Mathematics - 2016-11-23 (page 2 of 7)

Kaj Hansen

05:10

Bonsoir tout le monde

Empty chat. A shame.

Mike Miller

h

Kaj Hansen

A wild Mike appears!

Hey there

Mike Miller

how's it going

Kaj Hansen

I'm doing alright. Just getting settled into my relax-before-bed routine of music and browsing main.

Null

@KajHansen

Kaj Hansen

05:24

Hey @Null

Have you had that exam yet?

Null

its in 3 month or so ;)

4 hours ago, by Null

Let $K$ be a field and $A$ a non empty set.
Show that $V:=map(A,K)$ with the following binary operations is a $K$-vectorspace:

for $f,g\in V$ is defined by: $(f+g)(a)=f(a)+g(a)$ for all $a\in A$
for $r\in K$ and $f\in V$, $r\cdot f$ is defined by: $(r\cdot f)(a)=r\cdot f(a)$ for all $a\in A$

Closure under scalar multiplication.This is given by $K$ being a field, since any element of the image of $f$ will be still in $K$ if multiplied by an element in $K$ by definition of a field.
Therefore a function multiplied by $k\in K$ will still have the same domain and codomain, thus it's in $map(A,K)$.\\

Distributive property

If $u,v$ are functions and $c,d$ are scalars, then the following has to hold for all $a\in A$:

$c(\underbrace{(u+v)(a)}_{\text{some element in $K$}})=cu(a)+cv(a)$

Since $K$ is a field, this has to be true.\\

@KajHansen could you look over the above?

Kaj Hansen

Looks good @Null. You still need to show $g \in V$ such that $f + g = f$ for all $f \in V$ (add. identity) as well as add. inverses

Null

@KajHansen i showed the addition stuff with a similar excercise where K was a ring (i am allowed to point there)

Kaj Hansen

Yeah, the arguments will look pretty similar to that very last line above

Null

not that hard as i expected

Kaj Hansen

05:37

mhm

Null

procrastinated this a little while hehe

Kaj Hansen

The eternal struggle

Null

@KajHansen do you get everything someone talks about in a lecture?

Kaj Hansen

NOPE

I try to follow along

But the majority of my understanding comes from reading and doing the homework problems post-lecture

Reading pre-lecture would probably help to some degree, but I was always bad about that, heh

Null

i only go to a lecture if its on a day i have to give my homework

which is tuesday and wednesday

my impression is, that its way too much

in school we had basicly for every topic 1 month

and topic meant something like $e^x$

i try to go the whole week to the lectures but im just a sleepy person :D

Kaj Hansen

05:48

I have trouble with that personally @Null

I can't stay awake in lecture

Doesn't have anything to do with the lecture itself. In general, I get sleepy if I'm sitting down too long. I'll do homework and such while pacing around a room

Adeek

@KajHansen I don't get too much from lectures if I copy everything that prof is saying. So, I just attend lectures and concentrate without writing anything.

Kaj Hansen

It's really annoying, but I guess that's just the way I am.

Null

@KajHansen maybe sitting in general is the wrong way to visit lectures

Adeek

@KajHansen I learn more though from books than from prof.

Null

maybe math should be taught by running in a hall lol

Kaj Hansen

05:51

@Null, it'd be wayyy better if I could stand up in the back and be able to stretch, pace, and such mid-lecture. But then I'd be that guy

I too learn better by doing than by watching, but watching has some value to me, so I always went to class

@Adeek, I never took notes as an undergrad. I just watched

I agree with you

Null

@KajHansen i find it tedious that a prof has to write his definitions on the board

they should be prerequisite to even visit the lecture

Kaj Hansen

haha @Null

That'd be an interesting way to conduct things.

Mike Miller

Sometimes definitions take time and discussion to parse!

Adeek

hm

Null

if they'd give me their plan i would learn the definitions beforehand. and then he could make interesting applications

Adeek

05:55

I am little bit confused about the following issue from lecture notes.

Null

i visited one lecture that was only definitions.

