point-set-topology

[bitter smelt]

(a,b) is the same as ]a,b[

[nimble portal]

Then why not just write (a, b)? O.o

[bitter smelt]

people have been known to have different preferences for notation

[abstract saffron]

French notation

[nimble portal]

Good point

The French

[abstract saffron]

Idk if the French came up with it, but it's pretty common in France.

[untold lily]

I see Tu

[nimble portal]

How do I show that it's smooth? An induction argument?

Repeated differentiation just gives you sums of powers of sec & tan

[abstract saffron]

compositions of smooth functions are smooth

[nimble portal]

sec and tan

[abstract saffron]

You need some more scaling, and that's it

[nimble portal]

Since sec & tan are cont. on the domain then the derivatives are cont

[nimble portal]

Where do compositions come in here? The derivatives?

[unreal stratus]

Idk why compositions are relevant here lol, it's more products

Are we talking about a) I assume

[abstract saffron]

because of b) mainly

but you don't need it actually

[civic verge]

Guys, can someone give me the definition of CONEXO set?
Other than the definition of path.

I have searched the internet, but ended up quite confused.

[winged viper]

what does conexo stand for

[languid patrol]

It's a norwegian telecoms company

[civic verge]

For my conex*, I have a set in which I cannot split, remove or divide.

[winged viper]

what does that mean

does anyone know how to prove that the "theta" space does not cover any other spaces (other than itself)? Hatcher says it is true in his answer to this math stack exchange post https://math.stackexchange.com/questions/1129047/does-there-exist-a-space-w-s-t-x-to-w-y-to-w-are-covering-spaces-when-z

the theta space is the same as a graph with two vertices and 3 edges between them

i can see why it's true if you assume that the two "vertices" are mapped to vertices, but im not sure how i can justify that assumption

[novel acorn]

do you mean convex set?

[quick bough]

what does it mean to have two cells in each dimension in this case

[languid patrol]

It means that there are 2 n-cells for every n

So like up to n = 2 you start with 2 points, and you draw 2 arcs between them to make a circle

Then you glue one disc on either side of the circle to make a sphere

You can do the rest.

[civic verge]

set conexo

[civic verge]

my interpretation is that you can not separate, cut or divide, but as to how the definition is I do not know because I have been quite confused with my fellow internet stuff and what my teacher explained.

to be CONEXO-, it has to fulfill this property?

$X=A\cup B\
A\cap B=\emptyset \
A=\emptyset \text{ or } B=\emptyset$

[gentle osprey]

Awuita Fria

[languid patrol]

Yes this is the standard definition for a set to be connexo

But A, B must be open sets, in our conexo topol spa

[civic verge]

B_{r}(a)-\left{ a \right}

it's conexo

true?

[stark fog]

translation for connected sets

[quiet summit]

i will have my first topology exam next 10 minutes. Any advice?

[untold lily]

give the correct answers

don't make mistakes

[civic verge]

lul

[civic verge]

I guess going slow to think and structuring your concepts well, is what I did in my multi-variable analysis exam.

[civic verge]

To be dis-Connected, they must meet the following properties

$X=A\cup B\
A\cap B\neq \emptyset \
A\neq\emptyset \text{ or } B\neq\emptyset$

[gentle osprey]

Awuita Fria

[civic verge]

??

[urban zinc]

can someone give an example of a euclidean ball that isn't regular? I'm having a hard time coming up with one

[odd flame]

very broad and probably bad question but i wanna use it to formulate better questions: what is the intuition behind CW complexs

tterra

sorry for ping idec about the question but ur back

that's a lie i do care but still good to have u around again

[umbral panther]

Inside a sphere, everything but a pole. This is an open disk but its closure is not a closed disk. There is no larger open disk containing it

[odd flame]

i was asking this in particular with regards to doing it for S2 v S1

[odd flame]

what's the justification for this

the answer to this post isn't particularly enlightening

[odd flame]

holy shit it's a fucking dunce cap

i cant help but feel like ive been trolled

,ti

[gentle osprey]

The current time for stμ₂dying is 03:48 AM (EDT) on Thu, 27/04/2023.

[odd flame]

i cant take this rn

[unreal stratus]

Wdym

[odd flame]

it's okay i figured that one out

doing the dunce cap rn

[unreal stratus]

OK sure

[odd flame]

feel like im being trolled by my prof

[unreal stratus]

How come lol

[odd flame]

just the fact that it's very late and im doing a problem about a dunce cap

in any case it's htpy equivalent to S1 right

so what does a CW structure on the dunce cap look like

[fierce otter]

Hello can someone help me out with this question?

[novel acorn]

#old-network physics server

[hidden crag]

Stop posting this in every channel

[fierce otter]

bro i need help

whats wrong with asking

[hidden crag]

None of the channels are appropriate for your question

This isn’t even maths

[languid patrol]

Isn't it supposed to be setup so that advanced channels are gated in some way?

[vocal wharf]

they are (in fact there are multiple mechanisms to ensure that inappropriate stuff does not get posted in advanced channels); maybe its sometimes not enough but its a tough balance to strike, i think what we have going works well enough most of the time

[high quarry]

Well its like 5% of the overall posts, it isnt that bad

Have seen worst

[eager vigil]

Hey, I've just started on Topology by Munkres and I'm doing the problem for chapter 16, but I'm finding them quite hard and myself making mistakes. However, some solutions and problems I'm also simply misunderstanding. For example, this one, I simply assumed L to be described as normal in R^2 and then took the intersection. The topology then became generated by either half-open line segments on L, or open line segments or single points. I could of course see that this could be viewed as equivalent to R_l, R, and R_d if one viewed the line as the real line with ordering, but it didn't seem like that was what the problem wanted in the first place. And then adding direction to the line as well...
Am I missing something since viewing the line as such didn't even occur to me, or is it simply something one must pick up along the way?

[alpine bolt]

What would be a good reference for understanding a Postnikov system?
(I do AG and it is coming up in the papers I am reading)

[languid patrol]

I mean it depends on what you want. Just the definition is easy enough to understand, it’s a way of step-by-step building up a space which is weak homotopy equivalent to a given one. What do you want to know about it?

[alpine bolt]

I really just want to see some examples

[languid patrol]

Maybe it’s good to do some K(pi, n)‘s where it’s trivial? In general it will be very difficult to write out the postnikov tower explicitly. You can build the spaces by gluing in disks to kill homotopy groups but it is quite difficult

[languid patrol]

To get an explicit description

[nimble portal]

Can you guys pick out some problems for me because the only one that I'm interested in is 1.5 lol

But I don't think one problem is enough, though it is a fairly short chapter

1.8 doesn't look as boring as the others either

Maybe 1.3

Uh

I thought the tangent space at p was the VS of all vectors tangent to p

not just "emanating from p"

[rancid umbra]

1.2 is interesting

1.6 and 1.7 as well

[nimble portal]

Why is 1.2 interesting?

This is a cool concept

Is the point of this that a function is analytic if its Taylor series is in its germ?

[rancid umbra]

1.2 gives an example of a smooth function that is not real analytic

[nimble portal]

Yes, the classic one

[gritty widget]

Check my profile

[nimble portal]

Hm I guess that one would be good since I don't quite understand why it's smooth

I thought the derivatives wouldn't be continuous at 0

[rancid umbra]

its infinitely differentiable, by part (a) using induction

its not analytic because it does not equal its taylor series centered at zero

[urban zinc]

ohhhh that makes total sense, thank you!!

[nimble portal]

Oh wait it's defined at 0

Lol

Ok yeah never mind I see why

I think that is a good thing to prove though

[rancid umbra]

a modification that i like is exp(-x^{-2})

[urban zinc]

is this Tu?

[nimble portal]

Yes

[urban zinc]

TTerra recommended to me that since the exercises in Tu are fairly easy, you should also check out some exercises from Lee's Intro to Smooth Manifolds

[nimble portal]

I see

[urban zinc]

Prob doesn't apply until you reach the later chapters when it starts talking about manifolds tho

[nimble portal]

Yeah that's what I was about to say

Content doesn't match up this early

[rancid umbra]

no, its just like recentering R^n with center p

[nimble portal]

But then here it says elements of T_p(R^n) are tangent vectors?

