I am not really convinced of the "zero set is open" thing, therein lies the muckery

That's actually correct.

I thought homeomorphism preserves open set

It's worth understanding in standard examples why sections are local homeomorphisms.

{0} isn't open

Is it ever 0?

Well, if it's never 0, we're fine, Alessandro.

Indeed, that's why I'm asking :P

@TobiasKildetoft I use quite a lot of universal introduction in my arguments, if you don't mind

Remember that the projection $\pi:\mathscr S \to X$ is a local homeomorphism.

So in the Alexandroff case you can construct a minimal basis by looking at each point $x$ and taking the intersection of open sets containing it, calling that $U_x$

I mean there's no paradox if it isn't 0 ever

@LeakyNun Why would I mind?

I think being able to do something of the sort might just be equivalent to the minimal basis

@TobiasKildetoft no idea

Like

Right, Balarka. But I start with a continuous function that's somewhere 0, just to foil you.

blinks

(I have a question on main with a nice answer asking for reasonable properties ensuring that a space has a minimal basis if you want to look at it after thinking about this problem for a while)

heya Fargle

For sure. I have to understand the "section of the etale space" business better to actually understand why you're claiming the zero set is open though.

> In topology, an Alexandrov topology is a topology in which the intersection of any family of open sets is open.

Like take the open sets which are the smallest open sets around some point

I'm interested.

So ideally the way to close that off is to give a continuous function on a connected space which is zero somewhere but not everywhere

If those form a basis you're good

I just thought I'd throw that out there for you, Balarka.

Oh @Leaky Alexandroff stuff is what I'm gonna do a reading course on this fall and it's dank

But, Semiclassic, what's the flaw in the argument?

@Daminark do you have any example of an Alexandroff space?

Any finite space, trivially

@Semiclassical Sure, the conclusion is most obviously garbage. That's not the thing.

@Daminark otherwise?

Hell if I know.

I'm not giving hints.

I'm just saying that I don't like to call a conclusion like that 'obviously garbage' unless I have an actual counterexample on hand :P

IT's trivial to give zillions of counterexamples.

Surely you can give me a continuous real-valued function on, say, an interval, that is not everywhere 0 or nowhere 0.

Okay so I think there's a direct correspondence between posets and Alexandroff spaces

At least $T_0$ ones

well, $f(x)=x$? :D

well, sure.

Like if you give me an Alexandroff space, do the construction that I did before

though part of the reason I'd want to bring up counterexamples is to figure out why the argument is inapplicable in that case

the zero set is not open

I'm not saying anything until Balarka and Eric think about it.

so this part would fail

For $x\in X$, let $U_x$ be the intersection of open sets containing it

And let $x\le y$ if $x\in U_y$

@Daminark I mean, example of such space

So I think you can reverse that

Hi @Ted

@Leaky: No that, too, is correct. Because I said that the projection map is a local homeomorphism.

So if you give me a poset

And everyone else :3

o i was tabbed out lemme scroll up and see what's going on

Okay that's only a preorder, actually, $T_0$ makes it into an order since $U_x = U_y$ iff $x=y$. So now you have a partial order. Reverse, gimme a partial order

Then define $U_x = \{y:y\le x\}$

And the generated topology is Alexandroff and $T_0$

Yo @Fargle

Salut, GTR.

salut Ted

j'ai déménagé à Saclay c'est assez désert

où ça?

Oh I see the argument for the third line now.

I have got to be more careful of how I take naps, if only because if I sleep in the wrong position it leaves my neck sore for the rest of the day

And that kinda sucks

@TedShifrin sur le plateau de Saclay, là où est situé Polytechnique depuis ~1974. Le gouvernement veut rassembler les grandes écoles et les universités à cet endroit afin de former une sorte de campus unifié (et remonter dans les classements internationaux)

Eek

ohhhhh, je ne connaissais pas ce nom-là ...

someone clue me in

Quand j'étais là, on disait Palaiseau, je crois.

hi MikeM ... clue?

you're arguing something sheafy

Ah, I gave a paradox. Please don't spoil the children's fun :)

Oh, that's it? I assumed some people were just confused. :P

I will quiet.

does anyone know a number that is proven to be rational, but the period of its decimal expansion is still unknown ?

actually i am confused, about how the conclusion on the third line is being made

@GabrielRomon this can't happen.

It's easy to find the period of any rational number

unless you mean computationally unknown

in which case throw at me a rational number with a very big denominator and I probably can't tell you the period of its decimal expansion

if i have a section of the sheaf, the zero set is it's intersection with the zero section

that's still closed to me. i don't get it

But projection maps the zero section locally homeo to the base.

