point-set-topology

[hexed steppe]

it is true that continuous functions map intervals to intervals

[alpine nest]

That's correct, with the caveat that the interval might be degenerate (consist of a single point)

[white oxide]

[a,a] seems fine to me

[alpine nest]

Yeah, but sometimes you're fine treating singletons as intervals, and sometimes you're not, so I thought I'd clarify that

[tribal palm]

f is defined as a function X to R

[hexed steppe]

what is X

surely X=R

[tribal palm]

then they consider f(J) when J is a subset of R ??

[alpine nest]

Yeah, the notation is a bit inconsistent, but X has to be a subset of R containing J, otherwise the statement makes no sense

[tribal palm]

in the statement just above, any connected space

[hexed steppe]

its a typo then

[alpine nest]

Almost certainly, yeah

[white oxide]

As long as we take X to contain a subspace J homeomorphic to an interval in R it ought to be fine, but probably just a typo

[tribal palm]

right, i was wondering if the direct image were supposed to be a preimage or smth but that wouldn’t make sense either

must be that they mean X = R then

[hexed steppe]

the parenthetical tells you

[tribal palm]

yeah sorry i’m slow, lol

[white oxide]

f(J) is the image?

[tribal palm]

yes

direct image

[white oxide]

I honestly never heard that wordage outside of some functors

[tribal palm]

whaat? what do you call them then?

[hexed steppe]

image

in english, "image of J under f"

[tribal palm]

i may be more than averagely interested in functions on the form P^n(X) to P^n(Y) induced from f: X to Y

[white oxide]

P^n(X)?

[tribal palm]

power set composed with itself n times

[white oxide]

Well? What do you want to do with them?

[tribal palm]

pretend they’re all the same mwahaha

[white oxide]

You can just apply the corresponding functor n times

[tribal palm]

oh? sounds like something i should get to sometime

[white oxide]

the power set functor or functors in general?

[tribal palm]

cat theory

[white oxide]

[tribal palm]

for practical use

they must be useful, seeing how often my prof manages to bring them up out of the blue in lectures

[white oxide]

I can personally recommend vakil's first chapter. Its really good, even though the book is for AG

[ebon galleon]

You'll see the most important definition is mathematics: a category

[tribal palm]

*insert jens duck think*

(it is hard being nitro-less)

[ebon galleon]

(you should at least learn categories, functors, and natural transformations, so that you recognize these (very common) terms in the wild)

[tribal palm]

thank you, i mean to

functors seem very intuitive

cats somehow less so simply due to how simple they are… and natural transformations i know absolutely nothing about

[white oxide]

They dont have that much "meat" to them, they are "simple" objects just like how categories and functors are. But they turn out to be the correct notion often

You should familiarize yourself with objects defined through their universal properties first or right after

[tribal palm]

and phew yeah path-connectedness is proving much easier than connectedness

now why are these separation axioms such a mess

[alpine nest]

Yeah, they are

[white oxide]

how fine is your scale?

[tribal palm]

what does that mean?

[white oxide]

how many T_i's are making your life hard*

[tribal palm]

5

unless i have lost my ability count (happens often)

[white oxide]

Okay that's a reasonable set

(me forgetting kolmogorov being a thing in 3.1s)

[tribal palm]

actually i might drop Viro for this and follow munkres for this part

don’t have time to cover this in much detail

just two days of preparation until my exam and i still have a lot to cover

[white oxide]

Oof, yeah there is still quite a bit to go over

[alpine nest]

All these axioms and still no use found for anything beyond T2

They have played us for absolute fools

[tribal palm]

ok it seems the important part is T4 => T3 => T2

[alpine nest]

Yeah, anything without T2 I wouldn't hesitate to call pathological

And anything beyond T4 is quite restrictive

[tribal palm]

making algtop folks sad in the process?

[alpine nest]

Algebraic topology is not real and can't hurt me

[tribal palm]

actually the algtop people in my reading hall are quite buff

[white oxide]

But but tychonoff cube 🥺

[alpine nest]

I mean, that was obviously a sweeping and not entirely serious statement.

But also it does seem that T2 is the property the lack of which tends to break things the most

[white oxide]

Still mostly correct tho

Want something more than T2? We had something for that, we called it compact hausdorff

[alpine nest]

I mean, metric

[pallid delta]

the important part is T6 and T2
And potentially that compact => T4 (iirc)

[tribal palm]

what is t6, lol

[white oxide]

What was T6

[pallid delta]

disjoint closed sets are inverse images of 0 and 1 by a continuous function

[alpine nest]

It's a perfectly normal space

It's right there in the name

[pallid delta]

while T4 only yields inclusion

[pallid delta]

T6 <= metric so that is cool

[alpine nest]

See, metric spaces are perfectly normal

[white oxide]

I mean, once you can make distance arguments most of the T's are fairly straightforward

[pallid delta]

yeah ofc, this is one it's never taught in a purely metric context

[tribal palm]

this is necessarily a stronger condition than the conclusion in the urysohn lemma right

[white oxide]

Why do we care about precisely separated btw?

[pallid delta]

that's T4

[white oxide]

Like once you have urysohns lemma your separation axiom craving should've been satisfied

[red yoke]

Mapping to [0, 1] I guess

Say the (*) here https://math.stackexchange.com/questions/3547820/neighborhood-deformation-retracts-vs-cofibrations

[opaque cloud]

But it is insanely beautiful 😍 and super useful

[merry geode]

Show me the beauty of algebraic topology, I am still trying to recover from infinite lens spaces

[alpine nest]

it is! I just do nothing anywhere related

and in particular nothing that would require me to gaze into the non-Hausdorff abyss

[opaque cloud]

don't you work regularly in non-hausdorff spaces

[alpine nest]

no, just nonmetric

but still T2

[prisma garnet]

where do infinite CW-complexes show up?

[prisma garnet]

chrew

AG is a plague that haunts the earth

[red yoke]

Hatcher has a section on K(G, 1) in chapter 1 I think

[tribal palm]

i hate when people allow unions of sets of different types of objects 😭

[white oxide]

wdym, they are all sets

[tribal palm]

sets are informal objects, and so i refuse to not also informally talk about types

[opaque cloud]

set theorists are fuming right now

[tribal palm]

let them deceive themselves all they want

all my homies know sets are not real, mwahahaha

that said, i am very fond of set theory

[alpine nest]

It's just a collection of objects, innit

[tribal palm]

a “collection,” a “object”

[pallid delta]

a class, maybe even a category

[gleaming warren]

boom

[alpine nest]

I was being deliberately informal 😛

[alpine nest]

The empty set is real and cool and my friend.

[obtuse meteor]

How do you feel about algebraic geometry

[obtuse meteor]

algtop ppl generally work with Hausdorff spaces

[tribal palm]

fake

[tribal palm]

my ignorance has been revealed (again)

[tribal palm]

just a string of symbols… Hilbert’s got my back, mwahaha

[alpine nest]

I've happily avoided any interactions with algebraic geometry until now

I mean I'm not dissing algebraic geometry, it looks interesting from a distance, but my path has carried me elsewhere

Something something Zariski?

[fickle elm]

There is a similar notion called separated in AG, and it is related to Hausdorff as both can be characterized with that the immersion into diagonal is closed

Non Hausdorff is usually not a problem as we mostly not use set theoretic product, but fibered product.

And by Grothendieck, non separated scheme is called prescheme.

[paper crystal]

hello guys, how can I formally write the definition of a topology? currently I do: given a set $\mathbb X$, $\tau$ is a topology on $\mathbb X\iff\mathbb X,~\emptyset\in\tau~\land~\forall O_1,...O_n\in\tau,~\bigcap\limits_{i=1}^nO_i\in\tau~\land~\forall{O_i}\subseteq\tau,~\bigcup{O_i}\in\tau$ but I don't like the $O_1,...O_n$ and ${O_i}$ notation

[gentle osprey]

Jow

[thorny agate]

You haven’t defined what n is

Also what’s the problem with O_I notation?

[quiet thorn]

Also don't use \land in actual writing

[knotty vine]

The question was about a formal definition

[thorny agate]

If only one read the whole context

[quiet thorn]

Yes, but I'm just saying in general

I don't think it's a useful or good habit to have

[thorny agate]

ofc

[quiet thorn]

Sorry if it wasn't clear that that was not directly related to the question

[paper crystal]

that's one problem, what should I define it as?

