Mathematics - 2018-03-20 (page 1 of 4)

Bob

00:00

I have posted several problems and my answers to then to be checked by the group. They all deal with the same topic. Is that a problem?

Ted Shifrin

Well, Fargle, I'm always happy to chat diff geo with you.

0celo7

@AlessandroCodenotti The undergrads here don't know what PDEs are.

They also don't know that there are seminars.

Alessandro Codenotti

Here we have an undegrad PDE course so everybody can dislike them

0celo7

We do too, but what you get from that class has little to do with PDEs.

Ted Shifrin

We had an undergrad PDE course at UGA, but it was mostly engineering-style and often badly taught.

I never taught that, although I taught a lot of the material in the year-long applied math class I taught in 1986-7.

Bob

00:01

PDE can be very important in engineering

Ted Shifrin

they're important in tons of things, including mathematics.

Fargle

@Ted I don't know how I missed this in earlier reads, but I like how the Frenet frame kinda just builds itself. It almost feels like cheating. That constant length iff orthogonal to derivative theorem is insanely powerful once you see you can just normalize stuff.

Alessandro Codenotti

True, our undergrad PDE course wasn't very good, mostly because we don't have the prerequisits to actually do a PDE course

I skimmed the PDE part of Brezis when studying functional analysis, that looked much more interesting but still not my cup of tea :P

0celo7

I think that's the worst part of Brezis, and I'm a PDE person.

Daminark

In my functional class I liked Hille-Yosida

Alessandro Codenotti

00:04

Interesting, I really liked the first 5 or 6 chapters (and I tried quite a few texts before deciding to study from Brezis)

0celo7

I still don't understand the dedication of the book. Something about viewing PDEs through the eyes of a functional analyst, but the treatment uses virtually no functional analysis.

Bob

what is functional analysis?

0celo7

@AlessandroCodenotti Brezis and Yosida are my go-to books for functional analysis.

@Bob The theory of infinite-dimensional topological vector spaces.

Daminark

Sobolev spaces were a bit iffy though, we didn't talk weak derivatives and instead described them as the completion of the compactly supported smooth functions, but then it wasn't clear when things were defined everywhere

Alessandro Codenotti

That's the second recommendation for Yosida in two minutes, I'll see if our library has a copy

Daminark

00:08

(e.g. Laplacian is unbounded so we weren't sure how to extend it from smooth functions to H^2)

0celo7

@Daminark You extend it as a map $H^2\to L^2$.

Then it's bounded.

Alessandro Codenotti

@Dami you have a new quarter starting soon, right?

Daminark

Stuff like that made it more confusing, also it didn't feel like functional as much as a long drawn out example that used functional at the end. So yeah I dunno

@Alessandro yeah I've got this week of break and then spring quarter starts

0celo7

@Daminark That's what people in geometry do because it's easy and defining weak derivatives on manifolds is a pain in the ass. A careful student then has to spend a lot of time checking that everything makes sense.

It's definitely an issue when working with vector bundles.

Ted Shifrin

@Fargle: Yes, the orthogonal group is very powerful. (I was fixing a martini and getting some appetizers ...)

0celo7

00:10

@AlessandroCodenotti It's on Springer.

Alessandro Codenotti

Which courses are you going to take? @Dami

Daminark

Right now the plan is complex analysis, algebraic topology, Galois theory, and combinatorics

Ted Shifrin

I'm glad to see you taking things besides mathematics.

Daminark

Very much so

Fargle

sweats

Alessandro Codenotti

00:12

@TedShifrin that's not a nice thing to say about combinatorics

Fargle

Sorry, that was just a really hot take.

Ted Shifrin

Huh? @Fargle

Fargle

Never mind.

Ted Shifrin

@Alessandro: I appreciate your sarcastic retort to my sarcasm.

Alessandro Codenotti

:P

0celo7

00:14

The oldest trick in the book -- don't define what you're talking about so any claims you make cannot be falsified.

Alessandro Codenotti

I'm actually going to sleep now though, bye!

