point-set-topology

[plucky veldt]

just 59 pages, I'll probably skim through it to hammer down the concepts

this is nice, should be a good complement to topology without tears

then off to algebraic topology we go

it seems to me like twt sometimes lacks the motivation behind the theory

[strange folio]

this looks super nice, I might use this for something

[strange folio]

I think the motivation behind the definition of a topological space is tricky, which is why it took so long for people to come up with the correct definition. I think it is better to think of point set topology as foundational, and let the motivation sink in as you appreciate how nicely all these other things fit within it.

[small obsidian]

Big thing to remember is that we didn't discover math in the order that we teach it

Sometimes it's about digging in to find "oh this is why we discovered this"

[sleek thicket]

I haven't really been convinced that our definition of topological space is the 'right' one. Ig it's useful because it's super general

And it's a good place to build on

(eg it'd be awkward to define manifolds or cw complexes or schemes or w/e without it)

[tough imp]

Something something site

something something Grothendieck

[small obsidian]

It's right in the sense that you can completely describe continuous mappings with the relatively simple rules of topology

But who knows what we're possibly missing? Haha

[strange folio]

oh yeah, who knows maybe even better definitions will be found

but I would say, the biggest thing I try to impart on students is that definitions are not handed down from above, and that if you're asking about definitions, you're asking the wrong question. We can define all these things to mean whatever we want. But there are a very small number of definitions that I would say that doesn't apply to, and topological space is one of them

(What I mean by that is that in point set topology, the definition is the content)

[sleek thicket]

That makes sense

Regardless of whether a better defintion exists/whether we'll use something else in a century, topological spaces are what is used in math right now

[strange folio]

[Though the fact that people disagree about the definition of compact is one of my biggest petpeeves.]

[sleek thicket]

Oh like compact vs quasicompact?

[tough imp]

do you include hausdorff.... oh no????

[sleek thicket]

I read that in Hartshorne/another AG book and got super confused, but I figured it would never come up in practice

then I talked to someone on here and we confused eachother

Because they were a native French speaker

And their topology class used "compact" to mean "compact+hausdorff"

[strange folio]

I vote for not including hausdorff, for what its worth.

[sleek thicket]

Me too

[strange folio]

though if you allow people to get away with sneaking "second countable" into the definition of manifold, I see how they think they can get away with this

[sleek thicket]

hey now

I like my manifolds second countable

Give me the baire category theorem or give me death

This actually came up with a comment of mine on reddit a couple days ago

I said a continuous bijection between manifolds is a homeomorphism

[strange folio]

oh yeah, I mean, I like the second countable there too. But then, you convince geometers they can get away with throwing random things in definitions.

[sleek thicket]

someone objected about the discrete topology on R^n

Ahh makes sense

[tough imp]

I can just google those lol

[sleek thicket]

What you said was write

Yeah I want to say grothendieck initially defined schemes as separated? I might be misremembering

[tough imp]

I mean

the thing we call a scheme

he called a prescheme

I don't remember what property it was but a scheme as he defined it was some nicer scheme for us modern ppl

It might have been separated

[strange folio]

hehe a friend of mine always argued we should call subgroups presubgroups

and then call normal subgroups just subgroups

[sleek thicket]

me and Mathemagician had a friend who said we should define subgroups as images of group homomorphisms and normal subgroups as kernels

[tough imp]

HAHAHA

yeah that was pretty awesome

[sleek thicket]

~duality~

[tough imp]

shamrock got a reddit reply about how to introduce normal subgroups

and the response was as a fiber of 0 in the quotient or something

It was an... interesting take

[sleek thicket]

I still don't know how to explain it to students

I think max did a good job

[tough imp]

Idk, I'd just be like

lmao when you conjugate it, it stay the same

now you might be asking wtf is this, who cares

Cuz now I can make quotient

Why do we care about this?

Oh time ran out sorry see ya

[strange folio]

I do homomorphisms first

and then define kernels

and then ask the question "what can be a kernel"

and show, by basically brute force, that you can't make a subgroup of order 2 in S_3 a kernel

[tough imp]

I just don't think on a first pass there's really even a particular reason why someone would care if something can be a kernel

[strange folio]

Well, the trick is: once you see something can be a kernel, you know a homomorphism

[tough imp]

Maybe this is lazy on my part, but I feel like your first introduction to quotients and stuff like that is just gonna be a clusterfuck and super confusing

[strange folio]

just for free

[sleek thicket]

Idk that sounds pretty defeatist

[tough imp]

I mean... I guess

[strange folio]

As an instructor, they don't need to understand every step you do. But they need to see your tone of "this makes sense and it is logical to go from here to here to here"

[sleek thicket]

this reminds me of when my manifolds course introduced immersed submanifolds

In like, week 2 or 3

The motivation shows up at the end of the second quarter of the course

[marsh forge]

I mean the inuition of quotients being kernels

[sleek thicket]

And it's very hard to justify until then

[marsh forge]

is like

the defining idea

[tough imp]

I guess it doesn't really help that as a student I just sort of... listen to lectures and just learn the stuff. I don't really care all that much about motivation for stuff since it was cool stuff and I like it. Examples also don't really help me so it's a weird position to be in the teaching role since what works for me doesn't work for a lot of people it seems

I guess it doesn't really help that as a student I just sort of... listen to lectures and just learn the stuff. I don't really care all that much about motivation for stuff since it was cool stuff and I like it. Examples also don't really help me so it's a weird position to be in the teaching role since what works for me doesn't work for a lot of people it seems

I guess it doesn't really help that as a student I just sort of... listen to lectures and just learn the stuff. I don't really care all that much about motivation for stuff since it was cool stuff and I like it. Examples also don't really help me so it's a weird position to be in the teaching role since what works for me doesn't work for a lot of people it seems

[sleek thicket]

wym max?

[strange folio]

Oh yeah, that's definitely a big level up as a teacher. You're not teaching to a classroom full of yourself.

[sleek thicket]

Oof, mood

I get very jazzed about like active learning and involving students

And in my first quarter TAing I went super hard on that

And then realized I was making students uncomfortable by calling on them before they'd had time to process stuff

[tough imp]

I don't know, I find that I like to see like applications. If you introduce the notion of a flat module, after you show that free modules are flat or something, seeing a proof that uses flatness to get something cool helps me a lot more than like showing 8 different modules are flat.

But some people really like to just see examples before any sort of application so I have to like

take a step back and be like "okay, do you guys want to see an example?"

[strange folio]

I remember my advisor telling me "Before you can udnerstand any algebraic geometry, you really need to get a death grip on flatness".

I still don't understand anything about flatness ;-P

[sleek thicket]

lol

[strange folio]

(I like to think I understand some algebraic geometry haha)

[sleek thicket]

It just means curvature zero, don't worry about it

[tough imp]

I feel like flatness for modules I get, but not for like the AG version of flatness haha

[sleek thicket]

What's the difference?

I don't know what you mean by ag version I guess

[tough imp]

Uhhhhh

[marsh forge]

see algebraic geometry fucked up by not including algebra in the geometry

topology understands that algebra is a tool

[sleek thicket]

Oh like flat ring/scene map?

