point-set-topology

[unreal stratus]

Good

[surreal burrow]

thanks potat

[unreal stratus]

Np hug

[surreal burrow]

aw that's so cute

[broken nacelle]

it's a bit hard to read tbh lmfao

[gaunt laurel]

Dang

[broken nacelle]

for one I don't think collapsing S^1 \times {x} is you what you want

since that's a longitudinal circle

try to write an explicit deformation retraction of the filled in torus to the central circle

hint: ||use the universal property of the product of spaces||

and since you didn't specify what gamma is it doesn't actually have to be aa^-1

it could just be a constant loop

[gaunt laurel]

Oh, I didn't know longitudinal and meridian had been given convention to one of the dimensions. Thanks

[broken nacelle]

also, you said that the induced homomorphism is an isomorphism twice lel

I think you can safely delete "in fact, since phi [...] [phi i gamma].

[gaunt laurel]

Ahh, yes, thank you

[broken nacelle]

here was my proof btw

[solemn oar]

As an aside: Please separate your arguments into individual sentences. It makes it easier to read.

[ebon galleon]

Run-on sentence moment

[high hill]

booo

[unreal stratus]

Brain hurties

I just find it funny after you called the other proof hard to read

[broken nacelle]

[coral pivot]

Germanify the proof, make it one word

[broken nacelle]

I rarely have to write proofs for other people to read

so I suck at writing them lmao

[coral pivot]

Practice writing em up

[hidden crag]

Follows by Inklusionsschleifeninjektivitätswiderspruch

👍

[broken nacelle]

o7

it takes so much time tho

[coral pivot]

Yes math takes time

U can't skip it!

[broken nacelle]

[high hill]

[unreal stratus]

Google translate is famously good for as-yet-unwritten words

[ebon galleon]

apparently you can just make up word in german

[unreal stratus]

That is why I wrote it as such and not as just like fake words xd

deutsche Sprache, inklusionsschleifeninjekitivätswiderspruchsagenerlaubende Sprache

[hidden crag]

Pretty much

[feral copper]

Hey, dumb question, but: if I have an involution t:X->X on a cellular space X, what are the conditions on t to have a t-equivariant cellular decomposition? That is, cells are either fixed by t (set-wise), or t(e) intersects e only on lowed-dimentional cells?

Do I take a cellular structure on X/t and lift it?

But: is X/t necessarily cellular?

Ig I just have to ask that t itself is a cellular map?

[umbral panther]

There’s probably a topological manifold that isn’t a CW complex that has a double cover that is

[feral copper]

But most likely the involution will not be cellular then

[umbral panther]

Right, this would be an example of an involution that cannot be made cellular

[feral copper]

I think it's reasonable to think that if t is a cellular map, then I have what I want? Idk, I'm writing a proof and in any case I'm using it for a smooth involution on a smooth manifold x')

But the proof I have in mind only requires to work with a t-equivariant cellular decomposition of X

[umbral panther]

Asking that the quotient be cellular isn’t obviously enough. You need the fixed set to be cellular and the inclusion of the fixed set into the quotient to be cellular. Then you can lift.

[feral copper]

Yes ofc the fix-point set needs to be cellular, then I complete its decomposition into one of X/t and then I lift

Yeah I think if t is cellular itself, then Fix(t) is cellular and the inclusion Fix(t)cX and quotient map X->X/t are cellular

[tiny ridge]

There's no known example of a manifold which isn't a CW complex so bad strategy

In fact every manifold of dim =/= 4 is cellular by Kirby Seibenmann's topological handle structures

Here's a better example. Take the Alexander horned sphere, and consider the complement in S^3, known as the Alexander gored ball. Take the topological closure and double it along the topological boundary

By Bing's theorem this is homeomorphic to S^3. Gives you an involution of S^3 which has fixed point set as the Alexander horned sphere

Not CW

[umbral panther]

Every 4 manifold that is not triangulable is not a CW complex

In higher dimensions there used to be a question of whether non triangular manifolds are CW complexes, but I believe it was recently shown that lots aren’t

[tiny ridge]

Every high dimensional manifold is homeomorphic to a CW complex, so that is not right

It seems incorrect that nontriangulable implies nonCW

The attaching maps of a CW complex can be terrible

I think you should be careful about saying confidently wrong things

[feral copper]

Triangulable =/= cellular structure

[tiny ridge]

Kirby-Seibenmann's topological handle decomposition is a very classical theorem from the 70s

[umbral panther]

People are too careful

[tiny ridge]

Fine difference between a mathematician and a postmodernist philosopher.

[fair idol]

Sorry stupid question.

Let X be a topological space and U an open set. If I have the space of A(U) of constant functions f:U->A with A an abelian group with the discrete topology does A(U) consists of constant functions U is connected?

I'm trying to prove A(U)\cong A

I know with the discrete topology every element {x} is clopen. I think there is a basic fact about preimages of connected components I'm missing. Is it just that the preimage of {x} must also be connected so the only maps are U-> {x}?

