Mathematics - 2017-03-06 (page 4 of 5)

Secret

17:03

Some other observations:
Number of closed sets
Base case: 1
Inductive case: k
Number of open sets that are not open rays
Base case: 0
Inductive case: k-1

But in order for the induction to work, we need k+1 instead of k-1?

DHMO

how is that relevant?

Aaron Hall

@DHMO ???

Secret

ok sorry nvm, it still works. Yeah I think I am stuck at some kind of thinking blind spot. I really don't see anything misbehaving in that induction proof

DHMO

@AaronHall I thought mods only come when a message is flagged lol

Aaron Hall

I have the room favorited... but you guys are kinda boring... :P

(that's just a joke!)

Eric

17:10

@Secret what is it that you're trying to do by induction

Secret

@Eric Proving that complement of union of open intervals equals to the union of closed intervals

Eric

a countable union?

Secret

yes

Eric

this isn't true though

The Cantor set is a counterexample

Secret

DHMO said it also does not hold for the rationals and I struggle to find anything wrong in the above attempted proof

48 mins ago, by Secret

Let the rationals $q_i\in\Bbb {Q}$ be enumerated by the naturals $i\in\Bbb {N}$

Base case {$q_1$}, complement $(-\infty,q_1)\cup (q_1,\infty)$ true

2nd case {$q_1$} $\cup$ {$q_2$}, complement $(-\infty,q_1)\cup (q_1,q_2) \cup (q_2,\infty)$ true

Inductive case $\bigcup_{i=1}^k$ {$q_i$} $\cup $ {$q_{k+1}$}, complement $(-\infty,q_1)\cup (\bigcup_{i=1}^k(q_i,q_{i+1})) \cup (q_{k+1},\infty)$ true

Induction to the naturals
$(-\infty,q_1)\cup (\bigcup_{i\in \Bbb {N}}(q_i,q_i+1))$
-> ???

23 mins ago, by Secret

Actually the $\omega$-th case should look like this:
$$\bigcup_{i\in \Bbb{N}} \{q_i\}$$
and then take its complement, but the problem is I don't know how to prove this since this obviously does not have the same expression as the base case and the successor case (what is said to be the inductive case above)

Eric

17:18

This induction is basically just showing that finite unions of singletons are closed.

I see you're trying some sort of transfinite induction or something?

the problem with trying to prove the limit case is that it just isn't true.

Secret

Ah, so that's where the induction fails, that's why I struggle to find that in the succesor and base cases

Eric

yeah cause all the finite cases are fine

Secret

In that case, I can also have a counterexample where the singletons form some bounded countably infinite sequence such as {1/n} starting with n=1,2,3...

Eric

Sure there are lots of counterexamples

Secret

then I get the neither set (0,1] from the countable union of said singletons

Eric

17:24

I just like the Cantor set as an example because it shows you that characterizing closed subsets of $\mathbb{R}$ is far less simple than just characterizing open sets.

Secret

17:36

I still yet to fully comprehend the cantor set, but I will soon

Hmm, so that means...
Since the irrationals cannot be a countable union of open intervals, it is not open. Now the complement of the irrationals is the rationals which is also not open because any open interval contains an irrational so it cannot be a union of open intervals, and singletons are closed and pairwise disjoint thus. It is not closed either because it does not contain all of the limit points such as the irrationals. Therefore the rationals is a neither set. Therefore the irrationals are not closed thus the irratio…

The irrationals is so weird, it is a neither set that cannot be expressed as unions of any type of intervals except uncountably many singletons

Eric

I mean there are also tons of closed sets whose only representation as a union of intervals is as uncountably many singletons.

Secret

It seems I still have a lot to learn about how closed sets behave

These exercise given by DHMO does help in pondering about these sets

I wonder... since the rationals are dense, can I find a neither set of the form [a,b,...,x,y] so that the open ends are in the middle?

(to be investigated later...)