[knotty vine]

Here's how I would do it
[ \forall C \subset \tau, \bigcup C \in \tau \land \bigl( \operatorname{finite}(C) \implies \bigcap C \in \tau \bigr) ]

[paper crystal]

${O_i}$ notation doesn't say what $i$ is, and I'm sure there's a more concise way of saying $O_1,...O_n$

[gentle osprey]

semer

Jow

[paper crystal]

I like that, but how can I not use the word "finite"

[knotty vine]

Define $\operatorname{finite}(C)$ to be something like [ \exists n \in \NN, \exists f, \text{$f$ is a bijection from $n$ to $C$} ]

[gentle osprey]

semer

[paper crystal]

you mean $\exists n \in \mathbb N, \exists f, \text{$f$ is a bijection from ${m\in\mathbb N:m\leq n}$ to $C$} $?

[gentle osprey]

Jow

[knotty vine]

Yes, often the numbers in N are defined to be sets of cardinality n

But this is the same thing ofc

[paper crystal]

oh, interesting fact

[knotty vine]

Yours is better tbh

(but im lazy)

[paper crystal]

also that still has words `xD, I can do $|C|\in\mathbb N$

[gentle osprey]

Jow

[knotty vine]

Right, but if you allow the cardinality function, then you might as well just use the finite predicate

Only using symbols doesnt mean formal ofc

[paper crystal]

btw, what's the most common notation for cardinality? on wikipedia I found $|C|,~#C,~n(C),~card(C),~\bar{\bar C}$, $|C|$ seems to be most common but I like $#C$ better

[gentle osprey]

Jow

[knotty vine]

# is decent but it looks kinda ugly imo

[paper crystal]

comeon man! xd

[red yoke]

*screams* IS DECENT

[knotty vine]

|C| is ok, unless you're also working with negative numbers

card(C) is always unambiguous

[paper crystal]

?

[knotty vine]

Cant mistake it for anything else can you

[paper crystal]

oh nvm, I thought u said "ambiguous"

[paper crystal]

wym? it could be misinterpreted as the absolute value of the elements of the set?

[knotty vine]

Maybe, or maybe C is a number

[paper crystal]

well, problem solved anyways -\._./-

ty

[hidden crag]

what if we're modelling card games

[paper crystal]

name_of_the_game(C)?

[tidal lynx]

I don’t understand how the part starting from “Said differently” is relevant. It would make sense for Frechet-Urysohn spaces, where the sequential closure of any set equals the closure, but sequential spaces are concerned about something different

[white oxide]

I mean, unless you like circularity you have to settle with primitive notions lol

[hexed steppe]

closure is different from closed

[tribal palm]

i don’t have a problem with it at all, i just like to remind people that sets are informal objects, mwahaha

because i am evil

[white oxide]

whats a formal object

formal definitions involving only formal objects only

[hexed steppe]

maybe you can explain why you think it would make sense in the frechet-urysohn setting? @tidal lynx

[tidal lynx]

So in any topology the closure of a set is precisely the “net closure” (a point is in the closure of a set iff it is the limit of some net of points of the set). I read the last sentence in my image as “sequential spaces are the spaces where we can relax the word net (in my previous sentence) to sequence”

But what I have described is a space where the sequential closure is always equal to the closure, a F-U space

I guess my misinterpretation is the “any topology can be described in terms of nets” part, which I seem to have incorrect

[hexed steppe]

what do you mean by this

[hexed steppe]

the screenshot you posted says nothing about closures, so not sure where you got that from

[tidal lynx]

I don’t know how to decipher “Any topology can be described in terms of nets, but sequential spaces are those spaces for which nets of countable length are sufficient to describe the topology”

[tidal lynx]

So this was my attempt at giving meaning to “describing a topology in terms of nets”

But I guess it’s wrong

[white oxide]

My course defined frechet-urysohn as: For any x in the closure of A, there exists a sequence in A converging to x.

Sequential spaces are:
A set A is closed iff all limits of sequences in A are contained in the A.

So in sequential spaces closedness can be measured using sequences, but limit points dont have to come from a sequence

[tidal lynx]

I understand that part

I just don’t see the relation to nets

Is it maybe that in a general topological space “a set A is closed iff all limits of nets in A are contained in A” ?

That’s just a guess

[white oxide]

Well, if you can test closedness without ever talking about nets, you can also talk about open sets. So you pretty much dont need to use nets at any step when dealing with the topology of the space

brain fart, fixed now I hope

[tidal lynx]

No like in the Wikipedia image it says something like “for general topological spaces you need nets but here for sequential spaces you just need sequences”

[tidal lynx]

edited

[white oxide]

Yeah

Any limit point can be found with a net, all limit point able to be found using sequences is the property that frechet urysohn spaces have

With limit points I mean points of the closure of some set

[tidal lynx]

And also, the first direction of:
“A set A is closed iff all limits of sequences in A are contained in the A.”
holds in general, correct?

so sequential spaces only really ask for the reverse direction

[white oxide]

yes

[tidal lynx]

Nice, thanks

[white oxide]

If you care for it, id like to remember it as thinking of sequences as like a subset of things you can do with nets. Frechet-urysohn is equality and sequential space means that it is a dense subset in analogy to extending functions into R

[tidal lynx]

Trying really hard to understand this analogy but am failing

[white oxide]

its a "english class analogy" type analogy

[tidal lynx]

Since sequential spaces are determined by how sequences in them converge, why wouldn’t you say equality for them

Edited sorry

[white oxide]

whereas frechet urysohn tells you ALL limit points can be obtained through sequences, this is generally false in sequential spaces. But sequential spaces have the defining property that just knowing a set contains the limits to all sequences is enough to know it is closed

[tidal lynx]

Agree

[white oxide]

The latter is somewhat reminiscent to how if you have a cts function f: R -> R, you only need to know how f acts on the dense subset Q IMO

[tidal lynx]

Will think about that

[hexed steppe]

a set K is closed iff every convergent net in K has all of its limit points in K

this is false in general if “net” is replaced by “sequence”

[hexed steppe]

oh you said that

[paper crystal]

what does a topology being finer than another mean formally?

[balmy field]

The finer one has more open sets

I.e. (X, T') is finer than (X, T) if T' contains T

[gentle osprey]

Jow

Jow

[knotty vine]

@paper crystal It's not about the number of open sets or their size, its about set inclusion

So t1 and t2 are incomparable, similarly t3 and t4. But t4 is finer that t2

[paper crystal]

alright, thanks man

[paper crystal]

alright, thanks man

[knotty vine]

alright

[paper crystal]

why isn't a topology necessarily closed under arbitrary intersection? it intuitively feels like it should be

[alpine nest]

Why? The intersection of infinitely many open intervals on the real line might well be a single point

And treating singletons as open is not usually what we do

[wispy veldt]

meaning all topologies on R are discrete

[paper crystal]

how?

[alpine nest]

I_n = (-1/n,1/n)

All of them are open, and their intersection is just {0}

[paper crystal]

oh shit, yeah, yeah you're right

well, that's ultimately what I came for, an example

so thank you xd

hey, @alpine nest if you don't mind me asking (and also pinging you lol), you seem to know a lot about a lot in maths, ae you a student, prof ... ?

[alpine nest]

Used to do mathematical research, retired from that but I still teach a bit

I only know a few areas of math though 😄

[paper crystal]

really!? what research did you do?

[alpine nest]

dynamical systems

Sometimes topological, sometimes measure-theoretic, so those are the areas where I feel reasonably confident

[paper crystal]

that's really cool, damn the internet is a great place isn't it

[alpine nest]

It is!

I remmeber when we didn't have it 🧓

Or well, we did have dial up

[paper crystal]

lmao

can you tell me a bit about what mathematical research is concretely about? doing "research" in mathematics has always been obscure to me, I mean I just recently discovered my love for maths (I'm more of a physics guy originally) and the way it's presented often disregards the research aspect, everything is just presented as facts to know, and not as really as man-made constructs

[alpine nest]

Well, what exactly it's about, depends on the area, but it's solving problems and answering questions.

Either your own which resulted from your previous research, or someone else's

You read things, you talk to people you find out what questions could do with being answered

And hopefully you have some ideas of how to answer them

So basically like in other fields, you look at the current state of knowledge in your section of the field, and you try to expand it

[opaque scroll]

Step one, familiarize yourself with what people in your field know/doesn't know and what the open questions are.

Step two, look at some examples and make some guesses or simple questions.