[unreal stratus]

If U is connected yes

Image of a connected set under a continuous map is connected

[fair idol]

Are the only connected components of a discrete topology on A (say A has atleast 2 points) the singleton sets?

[gritty widget]

try to prove it

[formal tide]

Is there a name for this property: For any point x, any two open neighborhoods of it intersect at some point other than x.

It appears when proving that set-functions from R->Y, with Y Hausdorff, have unique limits

a sufficient conditions is that points are not open but doesn't seem to follow from any of the standard separation/countability axioms

[unreal stratus]

isn't this just equivalent to points not being open?

[opaque scroll]

The intersection of two open neighborhoods is open

[formal tide]

oh right lol I'm dumb

[floral bear]

what is the etymology of "homology"?

is it a reference to the quotient group, where homologous chains are made "the same"?

[queen prism]

https://mathshistory.st-andrews.ac.uk/Miller/mathword/h/ homology

i don't know what any of this means it's just a nice website

[floral bear]

but.. why the algebraic top definition?

why is that referred to as homology?

[empty grove]

Because you say that 2 chains are homologous when their difference is a boundary

[empty grove]

This one

I assume

[umbral panther]

1749! Wow, that’s way earlier than would have thought. Was it used in biology or chemistry by then?

[arctic relic]

Reviewing notes, have no idea what X_+ means in the context of CW complexes. So if X is a finite CW complex and X_+ is embedded in S^n, the spectrification of X is equivalent to the Spectrification of C(S^n/(i:X into S^n)[-n]

Well D/(spectrification of X_+) equalts that thing

I meant like DLD of X+ is equal to LD of C(…)

Where D is a prespectra, L is spectrification

Yeah i just dont know what X+ means in this context

[umbral panther]

Probably the disjoint union of X and a point, considered as a based space where the base point is the new point. (This is also the one point compactification of X, but that’s probably not very relevant)

[arctic relic]

So why is there this criteria of based X being embedded into a sphere vs just X on it’s own?

I guess the functor is literally from based to spectra

So it’d be fucking stupid if the space wasnt based.

Just like me

[unreal stratus]

I think the one point compactification thing is from like Atiyah duality right

[arctic relic]

maybe, i think there you remove a point from S^n so it’s R^n then just manipulate and do the one point compactification

Maybe im misremembering

[umbral panther]

Yeah, I didn’t pull one point compactification out of nowhere. The one point compactification of a vector bundle is the Thom space. But this X is supposed to be a finite complex, so it’s already compact and the compactification is just adding a point

[unreal stratus]

I am aware yeah

[heady skiff]

would the frontier of ${(x, y) \in \mathbb{R}^2 \mid 1 < x^2 + y^2 \leq 2}$ be $x^2 + y^2 = 2$ and $x^2 + y^2 = 1$?

[gentle osprey]

okeyokay

[queen prism]

frontier is boundary?

[heady skiff]

ye

i drew a picture and yea seems like it

[queen prism]

I mean I think it should be that
I don’t wanna prove it but it seems that way

[heady skiff]

i'm gonna come back and prove it rigorously but for now the problems are asking me to identify the boundary, interior and closure of sets

[heady skiff]

lol i feel you

the interior of b) is just E^2 with both axes removed right lol

[ebon galleon]

what the fuck is this book

E for R, frontier for boundary

[heady skiff]

armstrong topology lol

[heady skiff]

that's what i'm saying

bro tryna be different

[queen prism]

I only knew about frontier because of Wikipedia

[ebon galleon]

anyways yeah what you've said is correct

[heady skiff]

boundary is so much more intuitive lmao

[heady skiff]

is it considered cheating to plug c) into desmos

lol

[gritty widget]

why would it be

it's not like you're writing a strictly monitored exam. go for it king

[heady skiff]

nah ykwim

like as in cheating myself 😖

cuz like what if he like puts some weird ass trig function on the test and we have to find the boundary and shit and i don't have desmos ykwim

anyways he prob won't

oh bruh wtf nvm it's not like a graph of sin(1/x) will help me anyways

forgot how wack this function is

[heady skiff]

lmao is this right for both b and c

i mean like visually it seems pretty clear

but they're suspiciously similar

[ebon galleon]

c) closure is correct but interior and boundary are not

[heady skiff]

hmmmmmm ok

i'll think about the definitions then

[ebon galleon]

you'll need more points in the boundary. It's what you have + something else

(and by complement: less things in the interior)

[heady skiff]

bruh ok looking at the graph of sin(1/x) it looks like it covers almost everything at the origin but i know zooming in super close that's not exactly true....

i have no clue how to see this rigorously, lol

i don't think looking at a graph will help either

[ebon galleon]

what do you mean percisely by "covers almost everything at the origin"?