[urban zinc]

The tangent space is the space of directional derivatives

Directional derivatives and tangent vectors are the same

[nimble portal]

._.

[urban zinc]

If you have a function defined in a neighborhood of p, you can take a directional derivative in any direction

Hence why the tangent space is isomorphic to R^n

I also got tripped up a bit at first haha

[nimble portal]

Then why is it called...tangent...space...? 😭

[rancid umbra]

its weird for R^n. the tangent space of a manifold that is not R^n, like a surface or a curve (e.g. a sphere or a circle) is easier to visualize and you will see the why they call it a tangent space

[urban zinc]

^

[nimble portal]

So for a manifold it aligns with my intuition for what I think of as tangent? lol

[urban zinc]

Yes

For example if you have the graph of a function with one input and one output, then the notion of tangent space just corresponds to tangent lines

If you have a surface, the tangent spaces are just tangent planes

[nimble portal]

But how about if you just take like

[rancid umbra]

what is the definition tu gives for tangent space?

[nimble portal]

S^1

You can just take a vector normal to the circle

[urban zinc]

Tangent

Not normal

[rancid umbra]

you have to specify a point as well

[nimble portal]

Yes

[nimble portal]

But the definition just calls it "a vector"

And the directional derivative cna be in any direction

So why can't we take the normal direction at any point

[urban zinc]

That is for R^n

Not a circle

[nimble portal]

There's a different definition of tangent space for R^n versus a manifold?

[urban zinc]

When you learn about manifolds you'll see that the circle is an example of one

[nimble portal]

Yes

[urban zinc]

R^n is also a manifold

[nimble portal]

Yes

That's why I'm confuzion

[urban zinc]

For now you might as well accept the definition of tangent vectors as derivations, even though it seems weird at first

It will come to make sense

The tangent space of a point on a circle can be identified with the line tangent to the point on the circle, to answer your question though

[rancid umbra]

ew why does tu define tangent vectors as derivations first

[urban zinc]

Also based, 1.5 is cool

[urban zinc]

I think he gives two definitions at first, one concrete and one abstract

[gritty widget]

do all of them

[nimble portal]

not if i want to keep reading

😭

[urban zinc]

If you can do it in under five steps, it shouldn't take long to do

Lol

[nimble portal]

It's not about long or not

It's about the result that I'm trying to prove, whether I think it's interesting or not

[urban zinc]

Ngl I read Tu without doing many exercises but don't follow my example :P

[gritty widget]

tu's exercises aren't great anyways

(this is in response to eric)

[urban zinc]

Yeah I wanna go through Lee ITM and ISM now

Since you recommended ISM earlier

[nimble portal]

ig I'll just do 1.2, 1.5, 1.8

or maybe just 1.2 and 1.5

[urban zinc]

And then I can learn Riemannian geometry

I feel like Tu is generally a laidback book

Nice for getting first acquainted with the terminology and ideas

[nimble portal]

Or maybe I read section 2 and just combine exercises from 1 & 2 🤔 2 looks a lot more interesting

[gritty widget]

i don't think you should be skipping whole sections for reading

[nimble portal]

I've read section 1

It just goes over smoothness & analyticity

[gritty widget]

what's an example of a smooth function that's not analytic?

[nimble portal]

e^(-1/x) is the prototypical one

BUt I read that if you function satisfies lim n -> 0 f(x)/x^n = 0 then you can get a smooth but not analytic function with 1/f(1/x)

That sounds interesting to prove :o

but hard

LOL

[urban zinc]

Btw if you want some intuition on why e^(-1/x^2) works, note that it's a Gaussian but you do a hyperbolic change of variables x -> 1/x to switch the function's value at zero and at infinity

[nimble portal]

[urban zinc]

And the Gaussian decays to zero very nicely

[abstract saffron]

Another way to see it is that the higher derivatives, altho zero at the point of our interest, blow up around the nbhd as the degree increases (or was it the bump function?).

This can't happen for analytical functions

[abstract saffron]

This is provable with definition of derivative, unless I'm mistaken. I remember proving it some time ago.

[nimble portal]

What whole sections am I skipping?

[gritty widget]

i never said you skipped any sections. your message made it seem like you had planned on not reading section 1

and, as you already said when i sent that message, you had indeed read section 1

[nimble portal]

Oh okok my bad

I think that's what I'll do then 🤔 finish Section 2 and then make a pset for myself out of 1 & 2

[gritty widget]

a small word of warning

what tu calls "taylor's theorem" is actually hadamard's lemma

in case you want to be googling things

that's a nice and useful application of calculus (i think it was you who said you liked calculus stuff)

[urban zinc]

I wonder why he calls it taylor's thm...

[nimble portal]

Indeed!

[gritty widget]

maybe i should have said "the specific thing he calls taylor's theorem goes by hadamard's lemma" lol

[nimble portal]

So all this is saying is that if f is smooth on a star-shaped nbhd of some point then it's equal to another smooth function in that nbhd?

So we can always construct an approximation (with remainder 0) for f within the nbhd

[abstract saffron]

Wait, which one?

[abstract saffron]

Huh, I do call it Taylor's theorem

Or was it Taylor-Hadamard?

[nimble portal]

Isn't it just first order Taylor's theorem

or maybe if you take lim n -> infty

[abstract saffron]

The domain gives the trouble tho

That's the difference

[nimble portal]

Why?

[nimble portal]

Also is my understanding correct here?

[abstract saffron]

Ok, it's pretty annoying that we didn't do it explicitly in class

[abstract saffron]

More subtle than that, it's not obvious that you can combine linearly the differentials from a basis to make something sensible.

[nimble portal]

Why not?

Just chain together lines whose partial derivatives match f at the point

[abstract saffron]

There are cases where this doesn't hold, especially when f is not continuously differentiable.

Smoothness is a very strong condition here. It allows you to do basically whatever you want.

You can construct functions whose partial derivatives exist but it's not differentiable

I think the exact statement is, if all the partial derivatives exist and are continuous, then f is differentiable. In particular, the Taylor's formula is true.

[gritty widget]

"then it's equal to another smooth function in that neighborhood" basically completely gets rid of the conclusion of the theorem

[nimble portal]

wat

😭

I thought that was the conclusion

We can find g_i so that f is equal to f(p) + sum(x^i- p^i)g_i(x)

[gritty widget]

now that's the conclusion

[nimble portal]

How is that different from what I said?

[gritty widget]

"equal to another smooth function" is uselessly vague

[nimble portal]

Why?

[gritty widget]

because you haven't said what the smooth function is

when you do, you have a useful statement

[nimble portal]

Sure, but it guarantees existence

Are existence theorems not useful?

Sorry if I sound sassy I just don't see why it's so useless

[gritty widget]

i'm saying you can be more precise than just "another smooth function"

[odd flame]

ttera

[gritty widget]

hey

[odd flame]

[gritty widget]

i was finishing up my semester so i took a break off the server

but now i'm done

[odd flame]

good to have you back

[nimble portal]

So the significance of the result isn't just that it can be approximated, but rather that we have an explicit formula to approximate it? Or are both of those significant and just stating the first is missing the bigger picture?

[gritty widget]

the explicit formula

[nimble portal]

So there's another theorem (probably simpler) that guarantees existence?

[gritty widget]

i don't know what that means

[nimble portal]

Like existence of a smooth function that's equal to f on the domain

[gritty widget]

that's a trivial statement. f is equal to f on its domain

[nimble portal]

LOL

Okay another function

well

[nimble portal]

uh if it's equal then it's f so we just get back to this

[nimble portal]

I know I'm saying it wrong but do you get what I'm trying to say though?