Ted Shifrin

That's how I taught for 40 years, 0celo.

Night, @Alessandro.

0celo7

It's how certain geometers write.

Daminark

But yeah this quarter is the first where I'm doing this. We'll see how it goes. If it's too much then later I'll modify the strategy.

Ted Shifrin

I was being sarcastic, yet again.

Demonark: My personal advice is to take advantage of your last bit of undergraduateness before becoming a grad student.

Demonark: Grad complex?

Daminark

00:16

I was considering linguistics but one section conflicts with Galois and the other with complex. That and econ are probably the main non-math things I'm interested in trying

Yeah grad

Ted Shifrin

Probably a bit math-heavy, then.

0celo7

@TedShifrin Did you get the feeling at UGA that seminar/colloquium attendance by grad students was low? Alternatively, that there was a lack of curiosity?

Daminark

Perhaps. If I were to let go of anything it'd be AT since that's likely auditable (I do wanna take grad AT next fall so I'm gonna have to learn the undergrad stuff somehow by then)

Ted Shifrin

Colloquium attendance was generally pitiful on the parts of both faculty and grad students. On the other hand, when I was a Berkeley grad student, I think I missed a total of 5 colloquia (when I wasn't traveling), and I was one of the only grad students (out of 400) to attend regularly.

Demonark: Really? Don't they teach grad AT for incoming grad students who don't know anything from undergrad?

0celo7

400 grad students??

Ted Shifrin

00:19

Yes, Berkeley has now shrunk. Down around 200 PhD students, I think.

The attrition rate was humongous in the 70's.

Daminark

Uh, depends on who teaches. Danny Calegari taught last year and he said he was gonna go very quickly and assume that people have had undergrad AT

Ted Shifrin

That's not fair to all the first-years from elsewhere (which is most of 'em), Demonark.

What's in undergrad?

Daminark

Depends a lot on who teaches, course description says "Topics include the fundamental group of a space; Van Kampen's theorem; covering spaces and groups of covering transformation; existence of universal covering spaces built up out of cells; and theorems of Gauss, Brouwer, and Borsuk-Ulam."

0celo7

What is the Gauss theorem

Daminark

But this rarely is what the actual class looks like I think

Ted Shifrin

00:22

Definitely not fair to assume all beginning grad students have had fundamental group and covering spaces, van Kampen, etc. Ridiculous.

Daminark

And I'm not sure exactly what background the grad students have. I think you can still follow even Danny's class without prior background, you'll just have to work damn hard. Also next year Weinberger is teaching and I'm not sure what he assumes

Ted Shifrin

Well, in principle, one can learn homology and cohomology to a large extent without knowing the fundamental group, but that's not appropriate for, nor fair to, the grad students.

Daminark

Maybe they only take students who know some basic AT?

I'm not really sure. I'm guessing that not everyone has the background but they just make it work very quickly

Ted Shifrin

Demonark, that would be ludicrous, although I'm sure some of the grad students have in fact taken AT as undergrads.

Daminark

Maybe. Also I'm not sure if he necessarily assumes what's in the course description, because that's not really a good representation of the class. Someone took it with Peter a year ago and then grad with Danny

She said undergrad covered 85% of grad but that the latter was more in depth, so while he did go through all of Hatcher, start to finish, it was hard to absorb it in its full depth if it was your first time, but possible if you devote a ton of time

Whatever the system is, it seems to work out alright, and it's one in place, so you just have to play the cards you've been dealt

Tanuj

01:00

Guys, can a parabola have a center (point where a chord is bisected)?

On partially differentiating the general equation of parabola wrt x and then y and then solving them together gets me to the conclusion that a parabola cannot have a center. Is it true? If yes then how?

Maneesh Narayanan

01:14

How do you define centre as a point where chord bisected taken?

consider the circle.

Do you able to get the centre of the circle according to the definition?

@Tanuj

Niing

Is the reverse of Symmetry of second derivatives, or Schwarz's theorem, also correct?

Larson calculus p.895

Tanuj

01:50

@ManeeshNarayanan Well it's not the other way round. Every chord is not bisected at the center, but only those that pass through it.