O wait, I think the cantor set might work

Liad

18:04

Hi . anyone can explain this :
$\dfrac{d \ ^ 2 f}{dx_idx_j}(x_0)$ , this means derivative first with respect to $x_i $ then $x_j$ ?

Anheuser

Hello

s.harp

@Secret have you read a little of the Bourbaki book?

MickLH

@Semiclassical Fortunately yes, it was just a little C++ script I whipped up so I just added dump-to-file on the end of the program and went to sleep while it ran again :P

Anheuser

Are most of you undergraduates, graduates, post graduates or professors?

Assistant professors?

What's the deal.

Semiclassical

Grad student here.

DHMO

18:19

I'm still in high school lol

Anheuser

Cool

Semiclassical

There's a retired prof who comes by a lot, but beyond that I think it mostly plateaus at the grad-student level.

Mostly because if you're a post-doc or active prof then you're probably too busy to be here much.

Anheuser

That makes sense.

Semiclassical

That, and there are a lot more math-interested undergrads than there are math profs.

Anheuser

I am 22 and about to start my undergrad in math

Do you think you learned an adequate amount in your undergraduate?

Semiclassical

18:21

By contrast, I imagine the MO chat has quite a different pool.

Well, I should point out that I'm a physics grad student.

Anheuser

ah.

The language of the universe.

Semiclassical

And as far as what I needed for grad-level classes, I'd say my undergrad was fine.

Anheuser

that's interesting.

Semiclassical

But for the stuff I'm interested in I end up having to educate myself a lot because what I'm interested in isn't stuff I've taken classes on.

I've never taken algebraic topology, for instance, but I do understand what the various groups are adn how to make some conclusions. (Though if you asked me to do an arbitrary example out of Hatcher's AT book, I'd probably be stumped.)

Alessandro Codenotti

There's quite a few undergrads too here (including myself)

Krijn

18:25

I think I'm a grad in the US system

Not sure

s.harp

@Semiclassical does MO have a chat?

Semiclassical

Pretty sure.

Yeah, here: chat.stackexchange.com/rooms/9369/mathoverflow

But it's not very well-frequented.

s.harp

like 5 messages a day

Semiclassical

Yeah, and no conversations of any length in quite a while.

So I guess there's not an -active- MO chat.

Krijn

Define "chat"

DHMO

18:31

@AlessandroCodenotti sei qui?

Alessandro Codenotti

Nope

I have to go very soon

DHMO

Feb 25 at 9:52, by Alessandro Codenotti

@secret Look at the topology on $\Bbb R$ with basis $\{[a,b),a,b\in\Bbb R\}$. Is it comparable to the euclidean? Is $[0,1]$ compact? Does the sequence 1/n converge? Does the sequence 1-1/n converge?

Do you mind giving me questions like these ^ but under the right-order topology of $\Bbb R$?

Alessandro Codenotti

I never thought about that one specifically so I'm not sure

DHMO

alright

CausingUnderflowsEverywhere

18:54

who here thinks they are smarter than me and can answer the following question?
$4 = 5$

DHMO

smart people don't generally go around boasting their smartness

CausingUnderflowsEverywhere

hehe, I did not say smart I said, one who $thinks$

Anheuser

you divided by zero

by expanding the polynomials and canceling like variables which had a value of zero

that's simple enough

Balarka Sen

19:12

@s.harp my impression is that the main MO chat is homotopy theory

Ted Shifrin

19:26

Greetings @Balarka, @Alessandro, @DHMO

Balarka Sen

Hi @Ted!

DHMO

hi

Daminark

Hey @Ted!

Ted Shifrin

heya Demonark

Daminark

How's it going?

Ted Shifrin

19:29

Just back from the beginning of about $8K of dental work :( ...

Daminark

Damn, that's a lot

Ted Shifrin

yeah, I continue to have to get work by previous dentists repaired at great pain and expense ... ugh. And I may end up losing another tooth, too. :( ... What's math around your neighborhood?

Goodnight @MikeM

Mike Miller

h

Ted Shifrin

Your iPad functional yet?