[bitter smelt]

It should be insightful that you cannot phrase your statement well. Maybe it isn't a very meaningful statement

[nimble portal]

This is so boring

It feels so unmotivated

Just vomiting definitions after definitions 😭

[abstract saffron]

sounds like freshman complaining about epsilon-delta

[unreal stratus]

Utterly based

[abstract saffron]

Not saying it's not boring, but a large part of being a math student is understanding why we need those definitions

[nimble portal]

No but like

I see reasons for why we use those definitions and what use it is to establish those concepts first (usually)

And I'm given something to motivate those reasons beforehand, at least to some extent lol

Reasons in plain English**

[novel acorn]

Lol
Well if ur reading Tu there's some motivation in ch 1

Tho like generally speaking most math books are like this
Physics ppl like to meme on that

[nimble portal]

Oh we're given motivation for why we want to study differential geometry/manifolds as a subject

But I've already justified that to myself

[abstract saffron]

And math ppl meme on Jacobi's notation the world is perfectly balanced.

[nimble portal]

I don't know why I care about Hadamard's lemma or why we define derivations or why I care about germs or algebras or whatever

Like I get that maybe that seems ridiculous to you guys

"Just wait for it, you'll see why we want those concepts later" or whatever

But it would be nice to be given some reason to care about them now lol

[abstract saffron]

That's because you skipped the whole multivariable calculus. Basically we have all the nice tools in real anal, and we wanna bring em to higher dimensions.

[nimble portal]

I'm given a reason for the purpose of the chapter, to "characterize tangent vectors so they generalize to manifolds", but I don't see how a germ or a derivation or bla bla bla relates to that

I've taken multivariable calculus

[grizzled ibex]

wtf is a germ

[abstract saffron]

No one'll ever give you that. You kinda have to figure it out yourself

Welcome to math world

[nimble portal]

That's so dogshit lol what

Wtf is the point of having someone guide you, teach you material, and so on

[abstract saffron]

They all try to give some motivation, but at the end of the day, you have to understand it yourself.

[nimble portal]

If they can't help you understand it at least to some extent?

Yeah the work is the student's to do, but the teacher's responsibility is to guide their learning

[abstract saffron]

"I can explain it to you, but I can't understand it for you"

[nimble portal]

Yes

But they're not explaining anything

That's the issue I have

[grizzled ibex]

Why should you care at topology at all? @nimble portal

[nimble portal]

A definition is not an explanation

[grizzled ibex]

just search for more authors foccused on construtivism

[abstract saffron]

They explain how they understand it, which is most of the time not how you'll understand it. And they're now too advanced to see it your way and why it's so hard to understand.

Anyway, this is more #math-pedagogy , not topology

[abstract saffron]

You sure? 😄

[nimble portal]

Yes

[grizzled ibex]

you always have a more general way of understanding something, i think what you need to perceive is that every definition in some way encapsulates a intuition about how things are desired to work

if you keep that in your heart

it'll be more soft to learn by those bourbaki styles

[abstract saffron]

Hmm, can you state Schwartz's theorem?

[trail charm]

equivalence class of functions in a neighborhood of a point

[nimble portal]

No :p

Can you give me a rough idea of what it's about and I can probably remember it? Lol

[grizzled ibex]

equivalence about what

i mean

homotopic?

[gritty widget]

/clairaut's theorem. it goes by a few different names

[grizzled ibex]

what kind of relation

[nimble portal]

Oh is that the one about symmetry of second partials

[trail charm]

identical under restriction

[abstract saffron]

Yes

[gritty widget]

yes feather. what's the statement?

[trail charm]

so like if u have a neighborhood W around a point p

[grizzled ibex]

domain restriction then?

[trail charm]

then f|W = g|W

then f and g are in the same germ

yeah basically

[gritty widget]

they're equal when you take the limit 🤓

[grizzled ibex]

that's useful for diff-top?

[trail charm]

germs make stalks which make shaves

sheaves*

is it sheafs?

[abstract saffron]

sheaves

[gritty widget]

sheaves

[abstract saffron]

Leaf-leaves

[trail charm]

yea

[grizzled ibex]

i think we call it fiber in my country

that's my biggest problem, notation and names.,

[nimble portal]

Agh I don't remember the exact relation it has to smootheness 😭 just something about smoothness of partial derivatives means you can exchange differentiation order and it'll be equivalent

[abstract saffron]

What? They're not the same thing, aren't they?

[trail charm]

i don't think fibers are the same

i thought fiber had something to do with the inverse of a function

[grizzled ibex]

Not in english

[trail charm]

the preimage

[grizzled ibex]

that's another fiber

[nimble portal]

preimg

[abstract saffron]

Pretty sure not in French either

[gritty widget]

that's true, but the hypotheses of the theorem are a bit weaker than smoothness

typically

[nimble portal]

Is it just continuity?

So for any C^2 function you can switch differentiation order?

[abstract saffron]

Strange, because if you understand why we need that theorem, then you shouldn't have any troubles understand Hadamard's lemma either.

[grizzled ibex]

germ is kinda a bad name imo

[trail charm]

germ comes from the analogy of cereal germ, stalk, sheaf

[grizzled ibex]

Ohhhhhhhhhhh

[grizzled ibex]

and thats how you explain math guys

[gritty widget]

if your function is k-times continuously differentiable, then you get symmetry of all of the kth partial derivatives. the theorem is typically stated for k = 2, since it's simpler and the general statement for all k follows easily

[abstract saffron]

wait you know there's something called tropical algebra

[trail charm]

tbh didn't even know the d2/(dxdy) = d2/(dydx) had a name lol

[grizzled ibex]

thanks god i didn't knew

[trail charm]

i think it's pretty funny

albeit a bit ambiguous

[nimble portal]

I don't understand why we need that theorem

[abstract saffron]

Every physics person ever

[trail charm]

but the sacrifice for the pretty funny

[trail charm]

real

proud to say i used to want to be a physicist

now i want to be a pure math :)

that's called reformed

[abstract saffron]

Ok, either you're a physicist, or you're a lousy math student

[nimble portal]

I'm an engineering student

[trail charm]

what's the difference?

[nimble portal]

Worse than both

[abstract saffron]

The worst of both

[nimble portal]

Indeed

[gritty widget]

is there gonna be any topology talk any time soon?

[nimble portal]

LOL Sorry sorry

[grizzled ibex]

hmm idk

do you guys actually like general topology?

[abstract saffron]

Yes

[grizzled ibex]

every time i come here there's only talks about algebra and differential

[trail charm]

i will probably flood this channel come summer

when the only thing i read will be topology textbooks

[white oxide]

youve found the light, or well, got away from the course about it

[grizzled ibex]

is it useful to describe a space with directed graphs?

any theorem about that?

[white oxide]

A theorem about... usefulness?

[grizzled ibex]

both

[grizzled ibex]

that would be more logic focused xddd

[white oxide]

I guess you might call simplical complexes describing space with a directed graph

or rather its higher-dimensional analogue

[grizzled ibex]

interesting

[white oxide]

mayb diversify your reading? Did you have experience with analysis and some algebra before?

[trail charm]

i mean im reading topology textbooks for my directed reading this fall

[grizzled ibex]

have oyou guys seen anything about topological games?

[trail charm]

i took analysis and algebra this semester

[grizzled ibex]

like banach-mazur

[trail charm]

and will take more next semester

[white oxide]

I saw the wikipedia articles, thats it

[trail charm]

it would be silly to jump into topology without analysis

[grizzled ibex]

😭

[grizzled ibex]

i did that

analysis is boring

very very very very boring

only had introduction to analysis in my grad

since my grad wasn't bachelor at all

[white oxide]

only had introduction to analysis in my grad
the fuck

[grizzled ibex]

it was crazy

[white oxide]

Do you have a different major or something

[grizzled ibex]

hmm everything is a mess here

they taught us algebraic structures together with analysis

[white oxide]

Doesn't sound that crazy

[grizzled ibex]

it's called introduction to REAL ANALYSIS

and they taught us algebraic structures

together with some basic stuff from real analysis

i don't remember whats called in english

i think it's licensor

[abstract saffron]

Ever heard of simplicial complexes?