@Secret Hi! Help needed.

Secret

I need a picture to see which chords are we talking about

Niing

if $f_{xy} \not= f_{yx}$ then $f$ will have two value at a point?

orbit-stabilizer

If $f$ is a function, it cannot map one thing in the domain to two things in the co-domain.

Niing

Yes, so it will be a surface which has vertical line, so it's not even a function.

I'm imagining the picture of the function when the theorem is false.

Like you've said, if the theorem were false then it will lead to a contradiction.

Secret

02:10

Is Schwartz theorem an iff?

If not, what is an example of f that has commuting mixed derivatives but is discontinuous?

Tanuj

@Secret well, I don't know if they even exist, so I can't draw the picture, but tell me this, can a parabola have a center?

Secret

@Tanuj I don't know what the centre will be like if you don't tell me what the chords on the parabola like

Because I will expect all horizontal chords to be bisected by the line of symmetry of the parabola

Niing

it's not an iff, I was just imaging the case when the conclusion is false...

Tanuj

@Secret nvm leave it. I don't think there is a center for parabola.

orbit-stabilizer

math.tamu.edu/~tvogel/gallery/node18.html

Secret

02:28

No, I am thinking about the counterexample to the converse of Schwartz theorem, where a function where mixed partial diffs are equal but they are discontinuous on the open disk

Niing

03:07

it's impossible...?

since the those diffs exist means they're continuous?

orbit-stabilizer

No

Eric Silva

@TedShifrin they go through the same material the grad course is just harder and faster, so it works as a first course I think just a hard one

Danny definitely didn't assume we knew what the fundamental group was day one but I guess the pace would've probably been much harder without kind of already knowing what was going on

Secret

what's an example of a function with equal but discontinuous mixed partial derivatives?

Niing

03:32

So a function $f(x,y)$ both diffs exist and the same doesn't promise that the function is continuous?

Secret

response does not exist

Niing

So I should not ask this question?

Is this a bad question?

Secret

No its just the chat is currently too empty and nobody is responding to our questions

user76284

03:47

Let $S$ generate a group $G$. Then $x \in G$ is in the center of $G$ if and only if it commutes with all the elements of $S$, right?

Secret

Room is currently empty, responses does not exist, no questions will be answered

Niing

Life is too short to be rigorous

Secret

> Don't turn the room into a graveyard of deleted messages.

lol

Consider the following:

Define a lugotheropen clastopheromen be the defino $X: Y \to Y{X \to E^{e^x}}$ . Now let westerine $W$ be the squaoshmorin $s = \int _S^{s} s^{s^{gib}}df^{dsdt}$ such that $G = M\circ s_kc$. Now, it follows that the qwe of the 45768 is casnsurate to the following

$$\int^{\int^{\int}}D \otimes S^{♥}ghj (s\cdot edt)d\tau_{\int sa \lim_{a\to z} dr}$$

Niing

I can prove it

Use horseshoe

I'm giving you the link about horseshoe

wait a minute

An evening in the life of a typical math PhD student

►

@Secret: Take a look

Secret

04:06

You cannot use the horseshoe lemma because the above pseudointegral is irreflexive :P

also typo:

$X: Y \to Y^{X \to E^{e^x}}$

Let $S$ be a sextent. The the following holds:

$S^{S^{S(ubetj \mathop{E}(Putin)+\text{Rus}\frac{@}{3})}}$

Niing

maybe you can try differentiation, when fail, try twice or more

Secret

ok, let's see :P

$$\frac{d}{d\tau}\int^{\int^{\int}}D \otimes S^{♥}ghj (s\cdot edt)d\tau_{\int sa \lim_{a\to z} dr} = \int^{\int^{|}} D \otimes S^{♥}ghj (s\cdot edt)d\tau_{\int sa \lim_{a\to z} dr-1}$$

$$\frac{d^2}{d\tau^2}\int^{\int^{\int}}D \otimes S^{♥}ghj (s\cdot edt)d\tau_{\int sa \lim_{a\to z} dr} = \frac{d}{d\tau}\int^{\int^{|}} D \otimes S^{♥}ghj (s\cdot edt)d\tau_{\int sa \lim_{a\to z} dr-1} = \int^{|} D \otimes S^{♥}ghj (s\cdot edt)d\tau_{\int sa \lim_{a\to z} dr-2ux}$$