@LeakyNun As an example, if I take the sum of all reciprocal 100th powers of integers, I know that this sum will be $\pi^{100}$ times a rational number.

@Semiclassical oh, in that sense

but if someone asked me what the periodicity of the decimal expansion of that rational was, I'd just shrug my shoulders

but is it computable?

should be.

i.e. is there an algorithm?

lemme ask Mathematica :P

plot twist in mathematics: what if it turns out that $e+\pi \in \Bbb Q$ is independent of ZFC?

numerator: 18919607563824425059045486613898744374540568306613387226677139240862279083039449‌​5422

denominator: 98152054207575147101081780593695534583273922607504040499304079879335823590807672‌​25644716670683512153512547802166033089160919189453125

@Semiclassical then it's known :)

yep

@LeakyNun That would be pretty wicked

Also really sad

Actually is that even possible?

@Daminark prove that it isn't ;)

no u

I find it hard to be anything but childish when it comes to questions about $e+\pi$ tbh

I dunno, you dunno, no one know

@TedShifrin Are you arguing something like, if a section $s$ is so that $s(p) = 0$ at $p \in X$, then the germ of $s$ at $p$ is zero, so $s$ has to be zero on an open set on $X$ around $p$?

Ah ... a key word appeared.

a wild keyword appeared!

"A wild germ appeared!" sounds like something out of a nerdy Pokemon parody

Assume there's a model of ZFC in which $e+\pi\in\mathbb{Q}$

either that or an infectious disease

Then in that model there exist rational numbers $p$ and $q$ such that $e+\pi = \frac{p}{q}$

But then that equality has to hold in ZFC anyway

so?

@TedShifrin I guess the point is that the zero set in the topological context need not be the same as the sheaf context. The global function $f : X \to \Bbb R$ corresponding to $s$ can be zero at $p$ without it's germ being zero at $p$. Is that it?

I mean in ZFC those numbers exist and are equal, so if it's consistent with ZFC that $e+\pi\in\mathbb{Q}$, it's true in ZFC

Yup.

Kewl.

@Daminark so?

It took me 10 minutes to actually understand the argument lol

Well that's it, it's provable in ZFC

Good paradox, thanks Ted!

@Daminark it being true doesn't mean that it's provable...

That's OK.

I mean give me those rational numbers

ok, if it's true then it's provable, and then?

Well, that equality holds in every model of ZFC so now the negation is inconsistent

right.

so?

There's nothing else to say. If it is independent, well it's true and its negation is false. So not independent

oh wait

I see

well, let's say that in our model, $e+\pi=\dfrac12$.

Who said that you can prove it?

(obviously its negation is provable here, since $e>1$ and $\pi>0$...)

I mean maybe generating those rational numbers might be a problem

But once it's true that $e+\pi = \frac{p}{q}$

That equality must hold everywhere anyway

but what if you can't prove that they are equal?

I mean presumably all that matters is its truth

Even if there's no process to show it, once it's true, it's true

@Daminark doesn't mean it's provable...

I mean my point is that provability shouldn't be relevant at all

but independence is about provability right?

If that equality holds, independently of our ability to prove it, in any model, then that equality should hold in all models

Is it though? I thought it was about truth

@Alessandro help

well independence is about consistency

P is independent if ZFC+P is consistent, and ZFC-P is consistent

I mean my point is that if ZFC+P is consistent, that should correspond to a model in which P holds

But then within that model, you get a statement whose value is yes

and then?

@daminark yo learn sheaf theory w me d00d

@BalarkaSen yo learn proof theory w me

nah that's boring

why?

i'm kidding. i don't like logic that much but i should learn it at some point

@LeakyNun Maybe you want to slow down and just focus on your school work first. =D

@Jasper school hasn't started

gotta go faster

faster faster faster

woah woah woah

gotta go woah

stop shitposting

@LeakyNun I would argue that sheaf theory is a lot more useful on average than proof theory.

@TobiasKildetoft you're probably right

@BalarkaSen teach me sheaf theory

@TobiasKildetoft why don't you ask us to prove that if $H \operatorname{ch} G$ then $\varphi[H] = H$ for every $\varphi \in \operatorname{Aut}(G)$ lol

@TobiasKildetoft I wonder a little about that, though only in a roundabout way

Gaisi Takeuti: Proof theory is a good book for proof theory.

nostro salvatore e qui

@Daminark what's the question?

namely, that thinking about proof theory is tightly linked with the question of how one formalizes proof

@LeakyNun i had a meme for that but it's actually a bit edgy so i won't post it

and that leads one to thinking about stuff like incompleteness theorem, and that's got a definite connection to stuff like the halting problem

how does one prove the axiom of regularity in $V_\omega$?

induction?