[white oxide]

My analysis courses did so too, but in their defense they start in the first undergraduate semester

[grizzled ibex]

nope i haven't study algebraic top yet

[abstract saffron]

Well, then I can summarise that yes, it's very useful

[grizzled ibex]

when you can teach the kiddos 😎

[white oxide]

they really arent

just do singular homology always 😎

[grizzled ibex]

it's because i liked nets alot

filters and nets motivated me alot

seeing how they simplify alot of theorems

[abstract saffron]

And that's how you get poor

Learn some simplicial homo, do TDA, and get rich today!

[white oxide]

unironically something in the back of my mind that motivated my algebraic top learning

[trail charm]

ngl alg top is kinda cool

[white oxide]

Idk how big the memes are, but TDA sounds really cool. Actually use your degree in a non-trivial way

[abstract saffron]

It still breaks my mind. I can give a lecture on what I know, but I sure as hell won't do any hard exercises

[white oxide]

TDA?

[abstract saffron]

It's growing actually

[trail charm]

me when i read textbook :)
me when i get to the exercise page :(

then i skip it :)
then i understand nothing in the next chapter and go back to do them :(

[white oxide]

Last time I checked its a "maybe, hopefully it might be niche thing"

Hopefully the air bubble wont burst any time soon if its actually growing

[abstract saffron]

They have a team in Cambridge doing those things, and it's growing fast iirc

It hasn't spread to other branches yet. In Bio it's still limited to medical stuff.

[trail charm]

i heard from a prof that math bio used to be the next big thing like 10-20 years ago

everyone was doing math bio but then it kinda just disappeared

[abstract saffron]

That's why I'm majored in Bio

Ok, minored in Bio, but whatever

[abstract saffron]

It's going, but very slowly. Thing is, you need quite a broad background

[white oxide]

is that why you occasionally read about applying category theory to biology

Always pretty surprising to see that domain listed out of all thing

[abstract saffron]

category theory to Bio?

Are they gonna apply homology functor to Virus and categorise them or what?

[languid patrol]

You thought the world’s bestiary of diseases was bad, now imagine if they were functorial 😨

[abstract saffron]

An exact sequence over category of Diseases 😄

The ultimate way from Birth to Death

[white oxide]

Can't wait for applying the colimit of a family of donkeys and horses to figure out what one obtains

[trail charm]

snake lemma'd snakes

[urban zinc]

It's not quite, Taylor's theorem (first-order) usually says that if (f) is differentiable at (p), for (x) in some neighborhood of (p) and (\xi) between (x) and (p),
[f(x) = f(p) + f'(\xi)(x-p)]
Notice that in the first-order case, this is just the mean value theorem

[gentle osprey]

boolean_satisfiERIC

[white oxide]

how did taylors theorem get in here

[urban zinc]

feather had a question about the beginning of Tu intro to manifolds

[bitter smelt]

(Which is diff top but meh)

[urban zinc]

also the conditions I wrote are not really right, but if you know the mean value theorem, I'm sure you can figure out the actual conditions lol

(differentiable in an open segment, continuous on the closed interval)

[urban zinc]

germs are just a way to describe the behavior of a function at a point, where you don't care about how it behaves away from that point

you might find it easier to understand if you've taken an abstract algebra course or something

(I dunno what your background is)

if you're having trouble finding the motivation, it's also possible you don't have the intuitive background yet, but also, the server is here to help answer your questions about it, although I think it'd be better to put it in #diff-geo-diff-top like mig said

[nimble portal]

Will do!

[urban zinc]

Good luck with your manifold journey :))) I'm also learning about manifolds

[gritty widget]

is this true for any X, Y with some arbitrary Topology?

because in the proof we assume X, Y are metric spaces

[unreal stratus]

This is a definition

[gritty widget]

oh nvm i found a contradiction

[unreal stratus]

in general

Like this theorem justifies how you define the topology on a subset of a topological space (without appealing to a metric)

[gritty widget]

X = {1, 2, 3}, O_X = {{}, X}
Y = {1, 2}, O_Y = P(Y)
E = {2}

therese no such G

[abstract saffron]

Yes dear, G = {2}

[unreal stratus]

Lol

[abstract saffron]

Oh wait

[gritty widget]

no, {2} is not open in X

[unreal stratus]

Okay perhaps I see he issue

[abstract saffron]

I didn't read the whole thing, my bad

[unreal stratus]

Okay so

This is certainly not true in general if you just give Y an arbitrary topology

I mean if it were true, then all topologies on Y would be equivalent...

Since they have no dependence on X

[abstract saffron]

You can't just specify O_Y. E is open in the subset topology, which is induced by O_X

[gritty widget]

wait is compactness a thing only in metric spaces?

[abstract saffron]

Nope

[gritty widget]

because we use this theorem to say something about compact spaces

[unreal stratus]

But the point is that if Y is a subset of a topological space X then the subspace topology on Y - the natural topology on Y - is defined by the theorem you just used

And then it matches up with the topology you get from restricting a metric on X, if X is already a metric space

[gritty widget]

in the theorem their metric spaces, not induced by anything just a subset

[gritty widget]

if the topologies were not specified(i.e. any), and Y subset X, then do we assume subspace topology?

[gritty widget]

hmm, but in my book its only defined in metric spaces, so will not assume things about compactness outside metric spaces

[abstract saffron]

Well, you can remove the word metric

It stands perfectly well

[unreal stratus]

Ye

[unreal stratus]

I mean this is Rudin right, not a topology textbook

[plain raven]

I work in mathematical biology

so we are not dead yet

[nocturne basalt]

andrew blumberg is doing pretty well for himself

[plain raven]

So is Kathryn Hess.

[tidal lynx]

So is diligentClerk

[urban zinc]

I am also doing mathematical biology

[high quarry]

It’s a field I would like to work in the future, is a discussion possible in a few days?

[shadow charm]

Let’s go Kathryn Hess praise time

[abstract saffron]

this

I'm very much interested

[umbral panther]

Kathryn, I didn’t know you spoke Danish!

… I speak Scandinavian

She speaks Scandinavian with a strong Swedish accent

[pseudo ocean]

Guys is there a topological space $X$ such that $\pi_1(X)=\mathbb{Z}^{27}$?

[gentle osprey]

messyinterval

[gritty widget]

(S^1)^(27)

S^1 x S^1 x ... x S^1 (27 of them)

because fundamental groups interact nicely with products

[pseudo ocean]

THat could prolly be made into a torus of... Uhh some dimension

With a leftover circle

[hidden crag]

That is a 27-torus

[pseudo ocean]

Huh? Wouldnt that be (S¹)^(27×2)?

Because each torus is S¹×S¹

[gritty widget]

"n-torus" means (S^1)^n

[pseudo ocean]

Ahh

[gritty widget]

the ordinary torus is, in this language, the 2-torus

[pseudo ocean]

Makes sense

[gritty widget]

re: your question, every group is the fundamental group of some topological space

[pseudo ocean]

While we're on the topic, how about R?

There any spaces with fund. Grp R?

[nocturne basalt]

you can have a wedge of S^1 for each generator and quotient out any relations by attaching 2-cells

might not be a very satisfying construction

[novel acorn]

You could construct an eilenberg MacLane space with that group

[fading vale]

You can get close even w some pretty strong restrictions on the spaces

Like any finitely presented group is the fundamental group of some smooth compact 4 manifold lol

[plain raven]

I'm not going to put this in my calendar but if I do happen to be hanging out on discord you can ask me

[high quarry]

Thanks 🙂

[trail charm]

why does orientability require positive-definiteness of the jacobian? specifically, why can’t it be negative-definite?

[gritty widget]

i think you should test what you just wrote against a few examples

also, positive/negative-definiteness aren't the same as positive/negative determinant

[pulsar pumice]

If the jacobian is negative-definite, the function reflects or mirrors the tangent space, which results in a change in orientation. To ensure that the function maintains orientation, the jacobian must be positive-definite.

[trail charm]

ah that makes sense thanks

[trail charm]

i meant positive/negative-definiteness as in the determinant is positive everywhere/negative everywhere

prolly should've used a better word

[gritty widget]

just say positive/negative determinant

[wet finch]

I posted this in the book section

Does anyone have a translated version of Freudenthal's thesis on his theory of ends
https://dspace.library.uu.nl/bitstream/handle/1874/7437/1930-freudenthal-dissertatie.pdf?sequence=1

But no one responded

[rough cedar]

the quotient topology is just like, a set is open in the quotient if its preimage under the qutotient is open?