[heady skiff]

lol

precisely, nothing

it obviously doesn't but like

[ebon galleon]

That's kinda maybe close to how you should think about it, depending on what you mean lol

[heady skiff]

if you zoom out far enough it looks like a blob

lol maybe

[ebon galleon]

Do you have a guess for which points need to be added to the boundary?

(Here, note that the boundary is also the closure of the set { (x, sin(1/x)) | x > 0 })

[heady skiff]

i haven't really thought of that yet, but i'll keep that in mind

sorry im just like also doing some reading for the class, i'll return to this problem fs tho

[tawdry widget]

I want to learn this result.

Can you recommend me references? “One point compactification of a vector bundle is a Thom space”

[umbral panther]

Well, it’s only true when the base space is compact

But it’s pretty much just the definition of the Thom space

I guess people usually don’t work with the Thom space, but instead work with relative homology

[tawdry widget]

I see, thank you.
Sorry yeah I just saw it required to be compact somewhere else too. I thought this was true for general paracompact base space B

[umbral panther]

My recommended book is, as always, Bott and Tu.

[tawdry widget]

Thank you very much!

[cursive tendon]

I'm exploring the idea of the "roundness" of a polytope. Here in a paper, it's roughly defined by the ratio between the size of a containing ball and a ball it contains

Given a bounded polytope Ax <= b, can anything be said about the smallest containing ball and the biggest contained ball? Can I compute them?

[red yoke]

Maybe #linear-algebra or #optimization ?

[cursive tendon]

yeah I'll post it in optimization, I'm reading about basis reduction methods in integer linear optimization

thanks

linear algebra is maybe more fitting

[somber phoenix]

On 2 implies 3... He shows nothing. He showed that if you add extra points to A, then A closure is inside A.. But that shows nothing

[queen prism]

A is closed since it is equal to its own closure

but A is just f^-1 (B)

so f^-1 (B) is closed

[ebon galleon]

He's not "adding points to A", he assumes x is in the closure of A, and shows it must be in A. Thus, we conclude clA is a subset of A, hence clA = A

[somber phoenix]

He takes f(A), then adds points to it, and notes that f(A closure) is inside it...

[queen prism]

ok point to the line where it stops making sense

[somber phoenix]

Right after "therefore, of x in A closure

[ebon galleon]

If f(x) is in B, then x is in f^{-1}(B)

So he showed that if x is in closure of A, then f(x) is in B. Thus, since A = f^{-1} (B), we get that x is in A

[somber phoenix]

That first thing is not necessarily true, at least not by the lines before

That he wrote

[ebon galleon]

THis is a correct proof

If you're confused on the line that shows f(x) in B, we are assuming that f(cl A) is a subset of cl F(A). So this all makes sense, since cl B = B by assumption of B being closed

[somber phoenix]

He says f(f^-1(B)) is a subset of B. Therefore, some Y in Y may be missing, because some x in X may be missing..

[ebon galleon]

This is correct

f(f^{-1} (B)) is always a subset of B. Work through the definitions to show this

It is not always equal, as you might notice by your comment. That is, there may be points in B that aren't in f( f^{-1} (B)), but this direction he used always holds

[somber phoenix]

The proof is trying to say something about sets in X by figuring out that they might not be the same in Y.

It just doesn't work. Atleast not with what he's given.

[ebon galleon]

Perhaps instead of trying to argue why this standard proof is wrong, you should try to work through the details until you can point out exactly where the mistake is

Because this does work

[queen prism]

ye it's fine

[ebon galleon]

And this makes sense to do because the sets we have chosen are preimages of sets in Y. So it's not unreasonable to do this

[somber phoenix]

I've pointed out where it doesn't work, where he makes assumptions which he has given no argument for why it works.

[ebon galleon]

Well what you thought you pointed out was wrong. Read the above

[real notch]

It works

[queen prism]

if x is in cl(A) then f(x) is in f(cl(A))
but f(cl(A)) is contained in cl(f(A)) by hypothesis
thus f(x) is in cl(f(A))
that is, f(x) is in cl(B)
but B is closed so f(x) is in B
thus x is in f^-1 (B), which is just A

which line does not make sense to you

[ebon galleon]

@somber phoenix please give this a read and see if any part is still confusing to you

[somber phoenix]

Hm ok, that works.

Thanks.

[tidal lynx]

I need your type of confidence

[somber phoenix]

You must like math too much. You don't spend enough hours on things you don't really care for or like (though, topology is not even nearly as boring as ex analysis, it can even have it's nice moments).