Mike Miller

yup

Maks

19:35

@TedShifrin I had the linear algebra exam

Ted Shifrin

Ah yes, @Maks, and did all your hard work pay off?

Maks

I did really good on the theory part, not so much in the practice :(

I have to wait until tomorrow to get my grade

Ted Shifrin

Computational skill is important!

Balarka Sen

Exams are stressful. I have chemistry soon and I'll probably flunk it

Ted Shifrin

I never stressed over exams. Weird.

Astyx

19:40

Hi chat

Ted Shifrin

Salut @Astyx

Astyx

Alors ce concert ?

Ted Shifrin

C'était très bien, alors.

Balarka Sen

@TedShifrin I am just stressing with that particular thing, because I am not very good at it

Ted Shifrin

Just practice and have a good attitude.

Daminark

19:45

Hey I'm back!

Ted Shifrin

You left without permission.

Daminark

And today we (quickly) set up a couple results on spectra and weak compactness of the unit ball in reflexive, separable spaces

Sorry :(

Balarka Sen

@DHMO should tell me how to do chem.

Daminark

As to exams, I think it is tricky because on one hand, stress can definitely make it harder to think straight and get it right

But on the other hand, when things go badly in exams, there is often a big problem

So it's hard to avoid insane stress

Mike Miller

learning about the Sen conjecture

Daminark

19:53

What's that @Mike?

Balarka Sen

google gives me stuff about Tachyon condensations

Daminark

Same

Zaid Alyafeai

@BalarkaSen, you still active here!

Daminark

Somehow I don't think that's what it is

Balarka Sen

@Zaid Oh, hi. Yeah, I am.

Ted Shifrin

19:54

He's a pigment of our imagination.

Balarka Sen

I'm an imagination of your pigment.

Daminark

Can pigments of our imagination be sen-tient?

Mike Miller

ipam.ucla.edu/abstract/?tid=13898&pcode=GT2017

conjecturing even b4 birth

Liad

define $f(x,y) = \dfrac{xy(x \ ^ 2 - y \ ^ 2)}{x \ ^ 2 + y \ ^ 2}$ for $x\ne0 $ and$ y \ne 0$ and $f(0,0)= 0$.
I need to prove that $\dfrac{df}{dx}$ exists at $(0,0)$.
This means i have to show that $lim_{t \to 0}\dfrac{f(t,o)}{t}$ exists?

Balarka Sen

that's not me man

Zaid Alyafeai

19:57

@BalarkaSen, so how are you doing ?

Balarka Sen

Alright, more or less.

Liad

@Daminark why?

Ted Shifrin

No, Demonark. It's right.

Liad

@TedShifrin ain't it trivial? $f(t,0) = 0$ for all $t$

Ted Shifrin

Yup.

Daminark

19:59

I said nothing

Balarka Sen

@MikeMiller Is this the string theorist Sen, or someone else?

Liad

@TedShifrin i thought at first that i need to look at $f'$ with respect to $x$ and see that it exists at 0,0. what's wrong with that?

Ted Shifrin

You need to use the limit defn

Liad

@Daminark yeah. you said nothing :-)

Eric

Hi folk

Daminark

20:01

Yo Eric

Ted Shifrin

Yo.

Eric

Did you get your announcement

@daminark

Daminark

Yeah, Schlag said he'd poll us on Wednesday about whether we wanted to do the bootcamp

Eric

Very cool

Daminark

Unfortunately, he said that people who aren't citizens/residents would have to attend without pay

Because NSF

Eric

20:02

Ah

Yeah I see

Daminark

I know at least two people who are getting messed up by this fact

Eric

I'm excited to know in what capacity he expects my group to instruct you guys

Daminark

But yeah, starting July 5th, so we could actually do the summer school in analysis

That'd be pretty cool

You're prob gonna do a lot of the diffgeo :P

Eric

I'd be pretty down to hold some lectures on differential geometry stuff

Daminark

Also lol now that we're at the end of the quarter, Schlag's starting to speed up a little bit

Nearly as fast as Soug sometimes

He's trying real hard to reach spectral theorem so today he blasted through things pretty quickly, and also skipped/sketched some results

Told us to carefully read Kolmogorov-Fomin

s.harp

20:07

what spectral theorem are you going for?