[gritty widget]

yes that is the definition of an open set in the quotient topology

[rough cedar]

I was not able to find a clear definition of the quotient topology

the wikipedia one is like 20 lines

sorry if it was a dumb question

[rough cedar]

sorry

[white oxide]

Tbf the first definition of the quotient topology on that article is a bit dense if you aren't familiar with

[white oxide]

Is there is a way of deliberately quotening a subgroup out of the fundamental group of the bouqet of circles by doing... something to the space or how does that work?

[languid patrol]

Yes, this is how you build spaces with only one homotopy group.

You just take the relations that you want to hold and glue in a disc so that that path is contractible

(then if you wanted to do this with higher homotopy you glue in more cells)

[white oxide]

How would you glue in a disc to kill that specific relation tho?

Like, choose a path and then identity the edge of the disk with said path?

[nocturne basalt]

you glue a 2-cell along the relation

i.e. if ab = e then you attach a two cell along the S^1 for a and then along the S^1 for b

[languid patrol]

Wait that makes a = b = e? Maybe I misunderstand what you're describing...

Are you gluing in one or two two-cells?

[nocturne basalt]

just one

maybe that was unclear

[languid patrol]

Just to me ig

[white oxide]

It did kinda sound they just glued a one disk to each of the circles of a and b, thereby outright killing them completely and giving you a free group of lesser rank

So given a bouqet of n circles with generators a_1, .... a_n , after gluing in a disk along a_j1 , .... a_jk the only paths affected are those that loop around the edge of the disk completely, that is, loops that have at least the components a_j1+ ... a_jk?

[cosmic socket]

is there an easy way to see/show that the multiplication operation on the loop space of a topological (compact) group given by either loop concatenation and pointwise multiplication are homotopic?

also is compactness required?

[solemn oar]

You can use a homotopical version of the Eckmann-Hilton argument. Basically homotope the pointwise product of loops so that each is stationary at some point of the loop. The product is then the same as the concatenation.
https://en.wikipedia.org/wiki/Eckmann–Hilton_argument

[cosmic socket]

thanks

does anyone know of a good place to learn/understand how homotopy quotients work?

like why when I do a homotopy quotient by a Z/2 action do I usually multiply by some EZ/2 before quotienting

[pseudo ocean]

Maybe try Hatcher's book on Alg topology?

[unreal stratus]

That book doesn't cover homotopy quotients ig

[pseudo ocean]

Rats.

[plain raven]

Daniel Dugger, A Primer on Homotopy Colimits

[pseudo ocean]

Im feel confused

Someone on reddit told me S1 and T2 are homotopy equivalent

But they have different fundamental groups

How tf—

Also it is mentioned as a theorem that the fundamental group of a genus-g surface is the free group generated by g-elements

Possibly Z^g

But T2 is the case g=1

So π(T2)=Z

But by van kampen, π(T2)=Z*Z=Z²

And as far as i know Z aint isomorphic to Z²

[plain raven]

This is incorrect.

If the torus was solid, they would be homotopy equivalent.

but the hollow torus is not homotopy equivalent to S^1.

[umbral panther]

What did they actually say?

There are lots of things caused the torus. Clerk suggested one that is homotopy equivalent to the circle. Another is C*, the nonzero complex numbers

[pseudo ocean]

Well the original post was the quotient C/Z[i]

Which was group theory

But some know it all claimed the resulting construction is homotopy equivalent to S1

Smh i was talking about groups not spaces

[cedar pebble]

C/Z[i] is a torus, this is S^1 x S^1

[pseudo ocean]

I was originally pertaining to the quotient group, not the quotient space

But ugh some phrasing in that meme i sent gave that one guy the wrong idea

[languid patrol]

Also the fundamental group of a genus g surface is not abelian nor is it a free group on g generators

Although I guess in the world where S^1 x S^1 \sim S^1 maybe the second statement would be true

Anyway it’s isomorphic to the “surface group” which is 2g free generators in sets of g disjoint pairs modulo the relation that the products of the commutators of the pairs is trivial

[next crystal]

Show that if $Y$ is path connected, the set $[I, Y]$ has a single element, where $[I,Y]$ is the set of homotopy classes of maps of $X$ into $Y$.

My proof:
Suppose $f: I \to Y$ is continuous, and $g: I \to Y$ is the constant map $g(x)=y$ for all $x \in I$. Since $Y$ is path connected, there is a continuous function $\varphi: I \to Y$ such that $\varphi(0)=f(0)$ and $\varphi(1)=y$. Let $F: I \times I \to Y$ be given by $F(x,t) = \begin{cases} f((1-2t)x), t \in [0, 1/2]\ \varphi(2t-1), t \in [1/2, 1] \end{cases}$. Then $F$ is a homotopy from $f$ to $g$, so $[f]=[g]$ and $[I,Y] = {[g]}$.

[gentle osprey]

michαel

[next crystal]

this should be right, but when constructing the homotopy F it took a lot of brute force trying to figure out what works, even though F wasnt that complicated

is there some sort of diagram/graph or something that can make constructing a homotopy easier?

When sketching out the idea for why the product on paths is associative, my prof drew a diagram like this which motivated how to construct r(t), but im wondering if theres something similar i could do for this problem?

[coarse night]

this is false

[next crystal]

i meant to say where [I,Y] is the set of homotopy classes of maps of I into Y

[coarse night]

nvm I thought you are doing rel ∂I

is a visual argument enough for you?

I haven't read your solution tho

[next crystal]

im trying to use something visual to make coming up with a formual for F easier

so yeah ig a visual argument would be enough if it helps me construct F

[coarse night]

say you have path a--b and c--d, try to split the [0,1] interval in three parts where first you map b ->a via the path a--b, then 2nd stage you move a to c via any path a--c (exists as path connected) and then pull back b along the path c--d

Another way would be

That retract is easy to find

[next crystal]

is ∂IxI the |_| part of the picture?

[coarse night]

Yes

That’s partial drawing supposed to be |_|

[next crystal]

i think i sorta get what youre getting at but im not sure how i'd actually apply this to this problem

[coarse night]

do you understand the map F: IxI → Y I just described?

if you want to find r explicitly then here's what you can do

[next crystal]

not fully
so i get that we are finding a retraction of IxI to |_| but once we map into Y what are a,b,c,d?

[coarse night]

But most arguments in AT dont require you to give an explicit map

[pseudo ocean]

Ahhhh i guess i misinterpreted the definition. Anyway is there a case wherein the group is isomorphic to Z^2n?

[coarse night]

so we have to show any two paths in A can be connected via a homotopy

so I am assuming the paths are a--b and c--d

[next crystal]

in Y right not A

[coarse night]

we have a path a--c as path connected

that's what I drew

[next crystal]

okay i think im seeing it now

trying to connect it back to my drawing with f,g, and phi

so a--b might be one function c--d the other and a--c connecting f(0) to y in my proof

[coarse night]

okay I didn't read your proof

mb

but hopefully the notations are clear now

[next crystal]

yeah i got that now

[coarse night]

assuming you call f the path a--b, g be c--d and σ be a--c, r gives r(x,y) = 0xt or tx0 or 1xt
send 0xt to f(t)
tx0 to σ (t) and
1xt to g(t)

this is more explicit description

but it's clearer from the diagram

[next crystal]

okay i see how that matches with the diagram

so if i were to construct the homotopy with this my homotopy would have three parts?

a t \in [0,1/3]
t \in [1/3, 2/3]
t \in [2/3, 1]

[coarse night]

you can try to write it down explicitly

[languid patrol]

Only when g = 1 and the surface is a torus. Then the fundamental group is Z^2

[coarse night]

[next crystal]

okay i'll try doing that

idk if in the end it'll be faster to do this or just try and figure it out the old way

[coarse night]

(but trust me, it's unnecessary)

[next crystal]

i doubt we'll have to do this on the final just because of time but there were hw problems where we had to so

[pseudo ocean]

For g>1 we have to abelianize the group?