[ebon galleon]

[real notch]

what did he mean by this

[quiet thorn]

what is ex analysis

[hidden crag]

Lmfao

[tribal palm]

the only maths i’ve had to do so far that i’ve found genuinely boring is plane geometry and linear algebra

admittedly i didn’t feel like i understood much of the linalg

i anticipate this changing sometime

[hidden crag]

more lin alg is always the answer

[ebon galleon]

[quiet thorn]

inject yourself with lots of drugs linalg

[narrow cairn]

fucking dogshit notation what the hell (for (2))

$\subset$ for $\subseteq$ is terrible

[gentle osprey]

most likely to honorable

[quiet thorn]

It's fairly common

And you very used to it

[narrow cairn]

i hate that so much

why is math notation so ambiguous across authors

can we please just standardize this shit

[queen prism]

i think math has bigger problems than that

[ebon galleon]

^^

[narrow cairn]

like what

[ebon galleon]

Anyways, you should never be writing in a manner such that the distinction between the two is entirely within the notation

Especially between those two

Much better to just explicitly write (in words) that this cannot be equality

[queen prism]

mainly the atrocious education in many countries, which gives people a very bad impression of what math is

good notation is good for learners tho

[narrow cairn]

do yall write A complement B as A \ B or A - B

[ebon galleon]

Usually A - B but sometimes A \ B

[narrow cairn]

i feel like if you write A - B for that you have to write A + B for the disjoint union of A and B

[ebon galleon]

[queen prism]

I prefer A \ B
because A - B is often used for {a - b | a in A, b in B}
and because (A without B) u B is not really the same thing as A

[ebon galleon]

I mean that is fairly common

[gentle osprey]

Ryx (Home for flowers)

[narrow cairn]

upside down capital Pi will never not be funny to me

mathematicians running out of greek letters so they just rotate one

imo we should use $\Xi$

[gentle osprey]

most likely to honorable

[queen prism]

you are a letter

[ebon galleon]

Anyways, complement is an internal construction, whereas disjoint union is external, so they're not particularly connected anyways

[narrow cairn]

sure

[median sand]

the Riemann Hypothesis.

[narrow cairn]

nah

[median sand]

ok then, whether 0 is a natural number and what positive means

[narrow cairn]

easy solution to both just remove 0 overall

no more 0

[ebon galleon]

no more abelian goops

[median sand]

GOOP in shambles

no more vagina-scented candles

[narrow cairn]

make the ZFCN axioms which is ZFC plus the No Zero Axiom:

∀x(x ≠ 0)

[bitter smelt]

How has alg-top come to this

[narrow cairn]

0 is a natural since it is in ω, and positive means >0

both are solved

give me a fields medal

[median sand]

Discord standards of discourse are painfully low

[median sand]

i am a representative the IMC, please send me your shipment/billing address along with your credit card information (collateral), i will arrange the delivery of the medal (for a small administrative fee of $1999)

[empty grove]

How do I pay?

[median sand]

with our proprietary cryptovalue and NFT, Grothcoin

[empty grove]

📈

[void tapir]

proper subset is written subsetneq or just e.g. let A subset B be a proper subset so it's not ambiguous

[narrow cairn]

what is $\subsetneq$

[gentle osprey]

most likely to honorable

[narrow cairn]

huh

never seen that

interesting

[paper wedge]

A is subset of B but not equal to B

[unreal stratus]

I like it

[onyx raft]

Very cool topology discussions in here...

[kind pollen]

Subset but not equal

[high hill]

its for fools who use \subset to mean \subseteq

(I think most people do this )

[coarse night]

⊆ & ⊊

[quiet thorn]

rare sight to see ryu not have a troll name

[high hill]

trolling ryc

[coarse night]

lol

[tribal palm]

i have recently taken to doing this, as both my prof and munkres do

[tough hamlet]

well then you are lost

[hidden crag]

yoo discoshrug

[tribal palm]

honestly it is very interesting to see how my notation changes almost subconsciously

[queen prism]

if you repeat a lie enough times it becomes the truth

but yeah you just get used to certain things after using/seeing them a million times

[tribal palm]

as if there is any truth in choice of symbols

[high hill]

time to use < for leq

[tribal palm]

[high hill]

and < with the thingy on bottom for le

[hoary breach]

use picatchu

[tribal palm]

\lneq

[high hill]

[tribal palm]

btw to stay on channel topic, i’m not sure if i felt stupid or felt smart as i wondered if every topology had a basis, then realized the topology itself makes a basis, so i can always just state let (fancy) B be a basis

[high hill]

yeah i think basis is a silly term

generating set suits my taste buds

[queen prism]

what would a subbasis be

[high hill]

ok time to run before i get pickaxed by the topologists

[high hill]

open generating - basis
generating - subbasis

idk lmao

[tribal palm]

i struggle to wrap my head around how every open set can be written as a union of basis element

i can do the proof, but is somehow just falls together in such a way i struggle to believe it

[high hill]

the defn of basis i prefer is
every open set is a union of basis elements (which have to all be open)

but ig ur using the other one to prove this

[tribal palm]

iirc the one given by munkres is every x in X is contained in some basis element, and for any two basis elements, their intersection contains another basis element

[high hill]

yh thats the usual one. its probably the most useful in proofs is why

[tribal palm]

this does sound much nicer, but it sounds more general

it’s so fascinating how many of these fundamental concepts can be defined in seemingly differet but nonetheless equivalent ways

[hoary breach]

A collection of open sets is basis iff given an open set O and x in O there is a basis element B s.t. x in B subseteq O is one way to think of the open set as union of basis, but yea if you're fine with vectors as lin combination of basis then this isn't really all that diff

[tribal palm]

not that diff minus uniqueness

[red yoke]

More like spanning set

[tribal palm]

will any (\mathscr B \subset \mathscr P(X)) with (\bigcup \mathscr B = X) make ((X, \mathscr T)) a topology?

