lol

Yo @Ted

Balarka Sen

Ah yes

Meow Mix

and then you can like use that polynomial error theorem thing

Alessandro Codenotti

Hi @Ted.

Meow Mix

9:14 PM

weierstrass i think

or is that smth different

MatheinBoulomenos

There are also substantial theorems in category theory that can be used to prove quite non-trivial results in various fields, e.g. Yoneda lemma, Freyd's adjoint functor theorem, Beck's monadicity theorem etc.

Ted Shifrin

That's different.

See exercises at the end of chapter 19, Meow.

MatheinBoulomenos

Does this answer your question? @Leaky

Leaky Nun

yes

quallenjäger

Is there anyway to approximate a Lipschitz continuous function by $C^1$ functions?

Balarka Sen

9:18 PM

Lipschitz functions are almost everywhere C^1 so there's that

Ted Shifrin

$C^1$ functions are dense even in $L^1$ functions.

Balarka Sen

Also that... a rather uninspiring approximation

Ted Shifrin

I mean, you can always approximate a continuous function on a closed interval by polynomials. So the answer is yes, abstractly.

Eric Silva

The point of Lipschitz functions is that they're basically just C^1 with better compactness properties

@BalarkaSen idk what this means

quallenjäger

@BalarkaSen Are you sure about it? As far as I know it does only hold for nearly almost everywhere

Balarka Sen

9:20 PM

@Ted I think the right question to ask is if $f$ is Lipschitz, can you $f_n \to f$ with $f_n \in C^1$ such that $f_n = f$ on the complement of a union of balls around the singular locus of $f$, of decreasing radii as $n \to \infty$

Eric Silva

C^1 a.e. doesn't obviously mean something

Balarka Sen

@EricSilva It's C^1 outside a measure zero set isn't it?

Or am I misremembering?

Eric Silva

It's equal to a C^1 function on a set of arbitrarily small but positive measure

quallenjäger

What about $g(x)$ being the indicator function of Fat cantor-set and $f(x)=\int_0^x g(y)dy$

Ted Shifrin

All: This is Rademacher's Theorem to which Balarka refers.

quallenjäger

9:21 PM

It's derivative is g(x) and not continuous isnt it

Ted Shifrin

no, Eric, measure 0.

You're thinking Lusin or something from measure theory.

Balarka Sen

Oh, I want differentiable, not C^1

I misremembered totally

Eric Silva

@TedShifrin I'm thinking of a corollary to Whitney extension

Ted Shifrin

ohhh, right, diff, not C^1.

Eric Silva

Rademacher Is differentiable, that wasn't my complaint

Ted Shifrin

9:23 PM

My apologies.

Eric Silva

Mhm

They're still basically C^1 though lol

quallenjäger

puhhh, I just sent out a report, on which I claimed that lipschitz continuity is not sufficient for almost everywhere continuity of the derivative.

Balarka Sen

That's what my intuition is, and that's what misled me

Eric Silva

The derivative only makes sense a.e.

quallenjäger

Well my problem is, I have a Lipschitz function (unknown) and I have a method to reconstruct $C^1$ function. Is there any known approximation method of Lipschitz function by sequence of $C^1$ functions?

Eric Silva

9:27 PM

I mean you can approximate any continuous function by a C^1 guy

quallenjäger

@EricSilva Is there any numerical method doing that?

Ted Shifrin

Convolution is the best way to understand this, but I'm not sure how quallenjäger wants his approximation to work.

Eric Silva

Probably, beats me though

quallenjäger

Ok I will check that. Thanks @EricSilva @TedShifrin

Eric Silva

The way I always think of it is by solving the heat equation with you function as initial data and going forward a little bit @Ted

quallenjäger

9:28 PM

1

Q: Mean-Value Theorem for set

Suppose $f:A\rightarrow \Bbb R $ is Lipschitz continuous and its derivative $Df$ is continuous on a closed subset $E$, $E\subset A$. Define $$\eta_\delta(a)=\text{sup}_{0<|x-a|<\delta \text{ and }x\in E}\frac{|f(x)-f(a)-Df(a)(x-a)|}{|x-a|}$$. I would like to proof the pointwise convergence for $\...

real-analysis analysis measure-theory

Ted Shifrin

Not sure how to implement that in practice, Eric :)

Eric Silva

Me neither lmao

I'm not sure how to write down a Lipschitz function that I can't obviously approximate either though

quallenjäger

Any idea to this? Does Mean Value theorem holds for something like $[a,b] \cap E$ with $E$ being closed.