Step 3, Try to answer those questions, read theory you think might be relevant, talk to people and hear what they think.

Step 4, write down what you figure out in an article, present it at seminar or conference, publish it.

Step 5, profit (?!)

[alpine nest]

However, unlike most other disciplines you don't do experiments or surveys or delving into sources, you mostly just have a good think 😄

[paper crystal]

"profit (?!)" lmfao

[alpine nest]

Sadly there are a number of occupations that are much more lucrative than academic mathematics

[paper crystal]

well now that I think about it like that, it's kinda pretty obvious `xD

[paper crystal]

and yet some of the most lucrative occupations rely heavily on mathematical principles, finance, engineering, tech ...

[alpine nest]

Yep

There is money in being good with mathematics, it's just typically not at universities

[paper crystal]

I'll take that as career advice xd

still a shame though

[formal tide]

I'm having a silly moment: Why is not the uniform topology the same as the box topology?
From https://math.stackexchange.com/questions/315524/open-sets-in-uniform-and-box-topology?rq=1 I see that (xi-e,xi+e) is not open, but then how does the e ball around x_i looks like?

oh, I think it is just sequences that don't get arbitrarily close to xi +- e

[plain raven]

Ah yeah.

That's an interesting distinction

[paper crystal]

what is this nonsense
let (X,d) be a metric space and A be a subset of X, prove that |d(x, A) − d(y, A)| ≤ d(x, y)

chatgpt says d(x,A)=inf{d(x,a): a \in A}

[red yoke]

Chatgpt thinks it's obvious from the definition

[paper crystal]

I can't tell if this is meant to be sarcastic

I did the proof with that definition and it turns out to be true

[hexed steppe]

do you have another definition in mind

[paper crystal]

nope

[hexed steppe]

what is the issue then

[cedar pebble]

good lord do not go to chatgpt for math stuff

[paper crystal]

I never use it as a primary source, and it can be pretty useful still

[cedar pebble]

you shouldn't use it as any source

it produces utter nonsense for mathematics

[paper crystal]

it doesn't seem to have done that in this case ...

ok ok, I'll never ever use it again, pinky swear

[alpine nest]

I mean, it will probably restate definitions and theorems corectly, but it's unlikely to produce a valid chain of reasoning because that's not the purpose of the model at all.

[paper crystal]

I didn't say I trust it 100%, but we can't just disregard it's usefulness completely

[alpine nest]

it has uses in many situations, but solving mathematical problems is not one of those, sadly

I sometimes throw stuff at it in hopes that it will spit out some keyword or name that I'd forgotten about and that's connected to the problem

since that's pretty much what it's designed to do - produce natural-sounding text based on associations

but associations are not logic, even if they sometimes do a plausible impression of it

[hexed steppe]

yeah it might prove useful for searching the literature

broadly construed

but yeah for correct definitions or theorems or proving things…lol

[alpine nest]

I think it's likely to correctly state major theorems since they must have occurred a lot in the training data

On the whole the main issue in ChatGPT (in practical terms rather than in societal terms) is that it's not reliable, and when it's wrong, it's often wrong in ways in which humans tend not to be.

So if you want to trust the output, it requires careful verification, which often defeats the point of using the tool in the first place

[hexed steppe]

yeah

the only people in a position to use it effectively dont need it

(though terry tao might disagree)

[cedar pebble]

conversely all the people using it are also the people confused by utterly basic analysic problems

[alpine nest]

I have to say LLM tools have been surprisingly helfpul when writing code, but then they work as a highly advanced autocomplete

[cedar pebble]

Tao is going into crank territory with his GPT nonsense lol

[hexed steppe]

yeah coding is different

it would seem

[cedar pebble]

same thing happened to Voevodsky when he stopped doing motivic homotopy theory and started being like "homotopy type theory is the future" and "maybe Peano arithmetic is inconsistent but that's also okay"

lmao

[hexed steppe]

idk at least tao is proving things still

even if hes using them to talk about lean

[alpine nest]

Isn't Lean a different beast since it actually implements logical reasoning?

[hexed steppe]

kevin buzzard’s legacy

well you see

[alpine nest]

Unlike LLM's which are probabilistic in nature

[hexed steppe]

tao wants chatgpt to merge with lean

[alpine nest]

Ah

[hexed steppe]

so it will correct its math errors on the fly

[alpine nest]

As in using lean to verify and correct ChatGPT's output?

[hexed steppe]

(idt he came up with this but he has advertised it)

like, make chatgpt automatically generate lean code with its answers, and also check them

and then generate a new answer if it doesnt compile

[cedar pebble]

yeah I mean Lean is something you have to actually program yourself

it's a nice tool

it doesn't make up random bullshit for you like GPT does

[hexed steppe]

i think tao acknowledges that its garbage atm

but i think he just likes the idea of a big autocomplete tool

[hexed steppe]

basically just this

[cedar pebble]

yeah I mean no LLM is going to be able to do this reliably

[alpine nest]

I think there's great appeal in being able to communicate with software using natural language for inputs and outputs, and ChatGPT and its ilk make it appear tantalizingly close.

[hexed steppe]

its like OEIS

but way worse

is there actually appeal among like, professional scientists

[alpine nest]

I mean, for me writing up the actual paper was always the least fun part.

So if I could just tell my work to a LLM and have it spit out the LaTeX, that would be great 😄

[hexed steppe]

do you mean

write in english and not typeset

or let chatgpt actually write

the problem with the latter is the scarcity of well-written math papers to train on

[alpine nest]

I was thinking (in terms of pure SF, I don't think that's remotely doable right now) that I could explain my result the way I would to a human, and it would write it up 😄

Although my pain is mostly with the typesetting and managing notation, writing out the English was fine

[hexed steppe]

thatd be great lol

[alpine nest]

Also it's not really something I struggle with, but it's just tedious

[hexed steppe]

yeah

[alpine nest]

also because I have ADHD, my papers always have a lot of inconsistencies in notation

[hexed steppe]

i mean writing is very annoying

[alpine nest]

I need to find and check out one of these tools that generate TikZ code based on human description or image.

Because TeX as such I'm mostly fine with, but TikZ is awful to use

Incredibly powerful and versatile, but awful to use

[cedar pebble]

It’s not that bad tbh

You just get used to it

[alpine nest]

Yeah, fair, I only use it occasionally

So by the time I get over the initial pain I'm back to not using it

[cedar pebble]

It is fairly painful for complicated pictures sometimes

As long as you keep things simple with clean coordinates it’s not bad at all

most painful thing I have ever made in tikzpicture

[alpine nest]

looks great though!

[red yoke]

Surely inkscape be pog

[merry geode]

How do you even

[pallid delta]

I logged on and thought I was looking at some biology

blood flow and stuff

[tribal palm]

amazing

[formal tide]

Suppose the topology T' on X is finer than the topology T. Can this be detected by sequences? I.e. is there necessarily a sequence xn such that xn -> c in T' but not in T'?

[red yoke]

No

E.g. Long line + point is finer than closed long line

But either way a sequence converges to ω1 only if it is eventually constant

The problem is that the topology around ω1 is so fine that it's not first countable, thus you can't detect limit points using sequences

A generalization of sequences called nets can be used though

Two topologies are the same iff they have the same convergent nets and each convergent net converges to the same point(s)

[formal tide]

Thanks, yeah felt that'd be the case but couldn't find an example

that previous question is motivated by:
Let X be a set, and suppose we have a function that assigns to each sequences xn in X a set L (the set of the limits of xn). There is the very natural topology on X that has U close iff U is sequentially closed. I was wondering if this has a universal property (the finest/coarsest such that...) but it seems it isn't the case?

[ebon galleon]

Yeah unless you have first countable spaces you want nets as Arki said (or filters)

[formal tide]

You've just shown it's not the finest such that xn->c iff c in L. I don't think it's the coarsest either

yeah I know nets/filters work

I just find very weird that this natural definition doesn't have a universal property

first time I've seen that I think

[ebon galleon]

I haven't played around with them, but it looks to be something about sequential spaces, so I would expect that this process is a (left? right?) Adjoint to the inclusion SeqSpace --> Top

[formal tide]

ohhhh it does seem like a possibility, though you need extra data to carry the sequences and limits

but something like that seems like it could work

[ebon galleon]

Well okay that would be if you applied it to the limits of an existing topological space. I agree it should be either coarsest or finest topology such that blah blah blah (has exactly those limits?) But I'm not sure which

[formal tide]

not finest, as Arki shows

but doesn't seem the coarsest either

that's what I found very weird

[tribal palm]

if X connected and Y disconnected, is X times Y always disconnected?