[marsh forge]

see algebraic geometry fucked up by not including algebra in the geometry

topology understands that algebra is a tool

[strange folio]

yeah like a flat morphism of schemes has fibers all of the "correct" dimension

[tough imp]

yeah the scheme one

just becomes a lot more confusing

[strange folio]

(I think)

[tough imp]

cuz sheaf, and then stalks

[strange folio]

yeah like a flat morphism of schemes has fibers all of the "correct" dimension

[tough imp]

and all that jazz

[strange folio]

yeah like a flat morphism of schemes has fibers all of the "correct" dimension

[tough imp]

cuz sheaf, and then stalks

and all that jazz

[strange folio]

(I think)

[tough imp]

cuz sheaf, and then stalks

[sleek thicket]

It shows up in the weirdest places in algebra

I was reading something in Hatcher last quarter and I realized that you didn't really need free, only flat and finitely presented

It's just like a randomly good condition

That shows up

[strange folio]

Yeah, I just never found that germ of an idea that made it click intuitively

I always feel like I'm reasoning purely formally about it

[sleek thicket]

Same

[strange folio]

Yeah, I just never found that germ of an idea that made it click intuitively

[sleek thicket]

This is a reason that teaching things is useful for me

As a student I can slip into "okay just manipulate this purely formally" for a lot of stuff

But if I'm explaining it to someone else, I need to come up with some amount of motivation

[strange folio]

Yeah, definitely

though teaching flat/projective/injective modules did not do that for me haha

[sleek thicket]

Lol

I don't have an intuition for projective/injectives. I have a hope that learning more about vector bundles might make projectives click but ehhh

A friend of mine was like "oh yeah projectives and injectives will make perfect sense when you learn about model categories"

She said the same thing about chain homotopy

[tough imp]

who is this?

[sleek thicket]

It wasn't a helpful explanation either time

Olivia

[strange folio]

and here is where if you have a certain kind of brain you go "oh this is a really natural looking universal property" and that's all the motivaiton you need.

[tough imp]

Oh lmoa

I heard apparently that at some talk Lurie was asked a question like

"what's the geometric intuition for this"

[sleek thicket]

Zeta I used to think I was like that

[tough imp]

and his answer was that it was a contravariant functor

[sleek thicket]

Then I realized i didn't understand things

[strange folio]

Yeah, I'm an arithmetic geometer, and I do not like to think about commutative algebra at all.

[sleek thicket]

and had no reference point

[tough imp]

I do not claim this is true, I just heard it secondhand

[honest narwhal]

Sounds like my undergrad nt prof

He was like yeah fuck commalg

[sleek thicket]

I successfully avoided learning commutative algebra

[tough imp]

Wtf no you didn't

[sleek thicket]

I stopped going to class in the last couple weeks of last quarter

And then the final was canceled

[tough imp]

You avoided learning it when we covered it in class

[sleek thicket]

yeah I don't know anything about primary ideals

[tough imp]

by having self studied it a quarter early

[sleek thicket]

I was lost on some of the dimension theory stuff

506 did things in a lot more depth

[honest narwhal]

I don't know any commalg rn

[tough imp]

join me in reading matsumura

[honest narwhal]

Prob should just A-M speedrun

Or Matsumura that too

[sleek thicket]

absolutely not lmao

[tough imp]

Not you

you're "busy doing important things like" "research" and your "job" and stuff

[sleek thicket]

I have better things to do like get confused about whether the symmetric group acts on the left or right

lol

that took me like 2 hours together

Getting very confused

[strange folio]

Yeah, commutative algebra is like dental work to me. Which is funny becuase I love so much algebra stuff. But I cannot write down a short exact sequence or a chain of ideals without falling asleep.

[tough imp]

Lol on our 504 midterm I didn't want to bother learning how to make semidirect products work on the other side

So like, I just flipped how it works and just stated it works

but I got full points

The key is to just redefine it whenever you want it on one side

🤔

[sleek thicket]

oop

[strange folio]

let me tell you, wheny ou're grading a proof, there is a really strong tendency to go "I could look at this for 5 minutes and realize it isn't quite right and take off two points, or I could just give it full credit"

[sleek thicket]

that doesn't work when I'm reading a paper

[strange folio]

and there is a really strong pull towards the latter.

[sleek thicket]

Yeah zeta that's a mood

Honestly it's worse grading code

I have done both in the same class

[tough imp]

I mean what I said was true, but I really didn't want to reprove the existence of semi direct products haha

[sleek thicket]

Mine numbing

[strange folio]

I really love the fact that groups are isomorphic to their ops but rings aren't.

[tough imp]

grading code?

Just run it and see if it compiles :^)

[sleek thicket]

yeah I have caught bugs the test suite doesn't

By reading student code really closely

Also yeah zeta that's pretty weird

My algebra prof claimed that semisimple rings were iso to their opposites

As a consequence of Artin Wedderburn

Turns out it's false even for division rings

Which is pretty cool

Actually iirc the counterexample involved some number theory (brauer groups?)

[strange folio]

I love that there are rings that are PLIDs but not PRIDs

[sleek thicket]

oof

[strange folio]

(love in the sense that I never want to be forced to think about an example)

[sleek thicket]

There's a lot of bad rings

In my REU we're dealing with operads a lot

And like, monoidal categories

[strange folio]

but yeah, number theory is like, a machine that generates ugly division rings

[sleek thicket]

And there's a lot of really really weak algebraic structures

Like things that aren't associative

We unironically had to think about magmas the other day

It sucked

[tough imp]

OOOO

That reminds me of my favorite fucking proof on proofwiki

[strange folio]

oh god, I'm terrible about not using associativity

the fact that the notation x^5 assumes associativity is sad

[tough imp]

https://proofwiki.org/wiki/Idempotent_Magma_Element_forms_Singleton_Submagma

[sleek thicket]

Lol

[tough imp]

By Singleton of Element is Subset:

By the definition of submagma, the result follows.

■

[strange folio]

How very profound

I bet this will be helpful in my next paper

[sleek thicket]

Yeah the like, most important thing I'm defining involves "non-associative monomials"

[strange folio]

omg I'm so sorry

[sleek thicket]

which are products of {x0, x1,...,xn} where you can use x0 any number of times and must use x1 through xn exactly once

but you need to keep track of parentheses!!!

Yeah it's baaaad

why can't the tensor product be strictly associative

[strange folio]

something something Catalan numbers?

[sleek thicket]

In a perfect world...

[tough imp]

Wait, lie algebras aren't associative right?

[sleek thicket]

The lie bracket isn't, no

[strange folio]

Lie Algebras aren't associative, but they're almost associative

[tough imp]

hmmmmm

[sleek thicket]

Actually

That's sort of telecast

*relevant

[tough imp]

I was debating taking the lie algebra course in fall, but if non-associativity is this trash

[sleek thicket]

There's an operad that defines semigroups

[tough imp]

do I really want to subject myself to it 🤔

[strange folio]

I would not worry about dealing with non-associativity in a lie algebras course

[tough imp]

I think I honestly should

[strange folio]

a lie algebra is very highly structured

[sleek thicket]

Or rather assosciative algebras

[tough imp]

Since I will never ever learn that stuff unless I take a course

[sleek thicket]

And it's dual in a precise sense is the operad which defines lie algebras

Which is kind of cool

[strange folio]

That is the course I most wish I had paid attention to in grad school that I did not

[tough imp]

that's pretty cool

[sleek thicket]

I didn't think I would be interested in taking it

(we're at the same school zeta)

but then people convinced me to

And I got super hyped

And then found out I don't have time

r i p

[tough imp]

Sandor told me that specifically because I'm not interested in it that I should take it haha

[sleek thicket]

lol

[strange folio]

taht sounds like having me as an advisor 😛

[tough imp]

I mean, that's a huge area of math

[sleek thicket]

I was told "if you care about geometry/lie groups you need to take this course"

And then they were relevant to my research this summer

[tough imp]

and I will never ever know any lick of it unless I took a course in it

[strange folio]

it's intimately connected with lattices and those show up all over the place in good maths

[tough imp]

honestly I've still managed to come this far

barely knowing the definition of a lattice

Literally the only thing I think of is a grid of rhombuses in R^2 when you say lattice lmao

[sleek thicket]

I tried to teach 398 students and you wouldn't let me

Perpetuating the cycle

[strange folio]

I mean, grid of rhombuses in C is already getting into super deep shit about elliptic curves

so really, even if that is your only picture, you're already in deeper than yo uthink

[tough imp]

🥴

Uhhhh ohhhhh

[strange folio]

hey elliptic curves are the good stuff!