[gentle osprey]

Jens

[red yoke]

B may not generate the same topology if that's what you mean

E.g. (n, n+2) for n ∈ Z does not generate the reals

[tribal palm]

oops i forgot to specify (\mathscr T = \left{ \bigcup \mathscr A \mid \mathscr A \subset \mathscr B\right})

[red yoke]

But if S_b is the collection of open subsets of b then the union of S_b over b ∈ B will indeed generate the original topology

[hoary breach]

B has to satisfy some nice properties

[red yoke]

Yes by definition of generated topology

[gentle osprey]

Jens

[tribal palm]

feels like cheating ngl

but thanku

come to think of it i don’t even need the condition that the union over the basis is equal to X, if we adopt the convention that intersection and unions over the empty collection of sets is equal to X and ø resp

[red yoke]

Then B is a subbasis not a basis

[tribal palm]

whaat

if i drop the condition?

[red yoke]

Cuz basis doesn't involve intersections

Oh wait B may not be a basis actually

You need finite intersections of elements in B to be a union of elements of B

For it to be a basis

[tribal palm]

well yeah i’m not generating elements in T via intersections in B, i’m only saying that ø,X in T even in if U B ≠ C

[tribal palm]

right, i’m not sure how to handle that

[tribal palm]

how can this then be taken as a def?

[red yoke]

This is equivalent to
∪B = X and finite intersections of elements of B is an arbitrary union of elements in B

[red yoke]

Hm but aren't you getting the X using empty intersection

[tribal palm]

oops, true

[tribal palm]

so using this we don’t generate T by arbitrary unions in B? but we simply take it as given that T is a topology and that every open set can be written as a union in B

[red yoke]

Yes

[tribal palm]

okok i think i’m following now

thank again

[tribal palm]

oh wow gamelin and greene has a very clear and concise discussion on bases for topologies

love this small book

[tribal palm]

and damn you peeps think it’s bad to use subset for subseteq? i just noticed this book sometimes uses subseteq but more frequently uses subset to mean subseteq

may have to do with there being two authors

[queen prism]

i don't like it
but i get the feeling it's the dominant notation

doesn't matter as long as you're consistent tho

[high hill]

most of literature does except possibly the most modern.

[unreal stratus]

I like it but am too paranoid not to use subseteq and subsetneq lol

[feral copper]

$\varsubsetneq>\subsetneq$

[gentle osprey]

Matplotlib

[tribal palm]

terrifying

where \nsubseteq at

[feral copper]

$\not\varsubsetneq$

[gentle osprey]

Matplotlib

[feral copper]

Ugly

[tribal palm]

[feral copper]

[feral copper]

This means either that the thing on the left is equal to that on the right, or that it's not a subset of it
Useless x')

[tribal palm]

i made a summary of my findings after the above discussion

@red yoke @high hill checks out?

oops in 3. it should obviously be x in X not x in T

[red yoke]

Assuming TFAE refer to 1-3

Yes

[tribal palm]

why is not 4 equivalent?

well,

[red yoke]

4 is equivalent to true

[tribal palm]

i suppose it is too specific?

[ebon galleon]

4 is a basis for a topology

1 states B is a basis for a specific topology

[red yoke]

When you say "the following are equivalent" you mean any of the following statements imply any other

1, 2, 3 imply each other

4 is just a true statement

[ebon galleon]

oh true, it's just definitionally true anyways lol

[tribal palm]

4 does as well, though i see what you mean, in general these will be different topologies

[tribal palm]

i mean we have at least (4) => (1) right?

[ebon galleon]

no

[tribal palm]

whaat

[red yoke]

You can make 1,2,3,4 equivalent if you do

4. B ⊂ P(X) and T = {…} and for every …

[tribal palm]

but T is not a topology without what follows ! (i am clearly confused)

oh

right

[red yoke]

Your 4 is a theorem, it's like the difference between

x divides y
x divides y if and only if there exists integer n such that nx = y
One is a statement, one is a theorem

[tribal palm]

i just don't need to use the "the topology"

[red yoke]

You can say "x divides y" and "there exists integer n such that nx = y" are equivalent

But you don't say "x divides y" and "x divides y if and only if there exists integer n such that nx=y" are equivalent

[coarse night]

[tribal palm]

ok

i am still confused

but is it right now?