Adeek

Let H be a hilbert space, $y \in H$ and T be a linear functional on H defined by $T(x) = <x,y>$ for every $x \in H$, then it is easy to see that $||T|| = ||y||$ why ?

so we define $||T|| = sup_{||x|| = 1} ||Tx|| = sup_{||x|| = 1} ||<x,y>||$

why is first of all that $\|T\| \leq ||y||$ ?

Kaj Hansen

I haven't studied this stuff in generality, but this makes sense for vectors in Euclidean space

Adeek

yeah

Kaj Hansen

Due to Cauchy-Schwartz

Is there some analog for general inner products?

Adeek

06:00

hmm

I think there is something like that for hilbert spaces

just a sec let me check lecture notes

Null

cigarettes or beer, thats the question

deostroll

I am trying to understand the history of laplace transform - how the idea came about...

Michael Anderson

1. First show that ||T|| <= y, by considering x=y.
2. Break down z = <z,y> / ||y|| + (z - <z,y> / ||y|| ) and look at its value under T.

That should be ||T|| <= ||y|| of course.

Adeek

but ||x|| = 1 how do you know what ||y|| = 1 ?

Michael Anderson

You can use linearity there, T \lambda z = \lambda T z

deostroll

06:04

From wikipedia, it is said, that euler investigated some integrals as solutions to differential equations...What exactly were those?

Laplace transform

In mathematics the Laplace transform is an integral transform named after its discoverer Pierre-Simon Laplace (/ləˈplɑːs/). It takes a function of a positive real variable t (often time) to a function of a complex variable s (frequency). The Laplace transform is very similar to the Fourier transform. While the Fourier transform of a function is a complex function of a real variable (frequency), the Laplace transform of a function is a complex function of a complex variable. Laplace transforms are usually restricted to functions of t with t > 0. A consequence of this restriction is that the Laplace...

Michael Anderson

Ugh and my breakdown should be
z = ( <z,y> / ||y|| ) y + (z - (<z,y> / ||y||) y )
I think..

Adeek

@MichaelAnderson For the (1) to show $||T|| \leq y$ how can we consider x = y when it is defined norm is taken over the things which have norm one ?

Michael Anderson

Consider x = y / ||y|| then.

With a trivial special case of y=0.

Null

2

Q: $\mathbb R^3$ minus a line is connected.

Let $S\subseteq \mathbb R^3$ be homeomorphic to $\mathbb R$. Prove that $\mathbb R^3 \setminus S$ is connected. I haven't been able to solve this, although my topology skills are pretty weak. My friend told me he managed to prove this using results from his "dimension topology" class. Although I...

Adeek

ok I will try that.

thanks @MichaelAnderson

Kaj Hansen

06:07

I'm looking at that too @Null

It should be possible to show that that space is path-connected

Michael Anderson

T (y/||y||) = <y, y / ||y||> = <y,y> / ||y|| = ||y||

Kaj Hansen

The space itself should be homeomorphic to $\mathbb{R}^3 \setminus \{(x, y, z) \in \mathbb{R}^3 \ | \ y = z = 0 \}$

Null

@KajHansen isn't it the same as saying a zylinder by a paper is still completly connected if you cut it, and the flat paper is left?

(note that i have no clue whatsoever haha)

Adeek

yes @MichaelAnderson

Kaj Hansen

I'm not sure @Null

Adeek

06:11

@MichaelAnderson I guess then we get from that side that $\|y\| \leq \|f\|$

Null

$R^3$ minus a line would look like a cube with a tiny hole in it visually or?

Michael Anderson

Now if z = A y + w
T z = <A y + w, y > = A <y, y> + <w ,y>
We choose w s.t. <w,y> = 0
T z = A <y,y>
Then since ||Tz|| <= ||T|| ||z||
we find that
|A| ||y||^2 = ||T||

Oops hit enter too soon.

Kaj Hansen

an open cube with a hole drilled through @Null

It's obviously path connected

idk how to make the argument rigorous though

Adeek

what is A @MichaelAnderson ?

Michael Anderson

|A| ||y||^2 <= ||T|| ||z||
An I think the final bit comes from the triangle inequality on ||z|| < |A| ||y|| + ||w||

you find A and w by considering the <z,y> and wanting <w,y>=0

Null

06:14

$R^3$ minus a plane would fail tho or?