[tough imp]

Honestly I always end up liking the stuff I think is gonna mega suck

[sleek thicket]

You know what we should do

next year

Over zoom

In the spirit of stuff we think will suck

[tough imp]

Doing some slight topological groups stuff in complex was actually fun. I still don't like reals tho. this is why I will never do number theory

[sleek thicket]

number theory reading group with Thomas

[tough imp]

Because what if I...

[sleek thicket]

hahaha yes

[tough imp]

start to... like it

[strange folio]

well I went to grad school saying "I will absolutely not be a number theorist"

and then look what happeend

[sleek thicket]

Lol

[strange folio]

I turned into a rainbow frickin zeta 😛

[tough imp]

I haven't slaved away and spent like two weeks solving hartshorne problems all day

just to not do AG

[sleek thicket]

I have no idea what I want to do in grad school tbh

[tough imp]

Oh wait, I need that for ANT anyway

🥴

[sleek thicket]

there's a lot of fun math out there

[strange folio]

anyway folks, I'm crawling off to bed, good night!

[tough imp]

G'night

Oh boy, what did we do to the topology and geometry chatroom

[strange folio]

hahaha poor thing

at least you said Hartshorn enough times, that should mollify it

[sleek thicket]

manifolds manifolds manifolds

[tough imp]

Oh yeah, #geometry baby

[sleek thicket]

vector fields

lie derivatives

[tough imp]

uhhhh

[sleek thicket]

uhhh foliations

[tough imp]

parallel postulate

[nimble jolt]

sard's theorem

[bitter yoke]

Falting's theorem

[nimble jolt]

Weyl law

[sleek thicket]

oh I love sard

it's my happy place

• Rank Theorem
• Sard's Theorem
• Whitney fun bag
• tubular neighborhoods exist
• stokes'
choose your fighter

I really do love smooth manifolds

[tough imp]

What the heck is Whitney fun bag

Is that just something you threw in or is that real

[nimble jolt]

sards is one of those crazy bang for your buck theorems

another that comes to mind is Baire's in real analysis

it just does so much work and goes into so many things.

I think he was putting together in a bag whitney embedding/approximation (at least).

[sleek thicket]

Yup

[tough imp]

Oh I see

fun bag is like an adjectvie here

[sleek thicket]

baire is good

Also for manifolds

this is why I want second countable in my definition of manifolds

[honest narwhal]

Sard and BCT are fucking excellent

[floral gust]

I searched whitney fun bag expecting it to be a theorem

got bag pictures ):

[dire warren]

Theorem (Whitney fun bag):

Around every point of a flow on a manifold, there exists an open neighbourhood of the point (called a fun bag) such that..

[thin jewel]

Can anyone who is familiar with Several Complex Variables tell me why Definition 2.1 in this paper is integer valued?

https://core.ac.uk/download/pdf/82749806.pdf

I can't seem to figure it out

[sleek thicket]

oh shit 22 you're right

I forgot about existence of flows/solutions to ODEs

And more generally frobenius

[dire warren]

I like the flow box theorem for flows on manifolds

[sleek thicket]

What's that?

[dire warren]

Around every non critical point of the flow there exists a neighbourhood (called flow box) and coordinates one the neighbourhood such that the flow is conjugate to that of the vector field (1, 0, ..., 0)

[sleek thicket]

oh, yeah that's a good lemma

even cooler is that it works for multiple vector fields

if X1,...,Xk are linearly independent at p and [Xi, Xj] = 0 for each i, j, then there's coordinates near p where Xi = d/dx^i

[dire warren]

Oh wow didn’t know about this

That’s legit lol

[sleek thicket]

Yeah you use the fact that commuting vector fields have commuting flows

Which is also great

[limpid copper]

This is known as Frobenius theorem. You can also reformulate it in terms of ideals of differential forms, which is even cooler.

[nimble jolt]

I think it's a bit easier than local Frobenius because in his version the fields actually commute.

[limpid copper]

Right, his version is the only technical part in the proof of Frobenius. Nonetheless it is not exactly the same. :/

[paper crag]

Can someone tell me how the '+' in the picture below is defined ? It is from https://arxiv.org/pdf/1701.04862.pdf paper. Also, I don't think if dim(M)+dim(P)< d the manifolds can intersect transversally, but still the paper tries to proof that they don't intersect transversally using some other logic in Appendix. Please help . I would be grateful

[elder yew]

I think the addition is just defined by a "translation" of each point lying on the manifold

e.g.

If E is a subset of the reals, then E + x is just the set of y + x, where y is in E

[paper crag]

And x is a constant ?

[elder yew]

Usually it is, but in your case it's a continuous random variable

[gentle osprey]

Jeevesh:

[elder yew]

That would be my best guess

[paper crag]

Also , can you tell anything about the second problem ?

[elder yew]

So transverse intersection, if I recall correctly, is when the tangent spaces of the manifolds don't "line up"

They aren't linearly dependent

To say that they don't intersect transversally is to say where they intersect, the tangent spaces are linearly dependent

(I'm a little uncertain here)

I have no clue how that would be proven using dim(M) + dim(P) < d. Can you do contradiction? Say they are transverse and try a dimension counting argument?

dim(M) = m, dim(P) = p

Tangent space chops a dimension?

[paper crag]

Sorry I got something little wrong in my head.

[elder yew]

It seems like you'd need more dimension in order for them to be linearly independent

That's what my gut says

[paper crag]

I hope this clears things

[elder yew]

what're the dimensions of

T_x(M), T_x(P), and T_x(R^d)

we know that dim(M) + dim(P) < d

But it should be the case that you'll get

dim(T_x(M)) + dim(T_x(P)) = dim(T_x(R^d)) = d

Which (hopefully) contradicts the above

Yeah dim(T_x(M)) = dim(M)

etc.

[paper crag]

$m+p \le d$ is a case for proving lemma 2.

When $m+p \le d$ we need to prove that the intersection of M+$\eta$ and N+$\eta '$ is empty with probability 1.

In case, $m+p \geq d$ , we can say by General position Lemma that their intersection will be empty.

[elder yew]

So now you need to show, by virtue of transversality, $$ dim(T_x(M) + T_x(P)) = dim(T_x(M)) + dim(T_x(P)) - dim(T_x(M \cap P))$$

[gentle osprey]

MoonBears-C-:

Jeevesh:

[elder yew]

Yet : $$ dim(T_x(M \cap P)) = 0; dim(T_x(M)) = dim(M); dim(T_x(P)) = dim(P) $$

[gentle osprey]

MoonBears-C-:

[elder yew]

substitute it all in

So they can't intersect transversally

[supple locust]

I know that for oriented smooth manifold M without boundary, there exists a smooth non vanishing n form. Does this also hold for if we consider manifold with boundary?