[ebon galleon]

still you run into the issue i pointed out

Also this is formatted wrong

[tribal palm]

oh, right that's a separate problem

[ebon galleon]

The parts (X, T) is a topology and B \subseteq P(X) should be brought out of the statements

[tribal palm]

i did initially but moved it inside as i wanted to include (4)

[ebon galleon]

Also (X,T) is a topological space, not a topology

[tribal palm]

but i realize now that might've been a mistake

[unreal stratus]

This seems like the sort of theorem where trying to remember it seems more hard than actually working it out if you need it lmao

[tribal palm]

yes i'm just writing down the many different definitions i see so that i can prove the equivalences when i've an overview

[ebon galleon]

Like you should have something like:
"Let (X, T) be a topological space, and B subseteq P(X). Then the following are equivalent:

1. B is a basis for T
2. For every U in T there is some family A subseteq B with U = {union over A}
3. For each x in X and U in T with x in U, there is some V in B with x in V subseteq U"

Because you want these to be in reference to a specific B and T

[tribal palm]

[queen prism]

omg this is so cute

[ebon galleon]

birb

[tribal palm]

[tribal palm]

why B subseteq P(X) rather than T? i am after all taking T as given

[ebon galleon]

either is fine

A priori they are different, but if B generated T then clearly B is a subset of T

[tribal palm]

even if B did not generate T, B would still have to subset T right

[ebon galleon]

if B did not generate T then all of the equivalent statements are false

so no?

Like P(X) is a basis for the topology P(X) but clearly is not a subset of any other topology

[red yoke]

I don't see a problem with this

[ebon galleon]

It's better writen this way

Because then it is more clear that we are considering a fixed T and B

[red yoke]

"TFAE" already makes it clear

[tribal palm]

we can have a basis B for T without the basis having generated T, in either case B subseteq T

[tribal palm]

i don't see how this is relevant, we'd still have B = P(X) subset T

[ebon galleon]

if it doesn't generate T it's not a basis for T so that doesn't make sense

It doesn't matter which you have. It is a correct statement either way so maybe I am just misinterpreting what you mean

[tribal palm]

by "is not generated by B" in mean T is not given in terms of B

[red yoke]

There is no notion of "precedence" in math, in any case X, B, T are objects that already exist

[ebon galleon]

Even so, it is cumbersome and does not look as good to start each line with "(X, T) is a topological space, B subseteq T".

[tribal palm]

as in, you can explicitly give T, and then give a B that generates T

[red yoke]

Touche

[tribal palm]

i very much agree with this

[ebon galleon]

And I would argue if this is not such a simple statement where we know the symbols in reference actually mean the same thing, you could run into issues with how it is interpreted

[tribal palm]

i took (4) out of the list as well and am left with the very neat

[ebon galleon]

idk what the proper way of wording what I have in mind is lol

[ebon galleon]

but yes, i think this looks much neater

[tribal palm]

now i just gotta figure out how to best tie the previous (4) into this

[red yoke]

But then you lose a part of (4)

B ⊂ P(X) and T = {…} where for every … and for every …
Implies
(X, T) is a topology and B is a basis of T

If you take out the "(X, T) is a topology" you weaken the statement

[tribal palm]

but we wouldn't have the other direction right? simply as the two statements referred to different topologies

[red yoke]

No they are still equivalent

T = {…} is not a definition for T

It is a statement involving T

[tribal palm]

hmm

i will just drop the T = {...} part then

feels weird though, as i don't even mention T in (4)

but i suppose it works as it's equiv to (2)-(3) which do mention T

[ebon galleon]

see the problem is still that 4 is not in reference to T. So while 4 might be equivalent to the statement "B is a basis" [in the sense that basis is just some collection of sets satisfying a few properties i think mentions in Munkres], it is not equivalent to the statemetn "B is a base for T"

[tribal palm]

that sounds weird to me, even if T is not referred to in (4) it is in reference to both (2) and (3) and by the introduction B subset T

[ebon galleon]

Take X to be R, T to be the usual topology on R

Take B to be {emptyset, R}

B is a basis. but not for the usual topology on R

[tribal palm]

yes but then the statements are just all false

[ebon galleon]

4 is true

1-3 are false

[unreal stratus]

Yes

[ebon galleon]

who is "Yes" in reference to lol

[unreal stratus]

4 not referencing the topology is a big problem lol

[hoary breach]

yes

[unreal stratus]

I agree w you ryx

[ebon galleon]

okay 💀

i was mega confoosed

[unreal stratus]

Also isn't 2 just the definition of a base

[ebon galleon]

2 or 3 yeah, depends on the source i think

[tribal palm]

oh riight

i didn't see how for all x in X there was U in B with x in U but of course in your example U=X in B

[ebon galleon]

and the only way you can have x in (U cap V) is if U = V = X with mine

In particular, since any topology is a basis (for itself), any topology strictly coarser than T satisfies 4 but not 1-3

[unreal stratus]

I ave always found thr way 4 is written in munkres or whatever a bit weird idk

[empty grove]

3 is also a weird way of saying B is a cover of X 💀

Oh not 3 I'm thinking about the definition of a basis in munkres

I guess 3 is also weird because it's a wordier way of saying that every open set is a union of elements of B

[tribal palm]

[empty grove]

But then that is just 2 💀

[tidal lynx]

given a topology and a basis for the topology, can any open set be written as a disjoint union of basis elements?