Akiva Weinberger

\heartsuit $\heartsuit$

Secret

Ah finally, we are bored to death here

Akiva Weinberger

\varheartsuit $\varheartsuit$

Aw dangit

Niing

04:21

I'm sad, but they look beautiful.

And still, they're easier than Riemann Hypothesis.

Secret

O btw, we did formalise "$dx^{dx}$" somewhat so "$\int^{\int^{\int}}$" is not very far fetched

mowwwalker

Heya, I'm a comp sci grad looking to start studying math recreationally starting from the top and trying to understand everything intuitively. I'm planning to start with geometry / trigonometry. Anyone have any recommendations for resources?

Secret

I knew nothing about reference materials, I pretty much learnt most of it from the chat

Secret

$$\int^{\int^{\int}}d\tau = \int ^{e^{\int \ln \int}}d\tau = e^{\int \ln (\int \ln \int)} d\tau$$

Niing

@Secret: by the way, what do you meant by the Figure?:????

Akiva Weinberger

04:29

Eli Maor has a book called Trigonometric Delights

I don't remember if it's any good though

@mowwwalker

Secret

@Niing o that. That's just some worldbuilding stuff taken from h bar

also typo:

$$\int^{\int^{\int}}d\tau = \int ^{e^{\int \ln \int}}d\tau = e^{e^{\int \ln \int} \ln \int} d\tau$$

and then we can expand $\ln (id+\int)$ with the Taylor series (too lazy to write down), integrate that, expand the result in Taylor series again, and then integrate that, and then expand the exponential twice.

The result, I believe, is an infinite series of integral operators which has the general form:

$$\sum_{k=0}^{\infty} a_k \prod_{\ell=0}^{\infty}(\int^{(-k)})^{\ell}$$

which I think convergence will be a big issue

Actually on a related note, does:

$$\lim_{n\to \infty}\int \underbrace{\cdots}_{\text{n}} \int f(x) dx^n$$ always converge for all $f$ integrable functions?

We knew it does not for e.g. $\sin x, \cos x$, and it converges for $e^x$

A more interesting question will be. Is the set of all integrable function always closed under $\int$ ?

Recall an integrable function is something where $\int f(x) dx < \infty$ for all $f$

we also have complications like $x^m$ for integers $m > 0$ because its iterated integral can grow without bound, yet each member is a polynomial and hence integrable

Semiclassical

If I want the Gamma function $\Gamma(x)$, I do $\int e^{-t}t^{x-1}\,dt$ with an appropriate integration contour

0celo7

@Semiclassical preach

Secret

04:45

@0celo7 0celo, are the counterexamples to the converse of Schwartz theorem, i.e. a continuous function with equal mixed derivatives $f_{xy}=f_{yx}$ but $f_{xy}$, $f_{yx}$ are discontinuous?

Semiclassical

If I sub $t=xe^z$, this becomes $x^x \int e^{-x(e^z-z)}\,dx$

With saddle points at $2\pi i \mathbb{Z}$

0celo7

@Secret Presumably, yes

Semiclassical

So if for some perverse reason I wanted to have $ze^z$ inside that exponent instead

Secret

Semi: You seemed to want to do some Lambert W stuff, hmm...

Semiclassical

Yuuup

I’m trying to see the algebra on mobile

If I back-sub to get that modified integral in terms of t, that’s...

$(e^{-t})^z=(e^{z})^{-t}=(x/t)^t$ instead of $e^{-t}$

Secret

04:55

hmm... $ze^z = e^{-t} \implies \ln z + z = -t \implies -\ln z - z = t \implies (-\frac{1}{z} - 1)dz = dt$, ok that does not work as $t^{x-1}$ becomes umanagable... hmm...