[sleek thicket]

I think the same proof as for the without boundary case should work

You have an orientation

This defines a convex subset of the space of top degree forms at each point

Use partitions of unity

Something like that?

[supple locust]

ya

[gentle osprey]

Schuams:
Compile Error! Click the reaction for details. (You may edit your message)

Schuams:
Compile Error! Click the reaction for details. (You may edit your message)

[dusk heron]

Let $g$ be a Riemannian metric on the open unit disk ${(x,y)\mid x^2+y^2<1}$ which is complete and conformal to the Euclidean metric $dx^2+dy^2$. Can the curvature of $g$ go to zero as we approach the boundary circle?

[gentle osprey]

gustavn64:

[dusk heron]

Intuitively it feels like the completeness should force it to look like the hyperbolic metric near the circle, but I don't know how to justify that formally.

[tough current]

would someone care to help me with this one

[gritty widget]

that belongs in #geometry-and-trigonometry

@tough current

[tepid nest]

haha jeez

[gritty widget]

after reading the discussion in that channel i think this would have better luck in r/cheatatmathhomework....

😐

aaand he left

[coarse kestrel]

Lmao

[gritty widget]

I don't know whether this is the right channel or not, but I will ask anyway

how can I prove this :"let C be a conic and A and B two points on C, and let D and E be points on line (AB), then there exists a projective transformation which fixes lines (AB) and sends D to E, A to B and B to A"

nope

#geometry-and-trigonometry 4

[plucky veldt]

well, this channel has geometry in the name as well

[tepid nest]

it has topology in the name too

[gritty widget]

dude how is that related to trigonometry

that's projective geometry

geometry-trigonometry and topology-and-geometry can probably be confusing if there aren't any rules for content laid out clearly

i wouldn't think projective geometry lies under "pre-university" (maybe its not a canadian thing to teach that in high school), so i see no issue with it going here since "early university" doesn't really have a category for it

I agree

[coarse kestrel]

Idk why I can’t find a projective geometry book that doesn’t involve algebraic geometry

To me it feels like you shouldn’t need algebraic geometry for projective geometry, but I might be horribly wrong

[gritty widget]

how is this related to our discussion

because "requiring algebraic geometry" is a pretty significant way to determine which channel this belongs in, id think

im not the person youre replying to though; i don't mean to put words in their mouth

[bitter yoke]

@coarse kestrel algebraic geometry is like the reason people care about projective geometry in math

[tough imp]

I don’t know if that’s quite right. I think projective geometry has historical intrigue outside of the AG context

From my understanding

[gentle osprey]

slimvesus:

[honest narwhal]

So fwiw even if proper containment is what holds the whole proof is still fine

I'll think about it some more but there's a chance it's errata

Yeah it's errata

https://sites.math.washington.edu/~lee/Books/ISM/errata.pdf

@sleek thicket you must answer for Lee's crimes

[sleek thicket]

lol

[honest narwhal]

This is not too many tiers below Weibel level

[sleek thicket]

Consider the relative sizes

Most of those are minor

[honest narwhal]

no u

[narrow oar]

Hey guys, my university just got a new course "intro to topology" and the prereqs for it are analysis 1 (which I've taken) and a course on rings and fields (which I've not taken). I was wondering if I would be completely lost taking the course on topology while concurrently taking the rings and fields course? I have some very basic understanding of rings and fields from analysis 1.

[sleek thicket]

we don't know

It depends on the course content

You'll probably be fine

But it's hard to say without seeing a syllabus

[narrow oar]

yea.. I would share it but it hasn't been released yet unfortunately

[dim meadow]

Oh dami I got an invite for zulip for agitocc

[bitter yoke]

I also got one

[dim meadow]

I guess we're not in the same group lmao

[tough imp]

I think you can request

Also, where did you get that

👀

just in an email?

[narrow oar]

@sleek thicket the topics covered are these:
Point set topology: definition, continuous maps, homeomorphisms, product and quotient topologies, Hausdorff topologies, connectedness, compactness and compactifications. Algebraic topology: paths, homotopies, fundamental group, universal covering spaces.
If this tells you anything let me know 🙂

[sleek thicket]

yeah you should be fine w/o rings and fields

But you need to know group theory

Like, you should know what a group action is

And what a group is

[narrow oar]

Ok, thanks for the heads up!

[honest narwhal]

@dim meadow sad reacts only

[plucky veldt]

I think point set topology has almost nothing to do with abstract algebra

[sweet wing]

that list has some algebraic topology🙃

[plucky veldt]

Yeah, but before you get to it, you should have more than enough time to brush up on fields and rings

Does it mean A has no points that are isolated in T or in the induced subspace

the first one is trivial as (X, T) has no isolated points if its a perfect space

[sweet wing]

i dont really see fields and rings appearing lol

prob isolated as in the subspace topology

[small obsidian]

Indeed algebraic topology is pretty strictly groups

[plucky veldt]

for the second one, in an indiscrete space all points are a limit point of every non-empty subset, so its a perfect space and every non empty subset is dense, including a singleton set which obviously has an isolated point in its topology, so wouldnt that contradict the claim?

I suspect I made a mistake somewhere but I can't see it

@sweet wing that guy said rings and fields is a prerequisite for his intro to topology course

[small obsidian]

Found the question haha

[plucky veldt]

okay?

did you find the answer?

[small obsidian]

From what I am reading, a subset is perfect if it contains all of its limit points (therefore all perfect sets are closed!) and if every point in the set is a limit point (implying it has no isolated points)

Bear with me I'm at work so I might respond slow haha

Now, just because a space is perfect doesn't imply all of its subsets are.

You made a perfect example - every indiscrete is perfect, but every singleton subset of it is not

Unless the space itself is a singleton I guess

[plucky veldt]

but every non-empty subset of an indiscrete space is dense

including singleton subsets

[small obsidian]

Oh haha. A point is isolated if the point itself is an open set

[plucky veldt]

right, so a singleton subset obviously has an isolated point in its space

which means its a dense set with an isolated point

[small obsidian]

Right, but that's fine, because the singleton has no isolated points

[plucky veldt]

how so?

[small obsidian]

There's no open singleton

So there's no isolated points in the indiscrete

Ergo no subset can contain one

[plucky veldt]

okay, but the singleton subset has an open singleton in its induced subspace

[small obsidian]

There's no induced topologies happening here. Just subsets of the topology

[plucky veldt]

okay

so the answer is trivial

if (X, T) is a perfect space then it has no isolated points, therefore no subset of it can have an isolated point

the question asks if A that is an open set or a dense set, then A has no isolated points

obviously it has no points which are isolated in T, but does it have isolated points in the induced topology?

[small obsidian]

That seems too simple, you're right. I'm messing this up somewhere

[plucky veldt]

i asked if the question is about induced topologies or (X, T), @sweet wing said its probably about the subspace topology

[honest narwhal]

Even if Hatcher in particular might not use ring/module theory, it does come up a decent bit. Field theory not so much

[sleek thicket]

I don't think rings come up until like, cohomology

modules too

Maybe that's just my journey in topology

[gritty widget]

can anyone give me a hint with this

i’m not sure how to start

I(x) is just the family of neighbourhoods of x

[tough imp]

This is just

Oh shit uhhhh

Let me think I thought this was just definition of closure

Since {x} = its closure

[gritty widget]

no lol that’s what I thought at first too

[tough imp]

Wait no it is

So the inclusion < is immediate

Since x is in all sets U right?