Kaj Hansen

It would

Adeek

I see @MichaelAnderson

Danu

@TedShifrin No idea... I didn't edit it or anything.

Mike Miller

Keep in mind that the $\Bbb R$ could be wildly embedded.

Danu

Hi @MikeMiller

Kaj Hansen

06:15

That's what I'm trying to reason through @Mike

Michael Anderson

<z,y> = <A y + w, y> = A <y,y> + <w,y> = A <y,y>
So A = <z,y> / <y,y> and w = z - A y

Kaj Hansen

If that's the case, it wouldn't necessarily be the case that $\mathbb{R}^3 \setminus S \cong \mathbb{R}^3 \setminus \{(x, y, z) \in \mathbb{R}^3 \ | \ y = z = 0 \}$ ?

Balarka Sen

It could be a space-filling curve.

Danu

Hi @Balarka

Mike Miller

@BalarkaSen It's embedded...

Balarka Sen

06:16

Hi @Danu

Null

@KajHansen if R^3 minus a line is still connected, then it is conneted minus 2 lines. so R^3 is still connectet minus infitly lines. therefore its connected minus a plane? (just drooling around)

Balarka Sen

@MikeMiller Ah I didn't read the original question.

Mike Miller

@Null Dude, just delete a plane. It's visibly not connected.

Kaj Hansen

Just because a property holds for all $n$ doesn't mean the property holds as $n \rightarrow \infty$

Good example: $D_\infty$ is not the symmetry group of a circle even though inscribed $n$-gons more-and-more closely approximate a circle

Mike Miller

It's a rather good discrete approximation for $O(2)$ though.

Kaj Hansen

06:18

Wait

No

Balarka Sen

That's not disconnected?

Kaj Hansen

I messed up @BalarkaSen

I was thinking of $\mathbb{R}^2$ and the property didn't extend

:(

Balarka Sen

Right.

If $\Bbb R$ is smoothly embedded in $\Bbb R^3$, that the complement is connected is easy enough to prove. (take a path, make it transverse). So perhaps one can cook up a proof which avoids some heavy machinery by smooth approximation, given a randomly embedded R?

I dunno.

Danu

Oh hey @MikeMiller I have a small question that came up during the lecture on symplectic geometry: We're considering the blow-up of a point in $\Bbb CP^2$

The picture I have of this is the following

Mike Miller

@BalarkaSen Nonsense. There are wildly embedded $S^1$s. What makes you think $\Bbb R$s will be better?

@Danu Ask Balarka.

Danu

06:23

So any point $p$ that is not the one you're blowing up uniquely corresponds to a line through $p$ and $p_0$, and by picking two reference lines $L_1,L_2$ you can write such a line as $\lambda_1 Q_1+\lambda_2 Q_2=0$ where $Q_i$ are linear forms defining $L_i$ and $[\lambda_1:\lambda_2]\in \Bbb CP^1$ is unique.

Balarka Sen

@MikeMiller Wildly embedded ones cannot be approximated by smooth embeddings, right?

Danu

So now we have a commutative triangle

I think I understand this for every point in $\widehat{\Bbb CP^2}-E$ where $E$ is the exceptional divisor. But what does $\hat f$ do for $E$?

@MikeMiller Uh.. okay? @BalarkaSen?

Balarka Sen

bar{CP^2} lives in CP^2 x CP^1.

Mike Miller

I know basically nothing about complex geometry.

Balarka Sen

f hat projects to the second copy.

Danu

06:26

@MikeMiller What's going on?

@BalarkaSen Hmm... Why?

I thought about $f$ outside $E$ as just the composition of $f$ with $\beta$.

Balarka Sen

By definition of blowup. It's (point, line) in CP^2 x CP^1 such that point $\in$ line.

Danu

@BalarkaSen Funny, that's not the definition I've seen.

Balarka Sen

What's your definition, @Danu?

Danu

@MikeMiller But I think you know a lot about symplectic geometry---and $\Bbb CP^2$ is Kaehler :D

@BalarkaSen Connected sum with $\overline{\Bbb CP^2}$

But

Perhaps I can think about your definition.