Semiclassical

So $\int (x/t)^t t^{x-1}\,dt$ I think?

Which, oof

Secret

$\frac{1}{t^t}$ does not looks like a nice thing to integrate. In fact, its special function has no name as far I remember

Semiclassical

That just looks grotesque

Yeah. Do not want

0celo7

tfw you discover federer has elliptic regularity too

I think this book contains all math

Silent

how to write bold vector in math.se sites?

orbit-stabilizer

05:09

what book

Secret

what happens if $e^{-t}t^{x-1} = ze^z \implies e^{-\frac{t}{x-1}}t = (ze^z)^{\frac{1}{x-1}} \implies e^{-\frac{t}{x-1}}(-\frac{t}{x-1}) = (-\frac{1}{x-1})(ze^z)^{\frac{1}{x-1}} \implies -\frac{t}{x-1} = W((-\frac{1}{x-1})(ze^z)^{\frac{1}{x-1}})$

ok that's nasty, nvm

Silent

Suppose $C\subset \Bbb R^k$ is closed in $\Bbb R^k$ and let $\pmb z\in \Bbb R^k$. If $F=\pmb z-C$ where $\pmb z -C$ is the set of all $\pmb{z-y}$ with $\pmb y\in C$. How to show that $F$ is closed?

orbit-stabilizer

So, basically a translation.

Silent

05:35

Is this argument correct for my above question? Since $\pmb y\to \pmb {z-y}$ and $\pmb {z-y}\to \pmb y$ continuous, we get a homeomorphism, hence $F$ closed too. @Secret

Secret

I think it looks fine, though I am still very shaky on my background on continuous functions in topology

in particular, I am not sure how to show that the preimage of z-y (which I suspect to be closed in the usual topology) is also closed

Silent

ok

Zee

05:54

Whose here

Secret

No one only none persons

Zee

All persons are none persons

Niing

07:35

Can a single point be called a region? If it is, the point is interior or boundary point?

Niing

08:03

It's not a interior point since every $\delta-neighborhood$ will contain points outside. And it's not a boundary point, since there is no interior point so it cannot contain points inside and points outside

Alessandro Codenotti

That's not how boundary points are defined

hungryWolf

08:41

@Secret /o

o/

sorry :p

Guys if I have a stackoverflow account wih say 5000, will I get 100 if I signup for math.stackexchange?

Tobias Kildetoft

@hungryWolf Yes

hungryWolf

@TobiasKildetoft thanks for the info man

Secret

09:31

Fun fact (not): A coliation does not work because it is incomplete, not dense ans does not converge to all points of view

Consider the following:

Take a piece of Tol, wrap it into a Dyson cloth

Compute its generaloicseries which is given as follows:

$$\int^{\int^{\int}} e^{x^{\heartsuit}}gskfdx$$

where the tetration integral is defined as follows:

$\int^{\int^{\int}}=e^{e^{\int \ln \int}\ln \int}$

abenthy

If $F$ is a number field and $p$ a prime in $\mathbb{Q}$, are there any embeddings of $F$ into the $p$-adic completion $\mathbb{Q}_p$ and if so, how many?

Secret

In particular, the chat is currently empty for the above question to receive the answer it deserves

TheNotMe

09:55

How do I show that $\epsilon_n = \frac{1}{2} - (\frac{1}{2})^n \sum_{k=0}^n \binom{n}{k}max\{\frac{k}{n},1-\frac{k}{n}\} \approx \frac{-1}{\sqrt{2\pi n}}$?

hungryWolf

10:10

my friend has created a chatroom, but it does not look like this one

it is more kind of orange

can I post the link here?

Shobhit

10:29

On wikipedia, under the topic second countable space, its given "lower limit topology on the real line is not metrizable". We have not studied metrizability yet, but i found its definition. I wanted to know how can we know if a space is metrizable or not?