My problem is, I have something called signature of a path. Signature is a sequence of numbers, which describes a path uniquely. I have constructed a method to reconstruct a $C^1$ paths from its signature and I would like to generalize this method to lipschitz case. I thought, it might be possible to find a sequence of $C^1$ paths, which I can reconstruct, and then let it converge to the Lipschitz-Paths that I am looking for. I was wondering, if there is any numerical method doing it.

Ted Shifrin

Do you really need Lipschitz paths rather than piecewise $C^1$ paths?

quallenjäger

Yes.

philmcole

9:40 PM

Hello

Ted Shifrin

Beats me.

quallenjäger

Ok, thank you for your help anyways Ted.

Eric Silva

I wish I could help but I know next to nothing about numerical stuff

philmcole

Is there a definition for a derivative of a function $f: \mathbb R \to \mathbb C$?

MatheinBoulomenos

just take the derivative of real and imaginary parts

philmcole

9:42 PM

Is this the definition?

quallenjäger

@EricSilva You've helped me a lot already. Thanks

I am reading the book of Federer and Morgan as you suggested.

philmcole

@MatheinBoulomenos There is actually an excercise in which I need to show that those are equivalent

quallenjäger

They really sucks

:D

Eric Silva

Lololol

quallenjäger

Making things unnecessarily complicated.

Ted Shifrin

9:44 PM

Just write down the usual definition but for the vector-valued function, @philmcole.

Heya @Antonios

Antonios-Alexandros Robotis

Hey!

Eric Silva

Morgan is good but it's a book for getting a flavor of things rather than for learning the technical things

Antonios-Alexandros Robotis

quick question actually, if you have a moment.

Eric Silva

@quallenjäger I don't think it's unnecessary, some delicate questions require delicate and technical theory. It's a pain to learn but the principle of conservation of difficulty says that to do something hard you kind of have to struggle somewhere

Ted Shifrin

Go ahead, Antonios.

MatheinBoulomenos

9:45 PM

@philmcole en.wikipedia.org/wiki/…

Antonios-Alexandros Robotis

Mostly just a terminology : in the book I'm reading, the vanishing set of some collection of polynomials in $k[X_1,\ldots, X_n]$ is called an affine algebraic set in $k^n$. It seems like some sources call this an affine algebraic variety.

In this book, an "affine algebraic variety" is a ringed space isomorphic as a ringed space to $(V,\mathcal{O}_V)$, where $V$ is an affine algebraic set, and $\mathcal{O}_V$ is the sheaf of regular functions on $V$.

I'm just wondering if there's any reason people opt for one way or the other.

philmcole

Thanks, I'll try it @MatheinBoulomenos and @TedShifrin

Ted Shifrin

Ordinarily a pedantic person thinks of a variety as a set of points together with a structure sheaf (that's the ringed space stuff).

MatheinBoulomenos

@Antonios-AlexandrosRobotis the first is easier to define, but it has not structure

Antonios-Alexandros Robotis

gotcha, so it's just a pedantic thing

Ted Shifrin

9:49 PM

Like you can say the unit sphere is a manifold ... but you've just described its set of points. It's taken for granted that it has some obvious smooth structure inherent from its embedding.

Balarka Sen

A variety is an $\infty$-topos

quallenjäger

@EricSilva Well, you are right. The theory in there is very generalized.

Antonios-Alexandros Robotis

yeah I figured as much, I think hartshorne defines the vanishing set as a variety, and I was viewing that as the "absolute truth" lol.

Thanks @TedShifrin @MatheinBoulomenos

Ted Shifrin

Well, in chapter 1 he is giving examples and motivations. Just wait for chapters 2 and 3 !!

MatheinBoulomenos

I think there's more to this than pedantry actually. The first definition assumes that you have already embedded your affine algebraic variety into some affine space. This is like defining a manifold as a subset of $\Bbb R^n$. The second defintion just assumes "there exists an isomorphism", so we haven't chosen an embedding yet

Antonios-Alexandros Robotis

9:52 PM

@TedShifrin For now I'm staying mostly away from hartshorne, but I think he does a nice job defining sheaves in the beginning of II. The definition of presheaves as a functor is nice.

@MatheinBoulomenos ah okay, so it's kinda like the difference between the embedded approach to manifolds versus the more "intrinsic one" whatever that means

Eric Silva

@MatheinBoulomenos I guess the first is only pedantry once you know better but a priori it might not be

Which I guess is important

MatheinBoulomenos

Of course the way I think about affine algebraic varieties is as an object in the opposite category of finitely generated reduced $k$-algebras :P (jk)

Balarka Sen

I think about rings as affine k-algebras

beat drops

Antonios-Alexandros Robotis

lol well thanks all, time to get back to readin.'

Balarka Sen