So which point lies in which line? :P

Null

$R^2$ minus a point is still connected then too?

Mike Miller

06:31

@Danu That only makes sense smoothly. It's not a definition that makes sense in complex geometry.

Danu

@MikeMiller Right. This is a course in symplectic geometry.

Null

@KajHansen i think the best example for it is: a set is finite if it has n elements. doesnt hold if n->$\infty$

Mike Miller

I mean, that's not quite a sufficient definition in the symplectic context either, since connected sums don't usually make sense there.

Balarka Sen

@Danu Ok. Fix a point $p_0$ you're supposed to blowup in $\Bbb{CP}^2$. Then consider the elements $(p, \ell) \in \Bbb {CP}^2 \times \Bbb{CP}^1$ such that $\ell$ is a line through $p_0$ in $\Bbb{CP}^2$ (there are $\Bbb{CP}^1$-many of them) containing $p$.

Danu

@MikeMiller Well, they make sense but don't preserve symplectic structure I guess.

Balarka Sen

06:33

This collection of pairs is precisely the blowup.

It's interesting to understand why the one you said is equivalent to this.

Mike Miller

@Danu Then it doesn't make sense symplectically!

Danu

@BalarkaSen Okay. I guess my earlier picture should be helpful with this.

Balarka Sen

Yep.

Danu

@MikeMiller It makes sense for smooth manifolds without extra structure---I'm okay with that for now. We might discuss problems in the symplectic category later.

Mike Miller

Oh. So you literally just say the word blowup in the smooth context.

Danu

06:37

@MikeMiller Well, we did also do the blow-up of a symplectic manifold and showed that it is also symplectic. But via a detour.

Mike Miller

Yes, it makes sense. All I was saying is that the definition of symplectic blowup is not literally "connected sum with $\overline{\Bbb{CP}^2}$"

Danu

@MikeMiller Okay, sorry for my sloppiness.

sloppyness? No.

Mike Miller

i

Danu

Weird-looking word.

deostroll

Hi. While watching a video about galton board, I wondered why is it that all the balls pile up in the middle and not so much towards the ends...

but that video also said, if you take samples of say shoe sizes, or, heights of students in a school...they'd follow the same pattern...

that of a normal distribution curve...

why?

Kaj Hansen

06:44

Clever @Null

Null

@KajHansen not mine actually ;)

what you think of the answers in the thread?

i can only understand the one with the circle, but i dont know if that is valid

Kaj Hansen

I don't know anything about homotopy or homology

Well, other than the definition of "homotopy"

Null

maybe in the future someone drills a hole in the universe to find out :-)

Danu

@BalarkaSen Okay, I think I'm having a hard time with this :P

Balarka Sen

@Danu Can you see the exceptional subvariety in my definition?

Danu

06:59

Hi @TedShifrin!

Ted Shifrin

@Danu: Aren't you supposed to be asleep? ... I ended up clearing all cache and memory for Chrome on my iPad and the link went away.

Danu

@TedShifrin It's 8:00 AM here. Not so bad?

Ted Shifrin

Hi @Balarka @Kaj

Balarka Sen

Hi @TedShifrin.

Kaj Hansen

Hey @Ted

Danu

07:02

@BalarkaSen You didn't specify, but I assume the projection to the first factor should be the blow-down map? Then I guess $E$ is the preimage of $p_0$.

Balarka Sen

@Danu Yes.

Danu

Actually, I can live with this definition. But I don't see why it agrees with mine @BalarkaSen.

Balarka Sen

'Cause there are $\Bbb{CP}^1$'s worth of lines passing through $p_0$. For points away from $p_0$ there's just one, that's why blowdown is isomorphism away from the exceptional divisor.

@Danu I was trying to give you a walkthrough...

Danu

@BalarkaSen Yeah, that's okay.

I guess it's generally okay because $E$ looks the same too.

Balarka Sen

Ok, what's a neighborhood of the exceptional divisor, in my definition?

Danu

07:13

Well... A bunch of $(p,\ell)$ such that $p$ is close to $p_0$ and $\ell$ the unique line through it :P So it's like a nbh of the preimage of $0$ when blowing up $\Bbb C^2$.