Secret

I don't think that question has a first principle answer:

Metrization theorem

In topology and related areas of mathematics, a metrizable space is a topological space that is homeomorphic to a metric space. That is, a topological space ( X , T ) {\displaystyle (X,{\mathcal {T}})} is said to be metrizable if there is a metric d : X × X → [ 0 , ∞ ) {\displaystyle d\colon X\times X\to [0,\infty )} such that the topology induced by d is ...

though it seems second countable is a requirement for metrizablility

Shobhit

this is what i got when i googled, i have been reading this. But there are many technical terms, separable, second countable,..., so i am stuck in a web. Tbh, these are very cool. @Secret

TheNotMe

10:46

asking again

50 mins ago, by TheNotMe

How do I show that $\epsilon_n = \frac{1}{2} - (\frac{1}{2})^n \sum_{k=0}^n \binom{n}{k}max\{\frac{k}{n},1-\frac{k}{n}\} \approx \frac{-1}{\sqrt{2\pi n}}$?

Secret

@Shobhit yeah, you just have to go slow, topology has a lot of terms wrapped in them. But if you are clear what limit points and open sets are, then you are halfway there

for me, I like to find the bases directly, which is one reason why I spend a lot of time in uncountable stuff

Shobhit

how do you find the basis directly? @Secret

Secret

Well, Leaky trained myself somewhat. Basically say for the lower limit topology, you knew what the open sets are, you then try to find a family of these open sets such that they generate the topology For example $\{[x,\infty)\}$ will give open sets that are half open intervals of the form $[a,\infty)$ or a disjoint union of them

Feb 26 '17 at 8:03, by Secret

@DHMO $(-\infty,a)\cup [a,\infty) =\Bbb{R}$ is clopen. $(a,b)\cup [b,-\infty)=(a,\infty)$ is open. $[a,b] \cup (b,\infty)=[a,\infty)$ is closed.

Shobhit

i thought you had a general method :P @Secret

Secret

I just try to identify the different kinds of open sets in the topology ,and then attack from there

Shobhit

10:57

yes, thats basically it

Secret

For example $[a,\infty)$ and $[b,\infty)$ are of the same kind

maybe I am thinking in terms of equivalence classes of open sets in a topology?

Btw, my PhD started just when I reached the chapter on function continuity in munkres, thus I won't say I am very confident on function continuity proofs

Shobhit

you were helpful, ty

Silent

Suppose $0<x<1$. Let $k$ is any integer, and $n$ any positive integer. Fix $k,n$. If $q$ is such integer that $qn\le k<(q+1)n$, then how to show that there exists positive integer $p$ such that $k<px+qn<k+1$?

Silent

11:32

@AkivaWeinberger, will you please look at this (above post)?

Shobhit

11:43

@Silent first see that $k$ and $q$ must have same signs ($q \ne 1$), then from the second inequality $k = qn + \alpha$ where $\alpha$ is $0,1,..,n-1$ (for the case $k>0$), substitute this in the inequality which you want to prove, you will end up with $\alpha < px < \alpha +1$ so now you just need to prove that there exists $p$ for all these $\alpha$ 's such that this inequality is true. I leave that to you.

This is just over my head, i have exam tommorow, you can try if this works, until someone smarter comes along.

Dattier

12:05

@Lozansky : the path you choose is very difficult for me so here with an another path is more easy : fr.wikipedia.org/wiki/…

but there exists an intersting hint, maybe that can give the solution

with Rodrigues formula

$R(x)$ polynomial then $D^n[x\times R(x)]=xD^n[R(x)]+nD^{n-1}[R(x)]$

@Lozansky : ok ?

A topology enigma :

$\mathbb T$ all the connex by arc of $\mathbb R^2$ and, with $\forall A\in\mathbb T,\forall a\in A,A-\{a\}$ isn't convex by arc.

Is it true that $card(\mathbb E)=card(\mathbb R^{\mathbb R})$ ?

Silent

12:21

@Shobhit Thanks for help. Best wishes for exam.

Shobhit

@Silent ty :)

Secret

what is "convex by arc", convex curve?