[red yoke]

Disconnected times anything is disconnected

[formal tide]

Not clear if this even is a functor, as the domain is type dependent; i.e. it takes a set and a subset of limits of that specific set

[formal tide]

except times the empty set!

[red yoke]

A scholar of the empty set!

[tribal palm]

thanks

[formal tide]

[red yoke]

I think it should be finest?

Since if the natural topology T exists and any other T' has the same limits then any closed set is T' is a closed set of T?

[formal tide]

Applying the proccess to the closed long line would not give you the point I think

[red yoke]

I think the natural topology should be long line + point

[knotty vine]

This is one of the reasons the empty space should not be considered connected!

[formal tide]

I'm a bit confused by

any closed set in T' is a closed set of T?

could you elaborate why?

it should be

any sequentially closed set in T' is a sequentially closed set of T?
Right?

[ebon galleon]

closed ==> sequentially closed

[formal tide]

yeah but not conversely

so his statement is stronger

[formal tide]

the nlab agrees with that, so I must agree too

[ebon galleon]

So you take this topological space X, you get a collection of sequentially closed sets (which includes all closed sets), adn then you form this new topological space on the same set, say X_seq, which has all of these as closed sets. So it's a finer topology.

Which means the map id: X_seq --> X should be continuous, which should tell us which sided adjoint this is

https://en.wikipedia.org/wiki/Sequential_space#Categorical_properties which agrees with what was written here

[formal tide]

ohhhh this makes it very clear

thanks!

[ebon galleon]

Wait....

Sequential spaces are cartesian closed????

WTF

[tribal palm]

if X is connected and infinite and F is a finite subspace, i’m struggling to show X x X \ F x F is connected

[formal tide]

wikipedia say that it must be with its own product

weird

[ebon galleon]

Yeah, it has to be the categorical product for it to be cartesian. So this should be (I'm assuming we're right that the right adjoint is this process you described) this process applied to the usual product topology

[formal tide]

Try a particular example, let X=R and F={0}, see how you can "go around the hole"

[knotty vine]

Cool, sequential spaces seem to be both nice and form a convenient category at the same time, dont see that often!

[ebon galleon]

Yeah it's cool I'm just a bit surprised I've never actually seen them called a convenient category before lol

[tribal palm]

i can see it visually but i don’t know how to show it formally 😭

[white oxide]

For finite products you can quickly work this out

Actually, maybe it isnt clear. I recommend convincing yourself that products are distributive over unions

[tribal palm]

right, since you’ll get a product including an empty set

[white oxide]

Actually I am overcomplicating this, fix a projection map and use it

[tribal palm]

oh even better

[formal tide]

yeah mb I was thinking of path connectedness lol

[red yoke]

I think you can make use of the "path" visualization anyway

Note that being in the same connected component is an equivalence relation

[tribal palm]

is there enough information to define a continuous surj from X^2 \ F^2 to {0,1} ?

for a contradiction

um, i phrased that incorrectly but hopefully the idea is clear enough

[red yoke]

Disconnected does indeed imply that

Though not sure how that helps much

[white oxide]

This feels like a problem you have to draw out

If it was path-connectedness this would be fairly straightforward, maybe a similar argument works for connectedness

[red yoke]

And that products of connected spaces are connected

[unreal stratus]

Oh lol Jens

[red yoke]

Then you can draw a "path" with "horizontal" and "vertical" "lines" representing the relation

[white oxide]

Actually, X x X / F x F is the quotient space here right

[unreal stratus]

This is almost exactly in smth i've been working on

Isn't it \ kerr not /

[tribal palm]

hm?

no

[white oxide]

yeah mb

[unreal stratus]

dw

Anyway

[tribal palm]

my problem is with the set difference

[unreal stratus]

Good question

Yeah Jens dw that's what i'm saying

[red yoke]

Sol: ||For any (x, y) and (x', y'), take a finite sequence of points in X²\F² starting at (x, y) and ending at (x', y') such that consecutive points have one coordinate equal.|| Then ||consecutive points are in the same component by considering X × point||

[white oxide]

It should suffice showing that the union {x} x X of X x {y] is connected, so maybe showing that any map to {0,1} would have to factor through X?

[unreal stratus]

Ah my solution is basically equivalaent to yours arki

I think mine is perhaps easier to see though - for a hint: consider the characterisation in terms of functions into {0,1} (with discrete)

oh lol what kerr said

[tribal palm]

omg

that’s amazing

[unreal stratus]

Basically if you draw it out

You get uh

if you have a function X x X \ F x F -> {0,1}

it's constant for each fixed x and y

So you can see it must be constant overall

But yeah that is actually almost identical to Arki's nvm sorry 😭

[tribal palm]

i don’t follow

[white oxide]

In hindsight it justs making sure that being in the same connected component defines an equivalence relationship

and then given two points x and y in X x X \ F x F just consider the horizontal slice containg X and the vertical slice containing y, then their common point forces x and y to be equivalent

X x {x_2} is connected after all, so anything in it is equivalent to x. Same for {y_1} x X with y

[unreal stratus]

This is cause it is X x {y}

Or same other way round

[unreal stratus]

Ye

[tribal palm]

ok thanks for all the help but my brain has shut down and i will have to return to this later

i think i should also do some auxiliary problems connectedness and products behave with each other first

and the thing with connected spaces sharing a point implies their union is connected

in either case my top exam tomorrow will not go well

rip me

[gritty widget]

Keep us updated

[tribal palm]

i expect to pass, but not with a good grade :S certainly below a C

which will be my worst grade in a maths course but oh well

[gritty widget]

:/

[tribal palm]

i’ll probably take the exam again my first semester in my masters

(assuming i can get that far)

[stoic eagle]

Anyone have a direction? It’s from Hatcher 3 manifolds book, right after the chapter on prime decomposition

I think I’m missing something really easy

[tribal palm]

ok so i have figured if X is connected and homeomorphic by f with Y, and if A is subspace of X such that X \ A is disconnected, then so is Y \ f(A)

but i am trying to find a way to show two connected spaces X and Y are not homeomorphic by means of seing how their connectivity differs after removing points from each

[unreal stratus]

Yes there are simple examples

[white oxide]

the classic example involves your, probably, favourite basic top space example

[unreal stratus]

Well isn't that immediate from what you've done

[tribal palm]

i just realized i haven't even tried doing this directly with R and R^2 yet, and skipped straight to trying to generalize the procedure

[unreal stratus]

Or do you just want examples

[tribal palm]

i think so too but i'm not sure how to explain why and state the result

[unreal stratus]

Well if f: X -> Y is a homeo then X \ A -> Y \ f(A) is a homeo

[tribal palm]

right

[white oxide]

I am confused, what are you trying to do?

[unreal stratus]

So in particular if you can remove a point from each with one connected and the other not

Then the original spaces are not homeo

As it contradicts what I wrote

[tribal palm]

right, so simply by contradiction

[unreal stratus]

Yes

So for example R vs S^1

[tribal palm]

i am just struggling with language, well, among many otherthings

[unreal stratus]

Which is nice as both are 1D

They locally look the same

[tribal palm]

that's pretty swag

S^1 \ point is still connected

[unreal stratus]

Ye

[tribal palm]

it is clear some spaces are in a way "more" connected than others, is there a term for this? like, perhaps it can be measured by the cardinality of the least subspace such that it's complement is disconnected

[unreal stratus]

Well

[tribal palm]

no because that would not be a topological property would it

[white oxide]

There is path-connectedness, and then there is stuff that spiritually related

[unreal stratus]

There are notions of connectivity but they are probably not quite what you want lol

[tribal palm]

ye i think i can construct a counterexample

[unreal stratus]

Basically you consider maps of spheres into your space and consider then they can be deformed to constant maps

And the "0-dimensional sphere" case is path connectedness

[tribal palm]

whaat

[unreal stratus]

(there are certain extra things with basepoints but yes)

But anyway the simplest example here is the fundamental group

[tribal palm]

potato i love you just so you know

[unreal stratus]

like for each space X with a fixed point p, you consider loops in X starting and finishing at p

up to "deforming them"

[white oxide]

potato corrupting the youth with homotopy theory

[unreal stratus]

that gives you a group

(you can do one loop then another, etc)

And this is often useful for studying spaces

so for example under this, R^2 \ a point is less connected than, say, R^2

Because in R^2 \ pt you can't just like deform a circle about origin to a point (it has non-trivial fundamental group)

[tribal palm]

homotopies seem a lot of fun, i wish i had focused more of my effort into topology earlier

[unreal stratus]

It's okay, you can make a lot of progress w stuff anyway once you start

[tribal palm]

i blame my taking complex analysis last semester for my current unpreparedness, lol

[unreal stratus]

some point set topology is necessary + it's good you're doing some

[tribal palm]

*insert jens duck pray*

[white oxide]

what duck deities are you praying to

[tribal palm]

the duck god, duh

[white oxide]

Monoduckeistic? Heretic

[tribal palm]

*insert jens cat evil smile*

[tribal palm]

does this top on X = [0, infty) have a name? T = {ø,X} u {(a, infty) | a >= 0}

viro calls it "the arrow," but wikipedia uses that name for the lower limit topology, which is not homeomorphic to this one

[white oxide]

uh

[naive trench]

Mmmm i recognise that topology but I dont think it has a proper name

[white oxide]

the canonical toplogy

oh nvm it misses the "right arrows"

[tribal palm]

no, the above is compact

wait which do you mean by the canonical topology here

the subspace from R?