Ohhhh that's what you said at the last part

With modding out by the commutating something

We're modding out by the comm subgroup

Which, by def, is the abelianization of the grp

[next crystal]

if you were doing this problem would you just ditch all the diagrams and kinda brute force figure it out

thats how i got the homotopy in my solution i feel like its harder to come up with but might be less time consuming since i dont have to think about this retraction/other stuff

[coarse night]

no just diagrams

[next crystal]

what if i force you to write a formula

[coarse night]

those retracts will become a common theme as you progress

[coarse night]

I wouldn't

[next crystal]

he's normally fine with us just describing retracts and stuff

and like in the hws after homotopy he's like just say its homotopic but in the homotopy section hw we have to explicitly construct a bunch

[coarse night]

you are supposed to give these kinda arguments anyway otherwise they almost get into impossible territory if you want to describe explicitly

[next crystal]

yeah ig if it wasnt something made to be really nice it would be almost impossible

[languid patrol]

No to form the fundamental group you just mod out by the product of the commutators, the result is not an abelian group. If you mod out by all of the commutators you do get Z^{2g}

[coarse night]

your problem is still doable but I wouldn't try

[next crystal]

probably should spend my studying time on more important stuff but i always get stuck up on these things

the one in my solution is right apparently but i really didnt have a strategy to find it i just tried stuff until it worked

[coarse night]

the one I gave should be your general strategy

[pseudo ocean]

Ah the latter part is the one im aiming for

[coarse night]

this is a more general theme as I said when you have cofibrations and you extend homotopies etc

[languid patrol]

That is the homology of a surface, not the fundamental group.

[next crystal]

we didnt do cofibrations in this course but i will keep that in mind

[pseudo ocean]

So there is no symbolically compact way of expressing the fundamental groups of genus-n surfaces? Damn.

[next crystal]

it was a point-set course with a little algebraic in the last 40% of the course

[coarse night]

okay

[next crystal]

anyways thanks for the help 👍

[coarse night]

👍

[languid patrol]

Yes there is! I explained it to you, it's $<a_1, b_1, a_2, b_2, ..., a_n, b_n>/([a_1, b_1] \cdot ... \cdot [a_n, b_n])$

[gentle osprey]

Topos_Theory_E-Girl

[languid patrol]

If you want to be really symbolically compact some people denote this $S^g$, the surface group of genus g.

[gentle osprey]

Topos_Theory_E-Girl

[coarse night]

*orientable

[pseudo ocean]

Right but i meant in terms of isomorphisms with particular groups such that i can label em as the fundamental group

[languid patrol]

Eh?

This is an isomorphism with a specific group.

[pseudo ocean]

Which one?

[languid patrol]

The one I gave a presentation for.

[pseudo ocean]

Well the fun grp yes

[pseudo ocean]

Obviously

But apart from that one

[languid patrol]

Yes well I don't really understand what else you want....

[pseudo ocean]

Im looking for other groups our particular construction is isomorphic to

We established it isnt abelian so not Z

[languid patrol]

There are infinitely many groups which it is isomorphic to.

[pseudo ocean]

One of which is...?

[languid patrol]

Change one of the symbols in the presentation to any unicode symbol

[coarse night]

lol

[languid patrol]

Then when you get done with plugging in unicode symbols you can use non-unicode symbols.

[pseudo ocean]

Mfw i use $\Xi$ to denote the fund grp of T2

[gentle osprey]

messyinterval

[pseudo ocean]

I watched pierre albins lectures on this and i shit you not he claimed the fundamental group of n=2 genus surface is isomorphic to Z⁴

I musta been hallucinating

[coarse night]

[pseudo ocean]

Well he claimed "when abelianized"

[languid patrol]

Maybe he just meant the abelianization

[pseudo ocean]

Well yep

[languid patrol]

But the abelianization is just the homology

Which is Z^{2g}

[pseudo ocean]

Ohhhh

[languid patrol]

but if I were you I wouldn't be satisfied with that

Z^{2g} has infinitely many groups which it is isomorphic to

You should find them all.

[pseudo ocean]

Hm

How bout gaussian integers to the g?

[languid patrol]

That's one but keep going!

I believe in you

[coarse night]

Like coker(Z → Z^{2g+1})

[languid patrol]

@coarse night can help you now in your quest.

godspeed

[coarse night]

[pseudo ocean]

...cokernel?

I uh

Im still yet to study homology trembles in fear

Also how about T^n? What are their fundamental groups?

[coarse night]

π_1( (T²)^g )

[pseudo ocean]

Ah

Where g is the dimension, yes?

[coarse night]

If you are done with the classification, let me know. I would love to see

[pseudo ocean]

Classificatiob?

*classification?

[pseudo ocean]

So that would be a group of 2g generators?

[white oxide]

C'mon, at least give them a tip like well-ordering the set of all countable sets to make this easier

[languid patrol]

I don't think that would help since there is no set of all countable sets.

[white oxide]

I thought you were being sarcastic

[languid patrol]

Haha I definitely was, the fact that there is a proper class of such things makes it pretty unlikely that you can say anything....

[white oxide]

Guess I need more credentials before I can joke about well-ordering a different proper class without getting fact checked 😅

[pseudo ocean]

What does that have anything to do with finding other isomorphic homologies of T^n?

[white oxide]

Actually... wouldn't a subclass of the isomorphism class there be the free Z-module over {(G,1),(G,2),(G,3),(G,4)) M,with G a group from the class of groups. But for G=M that free Z-module is build up from itself 🤔

[white oxide]

What do you mean by "other homologies" ?

Choosing different representatives for the generators?

[coarse night]

who's gonna tell him

[pseudo ocean]

Yes...?

[pseudo ocean]

Well i meant groups tho

[white oxide]

If your representatives are n-cycles, then you will usually have uncountable many ways to chose them. If you are looking at finding all isomorphic groups, consider finding all the groups isomorphic to this particular trivial group:
{empty set}

[obsidian spoke]

I am going through munkres topology book and one of the exercises asks us the following: Let ${T_{\alpha}}$ be a family of topologies on $X$. Show that there is a unique smallest topology on $X$ containing all the collections $T_{\alpha}$, and a unique largest topology contained in all $T_{\alpha}$.

My question is when he says "largest topology" does he mean the finest topology? Or is it on cardinality? Similarly when he says "smallest topology" does he mean coarsest?

[gentle osprey]

FamilyFriendly

[gritty widget]

smallest/largest with respect to inclusion

"smaller" is coarser, "larger" is finer

[obsidian spoke]

oh ok, thank you for clearing that up

for the largest topology contained in all $T_{\alpha}$ would it be the intersection of all the topologies, $\bigcap T_{\alpha}$? Because suppose there was a finer topology, $T$, contained in all $T_{\alpha}$ then there is an open set $U \in T$ such that $U \in T_{\alpha}$ for every $\alpha$ but $U \notin \bigcap T_{\alpha}$ which would be a contradiction

[gentle osprey]

FamilyFriendly

[gritty widget]

a proof by contradiction is unnecessary but yes it'll be the intersection of them all

[abstract saffron]

Let the boy have his moment

[lavish geyser]

ok but

why is a closed set not the opposite of an open set

[gritty widget]

sets can be closed and open

a set which is not closed is not necessarily open

a set which is not open is not necessarily closed

etc

[lavish geyser]

https://www.youtube.com/watch?v=SyD4p8_y8Kw&ab_channel=Benjatastic

[abstract saffron]

we define a closed set as one whose complement is open

Under some conditions, one can show that a non-empty set that is not the whole space must be either open or closed.

But it doesn't hold in general. Topo can get pretty wild.

[lavish geyser]

oh ok

thanks

can someone explain pls

i dont understand

[plain raven]

Do you understand the definition of lim S?

[lavish geyser]

ye

what if S is just a sequence (x_n) and x

then lim S is just x

can the sequence be just x

so (x) converges to x?

but then lim S is always S

[deft bear]

so uhh mildly curious, can you prove JCT with solely point-set?