All U are nbd of x

So they contain x

[gritty widget]

wait but nbhds in I(x) could be open

[tough imp]

Doesn’t matter

They contain x

So their intersection contains x

[gritty widget]

well yes

oh yes ok I see

[tough imp]

Yeah so {x} is a subset of it

[gritty widget]

but the other direction

[honest narwhal]

Take any point that's not x

It's a finite subset so it's closed

The complement is an open neighborhood of x

[tough imp]

Oh Dami here with the omega plays

I realized my proof of the other direction fell through so I’m glad u came in

[gritty widget]

ok ok thanks I’ll try finishing it off

[honest narwhal]

So that should give T1 => {x} = blah

I'll let you try the other direction first yeah

[plucky veldt]

this is actually how you could construct the closure of {x}, yeah

personally, I'd first prove that every finite subset is closed iff every singleton is closed

that simplifies things a bit imo

and that is a definition of T1 that I'm familiar with

[small obsidian]

@plucky veldt
I've done a bit of investigating with a few others into the question and I think it's typo'd.

If A is merely a subset, then my argument works and no subset has isolated points, simply because there are none.

If A is the subspace topology, then you made a counter-example. Take the indiscrete space. The singleton is dense, and its subspace topology trivially has an isolated point.

[plucky veldt]

Okay, so what was the author's intent when writing the exercise?

[small obsidian]

Not sure yet! Haha. Still polling people so maybe something will come up

[plucky veldt]

@small obsidian thanks for investigating this, I was pretty puzzled by this so I'm glad I'm not the only person who wants to get to the bottom of this

[nimble jolt]

What is the exact question? @plucky veldt @small obsidian ?

[plucky veldt]

the one I posted above

[nimble jolt]

https://cdn.discordapp.com/attachments/496785752936546304/724240805123850280/Screenshot_20200621-083330_ReadEra.jpg this one?

[plucky veldt]

yes

specifically (iii)

[nimble jolt]

okay, will think about it at some point today.

[small obsidian]

Something that seems loosely related: If a space has an isolated point, any dense subset will contain it.

[plucky veldt]

well, an open subset of a perfect space indeed can't contain any isolated points in its subspace topology, since any set open in A is also open in X, so X would have an isolated point otherwise

but a dense subset? hmm... someone messed up here

[plucky veldt]

for a little less trivial example, if X = {1, 2, 3} and τ = {∅, {1, 2}, X}, then it is a perfect space, {2, 3} is a dense set which has a singleton {2} in its subspace topology

[sweet oasis]

is the function f: Y -> {a} x Y, such that f(u)={a} x u good enough to show the homeomorphism between {a}xY and Y?

like it feels lacking since it seems like i could have used anything to show that any other thing is homeomorphic

[plucky veldt]

I'm assuming that the topology is defined so that if U is open in Y, then {a} x U is open in {a} x Y

in that case, yes, your function is a homeomorphism

a homeomorphism is a bijection that preserves open sets both ways

and no, not every continuous function between spaces is a homeomorphism

it's rather obvious that your example is a homeomorphism, and not an interesting one

[elder yew]

But it is a useful one!

[sweet oasis]

its super useful and important

like anything else in topology that seems obvious

[plucky veldt]

@nimble jolt what conclusions did you come to after thinking about it yesterday

[elder yew]

https://math.stackexchange.com/questions/3667500/distortion-of-the-unknot

I had this question on distortion of the unknot

I emailed Gromov and he said he forgot, so if anyone's got anything I'd love to see it

[dim meadow]

Hey @elder yew where exactly in the paper does he state that?

[elder yew]

Page 114 of Filling Riemannian Manifolds

First example

The citation is for "Metric Structures for Riemannian Structures"

Pages 11-12, first remark

[dim meadow]

It says the equality only holds for circles

Not for unknots

So I looked at the proof in metric structures for riemannian manifolds

[elder yew]

I'm loosely calling unknots circle

[dim meadow]

You are or the book is?

[elder yew]

Well that's how knot theorists do it

Or at least the ones I've been citing

[dim meadow]

I think he might mean a literal circle

I looked at the proof in metric structures

And I was able to show r(s) = d(c(s+l/2), c(s)) is constant

And is l/(2dil(g))

So maybe that's enough to show that the curve c(s) is a circle

[elder yew]

https://arxiv.org/pdf/1010.1972.pdf

John Pardon cites it

Despite the simplicity of (1.1), very little is known about the distortion of knots, especially
if one is interested in lower bounds. Gromov showed that for any simple closed curve γ, we
have δ(γ) ≥
1
2
π, with equality if and only if γ is a circle, thus determining δ(unknot)

Sorry fraction didn't render from copy pasta

This is what Gromov said when I emailed him

[dim meadow]

Hmm maybe he defined it to be the min over the isotopy class?

[elder yew]

"My recollection is that a closed curve C of length 2\pi in the
Euclidean space, where the Euclidean distances between opposite
points c, c'\in C (with arc distances in C are equal to \pi) are all
\leq 2, is isometric to the unit planar circle, because the map of $C$
to the unit sphere for $c\mapsto (c-c')/2 is distance non-increasing.

But, probably, you had in mind distortion<\pi, where I don't remember
what happens."

[dim meadow]

Yeah gromov just talked about a regular circle

Not the unknot

[elder yew]

But the way distortion is defined in Pardon

[dim meadow]

Check the definition of distortion in your paper

Probably he defines it to be the min over the isotopy class

[elder yew]

Is that distortion of the unknot is the distortion of a regular circle

$$ \delta(K) := \inf_{\gamma \in K} \sup_{x, y \in \R^3 } \frac{d_{\gamma}(x,y)}{||x-y||} \geq 1 $$

[gentle osprey]

MoonBears-C-:
Compile Error! Click the reaction for details. (You may edit your message)

[elder yew]

x,y in R^3

[dim meadow]

Is K an equivalence class of knots?

Up to isotopy

[elder yew]

K represents smooth ambient isotopy class of knots

[dim meadow]

Then it makes sense

[elder yew]

Sorry I had this for a final project I turned in a month or so ago so some of the details aren't fresh

[dim meadow]

So in any case I'll go through the process he talks about in the paper

Cause I think it's interesting

Assuming dil(g) = pi/2, we get a lot of our inequalities turned into equalities almost everywhere

[elder yew]

Hrmm

[dim meadow]

Like 4/r(s)^2 = (4dil(g)/L)^2 is true almost everywhere

Since r(s) is continuous it's true everywhere

So r(s) is constant

[elder yew]

so it's a circle...

Hrmm all this stuff is very mysterious

[dim meadow]

It's not so mysterious

Well the way gromov writes is pretty mysterious

But the actual stuff isn't so crazy

There are a lot of details he mentions briefly which you have to pay close attention to

Like for example the norm of the gradient of c is 1

[elder yew]

Thanks!

If you want you can post on math stack and I'll green light it

[dim meadow]

Lol sure

[tepid jetty]

Hello! can you guys suggest an approachable introductory text for topology? My math background is not super strong, but I want to learn it to better understand some functional programming stuff I want to learn later on.