@LeakyNun I don't think that can happen

@AlessandroCodenotti why not?

so insofar as proof theory is linked to questions of computability, proof theory is important

....doesn't mean I can bring myself to care about it, though.

@Semiclassical hmm

so proof theory is more useful in CS than in mathematics

Just a feeling

@AlessandroCodenotti alright

Today I have been feeling tired but I cannot sleep.

@AlessandroCodenotti

@Balarka I may very well do so

@LeakyNun I'm pretty sure by any reasonable metric, proof theory is a part of math itself, just that the areas to which it applies are less other areas of math that aren't obviously related and more CS theory

I just peg it at the intersection

Computability theory too

induction on the rank of sets, I'd guess?

Though I think I'm prob gonna be more of a model theory person

model theory >>> proof theory

(well I know nothing about either, but model theory feels more interesting)

Hey guys/men and women... may I ask you all a personal question? (Sorry for my interuption). Why did you choose to study math (I suppose most of you did or are studying it).

*interruption

@AndreasAlmgren interest

Do you find it beautiful?

@LeakyNun

some of it is beautiful.

but ultimately, "mathematics is beautiful" is just a phrase that mathematicians tell others but that they don't believe in

@LeakyNun elaborate

@Abcd I like maths, so I study it.

Hmmm... I think that it is like poetry?

@LeakyNun why do you like it? How do you find it beautiful?

... in a more visual way.

@AndreasAlmgren there do exist people who believe that mathematics is an art.

@Abcd the beauty that everything can be proven (inb4 independence)

as I said, some of it is beautiful.

I would agree with them... but it is more complicated than art.

"mathematics is beautiful" is more like a gimmick to me

they say it but they don't really mean it

You always think that is the case?

@AndreasAlmgren I tend to.

Anybody else who want to share why they chose math?

I mean, I can show you examples of how mathematics is beautiful, but I don't really believe in that statement.

Can I ask a question? or does this chat is only closed to a specific topic?

@Mour_Ka you can ask any question.

It is not

(but you aren't guaranteed to get an answer :P)

@AndreasAlmgren why do you study maths?

great, I actually want to know what are the possibilities of getting -ve contributions (votes or comments) removed from a question I asked?

What if I want to remove the question at all with all its negative votes or replace it?

which question?

I actually deleted it, can I show it to you if it is deleted?

you can.

wait, I don't have 10k rep yet

haha ok tyt

you might want other people's attention then

I see upload key here

or I can rewrite the formula with the comment and you tell me what do you think

sure

@Daminark start reading forster :)

can I write latex here?

@Mour_Ka yes

Haha, I do not. I haven't finished highschool yet, or "gymnasium" as it is called in Sweden. But I think I will study math after finishing highschool. I find it beautiful, even the most intricate of details in it can be very beautiful. This may sound VERY strange to some ears, but math could be compared to love sometimes. It is interesting from time to time, and it is peaceful to work with math. It can also be very challenging, which I like @LeakyNun

but you would need to click a link in the room description to render it in your client @Mour_Ka

seems double dollar signs dont work here

what is the bull's eye?

so it does render at my side

ah so you see the formula correct?

yes

gr8

@AndreasAlmgren for example?

For example what, @LeakyNun?

math is useless trash

@AndreasAlmgren an example of beauty

You useless cunt, @BalarkaSen xD

no u

I think we both would fit in perfectly in kindergarten

i know swedish curses, back off

Oooh I'm scared I got IKEA

@LeakyNun, it could be as simple as a summation formula. It is not only art, the summation formula in itself is conveying a mathematical meaning.

hmm

The problem here seems to be that he don't know what argmin means. He want A to be the same size of formula and I got -ve vote. Even by replying to his comment asking what did he meant. I actually sent that the formula is from a published paper and send the link. I flagged the question for a moderator to have an intervention then I got another -ve vote in 1 minute. So I deleted the question.

I do believe that there exist people who mean it when they say that "math is beatiful". It is not something sterile, not only a tool to be picked up at certain times.

@Mour_Ka, this place can be a frustrating hell at times.

(MSE)

It is fine, I just want the question to be correct if it is not. I also want at least a guidance that I can start from. Seems I will ask people in office after all. But Is there a way that I can involve a moderator to help me remove the question totally and its negative contributions (votes and comments)

math is fun. the metaphilosophy behind why we study math is a tiresome question that has come up multiple times

I think that I remember a way to get in contact with a mod... wait a sec and I will see if I can find anything

@BalarkaSen cannot agree more

things happen without a reason

I study math, period.

my trash comment was, beyond making a joke, making the point that i don't really think the reason behind why i do math is really interesting

Boring.

i think it's trash, but i still do it

i enjoy trash

I suppose that can explain your smell...