[white oxide]

the (a,infty) and (-infty, b) generate the canonical top on R, I just didnt see that the latter was missing

[tribal palm]

oh ok

[tribal palm]

tf is this terminology lol just randomly deutsch mittem im Satz

[naive trench]

What?

[unreal stratus]

Yeah lol it's funny

also slightly annoying to my eyes since it should be im Kleinen ig

[lyric hedge]

Hi, I am currently reviewing topology problems to prepare for an exam. So lets assume I have an ideal point x* in a unit square, which is not in the left corner. So it is somewhere in this unit square, but when I compare two spots that are the same distance to the ideal point I prefer the one closer to the lower left corner. Disclaimer I am not a math student so I am not used to proving things, so sorry in advance. So I wanted to check if the preference relation is rational in a) and in b) if it is continuous. Since for rationality I need Completeness (so every point in this unit square has to have a preference) and transitive (that the preference has to be consistent, so e.g. if one point A is closer to the ideal point than another point B, if C is introduced and further away than B it should be less preferred than A). So first of all are my proofs somewhat correct?

Moreover if rationality and continuity is given then there should exist a utility function correct? If so how would it look like/ how would I formulate it?
Also is there a difference if the choice is given by real numbers (for the distance) or by pixels? Thanks for the help in advance!

I had posted it in a help channel but was asked to try to find help here, so sorry in advance if that is not the correct way to do it.

[brittle rapids]

what is an ideal point

[lyric hedge]

Imagine I want to place an ad on a website and my ideal point would be where I place it on the screen

[brittle rapids]

what is a rational relation?

[lyric hedge]

well rationality is given when the decision is complete and transitive

[unreal stratus]

I don't really understand what you are saying

Or how it has anything to do with topology

I assume these are economics terms?

[brittle rapids]

completeness? do you mean reflexivity?

what is a continuous relation?

[lime sable]

maybe "complete" is anti-symmetry?

or totality

[lyric hedge]

So completeness is for any x,y in X : x > y or y > x, whereas reflexive means for any x in X : x has a relation with x

[brittle rapids]

wait i kind of understand

[lime sable]

https://en.wikipedia.org/wiki/Rational_choice_theory

These preferences are assumed to be complete and transitive. Completeness refers to the individual being able to say which of the options they prefer (i.e. individual prefers A over B, B over A or are indifferent to both). Alternatively, transitivity is where the individual weakly prefers option A over B and weakly prefers option B over C, leading to the conclusion that the individual weakly prefers A over C. The rational agent will then perform their own cost–benefit analysis using a variety of criterion to perform their self-determined best choice of action.

[lyric hedge]

And transitivity would be for any x,y,z in X : [x >=y and y>= z] => x>= z

[lime sable]

so it's a partial order or a preorder i think

[brittle rapids]

i think linear order

[lime sable]

is "indifference" <= and >= or incomparability

[lyric hedge]

indifference

[lime sable]

yeah this article says total order

wait

"total order minus some assumptions"

great

[brittle rapids]

in detail the relation is: given a pair of points p, p'; p > p' if d(p, x*) > d(p', x*) or [d(p, x*) = d(p', x*) and d(p, 0) > d(p', 0)]

think lexicographical

[lime sable]

ok it's a "total preorder"

[lyric hedge]

and a preference relation is continuous if a>b implies that there are neighborhoods of a and b (B_a and B_b) such that x >y for all x in B_a and y in B_b.
and if {(x,y) s.t. x >= y} is a subset of X x X is a closed set: I.e. if {(a_n, b_n)} is a sequence of pairs of elements in X for which a_n >= b_n for all n and a_n -> a and b_n b, then a>=b

[brittle rapids]

ohhh

[lyric hedge]

yes

So it is related to topology no?

[brittle rapids]

ya

[brittle rapids]

did you show the first part?

[lyric hedge]

Maybe that is of help, so for it to be a total order it should also be antisymmetric

[brittle rapids]

it's almost antisymmetric, for any point there is at most one other point that shares the same preference

[lyric hedge]

I hope so that I did this in part b

I hope this makes any sense

[brittle rapids]

under the second definition you gave, and assuming i understood the relation correctly, i don't think the relation is continuous

[lyric hedge]

I never had to prove things properly before since as economics students we always take things as granted (shame on us), so now im struggling quite a bit when having to write proofs like these to even get a starting point. Not to speak of the notation which is probably disgusting

Oh okay

[brittle rapids]

by "distance to the origin" we mean the euclidean one right

[lyric hedge]

yes

[brittle rapids]

like sqrt of sum of squares

[lyric hedge]

yes

Sorry this is a problem from a problemset not related to this but there is the definition for d included

[brittle rapids]

if i interpreted the relation correctly, blue is the set of points whose preference is higher than p, right

bundle?

[lyric hedge]

I would interprete it that the the ones in the lower left quadrant if you would devide the unit square on the ideal point are the preferred ones no and not the whole rest
Since when two points having the same distance I prefer the one closer to the lower left corner as they like to call it

[brittle rapids]

take like (0.1, 0.1)

[lyric hedge]

Bundle is is here I prefer some x_1 and some x_2 so basically the point

[brittle rapids]

then p is closer than (0.1, 0.1) to x* right

so p is more preferred

[lyric hedge]

Hm yes that’s true

[brittle rapids]

also why would we see quadrants? aren't all our distances euclidean? (aka we should only see circles, and intersections of circles etc.)

[lyric hedge]

Makes sense in my foggy mind I chose the easy way of dividing it but you are correct

[brittle rapids]

i feel like i might be missing the important context but unfortunately i don't know economics

[brittle rapids]

if we fix p, and consider a sequence of points (in the blue set) that approaches the dashed part of the circle, this says that the relation is not continuous

because their limit will be more preferred than p

[lyric hedge]

but i mean the biggest circle you could draw would be one around x* that is going through 0,0 no?

The ideal point is not in the lower left corner. When comparing any two spots, you prefer the one that is closer to the ideal point. When two spots have the same distance to this ideal point, you prefer the spot closer to the lower left corner. This is all what the problem says

[brittle rapids]

right, and a spot where we'd prefer p over is exactly a spot in the blue region

ok so the region is drawn correctly

[brittle rapids]

sorry what's the point behind this?

well it might not be, like if x* was really close to 0,0 then we could draw a bigger circle

[lyric hedge]

sorry ignore it

the ideal point could also be close to the top right which makes the point behind my thought irrelevant

[brittle rapids]

does this read btw

just making sure, the ideal point is the same for all spots right

[lyric hedge]

So basically we would then experience a "jump" since we prefer p until some point of the sequence reaches the dashed line correct?

which makes it not continuous

[brittle rapids]

well it'd never actually reach the dashed line

but it would approach it

[lyric hedge]

sorry for this sloppy expressions but i don't know yet how to express it in math terms

[brittle rapids]

no worries

[lyric hedge]

I see but how would I prove this I was so happy with my solution haha

[brittle rapids]

btw the first definition of continuity also fails; e.g. if we choose a point q on the solid part of the circle, then p is preferred over q. but any neighbourhood of q contains points that would be preferred over p

[lyric hedge]

True

[brittle rapids]

i just don't understand what's being written with the implications, sorry ;(

i think usually math people like using words

because it gives more context

[lyric hedge]

Oh really I thought they like all these fany notation thingies

I tried to verbalize the proof but i got lost in the middle so if i would do a proof by contradiction so: Show that a rational preference relation on X is continuous if a is preferred over b implies that there are neighborhoods of a and b such that x preferred over y for all x in B_a and y in B_b. would be the first proof and the second one would be to show that on X is continuous if {(x,y) s.t. x weakly preferred over y} subset of X x X is a closed set...." and then show through the assumption that they are not continuous correct?