[frigid patrol]

I like Elementary Topology by Oleg Viro

[marsh forge]

What kind of topology @tepid jetty

Do you know any yet? What kinds of things do you want to know

[tepid jetty]

I don't know any at all, just basic stuff from youtube videos.

Need a jumping off point.

[marsh forge]

Does anyone know of reference for the fact that the bar complex actually gives you the right classifyng space

i can only find it asserted lmao

[plucky veldt]

@tepid jetty Allen Hatcher's Notes on Introductory Point-Set Topology for a bird's eye view or Topology Without Tears for a more detailed treatment

if you just need it for programming then Hatcher's notes will probably suffice

although I would recommend reading a proof that (a, b) is homeomorphic to R, which Hatcher hasn't yet included in his notes

[tepid jetty]

Thanks!

[heady epoch]

can someone help me?

e.g. verify (a) and give a hint on (b)?

[small obsidian]

Ya

[heady epoch]

so I asked on stackexchange

no helpful comments so far 😦

https://math.stackexchange.com/questions/3734150/surface-metric-in-normal-coordinates-and-gaussian-curvature?noredirect=1#comment7677129_3734150

[gritty widget]

can anyone give me a hint for the backward direction

it's probably just stupid but someone said bump functions and i have no idea what that is q_q

[dim meadow]

Do you know what partitions of unity are?

Also where is chi coming from and going to? Where is f coming from and going to?

@gritty widget

The phrasing of the problem is a little ambiguous

[honest narwhal]

\chi:M->N

f:N\to\mathbb{R}

They're using the convention that "function" = codomain of R

I see why someone was thinking of bump functions here but I don't think they're necessary

Or wait no I get it

If $\chi:M\to N$ isn't smooth, then we pick a chart $(U,\phi)$ in $M$ and $(V,\psi)$ in $N$, let's say $\chi(U)\subset V$, such that $\psi \circ \chi \circ \phi^{-1}:\phi(U)\to \psi(V)$ isn't smooth

[gentle osprey]

Daminark:

Daminark:

[honest narwhal]

How do we extend psi to a smooth function on all of V? That's kinda where the bump function business kicks in.

[gentle osprey]

Daminark:

[honest narwhal]

This is super useful for manifolds because it basically lets you kinda fuck with shit strictly locally

So what's the idea? We have V in N, we can find a ball in it's image, a smaller ball in that

Or okay well psi doesn't land in R but you can handle that by just projecting, some projection of psi is not smooth if psi isn't

But yeah so get psi':V->R

Choose W_1 \subset W_2 open subsets of the image of psi'. Then find some function g that's 1 on W_1 and 0 outside W_2.

Then g times psi' (pointwise multiplication) actually extends to a function on all of N, because you just say it's 0 outside V

That's smooth still because g times psi' is already 0 on a neighborhood of the complement of V

Or I guess it depends on how you choose W_2, at least it's 0 toward the boundary of V

The upshot here is the earlier argument, that I know psi chi phi^{-1} isn't smooth, well alright some component psi' of psi will satisfy psi' chi phi^{-1} isn't smooth as a map to R. Then using bump functions I can actually refer to psi' as a function on all of N. And then psi' is our f

[dim meadow]

Yeah that seems like a reasonable argument

I guess the big fact you are using here is that a map from R^m to R^n is smooth iff all of the coordinate functions are smooth

Which is the only thing you really need calculus to prove

[gritty widget]

oh jesus

wait a second

[honest narwhal]

Focus on my last message, it's got what you want, the previous stuff was me explanation bump functions and all

[gritty widget]

oh i'll read it anyway ty

[dim meadow]

And yeah dami that is obvious because literally you can construct the Jacobian from the derivatives of the coordinate functions

[gritty widget]

oh this is really detailed ty n.n

[dim meadow]

@gritty widget is this Lee?

Or some other book

[gritty widget]

no it's just lecture notes lol

[dim meadow]

Oh okay

Maybe you should go over stuff about bump functions and partitions of unity

[honest narwhal]

That would be good probably

[gritty widget]

is that in lee?

[dim meadow]

Yes

[gritty widget]

ok i'll check it out ty

[honest narwhal]

(Also think about why bump functions aren't a thing in complex analysis)

[dim meadow]

At the very beginning if I recall correctly

Lol

[gritty widget]

great thanks all

[dim meadow]

That's because if a holomorphic function on a connected open set is 0 on an open ball it is the constant 0 function

[fringe glen]

no

[gritty widget]

[magic fractal]

Can someone help me with this?

[hexed holly]

which part?

for i) you want to show that if f_1, f_2 is another orthonormal basis, the value of the jacobian stays the same

for i) you will need to remember the condition under which axb=0

and IVT probably

[median kite]

Can someone help in this??

[magic fractal]

Yeah i got it... thanks @hexed holly

[gritty widget]

did i prove this right

i feel like i didn't

[plucky veldt]

an intersection of neighborhoods isn't necessarily an open neighborhood, but that doesn't matter match for the purpose of this problem

this seems correct, although I'd change some notation

[gritty widget]

but this is a finite intersection right?

[plucky veldt]

personally, I'd prove it by induction I think

it is a finite intersection, and a finite intersection of neighborhoods is always a neighborhood, but it isn't necessarily an open neighborhood

[gritty widget]

oh right our book defines neighbourhoods as being open

[plucky veldt]

thats stupid

and what's the point of saying "open neighborhood" then

[gritty widget]

yes sorry you’re right that’s redundant

why would that be stupid

maybe I’m misunderstanding hold up let me take another look

[plucky veldt]

so the terms neighborhood and open set are synonyms?

[gritty widget]

no that’s not what I mean

[plucky veldt]

I'd say that a neighborhood N of p contains a point of A by definition of p being a limit point, and if N contains n points of A then it contains n + 1 points of A, since we can take an intersection of all neighborhoods of p that don't contain one of those n points with N and that would be a neighborhood of p in N that contains another point of A, so N contains an infinite amount of points of A

[fervent citrus]

I would tend to believe this is what most people are used to

[plucky veldt]

wikipedia defines it differently

[gritty widget]

ok right I’ll keep that in mind

i guess I just tried to mirror the proof from analysis in metric spaces

[plucky veldt]

@gritty widget it's not clear to me why you think you proved it

also interesting corollary: if a hausdorff space is finite it's discrete and therefore has no limit points

[gritty widget]

oh nice

seems like I misunderstood but didn’t you say ‘this seems correct, although I'd change some notation’

[plucky veldt]

your steps are correct and are necessary for this proof, but I'm not sure why you think your last sentence proves the claim

there's something lacking that would make me sure you know what you're doing

[gritty widget]

oh you mean I should add a sentence saying this is a contradiction

how would you change the notation?

[plucky veldt]

well I think you wrote (A∩U) - {p} twice unnecessarily, also it doesn't matter that the resulting neighborhood is contained in U, what matters is that it's disjoint from A

also hausdorff implies that there's a neighborhood of p that doesn't contain q, you can have one that is a subset of U but you'd have to take the intersection with the one that's guaranteed by hausdorff, I think

[gritty widget]

oh right, this has been really helpful, thanks n.n

[plucky veldt]

and then the intersection of all those is a neighborhood of p that is disjoint from A, so p is not a limit point of A

[plucky veldt]

sorry it's not necessarily disjoint, if p is in A then it intersects with A, but that intersection contains only p

[willow spear]

Can someone help me understand this definintion of a continuous function (Topology Context)

This is from Hatcher's Basic Point-Set Topology Notes

[honest narwhal]

Your definition in mind is delta epsilon right?