Daminark

Compact operators on separable Hilbert spaces

s.harp

With a friend of mine I am working through von Neumanns book on QM, where he does it for unbounded operators

Balarka Sen

lol

s.harp

its pretty fun, but the language he uses is horrible

(and sometimes he says things which are not right)

Daminark

Eek

Zach Hauk

20:13

Good news

Wait, bad news

I don't know.

Krijn

@BalarkaSen How good is your algebraic geometry?

Mike Miller

@Balarka I dunno

Zach Hauk

II still need to convince my parents to let me apply

Balarka Sen

@Krijn I know a thing or two about varieties.

Krijn

@BalarkaSen In the language of schemes and sheafes?

Balarka Sen

20:14

Nah

Zach Hauk

But, two things.

1. It's more likely I'll be accepted if I apply this year and don't go, even though I'll still have to apply
2. I can have Ted write a recommendation (if he wants)

Sad part:

I was busy doing other things (and waiting for her to reply) that I didn't get the essay OR the problems done besides what I already had

Balarka Sen

that's why you do the essential stuff before fretting over unimportant things

Liad

@TedShifrin about the function i asked you. if i want to find $df/dx$ for a point which is not $(0,0)$. i just make the derivative with respect to x, and fix y. what is the explanation that i can do that?

CausingUnderflowsEverywhere

hey @ZachHauk Im going to fail the test I have today

Zach Hauk

I'm not going to make the application deadline

CausingUnderflowsEverywhere

20:25

how young are you?

Zach Hauk

13

Sha Vuklia

Hi guys. My books says that a homogeneous function is completely determined by its "restriction" to/on $S^{n-1}=\{x\in\mathbb R^n\mid \mid x\mid=1\}$

CausingUnderflowsEverywhere

oh lovely

Zach Hauk

what are you having issues with?

CausingUnderflowsEverywhere

what level math are you taking right now?

Sha Vuklia

20:26

but what do they exactly mean by restriction?

because they show the following:

Zach Hauk

i'm in geometry because my teachers won't let me skip any levels

besides the one i already sklipped

Daminark

@Sha Restriction is of domain

Sha Vuklia

Daminark

So you con consider only what the function does to the unit sphere

Sha Vuklia

ahh

right

yea yea, makes sense! thanks:)

CausingUnderflowsEverywhere

20:27

you should take online math classes at the same time as your day ones if you have

Daminark

And it is true that everything else is determined in the case of a homogeneous function, since everything is a multiple of something on there, and then you can just pull out the constant

Sha Vuklia

yup got it:)

Balarka Sen

The point is R^n - 0 is S^(n-1) x R

(fill in with spheres centered at origin, of varying radius)

CausingUnderflowsEverywhere

haha when did I start writing essays was it in the 8th grade..

Krijn

@ShaVuklia Is this Dutch?

Sha Vuklia

20:31

@Krijn yes it is!

omgoodness!

you're Dutch too!:D

CausingUnderflowsEverywhere

@AkivaWeinberger wait so are you saying that the limit of 0.1, 0.001, 0.0001 , ... is 0 because after 0/0 we cant provide any rational numbers that can be smaller than any value in the sequence?

Krijn

Jazeker ja

Sha Vuklia

Hahah, grappig!

Maar je bent al een masterstudent:) Ik zit pas in mijn eerste jaar:(

Zach Hauk

@CausingUnderflowsEverywhere no

Krijn

@ShaVuklia Als ik mocht kiezen zou ik liever nu in mijn eerste jaar zitten

Geniet ervan :)

Astyx

20:33

Rehi chat

Sha Vuklia

@Krijn Haha, ja, ik heb er al twee "eerste jaren" opzitten XD (ben twee keer geswitcht van studie), maar het blijft idd leuk om eerstejaars te zijn :P

Is that a salute @Astyx? XD

ohhh

re-hi XD

Astyx

Yes :)

Sha Vuklia

Long time no chat! How have you been?