Dattier

sorry it's a mistake, "connex by arc"

Silent

@Shobhit I am sorry, but can't show that there is $p$ such that $\alpha < px < \alpha +1$.

Shobhit

12:43

@Silent this inequality is equivalent to $p \in (\frac{\alpha}{x}, \frac{\alpha +1}{x})$, so all you need to show is that there is always an integer in this interval. Can you do now?

Silent

@Shobhit No! suppose $(1/5,2/5)$ does not have an integer in between them, can't do!

Shobhit

you are forgetting the range of x @Silent

Silent

@Shobhit Oh! $p$ exists because $\frac{\alpha} x<\frac{\alpha} x+1<\frac{\alpha} x+\frac1x$, because $0<x<1$, right?

Thank yo.

Shobhit

@Silent perfect. Np :)

@Silent I think you should argue that since the length of the interval in which $p$ belongs is $\frac{1}{x} >1$ so it contains an integer.

Silent

Ok!, i will do that.

@Shobhit

Lozansky

12:58

@Dattier Yes, that's what I got so far, namely $(n+1)P_{n+1} = \dfrac{n+1}{2^n n!}[xD^n((x^2-1)^n)+nD^{n-1}((x^2-1)^n)]$. So you see it's that last term that needs fixing

Dattier

@Lozansky : the path you choose is difficult, do without the Rodrigues fromula

like here : fr.wikipedia.org/wiki/…

Lozansky

Via fonction génératrice?

Dattier

yes

in this link the solution is give

do you speak french ?

@Lozansky

Akiva Weinberger

@Vrouvrou speaks French (@Dattier)

Secret

55 mins ago, by Dattier

$\mathbb T$ all the connex by arc of $\mathbb R^2$ and, with $\forall A\in\mathbb T,\forall a\in A,A-\{a\}$ isn't convex by arc.

Lozansky

13:10

@Dattier Très peu

Secret

$\Bbb{T}$ is the topology where all open sets are connected curves in $\Bbb{R}^2$ except for a point?

Dattier

@Lozansky : it does not matter, did you understand the proposed solution?

@Secret : no, the topology is fixed it's the canonical one of R^2

Secret

ok, usual topology of \Bbb{R}^2$ and you are considering all the connected components on it except for a point?

Lozansky

@Dattier Yes, I got it. But without a GF, it would be difficult no?

Dattier

@Secret : no, T is a set of part A connex by arc of R^2, and A-{a} is not connex by arc (forall a in A)

@Lozansky : GF ?

Lozansky

13:19

@Dattier Generating function

Dattier

ok

Maybe you can do that with the definition of Legendre Polynomial

@Lozansky

Show by induction : $\int_{-1}^1 P_m(x)P_n(x)dx=\frac{2}{2n+1}\delta_{m,n}$

Secret

this set is so weird

o wait what I am doing, a curve in $\Bbb{R}^2$ already fits the criteria of $A$

Dattier

@Secret : this set contains the open segment so card(T)\geq card(R)

@Secret an open curve. (for exemple [0,1]\times {0} is not in T)

but ]0,1[\times {0} is in T

Secret

Yeah, since $A$ is connected, but $A-\{a\}$ is not for all $a \in A$, it follows $A$ has to be an open curve in $\Bbb{R}^2$

So I think $T$ is like this:

So that means, we are counting the number of open curves in $\Bbb{R}^2$

(computing...)

Dattier

I think he misses you a lot

what about the open cross ?

@Secret

Secret

13:34

ah right, cause A-{a} only need to be disconnected, not disconnected with 2 components

So T has to be all star shaped sets in $\Bbb{R}^2$ I think...?

(no, wrong terminology, let me think again)

Air Conditioner

Lately, I've seen some posts that aren't in English, and they almost immediately get comments saying something along the lines of "We use English only here". The top answers to this meta post (from 2011) seem to mostly agree that posting in a different language is okay. Has something changed in the attitude toward these questions in recent years?