If so, you can show these are equivalent. The idea is that, okay let's say you satisfy delta-epsilon and let O be an open set in R. We wanna show f^{-1}(O) is open. Choose x in f^{-1}(0). Then f(x) is in O, which is open.

Means we can choose some ball of radius epsilon around f(x) that's contained in O. Now the delta-epsilon stuff kicks in

There exists some delta such that |x-a| < delta => |f(x) - f(a)| < epsilon. What does that mean? That means the ball of radius delta around x maps inside the ball of radius epsilon around f(x), and thus inside O

So the ball of radius delta around x is a subset of f^{-1}(O).

What have we done? We've shown that for every x in f^{-1}(O), there is some ball B such that x\in B \subset f^{-1}(O)

But that's just the def of being open

Now assume we know f^{-1}(O) is open whenever O is open. Fix a point x and choose epsilon > 0. The ball of radius epsilon around f(x) is open, call that O

f^{-1}(O) is open. Well, x\in f^{-1}(O), so by def of open we can find some ball B, call its radius delta, such that x\in B \subset f^{-1}(O)

[willow spear]

Wdym by ball of radius

[honest narwhal]

But okay, let's say |x-a| < delta. Then a is in B. So it's in f^{-1}(O). So f(a) is in O, which is a ball of radius epsilon around f(x). Which means |f(x)-f(a)| < epsilon

Oh replace the word ball with interval

[willow spear]

Okay...why do we have to prove using the inverse function?

I’m pretty new to topology...I’m used to the def taught in ap calc. with epsilon

An open set is just the collection of open intervals right?

See I’m also confused with the ball of radius thing

What is that?

Like I’ve taken ap calculus

I’m in high school

So what would be covered in an analysis course... how long would it take?

[gentle osprey]

slimvesus:

[willow spear]

Analysis is just calc right?

Oof ok

[ivory dragon]

well it is worth noting that

[willow spear]

So do you think I can just continue with Hatcher?

[ivory dragon]

some schools call "intro analysis" something like "advanced calculus"

and in continental europe the terms tend to be synonyms

kinda

[willow spear]

Are u in university?

Is topology without tears a better alternative?

Oh okay

[nimble jolt]

I think either munkres or hatchers notes would be better than topology without tears most likely.

[willow spear]

Homestly it’s the notation that so annoying and confusing

[nimble jolt]

Example? It may just be standard in the subject rather than something hatcher specific djoker.

Doable but I think the other way makes more sense pedagogically.

[ivory dragon]

topology will feel pretty unmotivated without analytic connections

like it wont be obvious why some definitions are the way they are

[nimble jolt]

at least a first course in analysis up to the point where open sets in R^n are defined and you see how slick it is to work with continuity phrased in this way rather than epsilon delta.

By that point in the course, the typical student is probably sick of epsilon delta.

And so the chance to work purely with set operations is probably going to be welcomed.

[ivory dragon]

point set definitions are really just "intro analytic definitions but less finnicky"

ehh jan

i think epsilon-delta is good "first practice" in inequality juggling

[nimble jolt]

epsilon delta is great motivation for the abstraction of metric spaces and then point set. Also it can't hurt to give students some basic practice manipulating inequalities.

yeah

[ivory dragon]

i'd accept the statement that eps-del stuff is overemphasized though

but i think its important pedagogically

[nimble jolt]

yeah and kinda tricky to teach well, a lot of students walk away with the impression that it is a valuable skill to be able to choose the delta well to make whatever exactly smaller than epsilon.

[ivory dragon]

oh man i remember undergrad analysis

[nimble jolt]

when usually you are extremely happy to just get an upper bound that is o(1) in delta.

[ivory dragon]

my choices of deltas were

ugly as shit

consistently some weirdass min{1, fuckyfraction}

or whatever

[nimble jolt]

haha

[ivory dragon]

but hey it works

[marsh forge]

I got burned once bc i didnt choose it quite small enough (some scalar)

So i started using delta/100000

For all of my proofs

[dim meadow]

Lol

That's gross

[plucky veldt]

I noticed almost no overlap between analysis and topology, except for the fact that the topological definition of continuity is equivalent to the one from analysis in the euclidean R

never heard about any of those before topology

openness only in the context of open intervals

[honest narwhal]

A lot of early topology is there to formalize basic notions in analysis

[plucky veldt]

those are already formalized, topology generalizes them

by treating R as one of its spaces

[honest narwhal]

Well it defines a notion that applies more generally but it was also kinda how they realized "the right" way to think about those concepts even in the setting of R

Plus analysis cares about way more general things lol

[elder yew]

I had a question on Hatcher's proof that the fundamental group of the circle is Z

It's the part where he shows his map phi is a homomorphism

[marsh forge]

what page?

[elder yew]

Uhh hold on lemme get it out

[marsh forge]

you can explain it too if oyu want

just thoight page would be faster

[elder yew]

Page 29

Fourth paragraph

"To verify..."

[marsh forge]

hmm sorry I'm not sure i see a to verify

[elder yew]

Top of page says The fundamental group of the circle

[marsh forge]

yeah

[elder yew]

"To verify that phi is a homomorphism, let $\tau_m : \mathbb{R} to \mathbb{R}$ "

[gentle osprey]

MoonBears-C-:

[marsh forge]

am i just blind

[elder yew]

go to the next page

It's literally right here on my physical copy

[marsh forge]

homomorphism doesn't appear untl page 34 according to my search

maybe he changed the proof?

[marsh forge]

oh wow

yeah thats just not there in the online version haha

[elder yew]

He does a translation on the lifts, then projects down

[marsh forge]

ok yeah same basic idea

whats the question?

[elder yew]

The translation takes you from ns to (n+m)s

Then when you do the group product you go to

[(n+m) + m]s

Are you gluing n+m to 0 and looping back around to m?

(on the helix that is)

Or is it only a loop when you project down

[marsh forge]

I'm not entirely sure I see what you are saying?

It's only a loop when you project

It's a path lifting of the loop

[elder yew]

Right. so he writes w_n = ns

And you translate by m

t_m w_n = (n+m)s? or ns + m?

[marsh forge]

the latter

[elder yew]

ns + m

[marsh forge]

which it why you can compose tmwn with wm

because the latter ends at m

and the former starts there

[elder yew]

Ahh yes that was my question

[marsh forge]

yeah he's usng that

[elder yew]

The notation isn't too clear

[marsh forge]

and implicitly the fact that \tau_m is not visible once you take the quotient

like R->R/Z is the map from R to the circle

you're working mod 1

so adding m doesn't matter

[elder yew]

Ok so this way it does form a loop when you project

yeah

Ok thanks for the clarification. It's been a while since I did anything like this

[marsh forge]

np

hatcher is notoriously much easier when you know what it is supposed to say and much harder when you dont

[elder yew]

I got it about a year ago after taking knots trying to rigorize what I had learned

But other things got in the way. Since I've done 3-manifolds and my reading comprehension of hatcher is much, much higher now

[marsh forge]

tbh i dont entirely blame hatcher, i think topological arguments are often like, very hard to explain until you see it and trivial once you do haha

[elder yew]

Yeah, every time I was at office hours it was an hour and a half discussion just explaining some weird geometric thing

and I got a headache, but the next day it'd be cleared up

[marsh forge]

hahaha

[elder yew]

I'm just doing chapter 1 of Hatcher

I don't need any of the fancy stuff

[marsh forge]

I kinda like it in a sadistic way

the moment when you finally see it

is worth the pain

there's an especially bad proof like that in hatcher when he proves results about K_m,n knots on the torus

[elder yew]

Yeah I've spent a long time staring at that

I need to go back through that and the wirtinger presentation stuff

Then I'll transfer back over to heegaard splittings

[willow spear]

How is this true?