Balarka Sen

Well, what does it look like? The exceptional divisor is a $\Bbb{CP}^1$, so a small nbhd of that would be a bundle on it... what's that bundle?

Null

is $\mathbb{C}^2$ even visualisable?

Danu

@BalarkaSen $\mathcal O(-1)$, I know that much already ;)

Mike Miller

you put a C on another C

2

Danu

^ this guy

Balarka Sen

07:16

@Danu I asked whether you can see it from the definition I gave.

If you can, you're done, more or less.

Because then blowup at $p_0$ is exactly obtained by replacing the point $p_0$ by a $\Bbb{CP}^1$ and a neighborhood of $p_0$ by a tautological bundle neighborhood of $\Bbb{CP}^1$.

That's precisely what everything looks like in $\Bbb{CP}^2 \# \overline{\Bbb{CP}^2}$.

Danu

Yeah, I guess I'm okay with it. Thanks.

Balarka Sen

ok

Danu

Simon & Garfunkel - Mrs. Robinson (Audio)

►

Back to grading quantum mechanics homework.

@TedShifrin Wanna help with a bit of complex geometry? :)

Null

07:54

what is a honor in math?

10

Q: Do I still need a PhD to do research if I have double honours in CS and Pure Math?

I'm planning to get double honours in CS and Pure Math. I'm confused what's the biggest benefit of Ph.D. will be for me? People say you learn to do research, but I can do research(as I have the appropriate background, CS and Math) sitting at home or in industry. Then what's the biggest benefit of...

publications phd research mathematics computer-science

Brody

08:20

@Null Sounds like this...

Joint honours degree

A joint honours degree (also known as dual honours, double majors, or Two Subject Moderatorship at the University of Dublin) is a specific type of degree offered generally at the Honours Bachelor's degree level by certain universities in Ireland, the UK, Canada, Malta, and Australia. In a joint honours degree, two subjects are studied concurrently within the timeframe of one honours. == Requirements == A joint honours degree typically requires at least half, often almost all, of the credits required for each of its respective subjects. The two subjects do not have to be highly related; indeed,...

Kaj Hansen

Hey @Andrew

Welcome back Mr. Sen

Null

@NaCl hi =)

Andrew Thompson

Hellu.

Balarka Sen

G'morning, @Kaj

Also, hi @AndrewT, @Brody.

Kaj Hansen

Night time

Brody

08:23

Hi @Balarka and all

Balarka Sen

I realized but I was confused what I should greet with, a good morning or a good night.

Brody

Curious... what's the meaning in Danu's blue name?

Kaj Hansen

haha

Balarka Sen

you mean the significance of him being blue?

Brody

...yep!

Balarka Sen

08:26

"he's just a little guy who lives in a blue world and everything he sees is just blue like him"

he's a moderator.

Brody

Thanks. I also had to look that up

Wow, so that's what that tune is..

Null

any heavy metal fans here?

@Brody hi, did you where yesterday not here?

Brody

@Null I suppose not, can't recall tbh

Alessandro

Kinda, more on the rock side though

Brody

How are you by the way? @Null

Kaj Hansen

08:33

I am

Heavily

Null

@Brody I'm fine as I am done with my homeworks in time. How do you feel?

Brody

@Null Must feel nice! I'm okay

Null

@Brody actually it is a little bit dampered as I did not get something specific our prof said. Well i guess that's how it is. But now I only have to worry how I pass my examen for the next 2 month. I think that's enough time to prepare :)

Brody

@Null Two months sounds great, just don't procrastinate oc ;)

By the multinomial theorem, the expansion of the square of a multinomial with $n$ terms is given by $$\displaystyle (x_1+\cdots + x_n)^2 = \sum_{|\kappa |=2} \left[ \dfrac{2}{\kappa !}\,x^\kappa\right]. $$

I'm wondering if there is a way to index the RHS sum linearly in $i\in\{1,\ldots, I\}$ fashion. The upper bound $I$ would be the $n$th triangular number $n(n+1)/2$.