ÍgjøgnumMeg

@AirConditioner I don't think so, there are just a lot of comments from people who never look at meta

Dattier

It's a good news (for me, my english is very bad)

Silent

Let $A\subset B\subset G$, where $G$ is group and $A$ nonempty, then centralizers satisfy this relation: $C_G(B)\subset C_G(A)$. Can something be said about relation between normalizers $N_G(A)$ and $N_G(B)?$

Akiva Weinberger

*It's good news

Dattier

13:42

that's the proof of "my english is very bad" lol

Akiva Weinberger

@Silent Remind me what a normalizer is?

Silent

ok

Secret

T is the set of all star domains and open curves and their union in $\Bbb{R}^2$

Akiva Weinberger

Oh, it's the set of things where $xA=Ax$?

Silent

@AkivaWeinberger $N_G(A)=\{g\in G: gAg^{-1}=A\}$

Dattier

13:45

@Secret : what about the open disc is a star domains, no ?

Secret

but open disk won't become disconnected if you remove any point, is it?

Silent

@AkivaWeinberger yes!

Dattier

@Secret : so T is not the set of all star domains and open curves and their union in $\Bbb{R}^2$

user193319

Dumb question: if $\varphi : E \to \Bbb{R}$ is a simple function that is equal to $0$ almost everywhere on $E$, then isn't $\varphi = 0$ everywhere on $E$?

Akiva Weinberger

In $S_3$, is $(23)$ in $N_{S_3}\{(12),(13)\}$?

Secret

13:47

ugh, how do i describe this shape?

Dattier

@Secret The question is about cardinality, just have enough to answer the question

it is not necessary to describe everything

Akiva Weinberger

I think $N_{S_3}\{(12)\}=\{e,(12)\}$ and $N_{S_3}\{(12),(13)\}=\{e,(23)\}$ @Silent

Secret

Well, T is gonna to have something to do with some union of $\gamma : [0,1] \to \Bbb{R}^2$. That means we need to compute $\text{card}((\Bbb{R}^2)^{[0,1]})$. Let's see...

Akiva Weinberger

$\{(12)\}\subset\{(12),(13)\}$ but $N_G\{(12)\}\not\subseteq,\not\supseteq N_G\{(12),(13)\}$

(No idea what's the best way to notate that)

Neither is a subset of the other

Silent

I think $(23)$ is in $N_{S_3}\{(12),(13)\}$. @AkivaWeinberger

Dattier

13:52

@Secret : I think your idea of working with starred ensembles is good, but you have to restrict yourself or only good together

user193319

Any care to answer the dumb question I posed above? I'm pretty certain it's true.

Silent

Oh, you already gave more info@AkivaWeinberger

Thank u!!!

Akiva Weinberger

@user193319 What about $1_{\Bbb Q}$?

Secret

$\Bbb{R}^2 = \mathfrak{c}$, $[0,1]=\mathfrak{c}$ thus we have $\text{card}(T) = \mathfrak{c}^{\mathfrak{c}} = 2^{\mathfrak{c}} = \text{card}(\Bbb{R}^{\Bbb{R}})$

Akiva Weinberger

I don't know what notation you use; I mean the function that is $1$ on the rationals and $0$ everywhere else @user193319

user193319

13:57

@AkivaWeinberger Hmm...That seems to be a counter-example...I am trying to show that if $\varphi$ is a simple function that is $0$ almost everywhere on $E$, then $\int_E \varphi = 0$.

Dattier

@Secret : why card(T)=c^c ?

Akiva Weinberger

@user193319 Well, $\int_E(1_{\Bbb Q})$ does equal $0$

It equals the Lebeg measure of $\Bbb Q$ which is 0

user193319

@AkivaWeinberger Wait! $1_{\Bbb{Q}}$ isn't $0$ almost everywhere, is it?

It's equal to $1$ almost everywhere, right?

Akiva Weinberger

It's 0 on all the irrationals, which is almost everywhere

It's only 1 on the rationals, which is almost nowhere

Very few numbers are rational

user193319

Whoops...I was thinking things backwards.

I still don't see how to prove that $\int_E \varphi = 0$ in general.

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