[abstract saffron]

Hmm, there's something called sequential characterisaton of a closed set

[unreal stratus]

There are some more elementary proofs like this https://www.maths.ed.ac.uk/~v1ranick/jordan/tverberg.pdf

[deft bear]

So the main method is show it holds for an easy shape, and show you can turn messy shapes into easy shapes

[stark fog]

the idea is to show that if you take any point of lim S, it is a limit point of lim S, hence it must be closed

[next crystal]

Let $p: E \to B$ be a covering map, with $E$ path connected. Show that if $B$ is simply connected, then $p$ is a homeomorphism.

What i have so far:
Suppose $e_0 \in E, p(e_0)=b_0$. Since $E$ is path connected, the lifting correspondence $\phi : \pi_1(B, b_0) \to p^{-1}({b_0})$ is surjective. Since $B$ is simply connected, $\pi_1(B, b_0)$ is trivial, so $p^{-1}({b_0})$ must be a singleton set. Since $B$ is connected, by a previous problem we did we know $p^{-1}({b})$ is a singleton set for every $b \in B$. \\ $p$ is continuous and surjective by definition. To show injectivity, note that if $\alpha, \beta \in E$ satisfying $p(\alpha)=p(\beta)=b$, then $p^{-1}({b})$ is not a singleton set.

[gentle osprey]

michαel

[next crystal]

all i have left to show is that p^{-1} is continuous, so if U is open in E then p(U) is open in B, but im not sure how to do this. Any hints?

[next crystal]

nvm i got it we showed p is an open map in class

[abstract saffron]

I know a very long chain of proofs that need some multivariable calculus, but yes, you can

[obsidian spoke]

are there topologies for which no basis exist?

[novel acorn]

No
Every topology is a basis for itself

[obsidian spoke]

ah true

this might be a dumb question but one of the conditions for a collection to be a basis is that if we have an element that is common in two basis elements then there is a third basis element that contains the common element and is a subset of the intersection of the previous two, can we just take the third basis element to be the intersection of the two basis elements?

[untold lily]

that's pretty much what a subbasis is

[obsidian spoke]

oh bruh, really?

[untold lily]

for a subbasis S, the set of all finite intersections of elements of S is a basis, which is sort of what you are describing

[obsidian spoke]

hmm thats interesting

in that case would every basis be a subbasis as well?

[gaunt linden]

Yes.

The word "basis" can be a bit misleading if you expect it to be like linear algebra where every vector can be made as a combination of basis elements in exactly one way. There's no such requirement for a basis of a topology; as long as each open set is a union of basis sets in at least one way, it's a good basis.

[obsidian spoke]

yeah I was trying to connect it to linear algebra, but couldn't think of an analogous idea to subbasis

[gaunt linden]

In linear-algebra terms, both the topological concepts are more analogous to "spanning set", but in two different ways. For "basis" we consider "span" to mean "everything we can get as sums of these" (where sums then becomes unions). For "subbasis" we think of the span of a set of vectors as "the smallest subspace that contains all these", and a set A is a subbasis for T if and only if T is the smallest topology that contains A.

[obsidian spoke]

These ideas are gonna take a while to settle in ngl

[gaunt linden]

Yeah, it takes some time.

[obsidian spoke]

ok so say I had a collection of topologies, ${T_{\alpha}}$ on some set $X$ then does $B = \left{\bigcup_{\beta \in I}U_{\beta} \left\mid\right U_{\beta} \in T_{\beta}\right}$ define a subbasis for some new topology? Since the union of all elements in $B$ is $X$?

[gentle osprey]

FamilyFriendly
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[gaunt linden]

Every set of subsets of X is a subbasis for some topology.

[obsidian spoke]

hmmm

in munkres he says that "A subbasis S for a topology on X is a collection of subsets of X whose union is X"

what i get from that definition is that if X = {1,2,3} then S = {{1},{2}} would not be a subbasis

unless I'm missing something

[gaunt linden]

Oh yeah, authors differ slightly in whether they require every point of X to be covered. If you define the topology it's a subbasis for as "the coarsest topology where all these sets are open", then you don't need the extra condition -- but if you define it as "all unions of finite intersections of subbasis elements" you may need it -- depending on whether you're comfortable with calling X itself the intersection of no subsets.

[obsidian spoke]

oh I see, so these are pretty much the same thing

[gaunt linden]

Eyup.

[obsidian spoke]

would this generate the coarsest topology that contains all T_alpha?

[gaunt linden]

Yes, but it's a bit roundabout. You could just take $\bigcup_{\alpha\in I} T_\alpha$ to be your subbasis.

[gentle osprey]

Troposphere

[obsidian spoke]

oh yeah you could

I didn't do that cuz I got it confused with that fact that U T_alpha may not be a topology

lol

[tough zephyr]

A proof I'm reading makes the claim that any connected subset of $\mathbb{R}^n$ is path connected via piecewise-linear paths. Yet, I see no reason how this is obvious and can't seem to find anything on this.

[gentle osprey]

Philka

[umbral panther]

It doesn’t seem true for the circle. Are you sure the hypothesis is connected and the conclusion piece wise linear?

[tough zephyr]

Actually now that I think about it I think it's necessary that the subset is open, since then you can show the set of points connected piecewise linearly to an arbitrary point is a nontrivial clopen subset of the connected set; therefore the whole set

[unreal stratus]

Yes

Also so like morally uhhh

If you know a subset U of R^n is path connected and open then given some path in U with image C, we know C is compact and admits some epsilon thickening around it within U

So we could use that to cut up C into pieces and replace it by something piecewise linear lol

But yeah easiest to just do what you dod

[gritty widget]

are these two diffeomorphic

here is a back side

[abstract saffron]

No

[willow viper]

Hello how can I show that the only continuous maps from T (trivial space) to H (hausdorff space) are the constant maps? I've tried with a map $f:X\to Y$ and $T=(X,\tau_1), H=(Y,\tau_2)$ saying that $U\in\tau_2$ open, so $f^{-1}(U) = X$ or $f^{-1}(U) = \emptyset$ which are both open and so f is continuous. But how can I show that they are constant? If that's what I wrote even makes sense?

[gentle osprey]

b3s4d

[gritty widget]

assume f is not constant and get a contradiction

"there exist x and x' in X such that f(x) is not equal to f(x')"

[willow viper]

Oooh ok let me try

[tidal lynx]

Here's a proof that we can define a topological manifold M in two different ways, one requiring the charts to carry open sets into open sets, and the other requiring them to carry open sets onto open balls (and hence also R^n by transitivity):
Suppose x is a point of M, and U is an open nbhd of x such that there exists a homeomorphism f from U into V, an open set in R^n. Then since V is open, there exists an open ball B centered at f(x) and contained in V. Now note f^-1 exists and is continuous, so we can restrict it to B to get our desired homeomorphism.

I wanted to confirm that the last step relies on two facts, and it's not just immediate some other way:

1. If f: U -> V is a bijection, then so too is f restricted to any subset of U
2. If f: U -> V is a continuous map, then so is f restricted to any subset of U (with the corresponding subspace topology on the subset and its image)
   or I guess they can be combined into
3. If f: U -> V is a homeomorphism, then so too is f restricted to any subset of U

nbhd -> nbhd
open nbhd -> open nbhd
open nbhd -> nbhd
nbhd -> open nbhd

And it looks like defining it in any of the above ways (X -> Y meaning the charts provide a homeomorphism between a set of type X in M to a set of type Y in R^n) are all equivalent by (3) above?

[tidal lynx]

---

Bumpp

[willow viper]

Hello, can I say that: Assume f is not constant, so $\exists x,x' \in X : f(x) \ne f(x')$. So in H $f(x)\in U, f(x')\in V: U\cap V = \emptyset$. But with $f^{-1}(X\cap X) \ne \emptyset$ this is a contradiction?

[gentle osprey]

b3s4d

[willow viper]

I don't think this is correct because U must differ from V right??

[gritty widget]

did you mean f^{-1}(U cap V)

f^{-1}(X \cap X) makes no sense

[willow viper]

Yes I've changed the U and V with the X

Because I thought that $f^{-1}(U\cap V) = f^{-1}(U) \cap f^{-1}(V)$

[gentle osprey]

b3s4d

[gritty widget]

this is true

[willow viper]

But I can't replace U and V with X at the same time right?