Danu

09:01

By the way @TedShifrin, that equation $\mathcal O(-Y)^*\cong \mathcal O(Y)\otimes K_X$ I was trying to get is wrong (it *looks* like adjunction, but these are not bundles on $Y\subset X$ but on $X$ itself). Of course it has to be since the map $\operatorname{Div}(X)\to\operatorname{Pic}(X)$ is a homomorphism, so $\mathcal O(Y)^*=\mathcal O(-Y)$ for sure (cc. @BalarkaSen).

The ingredient I was missing was a special case of Serre duality: $H^q(X,E)=H^{n-q}(X,E^*\otimes K_X)$, applied to $E=\Omega^p\otimes \mathcal O(-Y)$. It's tricky because when I see $H^q(X,\Omega^p\otimes E)$ I immediatel…

Brody

09:16

Btw, $\kappa$ denotes any $n$-tuple whose elements are taken from $\{0,1,2\}$, respecting multi-index notation. We are summing over the multi-indices satisfying $|\kappa |=\kappa_1 + \cdots +\kappa_n = 2$, and it's known there are $I=\dfrac{n(n+1)}{2}$ distinct such $\kappa$. I'm just wondering how I'd symbolically write out such a mapping, as I have at least one in mind.

Balarka Sen

It's probably not possible to enumerate the partitions of some number explicitly, although obviously there are lots of such enumerations.

Brody

09:28

Responding to me or someone else @BalarkaSen?

Balarka Sen

It was a response to you, @Brody

Brody

@BalarkaSen Okay. My bad, wasn't certain and really didn't want to assume (lol)

Balarka Sen

It's fine. I should have pinged you.

Brody

lol no worries @Balarka. I was trying to avoid a potentially awkward situation (by being awkward)

Because of the $|\kappa |=2$ restriction, we know our $n$-tuples must contain exactly one $2$ and the rest zeros or exactly two $1$s and the rest zeros. One mapping scheme is to have the first $n$ natural numbers map to the multi-indices where the first position is taken by $1$ or $2$.

Then the next $n-1$ natural numbers map to the multi-indices where the second position is taken by $1$ or $2$, preserving distinction from the first $n$ mappings. Then, the following $n-2$ to the next block of multi-indices, distinct from the previous $n+(n-1)$ ones. And so on. Is there any standard(-ish) way to "encode" this sort of function?

Balarka Sen

Shrug. No idea.

Boni Lindsley

09:47

Isn't that the just the number of ways to add standard basis vectors together?

Brody

In retrospect, I maybe didn't need all the background crap. Could've just asked how to--with some conditions--assign natural numbers to tuples, heh

Fargle

Hello night-chat.

Brody

@BoniLindsley Hmm yeah. In this case, the multi-indices are just $\mathbf{u}+\mathbf{v}$ with $\mathbf{u},\mathbf{v}$ from the standard basis for $\Bbb R^n$

That should help define a function

Hi @Fargle

Fargle

How goes it @Brody?

Brody

@Fargle Okay. Think I'm coming out of a head cold only to sink into something else

Hbu?

Fargle

09:59

@Brody Huh--coming into a head cold, myself. Working through Rudin to mixed success.

Kaj Hansen

What chapter @Fargle ?

Fargle

@KajHansen 2 and 3.

Brody

@Fargle Must be that time of year. And you're doing the analysis text?

Fargle

There are some exercises I still want to do out of chapter 2 but I also want to read ahead.

@Brody Indeed. Working the exercises that seem interesting.

Balarka Sen

Rudin is lots of fun.

Fargle

10:06

I really love the exercises. They're generally really elucidating.

LeGrandDODOM

130 close votes awaiting review... 2 years ago there were 30 at most... what happened ?

Balarka Sen

There's a nice one in there: can you find a perfect set in R which does not contain any rationals?

Fargle

@BalarkaSen Yeah, that one is still screwing me a bit.

Kaj Hansen

That topology chapter is fun

Balarka Sen

Yep, @Kaj. The exercises, mostly, though. The theory bit was too terse for me.

Kaj Hansen

10:09

That's what I meant

Balarka Sen

Gotcha.

Fargle

@BalarkaSen My gut tells me "no", but I'll have to see why.

Balarka Sen

:)

Kaj Hansen

@LeGrandDODOM, more questions being asked?