[brittle rapids]

i'm finding what you wrote very difficult to read

the first proof? i thought that was the definition of continuity

same for the second

[brittle rapids]

ok so this confirms my suspicion

they are equivalent definitions

[lyric hedge]

Ooooooh

So I choose the one that is easiest to show that it is not continuous

[brittle rapids]

i mean

i think i showed both for you

oops

[lyric hedge]

I see you would use the graph and simply use the verbal description to prove it

[brittle rapids]

you can translate it into a more formal proof, i.e. make the sequence explicit, show that the point the sequence converges to is really more preferred

[lyric hedge]

Man thank you so so much, really your help was much appreciated

[brittle rapids]

btw i think it's worth it to show that the two definitions are equivalent

it helps gain familiarity with moving between X × X and X

[lyric hedge]

hm might be a good task to study thanks for the tip

Again thank you for helping me out and dealing with my sloppy math 😄

[brittle rapids]

hints: ||a isn't preferred over b if and only if b is strictly preferred over a|| ||open sets and closed sets are complements.|| ||what is the definition of an open set in X × X?||

[lyric hedge]

Thank you will definitely come back to that later on 😁

[west brook]

Is the fiber bundle commutative?

[somber falcon]

Not immediately sure what that means, are you sure you're using the right term?

[west brook]

Can you always interchange the fiber and base space?

[somber falcon]

Well unless it is a trivial bundle, it's not really a product. The way the bundle turns out is going to depend on automorphisms of the fiber, and if you switch the two, you're going to have a different automorphism group, so different gluings will be possible.

Consider the mobius bundle, a line bundle over the circle. If you "commute" and make a circle bundle over a line, and it doesn't seem possible to do this without it being a trivial bundle (naively...maybe somebody will show a non-trivial circle bundle over a line).

You can think of a fiber bundle as two things (roughly): a collection of product neighborhoods where factors are the base and fiber, and a collection of maps from the fiber to itself that show how to glue the product neighborhoods together.

[lime sable]

the line is contractible, so all (topological) bundles are trivial

[tribal palm]

wow almost all topology books i’ve looked at has introduced the same topological ideas from different starting points

i love this one:

[tribal palm]

someone plz critique my two proofs here, i am not sure if i am missing something

[white oxide]

They are fine

Maybe the second one is a bit too wordy

Like you could've looked at the pre-image of 0 and 1 and by connected argued one of those must be empty

[tribal palm]

i generally prefer that over too little words

right that would also work

i like how i can evade any mention of symmetry in my first proof though

but i feel it's a bit more handwavy than i prefer

[white oxide]

What symmetry

[tribal palm]

in S^0, you can interchange the two elements, lol

[tribal palm]

i suppose you achieve the same by just saying as you do here that one of the two has to be empty

[white oxide]

So?

[tribal palm]

typically what you do then, "is suppose without loss of generality" ...

[white oxide]

I still don't know what you mean by symmetry in the first proof

The image is connected, so you need a non-empty connected subset of S^0 as your image, this forces you to have your image to contain only one element

[tribal palm]

yes, and there is a symmetry in "image with only one elements," as it contains two possibilities which are symmetric

[white oxide]

I am not sure what you are getting at

Yeah, there are two choice functions and they act exactly the same

[tribal palm]

X Hausdorff iff the diagonal in X x X is closed, is so weird to me

and wow it generalises to X/~ Hausdorff iff ~ is closed in X x X

that's so swag

[naive trench]

What?

[red yoke]

Rare point-set be pog

[unreal stratus]

It is super useful

[tribal palm]

what what?

[unreal stratus]

Nice application: if f:X -> Y is a (continous) map, Y Hausdorff and G := {(x, f(x)} is the graph, then G is a closed subset of X x Y

Good exercise and has a v short solution using what you just mentioned

[tribal palm]

whaat

[naive trench]

The statement

[unreal stratus]

This is useful for example because you immediately know that, like, the graph of a curve is closed lol

[tribal palm]

i did not know continuos images from hausdorff spaces were always closed

[unreal stratus]

So many things you can draw in R^2 will be closed lol

They aren't

That's not what mine is anyway

[tribal palm]

oh, of X x Y, right

[unreal stratus]

Yeah

An important criterion to know for a map to be closed is that the domain is compact and codomain is Hausdorff

[tribal palm]

um can you restate in less condenced language

[unreal stratus]

Sorry

[tribal palm]

aha

thank you

i have just over an hour until my exam starts, and it is abundantly clear i have not looked enough at separation axioms and compactness

rip

[white oxide]

The diagonal being closed criterion is so good that AG uses it as it's own version of "hausdorffness", which I find kinda cool

[tribal palm]

oh actually i knew for metric codomains

[unreal stratus]

Yeah nice

Yeah the diagonal being closed is nice as it doesn't mention elements or anything

[white oxide]

I didn't think of it that way but yeah

Closedness is much more "point agnostic"

[wispy veldt]

Another usefull criteria i have been using lately is f is cont on compact set E iff its graph in E is compact

think i was trying to bound integral phi(x,1/x) where phi smooth and has compact support in R^2 and it ended up being usefull

[unreal stratus]

Oh interesting

[formal tide]

how does R/[0,1) looks like? I.e. [0,1) identified to a single point

[unreal stratus]

Hm it should be the line with two origins

Like R/[0,1] would literally just be crushing down [0,1] to nothing

so you just get R

and that is just R/[0,1) after you identify the point [0,1) with the point {1}

[opaque scroll]

Is not quite the line with two origins, since a set like
(-1, 1) is open, but looks like (-1, 0] in the quotient

[unreal stratus]

Yeah good point

[opaque scroll]

But I guess you can describe it as R with a special point x, such that the open sets are open sets of R or x with an open set of R that has 0 as the limit of an increasing sequence

Or at least contains a neighborhood of 0 intersected (-1, 0)

[tribal palm]

i am confident in my answers to exactly 50% of the exam

but then again i only answered 55% of it

so not much to hope for

the last 30% were clearly modelled on/ along the lines of the proof for the Urysohn lemma, which i am not at all famliar with beyond “you use some dense set” and that i consequently had absolutely no idea how to answer :S

the last 20% i am missing i spent like two and a half (out of four) hours on, getting nowhere, i suspect i missed something really obvious ehehe

there was this problem (10%): X and Y both connected, compact Hausdorff spaces, and some surjective f: X -> Y continuous with for each x in X some nbhd Nx of x with f(Nx) a nbhd of f(x), show that for each y in Y, f^-1{y} is finite

[tribal palm]

oops forgot a probably crucial assumption: each f|Nx is an imbedding

[white oxide]

This gives f^-1{y}, by injectivity, the discrete topology

[tribal palm]

oh

for each f|Nx you mean

but one can also show f^-1{y} is compact

and take {Nx : x in f^-1{y}} as an open cover

[white oxide]

Yeah each Nx only hits a single element in the fiber

So it has the discrete subspace topology and is compact. Hence it must be finite

[tribal palm]

thank you very much Kerr, and lol i was certain there was something very simple i was missing

i’m not sure how i even managed to miss this 😭, i set up everything else i needed, that is, i found the finite cover of Nx’s of f^-1{y}, and i managed part before to show that f is surjective, which was originally not an assumption

i suppose i just forgot to remind myself f|Nx imbedding means it makes a bijection with it’s image

[fathom steeple]

dosent this prove the stronger statement that every function f:D^n->D^n has a fixed point where x is in D^n-S^(n-1)

since if every point on the boundary was a fixed point, and were the only fixed points, there would still be a contradiction?