[gritty widget]

.

[willow viper]

Yes hm ok I see it now

[gritty widget]

why is f^{-1}(U \cap V) non-empty

[willow viper]

maybe like this: f^{-1}(U cap V) = f^{-1}(U) cap f^{-1}(V) = X cap X = non-empty??

[gritty widget]

that is correct

and this gives a contradiction because?

[willow viper]

Because U cap V must be empty

[gritty widget]

so the pre-image is also empty. there you go

[willow viper]

But I don't understand

Why can I choose X for V and U

I thought they are disjunctive neighbourhoods

[gritty widget]

are you asking why f^{-1}(X \cap X) makes no sense or are you asking why f^{-1}(U) \cap f^{-1}(V) = X \cap X

[willow viper]

The first one I get

But for example I choose x, x' from U and V so U and V must be seperate neighbourhoods from the beginning but then I choose U and V to be equal

[gritty widget]

no

you're not choosing U and V equal

f^{-1}(U) and f^{-1}(V) will be equal. the pre-images, not the sets themselves

[willow viper]

Oooooh

Ok true my bad

thank u

[gritty widget]

why are f^{-1}(U) and f^{-1}(V) both equal to X

[willow viper]

well they can also be equal to the emptyset or a combination I suppose?

[gritty widget]

no

[gritty widget]

you wrote, essentially, f^{-1}(U) = X here. why is this true

i am not saying it is not true

i am saying you haven't justified why it is true

[willow viper]

f^{-1}(U) \subset X and f(X) \subset U <-> X \subset f^{-1](U) so f^{-1}(U) = X

[gritty widget]

f(X) \subset U
why

this doesn't make sense to me at all

[willow viper]

Ooooh

[gritty widget]

remember you're using the trivial topology on X

what are the open sets of X

[willow viper]

you're right only f(x) \subset U

[gritty widget]

\in

[willow viper]

not the whole X

well X and the emptyset

[gritty widget]

ok

f^{-1}(U) and f^{-1}(V) are open sets of X

so what can they be

[willow viper]

emptyset cap X
emptyset cap emptyset
X cap X??

[gritty widget]

those would be the possibilities for f^{-1}(U) \cap f^{-1}(V)

so are either of f^{-1}(U) or f^{-1}(V) empty?

(neither is empty. so both equal X)

[willow viper]

Right because that's what brings the counter example

[gritty widget]

justify

[willow viper]

But what is there to justify

[gritty widget]

that they are non-empty

[willow viper]

Well they have to be to satisfy f^{-1}(U) cap f^{-1}(V) != emptyset

[gritty widget]

you are proving this

by showing that f^{-1}(U) and f^{-1}(V) are both equal to X

from which you will deduce that the intersection is not empty

[willow viper]

Oh okay so I can't just choose them

[gritty widget]

you are showing that they are equal to X by showing that they are not equal to the emptyset

this works because you put the trivial topology on X

you cannot just choose them but this is not a problem

so why are they non-empty

[willow viper]

Ok

Because f maps the emptyset to the emptyset

So f(emptyset) != U

[gritty widget]

x is in f^{-1}(U)

so it is nonempty

[willow viper]

Oooook

then it's easy because x is not an element of the emptyset

[gritty widget]

interesting way of saying "it contains an element so it is nonempty"

[willow viper]

Oh ok that also works xD

[willow viper]

hold on, then here the fourth line is basically wrong because f^{-1}(U) != \emptyset

[gritty widget]

?

you said "or"

and you got one of the possibilities

you also seemed to conclude from that that f is continuous, but you are supposed to assume this, not prove it

you need to know that f^{-1}(U) is open in X to say it's equal to the empty set or to X. you need continuity of f for this

[willow viper]

Oh that makes a bit more sense now

[obsidian spoke]

Im confused as to whether or not the set $(0, 1) \setminus K$ where $K = {\frac{1}{n} \mid n \in \mathbb{N}}$ is open in $\mathbb{R}$ with the standard topology

[gentle osprey]

FamilyFriendly

[obsidian spoke]

I believe that it is open

since we just have it like (1/2, 1) U (1/3, 1/2) U ...

which is just a countable union of open sets

[gritty widget]

any union of open sets is open

[obsidian spoke]

so this set would be open then?

[gritty widget]

seems like it to me

[obsidian spoke]

ight thank you

[willow viper]

Hello, I'm trying to prove that every element of B_1 is an open subset of Rn, but I'm really not sure how.. B_1 will look like B_1 = {{y1}, ..., {yk}}, but then they already are open subsets of Rn??

[bitter smelt]

huh?

I am not sure what you are saying but you usually do this by using Hausdorff's criterion

then find nested neighborhoods

[willow viper]

I was trying with this

[bitter smelt]

its hard to write a cube as a union/intersection of balls

you'll want to use some theorem, probably

[willow viper]

Okay

[willow viper]

I don't know this explanation came up followed by the example I've shown

[bitter smelt]

you mean the question came immediately after the definition?

[willow viper]

basically yeah

[bitter smelt]

realistically, just prove this. its easy

[willow viper]

that's it

[willow viper]

Okay I'll try to

[bitter smelt]

You can also use 2.40.

(Its actually just the same thing)

[willow viper]

it did look a bit familiar to yours

but I wasn't sure

maybe because I didn't understand it

[bitter smelt]

You can also use 2.40 to prove the critereon

(and the criterion is a generally useful thing to know)

using 2.40 directly for 2.42 is just a little more annoying and will involve actual numbers and such

whatever you prefer

[willow viper]

okok I'll try it with the 2.40

[vast estuary]

the identity map is continuous, and the product of two compact sets is compact, so the graph of the identity map is also compact?

am i missing something? this is a qual problem and seems too good to be true

i suspect something goes haywire if X is not hausdorff or something

[lime sable]

definitely a weird question

[red yoke]

Is not a hard question

Just don't accidentally assume X is Hausdorff or sth

[high quarry]

well, there is a result that shows that the diagonal of X is closed in XxX if and only if X is separated (can you prove it?), here we have something a little similar, but with compacity :>

[lime sable]

isn't this only a thing in AG

i was thinking about that

hausdorff (separated X -> *) iff the diagonal is closed, compact iff universally closed (pullback of Y -> * along X -> * is closed, where * is the point)

in AG, the diagonal is quasi-compact as a map iff quasi-separated

[abstract saffron]

[lime sable]

but none of these are relevant to the situation at hand lol

[abstract saffron]

Chill, it's point-set topo

[lime sable]

yeah i saw diagonal though

isn't it weird for that kind of question to be on a qual?

[languid patrol]

This is sort of right but it has nothing to do with the product of compact sets being compact, all you need is that the map $x \to (x, x)$ is continuous.

[gentle osprey]

Topos_Theory_E-Girl

[abstract saffron]

Was about to say didn't it follow from closed graph thm.

[high quarry]

(AG? XD)

(no it's quite general, but my point really was to show that we needed the space to be Hausdorff, or else we would get something not closed, and thus not compact if everything is, lets say, metric)

[lime sable]

not sure what you mean. usually people say "hausdorff" instead of "separated" outside of algebraic geometry for the analogous condition, but the result of the problem follows because continuous images of compact sets are compact

metric spaces are automatically hausdorff so that counterexample doesn't work

[high quarry]

wait, you are right, i'm stupid ; I'm looking for a "set" of non-hausdorff spaces but such that any compact spaces of this set is closed

[high quarry]

I think it's quite langage-dependent ; in my french courses, we used both. Also the notion of compact in french include separability (the english compact is called *quasicompact [changed])

[lime sable]

ah, my mistake

[languid patrol]

Yes it is a common confusion when reading French texts on topology, I've seen many people ask questions like "isn't there a hypothesis missing here or do I not understand the grammar?"

[umbral panther]

If the space is compact but not Hausdorff (often called quasi-compact) then the diagonal is not closed. But it is still compact! The image of a compact set under any continuous map is again compact