Oops any connected set fails at this by definition

[coarse night]

what 4 2 3 what are you guys talking about?

[tidal lynx]

this

[coarse night]

not disjoint, just union

[tidal lynx]

yea ik I just wanted to see if that stronger thing was also true

[ebon galleon]

A topology is a basis for itself, and it trivially holds when we take such a basis that any open set can be writen as a disjoint union of basis elements regardless of whether the set is connected or not (namely: the union of a single set: itself)

[tidal lynx]

This is what Munkres does:

• He defines a "basis for a topology on X" as a collection of subsets of X that satisfies condition (4).
• He then defines the "topology generated by the basis" as the collection of all subsets of X such that U is in the collection iff for all x ∈ U there exists a basis element B such that x ∈ B ⊂ U.
• He then proves that the colletion above is actually a topology on X
• He then shows the collection above is equal to the collection of unions of basis elements (so basically condition (2))
• He then proves that if you are given a topological space (X, T), and you are given a collection of open sets B ⊂ T, then if B satisfies condition (3) it is a basis for the topology (in the sense that the topology it will generate is exactly (X, T). Or equivalently, by the fact above, the collection of unions of elements from B is exactly T)

I'm curious as to what characterization of basis condition (1) is intended to give

[ebon galleon]

T is the intersection of all topologies containing B

[tribal palm]

haha that was what set me off with this in the first place, my books didn’t agree on which statement to take as the definition

i’m beginning to realize it might be a bad idea to use many different textbooks when they all define things differently, what is especially painful to me rn is how they use the notion of neighbourhood differently

[tidal lynx]

I think basis/neighborhood are the main ones

[tribal palm]

whew ok that’s reassuring

[tidal lynx]

Maybe limit/accumulation point of a sequence too

[tribal palm]

so if i can just get a grip of this i can actually get down to business

[tidal lynx]

OTOH "the closure of a set S is the intersection of all closed sets containing S" is an amazing definition

[ebon galleon]

this is also an amazing definition

[tribal palm]

folland mentions something like that in passing

[tidal lynx]

which one

[tribal palm]

real analysis

[tidal lynx]

no like mentions the closure def? or smthn else

[tribal palm]

this def, that is

[tidal lynx]

Oh

[queen prism]

i like that

[tidal lynx]

It's an exercise to Munkres to prove that

I asked abt it here like a week ago lol

Well not exactly that but a variant of it

[ebon galleon]

it's good because it works for so many different structures

[tidal lynx]

Like what

[ebon galleon]

cl A = (intersection of all substructures containing A)
Topological spaces and their many generalizations/alternatives. Ideals in a ring. Subgroups in a group.....

[tribal palm]

gamelin and greene say S subset X is open iff S is a neighbourhood of each of its points… it really makes no difference here whether you require nbhds to be open or not does it?

[ebon galleon]

Well if you require neighborhoods to be open, this is trivially true

[tidal lynx]

Oh I see wym

[ebon galleon]

The point is that if we take neighborhood to be "N is a neighborhood of x if there is an open set U with x in U subseteq N", these are equivalent

[tidal lynx]

Like "the field extension generated by a set of elements S is the intersection of all fields containing S..." or smthn

[tribal palm]

so if you require nbhds to be open there’s really not much reason to talk of nbhds instead of just open sets?

[queen prism]

i mean it's convenient

[ebon galleon]

They're really two different notions of neighborhood imo

The usual one is that "a(n open) neighborhood of x is an open set containing x" is really just shorthand, since this is a common phrase

[tribal palm]

right but this is logically more specific than the more general sense of neighbourhoods

i have seen some books define the topology in terms of a function X to P(X) that assigns to each x a set of neighbourhoods

[ebon galleon]

The (not necessarily open) neighborhood in context here is really more of a historical note. You can define topologies in terms of a collection of (not necessarily open) neighborhood at each point - a set N(x) subseteq P(X) for each x in X - which satisfy a few requirements. Given this notion of neighborhood, you can define what an open set is: it's a set that's a neighborhood of each of its points. What this statement is about, is that these two definitions of topology are equivalent: Their notion of neighborhood and open sets are consistent between the two of them

[tribal palm]

right right

what is the most common use of the word in the modern literature?

nbhds = open nbhds ?