[umbral panther]

A function from D^n to D^n with only a fixed point on the boundary is the constant function that lands on a point in the boundary

[fathom steeple]

why

[hexed steppe]

any constant function has exactly one fixed point

[fathom steeple]

bw claimed the converse

[knotty vine]

I dont think this is true? Take the interval [0,1] and the function x/2

[opaque scroll]

Yeah, in general just moving everything closer to a single point will only have that one point as a fixed point. Wheter that be on the boundary or not

[knotty vine]

I think what @hexed steppe said is actually a response to your original question, instead

[fathom steeple]

hm, how

[fathom steeple]

my point is that if f only has a fixed point on the boundary, the construction in this proof still gives a retraction

[umbral panther]

By “a function” I mean an example of a function. A counter example to what was claimed

[hexed steppe]

no

[knotty vine]

Oh right I see

[hexed steppe]

i was answering the question about bw’s example

[knotty vine]

Yeah I misunderstood bw, I though you meant "any function"

[hexed steppe]

this is not a realistic scenario

or rather

the proof shows that the set of fixed points cannot be S^{n-1}

[hexed steppe]

but that is way weaker than your claim here

[fathom steeple]

if there are no fixed points in D-S, the map given in the proof is a retraction?

[knotty vine]

No because it's not continuous!

Try it with the examples we gave

[hexed steppe]

yes which is a contradiction

[knotty vine]

I think the question was: why cant we prove that the fixed point must in fact lie in D\S by modifying the original argument to say "Suppose that f(x) =/= x for all x in D\S"

The problem being that the g so constructed may not be continuous

Of course we already knew that the (possible unique) fixed point can lie on the boundary as well (using any of the examples)

Now that I think about it, if a function D^n -> D^n has only a finite number of fixed points, must it always be an odd number?

[fathom steeple]

you're saying that my strong claim is true then?

[hexed steppe]

what

no

you have already gotten a counterexample for it

[hexed steppe]

maybe

its not true for S^1

[knotty vine]

Obviously since S^1 may have no fixed points at all, i.e. an even number

[fathom steeple]

so if there is a fixed point in the boundary but not in interior, then the map given in the proof is surjective, the identity on the boundary, but not continuous? is this right

[knotty vine]

Yep

[fathom steeple]

is there an intuitive way to see why its not continuous

or do i need to do the calculation

[umbral panther]

F(x)=x^2 has two fixed points, 0 and 1

[hexed steppe]

why is it surjective

[fathom steeple]

sorry, range is boundary

thats what i meant

[knotty vine]

The easiest to see is D^1 with a constant map to one boundary point

[knotty vine]

Yeah surjective to S^{n-1}

[hexed steppe]

no that is not correct

i dont understand what oyu are even asking

the map cannot be defined

if there is a fixed point

it is a proof by contradiction

[fathom steeple]

i thought if there is a fixed point it gets sent to itself

since the "ray" between f(x) and x is just the point

or i guess its not even a ray

[knotty vine]

Yes it can if the fixed point is only on the boundary, just define g(x) to be the identity on the boundary, have some imagination!

[hexed steppe]

no that is a completely different situation than a ray

[hexed steppe]

???

yeah obviously this wont be continuous

[knotty vine]

Yeah that was the question, no?

[hexed steppe]

i genuinely dont know what the question is

setting g to be identity on the boundary

[fathom steeple]

my question was based on the assumption that the ray between x and f(x) is x if x is faixed p[oint

[hexed steppe]

is completely different

from fixing a single point

[knotty vine]

In the proof we assumed that f has no fixed points, in order to derive a contradiction

[fathom steeple]

right, but i was thinking that the same proof implied that there had to be a fixed point in D-S

so assuming that there are no fixed points in D-S, but maybe on S

[hexed steppe]

you cannot get a contradiction because there are continuous maps which fix only one point on the boundary

sure you can define g(x) = x for x the fixed point

[fathom steeple]

i see

[hexed steppe]

but like

then what

[knotty vine]

Yeah, a modified argument can't go through because g may not be continuous

[hexed steppe]

you dont really know anything about this g

[hexed steppe]

you should really try to prove the map here is continuous

that will probably clarify a lot

[fathom steeple]

do you do it explictly

with coordinates

[hexed steppe]

yes that is what analytic geometry usually connotates

[fathom steeple]

oof

[knotty vine]

Define g via the ray method on the interior and the identity on the boundary (this defines the same map as the original argument in the case that f has no fixed points at all). If f has a fixed point on the boundary, then g is not continuous, hence the argument doesn't work

[hexed steppe]

yeah maybe it would help to do an example

[fathom steeple]

so the equation of the ray is f(x)(1-t)+tx, t>=0

[hexed steppe]

consider f a constant boundary-valued map

[fathom steeple]

how do i explictly find when this intersects the boundary

cleanly

[hexed steppe]

try defining g

and like draw a picture

[fathom steeple]

g?

[knotty vine]

The function in the argument

[hexed steppe]

im not responding to your analytic geometry question lol

just think about that

for more than 2 seconds

and see if you cant figure it out

i am instead trying to outline an example

which might further illustrate why g ought to be discontinuous in the case where f fixes a point on the boundary

[knotty vine]

The easiest is on D^1

[hexed steppe]

yeah

do you mean D^2

[fathom steeple]

what is the difference between f and g in your example

[knotty vine]

No

[hexed steppe]

oh i like D^2 better lol

[hexed steppe]

i dont understand

f is the constant map

and as semer said

g is the map in the argument

[knotty vine]

On D^1 you dont have to think about annoying intersections with a circle

[hexed steppe]

well anyway

definitely do one you can draw

[fierce bobcat]

guys can yall rank
algebraic geometry, differential geometry, algebraic topology, differential topology
by difficulty

[somber falcon]

It doesn't really work like that imo

They are all very difficult if you learn them thoroughly, and not so much if you just want to get by

There is a ton of overlap material too

Different people will excel at one or the other too

Some people love geometric arguments but hate abstract topological ones, some vice-versa. Depends on how your intuition works.

If you like coordinates and calculus and analysis, differential geometry can be more like that. If you prefer clean abstractions and abstract algebra and logic, algebraic geometry and topology are more like that. Differential topology mixes both.

There are subflavors of each, so it would depend on your instructor too. Some people teach diff geo like it is basically an analysis class, some people avoid coordinates like the plague. It depends.

They are huge subjects.

[fierce bobcat]

thx for the in depth explanation well written imo

like im a noob at math and i 100% understood that (unlike edit:any analysis textbook)

[somber falcon]

If you are looking at undergrad classes, it's common for the undergrad differential geometry course to follow something like Do Carmo's 'differential geometry of curves and surfaces', and it's taught like an extension of multi variable calculus where you do more interesting geometry. You could check that out for the flavor.

What "algebraic topology" means as an undergrad class can vary a lot by institution, so you could look up whatever the course text is.

Algebraic geometry is typically not offered at undergrad because you need a lot of abstract algebra just to understand the basics afaik.

[balmy field]

Rudin is evil!!!

There are good reference books and there are bad ones

[rancid umbra]

rudin is a fantastic second read after you’re more familiar with the material

[balmy field]

[proud pawn]

Love him or hate him he spitting theorems

[alpine nest]

Every Rudin book:

I quite like them as reference, but I really wouldn't recommend any of them to someone learning the subject for the first time.

[naive trench]

AH yes my fav book

[balmy field]

So true lmao

[lament vortex]

i was taught analysis from rudin
but i looked at the book maybe 3 times the whole semester
prof actually took care to elaborate on things and i learned a ton

[somber falcon]

Spivak's calculus is my favorite first encounter

I found an error in it once and he included it in the errata. This is the most famous I will ever be lol

[tribal palm]

that's swag

[merry geode]

Wow

[somber falcon]

it happens way more often than you'd think

[merry geode]

Still, great feat

[quartz crater]

Hello, im taking a “multivariable analysis” course as they call it in Sweden. We’ve been introduced to very basic terms from topology regarding sets in the (Euclidean) space R^n, like compactness, openness, etc. and the definitions use neighborhoods/open balls and set theory.

That is, we haven’t explicitly been introduced to metric spaces or topological spaces so im saying all this as a disclaimer(?) if I don’t understand something, since I couldn’t find an appropriate channel to ask this in

I have a question regarding an infinite union of closed sets, specifically, $$\cup_{n=1}^\infty [\frac{1}{n},\frac{n-1}{n}]$$

[gentle osprey]

Aslan

[quartz crater]

Why does the resulting set become open? $(0,1)$

[gentle osprey]

Aslan

[naive trench]

Thats (0,1)

Oh

[red yoke]

Well do you agree that (0, 1) is open

[quartz crater]

Yeah

[red yoke]

Arbitrary unions of closed sets need not be closed

[naive trench]

Yep

[red yoke]

Although finite unions are closed

[gentle girder]

are you asking why the union is (0,1)?