Balarka Sen

A person told me that it took him years to come up with a proof by his own. And that involved hyperbolic geometry.

Kaj Hansen

10:11

When I joined, there were like 300k questions asked to date

And now there are 700k

Alessandro

That sounds obviously false, so it's probably true :P @Balarka

Fargle

@BalarkaSen That's...heartening.

Kaj Hansen

And I only joined ~3 years ago

Balarka Sen

Which sounds obviously false, @Alessandro?

Alessandro

"There is a perfect set containing no rationals"

Balarka Sen

10:13

I'm not spilling the beans out for you!

Fargle

Good--I don't want it spoiled.

Alessandro

I don't want you to

It was just my first reaction

My intuition is known to be always wrong though

Balarka Sen

Ah, ok. Yeah, it does sound obviously false.

Kaj Hansen

Monte Hall is obviously false

Brody

Apparently, the best heuristic evidence against a conjecture is if @Alessandro thinks it's true.

@KajHansen Didn't Erdos presume it false for a while?

Kaj Hansen

10:17

@Alessandro, simple fix: create a new intuition, say intuition-prime, where every statement $x$ is $\neg y$ for a statement $y$ in current intuition

Yeah @Brody

He didn't believe it until shown experimental results

Balarka Sen

On the other hand, the Jordan curve theorem sounds obviously true

I mean it's true but not obviously so

Kaj Hansen

And it is :)

DHMO

@Alessandro I thought R\Q is a perfect set

Kaj Hansen

lol

Alessandro

That's not closed

Brody

10:19

pssst @Balarka, I say yay to the question-ay

Kaj Hansen

$\mathbb{Q}$ isn't open

DHMO

fair enough

Balarka Sen

@KajHansen On the other hand you could take the 3-dimensional variant of Jordan curve theorem.

That's obviously true, but actually neither obvious nor true

Kaj Hansen

How do you define a closed surface?

Balarka Sen

It just says "an embedded sphere in R^3 disconnects it"

by an embedded sphere I mean a subspace homeomorphic to the sphere.

Kaj Hansen

10:22

IT SHOULD!

Balarka Sen

yep. but...

Kaj Hansen

You see

This is why we need to use ZF, not ZFC

I don't know if this actually uses choice, but it seems like the kind of trickery that arises from choice

Balarka Sen

I actually messed up a bit. It does disconnect; just that one of the components is not a topological ball.

Kaj Hansen

Oh

I'm more okay with that

Balarka Sen

Jordan's theorem does hold in higher dimension. Just not the Jordan-Schoenflies theorem.

I forgot all these. Sigh.

DHMO

10:28

any fun/useful questions on non-standard analysis (infinitesimal)?

@balarka are you familiar?

Balarka Sen

no

Kaj Hansen

Is NS analysis actually useful?

Does it help with anything regular analysis can't?

It's My Turn (1980) Snake Lemma

►

Balarka Sen

hah

i like the dude's objection though

Kaj Hansen

it's NOT well-defined

Brody

Clayburgh was beautiful. Think she had any idea what she was talking about? (not knowing if she has any mathematical background)

Kaj Hansen

10:45

She didn't @Brody. I went to a talk by Benedict Gross from Harvard who recounted coaching her for that part of the movie.

Alessandro

@Balarka my intuition was wrong, as expected, I think I actually know such a set, there's a pretty famous one (after throwing away some rationals)

Kaj Hansen

Cantor is perfect, but it has rationals, if that's what you're thinking of @Alessandro

Balarka Sen

It's plenty famous, yep

Alessandro

I think it still works after throwing them away

Kaj Hansen

Interesting

Brody

10:51

@KajHansen Interesting, thanks. Her performance was pretty convincing imo

Alessandro

It surely stays closed, and I'd be very surprised if some point becomes isolated

Kaj Hansen

It really is

Balarka Sen

goes to mute mode

Alessandro

Actually I'm not that sure thinking twice about it

Lozansky

Will $$ \int_{\delta < |s| < a } f(s) ds$$ be two integrals?

For some $\delta > 0$

Balarka Sen

10:59

Yep.

Lozansky

Thanks :)

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$Mathematics$ Mathematics