[gritty widget]

A is closed because it contains its limit points

so the complement is open

[sleek thicket]

what's your definition of an open set?

[gritty widget]

alternatively, if you pick a point x in the complement of A, then it shouldn't be too hard to find an open interval containing x disjoint from A using the archimedean property

[willow spear]

what does teh phrase "together with its limit 0" mean?

[marsh forge]

Union

Sequence union 0

[gritty widget]

can anyone give me a hint for the backward direction of (b)

ty n.n

[marsh forge]

assume its not in the interior and build a sequence

[gritty widget]

ok I’ll try ty

[marsh forge]

lmk if you want a bigger hint

[gritty widget]

ty

[gritty widget]

is this fine

[gentle osprey]

val:

[gritty widget]

@marsh forge sorry for the pung

[marsh forge]

yeah this is the idea

[gritty widget]

ty

[plucky veldt]

you haven't quite proved that Ui ⊆ A

a better way in my opinion is that if x ∉ Int(A) then we can construct a sequence that is not in A

and an even better way would be to show that (b) follows easily from (a)

[gritty widget]

wait why did i not prove U_i\subseteq A?

@plucky veldt

[plucky veldt]

which Ui is in A

[gritty widget]

all but finitely many right?

[plucky veldt]

how do you know that

[gritty widget]

because every sequence in X is eventually in A

*every sequence which converges to x

[plucky veldt]

how does that follow

[gritty widget]

well x_i is just an arbitrary element of U_i

and for all but finitely many i, x_i must be in A

since x_i was arbitrary U_i \subseteq A

i mean that might be wrong

[plucky veldt]

which Ui ⊆ A

[gritty widget]

all but finitely many

i mean i think

but a correction would be helpful

[plucky veldt]

I'm not asking how many

I'm asking which

[gritty widget]

does it matter?

we know they exist

@plucky veldt sorry was eating

[plucky veldt]

how do you know they exist

if they exist then surely you are able to provide an example of one

or are you not

[gritty widget]

huh

do you want me to construct an explicit example

@plucky veldt

[plucky veldt]

yes

[gritty widget]

you could do something with 1/n probably

but why

is this proof right or not

[plucky veldt]

you say that Ui ⊆ A but you didn't prove it

[gritty widget]

i don't know what to say

right so we constructed (x_i) by taking an arbitrary member of (U_i) for each i

so (x_i) is eventually in A, which means that there is some j\in\bN with the property that i\geq j has x_i\in A

since our choice of x_i was arbitrary, every i\geq j has U_i\subseteq A

what is the problem with this argument @plucky veldt

[plucky veldt]

since our choice of x_i was arbitrary, every i\geq j has U_i\subseteq A
@gritty widget WHAT

[gritty widget]

each x_i is an arbitrary member of U_i

[plucky veldt]

since our choice of x_i was arbitrary, every i\geq j has U_i\subseteq A

what is j

[gritty widget]

so (x_i) is eventually in A, which means that there is some j\in\bN with the property that i\geq j has x_i\in A

[plucky veldt]

what if I take another sequence

which has a different j

[gritty widget]

it doesn't matter

the point is j exists

[plucky veldt]

of course j exists

but it isn't true that Ui ⊆ A for i ⩾ j

because who said there is no another sequence

where the j is bigger than the j from the previous sequence

so you necessarily have points in some Ui that aren't in A

[gritty widget]

i see what you're saying

hmm

so what was wrong with my reasoning

well i'm still not entirely convinced it's wrong

[plucky veldt]

everything you wrote was true, but you didn't prove it

[gritty widget]

how would i fix it then

could i have a hint

[plucky veldt]

There are better ways of proving the lemma, but if you want to go down this path, you have to show that there is a Ui ⊆ A

it is true that there is a N such that Ui ⊆ A for i ⩾ N, but what's N

[gritty widget]

idk

hmmm

actually no i have no idea

[plucky veldt]

it's sup{j: x_i ∈ A for i ⩾ j for all sequences {x_i}}

now we know there is a supremum

[gritty widget]

how do we know it exists

[plucky veldt]

because otherwise it would mean that for every n ∈ N there's a sequence {x_i} so that x_i ∉ A for i ⩽ n

in particular x_n ∉ A

so if we take x_n as a sequence, we have a sequence convergent to x that is outside of A, which contradicts our assumption

[gritty widget]

hmm

you mentioned that this follows from the first part

[plucky veldt]

and still, x_i ∈ U_i for all i

[gritty widget]

right ok that makes sense

how does the first part imply this?

[plucky veldt]

x ∈ Int(A) is equivalent to saying x ∉ cl(X \ A), which is equivalent to saying x is not a limit of a sequence in X \ A, which is equivalent to saying every sequence converging to x is eventually in A

[gritty widget]

wow nice

that's cool

thanks for all the help again

[slow urchin]

hi folks

would a course in Analysis and a course in Abstract Algebra be useful/essential for getting through either Munkres or Lee?

[coarse kestrel]

Basic analysis would help for Munkres

And algebra for later chapters

[slow urchin]

i am a math noob, and am very slow. so you gotta pretend i'm lacking in maths maturity.

oh, okay, algebra, ONLY for the later chapters.

thats interesting.

[coarse kestrel]

Yeah the second half is called "Algebraic topology" for a reason

[slow urchin]

tyty. i will pass this info along.

One last question.

Which book is the "lower level book" Lee or Munkres?

[coarse kestrel]

Munkres contains all the basic topology stuff

Lee doesn't cover basic topology but more differential geometry stuff, but Munkres has algebraic topology, which is quite different from differential geometry

[honest narwhal]

He might've meant Lee Topological Manifolds

[coarse kestrel]

Wait by Lee I assumed you're talking about Lee's Introduction to Smooth Manifold

ah

[honest narwhal]

If he's comparing to Munkres that's the only thing that would've made sense

[coarse kestrel]

True

[slow urchin]

He might've meant Lee Topological Manifolds
@honest narwhal

I'm sorry, i should have been clearer, i didn't realize there were two books by lee.

[honest narwhal]

So you meant topological? In hindsight actually there are also two books by Munkres lol

So you could've meant: Munkres Topology/Lee Topological Manifolds, or Munkes Analysis on Manifolds/Lee Smooth Manifolds

[slow urchin]

oops.

[honest narwhal]

I imagine former? 😛

[slow urchin]

the most intro level of either from either author.

[honest narwhal]

So, for Lee that's definitely topological manifolds

For Munkres... idk if either is "more intro" it's sort of a lateral shift

[slow urchin]

yes, Lee Intro to Top Manifolds.
and, Munkres it'd be Topology.

[honest narwhal]

Of the two I'd prefer Lee

[slow urchin]

its more noob friendly?

[honest narwhal]

And for that you'd want some basic algebra but you wouldn't need that much, mostly basics of group theory and all