[quartz crater]

I would naively take the limit

But don’t quite know how to reason it down to being open

From taking the union that is

[gentle girder]

you should take any element of (0,1) and show that it's in the union, and any element of the union and show that it's in (0,1)

and then since (0,1) is an open interval, it's open

[gentle girder]

if you find a simpler form for your set, then it'll be clear that it's open

[quartz crater]

What if I don’t know that it’s (0,1)?

[gentle girder]

well you should reason that you get (0,1), and then prove that the union is (0,1)

use your geometric intuition, you basically will get everything in the interval without the endpoints, but you won't get 0 or 1

alternatively, you have a sequence of closed intervals $[a_n, b_n]$ where $a_n$ is decreasing and $b_n$ is increasing, then the union of all such intervals will be $(a,b)$, where $a = \text{inf} (a_n)$ and $b = \text{sup} (b_n)$ as long as $a_n$ and $b_n$ are not eventually constant

[gentle osprey]

smay

[gentle girder]

imagine the interval expanding to contain more and more things

(if $a_n$ attains it's minimum then you get that the lower endpoint is contained in the uinion)

[gentle osprey]

smay

[quartz crater]

Reason I’m unsure about my intuition is for example when such a union is closed(?), so I feel uneasy going with my gut feeling unless im missing something

[gentle girder]

this is why math has proofs, our intuitions are often not good enough

[red yoke]

It might help to check the endpoints specifically

[gentle girder]

if you intuit something, try to prove it and if you can't, or find out that what you said isn't true, then try to update your intuition

[red yoke]

Is there any [1/n, (n-1)/n] containing 0?
No
Is there any [1/n, (n-1)/n] containing 1?
No
Is everything in between contained in some [1/n, (n-1)/n]?
Yes

[gentle girder]

this is also why you draw pictures when you're given problems

so that you know what questions to ask

[quartz crater]

I’m sorry if it seems I’m ignoring some of your replies just trying to respond to at least one of you while slowly reading your replies

[gentle girder]

you should have some suspicions every time you draw a picture, and those are all avenues of possible solutions, some might get shut down, but you can always go back and try another one

it's all good

take your time

[quartz crater]

Thanks for the help i appreciate it

So i should be wary since theres no n s.t 1/n = 0? or resp. (n-1)/n =1 ?

[gentle girder]

yes that’s right

the union is the thing that “contains all the elements in all of the sets”

[gentle girder]

it’s actually a bit different

[quartz crater]

Hm

[gentle girder]

1 being in the interval happens to be equivalent to "(n-1)/n = 1 for some n" because you are suspecting that the upper endpoint of the union is 1. The point is that 1 is not in the interval becuase none of the upper endpoints are greater than or equal to 1 (resp. lower endpoint not being less than or equal to 1)

but like you have the right intuition anyways so I shouldn't actually be pedantic about it

[gentle girder]

but yes this is the correct intuition to have

[quartz crater]

I dont mind it, if anything i appreciate it!

[quartz crater]

If its possible could you expand on that last bit?

I think im not fully getting it

[gentle girder]

draw a picture of the first 3 sets, and notice that there is space in between the upper endpoints. The union will contain all those values that you're shading, even the irrational ones. But does that mean that there's some n such that we have r = (n-1)/n for all r in (0,1)? no, the point was that r was contained in at least one of the intervals

which is to say that r was less than (or equal to, but it doesn't have to be equal) the upper endpoint and greater than the lower one for at least one interval

1 is not in the union because it doesn't meet that condition for any such set, since our upper endpoint is going to be less than (not just "not equal to") 1 no matter what n is.

[quartz crater]

I think im missing something, i just cant follow the arugment you're making :(
Just to make sure, is it the "red space" youre talking about here?

[gentle girder]

sorry what I meant was draw the first 3 nontrivial sets lol, it'll be more illustrative, draw out like, [1/3,2/3], and [1/4,3/4] and notice that there is no such n satisfying 1/n = (any of the points in between 1/4 and 1/3)

and notice also that there is no n satisfying (n-1)/n = (any of the points between 2/3 and 3/4)

but all of those points are also going to be contained in the union anyways

[quartz crater]

Hm, so taking it to the extreme is your point that there will will be no point inbetween 1/n and 0 for large n, and since we're never actually reaching 0 there will only be abritrally close rationals to it?

[brittle rapids]

consider $\bigcup_{0<r<1}[0,r]$

[gentle osprey]

quasi_semi_group

[brittle rapids]

you would agree that this is [0, 1), right

[quartz crater]

Sure

[brittle rapids]

why

[quartz crater]

r never reaches 1?

[brittle rapids]

well i mean

why is every point in [0, 1) contained in the union

[quartz crater]

This feels weird now lol, it was a bit easier when there was discrete steps like in the union above but now i cant even begin doing anything exepct naivley doing an uncountable union(?)

[brittle rapids]

try to think of the reason

forget the "approaching" nature of the union from earlier

[quartz crater]

Where do i start for example in this case?

[brittle rapids]

wdym

well ok, take 0.99

[gentle girder]

yeah, unions of sets can be unnatural to think about at first, but the point is that there is a good definition of union that makes it more clear, you should recall the definition of a union to start and then we can do an example

[brittle rapids]

why is that in the union

[gentle girder]

like what is a union

[quartz crater]

Alright by definition a "single" union if i recall correctly is just the set of all say x's such that x is in some set A or in some set B

[alpine nest]

a miserable pile of elements

[gentle girder]

okay yeah, and what about the union of any number of sets

[alpine nest]

By "single" you mean the union of two sets, and that's correct

[quartz crater]

While thinking of the def. i've only seen one for families and not one like above where r can range in R

so like U_(i in I) where I is an index set

[gentle girder]

they are the same actually

[alpine nest]

Well, that's what happens here, except your index set is (0,1)

[quartz crater]

Hm

Oh

[gentle girder]

U_i in I is just taking the union of the family A_i

[alpine nest]

For every index i you have the set A_i, and thus you have a family of sets, one for each value of the index.

And you take the union of that family, i.e. the set of elements which belong to at least one of them.

[quartz crater]

Oh okay

[alpine nest]

I.e. the union consists of all x for which there exists an i such that x is in A_i

[quartz crater]

So like @gentle girder mentioed somehwat i think my issue now may stem from the def. of taking union of a family

[gentle girder]

yep

[quartz crater]

Havent really thought about it at all

[gentle girder]

it's just the generalization of what you said earlier

it's the notion of "or"

[brittle rapids]

if you tried to rephrase this for a large index set, you would end up getting the same definition :)

[gentle girder]

a hint: ||you already have the union over at least one family, A U B is the union over the two-element index set {0,1}||

[quartz crater]

So A_0 or A_1 or ... etc? which i guess is this right^

(btw i really appreciate the help on this I think im learning something lol)

[gentle girder]

yes, but if you want to phrase it to accomodate uncountable index sets it's more like ||"all elements that are contained in at least one of the A_i"||

[alpine nest]

Well, that works if your set of indices is the natural numbers.

[quartz crater]

Hm i see

Whats so subtle about rephrasing it like that?

[alpine nest]

Well, what about this one?

[gentle girder]

you can't list an uncountable set

[quartz crater]

Oh right!

damn

[gentle girder]

so having it as a list can't refer to anything indexed over some uncountable set!

but usually these kinds of statements are generalizable

[strong swift]

what does Hatcher mean with Hom(-, G)?

[quartz crater]

Damn

[white oxide]

You plug in your objects into where the - is

[gentle girder]

hom is a functor from chain complexes to cochain complexes since it's a contravariant functor R-mod -> R-mod (or abelian groups if R = Z)

[quartz crater]

I would never take up on this subtieltily (cant spell) if someone mentioed it (like now)

[white oxide]

And get maps given by pre-composing

[brittle rapids]

you don't really need to worry about it, as long as you know what you're doing

[gentle girder]

you are allowed to be told about this subtlety, you can usually list stuff and everything will be all right, people will know what you mean

[brittle rapids]

for instance, it's important to learn what an argument based on countability looks like

and to know that you can't use it if your index family is not countable

[quartz crater]

Cool I see

[brittle rapids]

anyways the definition is good and all, but it's good to get practice of moving between your geometric intuition and the abstract definition

[quartz crater]

In this case picking r=0.99 works because theres exists some x such that x \in [0,0.99]?