Balarka Sen

I'm gonna favorite that comrade matrix question

Astyx

I've had my ups and downs, and you ? :)

Sha Vuklia

20:37

Same! Actually, my entire (last) week seemed like a down, but the weekend and today are an up, so that's great:)

CausingUnderflowsEverywhere

awww what happened? D:

Astyx

For my part, my evenings were downs whereas my days were ups, but this evening I'm feeling good

I'm glad you're feeling better !

Sha Vuklia

Thank you! Same to you ofc:)

My book says that $\{rx\mid r>0\}$ (for $x\in S^{n-1}$) are 'half lines'. I'm assuming I can translate that to ray/half-infinite line. But in this case the line doesn't really have a beginning point (it doesn't start at 0). So are they just clumsy or am I confused about the definition?

I would have thought that this is a half line:

$\{rx\mid r\geq0\}$

Astyx

Well it's half a line minus a point

But when you come to think of it, if one half has the origin, the other half does not, right ?

Sha Vuklia

that is true :P

Balarka Sen

20:45

Yeah, they are actually (0, infty) not [0, infty)

But it's not technically a line, since it's not a vector subspace of R^n. "half-line" is the best word I'd use to describe it too

s.harp

Has anybody here played a bit with fractional derivatives?

Liad

i want to show that the function $f(x,y) = \dfrac{xy(x \ ^ 2 - y \ ^ 2)}{x \ ^ 2 + y \ ^ 2}$ for $(x,y) \in R \ ^ 2 -\{0\}$ and $f(0,0) = 0$ is continuously differentiable. i showed that $\dfrac{df}{dx}(0,0) = \dfrac{df}{dy}(0,0) = 0$. now at every other point i also have an expresssion for $df/dx $ and $df/dy$. now to show that $f \in C \ ^ 1$. i need to show that $Df$ is continuous at $(0,0)$ correct ?

Krijn

@BalarkaSen Ray, maybe?

Balarka Sen

Ray usually has an end.

@Liad you first need to show Df exists

Liad

@BalarkaSen as i said, i have $df/dx$and $df/y$ at each point

Balarka Sen

20:50

Doesn't suffice.

Liad

they need to be continous

Balarka Sen

Yes.

Liad

then i will get that f is continuously differentiable ?

if i have for example $df/dx$. how do i show it is continuous at $(0,0)$ ?

Balarka Sen

And in particular that f is differentiable, sure.

Liad

i have an expression for $df/dx(x,y)$ i just not sure how to show it is continuous

Ted Shifrin

20:52

@Liad: To respond to the question you asked, definition of partial derivative.

You need to show partials are continuous to show $C^1$.

Liad

yes this is what i thought. but in every point that is not $(0,0)$ this is trivial

Ted Shifrin

At 0 you need some estimates on the partials to show the limit is 0.

Liad

and i also have $df/dx(0,0) = df/dy(0,0) = 0$

now if i want to show that , for example , $df/dx$ is continuous at $(0,0)$

Balarka Sen

The values of the partials at 0 doesn't suffice. You have to show that they are continuous by hand.

Ted Shifrin

You need to show the limit is 0.

Liad

20:56

limit of what?

df/dx is a function of x right?

Balarka Sen

as a function of x and y both.

Ted Shifrin

Of the partial derivative as a function of $(x,y)$.

Liad

but df/dx is a function of one variable , doesnt it?

huh.

Balarka Sen

No!

Ted Shifrin

You keep writing d, but it's $\partial$.

Liad

20:57

ops. i got confused because we derivative it like $y$ is constant :P

so i need to show that $\dfrac{\partial f}{\partial x}(x,y)$ is continuous at $(0,0)$

@TedShifrin how is it now :-)

Ted Shifrin

Good :)

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$Mathematics$ Mathematics