I get paid for it, best reason ever :^)

im covered in rotten stench d00d

got a problem with it

Final question for today

@Mour_Ka, I'm on my mobile unit and cannot find a mod chatroom but I think that there are people there who are mods and have chatrooms

Theorem : Suppose $M^n$ is a smooth manifold with boundary and $p \in M$. If there exists a smooth chart $(U, \psi)$ for $M$ such that $\psi[U] \subseteq H^n$ and $\psi(p) \in \partial H^n$, then for every other smooth chart $(U, \phi)$, with $\phi[U] \subseteq H^n$, we have $\phi(p) \in \partial H^n$

@Perturbative Think in terms of charts. What does a boundary chart look like?

@Perturbative Consider $\operatorname{Aut}(M)$ and $\operatorname{Hom}(N)$. Clearly, they are isomorphic, now, use sheaf theory to show that $\operatorname{Int}M$ and $\delta N$ have the same section. After that, realize that I have absolutely no idea what I'm talking about.

top 10 nlab fails ever

yeah but do witness something like that before or you know if moderator would help?

@BalarkaSen example of compact $X,Y \subseteq \Bbb C$ such that $\delta X \cong \delta Y$ but $X \not\cong Y$ :P

@BalarkaSen Well a boundary chart $(U, \psi)$ around $p \in \partial M$, is gonna map a neighbourhood $U$ of $p$ to $\psi[U]\subseteq H^m$ with $\psi[U] \cap \partial H^m \neq \emptyset$

@LeakyNun Disk and torus with a disk removed.

$$-\log (1.8* 10^{-5})= 5 - \log 1.8$$ Please verify if you can.. I don't know logs properly, will study them in Maths later. Right now I am dealing with $pH$ calculation problems so I need basic log knowledge.

@BalarkaSen :O

@Perturbative Right. So locally the boundary looks like $\Bbb R^{n-1} = \partial \Bbb H^n \subset \Bbb H^n$. If $f$ maps $p \in \partial M$ to $f(p)$ diffeomorphically, what must a chart around $f(p)$ look like?

wait, why aren't they homeomorphic?

@BalarkaSen In essence what I tried to do was complete a commutative diagram for $F$, and handle the cases for the dimensions of $M$ and $N$ respectively

@Abcd $\log(1.8 \times 10^{-5}) = \log(1.8) + \log(10^{-5}) = \log(1.8) - 5$

right

@LeakyNun Thank you very much for replying to my queries.

@LeakyNun Different fundamental groups, say.

Fixed, @Perturbative

@BalarkaSen right

$\pi_1(B_2) = \{e\}$ right?

Ya it's contractible

$\pi_1(T-B_2) = \Bbb Z^3$ right

Ohhh wait, dammit forgot that $M$ and $N$ can only be diffeomorphic if they have the same dimension

@Leaky, no, that's the free group on 2 letters.

Is topology on $\Bbb Q$-vector spaces interesting?

@BalarkaSen what's a characterisation of spaces with abelian first-homotopy group?

idt there's one. topological groups (in general h-spaces) have abelian $\pi_1$

so there's that

but torus is abelian?

yeah torus is a lie group

nvm, $\Bbb Q$ isn't even complete

Salut @GabrielRomon

salut

@BalarkaSen There would be two possibilities for a chart around $f(p)$, $f(p)$ would be the domain of an interior chart or the domain of a boundary chart, either mapping to $\mathbb{R}^n$ or $H^n$ with nonempty intersection of $\partial H^n$ respectively

As-tu lu les reponses sur ta question de la periode des nombres en $\Bbb Q$?

I'm a bit sleepy, especially after making that error thinking that $M$ and $N$ can be diffeomorphic if they had different dimension

But thanks! @BalarkaSen :) I'll complete the proof on my own

Night everyone!

@Perturbative Right. But since $p$ is a boundary point in $M$, you can take a boundary chart $(U, \varphi)$ around $p$ which maps $\varphi : U \to \Bbb H^n$ diffeomorphically to an open set on the closed upper half plane hitting the boundary.

With the diffeomorphism $f : M \to N$, you should be able to produce a boundary chart for $f(p)$.

G'night!

@BalarkaSen our timezones are near lol

what's the time there

Thank you guys :)

@BalarkaSen wait, not very close

but it's 03:57 here

which should be 01:27 there

yeah you're way ahead of me