[ebon galleon]

yeah open nbhds

[tribal palm]

excellent, thanku

i’m curious to have a look at the N(x) approach sometime though

so let me attempt to rephrase this definition in gamelin and greene without using the terminology of nbhds:

a point x in X is adherent to S subset X if all open U containing x meets S (so U n S ≠ ø)

or if all open nbhds of x meets S

[ebon galleon]

adherence

[tribal palm]

i’m a sucker for classification of points in terms of a some subset of a space

makes the def of the closure very simple

[ebon galleon]

adherence

i was unaware of adherence until i learned of convergence spaces

[earnest ibex]

In this proof, how do we know that the alpha_i are in finite number?

[opaque scroll]

That's the definition of the product topology

[ebon galleon]

By definition of the basis for the product topology

[earnest ibex]

Oh I get my confusion, in Munkres that isn't the definition of product topology, but Theorem 19.1

Thank you both

[opaque scroll]

Do they define it with universal property or something?

[ebon galleon]

Munkres turning definitions into theorems 🗿

[ebon galleon]

Probably take a subbasis to be inverse image under the projections of all open sets

Then you can show this gives the usual basis we think of

[earnest ibex]

Exactly that

[ebon galleon]

Because then it's just a simple statement about finite intersections of sets of the form π_α^{-1}(U)

[earnest ibex]

I'll give it a better look, thanks

[unreal stratus]

Okay not expecting a full answer lol but does anyone have any ideas for how one might study the map $T^2 \to T^2 \land T^2$ on homology?

[gentle osprey]

potato

[unreal stratus]

I'm actually interested in taking mod 2 coinvariants on the last bit too lol

I imagine this is the sort of thing where you should just find more-or-less explicit cycles and see what happens but idk

Or break it up into T^2 -> T^2 x T^2 and then the quotient map lol

[tribal palm]

that looks like something else than what i’m thinking of

[obtuse meteor]

Hm I'm not exactly familar with this map where does it come from?

like is this just like diagonals and then smash or what

[unreal stratus]

Yes

[languid patrol]

It’s not so bad, the diagonal map has the obvious effect on homology, then the inclusion of T^2 x {} \cup {} x T^2 is a cofibration so the reduced homology of the smash product is the relative homology of the product with respect to this subspace

[unreal stratus]

Yeah that's what I had too tbh should just keep going

With the actual compoot

I wi another gi

[languid patrol]

Especially for T^2 what happens is pretty simple

But you can do this for very general spaces as well

Yes you have that this is dual to cup product over a field, but more than that in this case the homology is very simple so it’s clear where the diagonal cycles go

[unreal stratus]

Yup agreed

[languid patrol]

In general yes the diagonal is about as hard to understand as the ring structure on cohomology, so it is not always simple

[unreal stratus]

Okay sure

Phew I wasn't being silly lol i just assumed you had an argument that works in general

I'm pretty sure the eventual map T^2 -> (T^2 smash T^2)/Z/2 ends up sending the degree 2 generator to another generator and everything else to 0

at least on like lol Fp homology which is what i'm doing

[languid patrol]

This is a calculation which is doable even over Z

[unreal stratus]

Yeah though I think the homology of that last thing i mentioned is a little annoying to determine right

[languid patrol]

Anyway everything is torsion free so it’s basically the same

[unreal stratus]

Ye

Oh my I'm a dumbass

I kept wondering why Chmonkey was typing but not actually saynig anything

[languid patrol]

Chmonkey all of a sudden learned alg top

[unreal stratus]

Well thank you

Anyway this is ultimately quite nice because uhhh

Basically the result I am trying to get involves a map X -> X smash X / Z/2

and it ought to be determined in terms of the cup product structure of X

which does indeed seem the case which is nice

:)

[feral copper]

Heyo! If I have a double branched cover X->S^4, is there a description of the generators of the second homology H_2(X;Z/2)?
I've tried to scratch my head around this, I somewhat managed something by hand where b_2(X;Z/2)=2, but I'm having to deal with arbitrarily large b_2...

I'm still struggling to understand where the second homology comes from. I accept that there is second homology, but Idk what surfaces would not be null-homologous above where everything is below.

[tiny ridge]

@feral copper Why is this surprising? Every 3-manifold is a branched cover over S^3, which has no first homology. But a loop which winds around the branching set degree many times can unfold to a nontrivial loop upstairs.

Similar things can happen with a codimension 2 branching set of a branched cover over S^4

[fathom steeple]

What does the "Each had a finite subcovering" thing meaning

A subcovering is a subset of a covering which still covers the set right? How is it being used in this context then

[ebon galleon]

If each B(x, epsilon_n) could be covered by finitely many elements in this open cover (i.e., is a subset of a finite union within the open cover), the since there are finitely many B(x, epsilon_n) that cover the space S, we could take the union of all these subcovers (finite union of sets, each with finitely many elemenets) to get a finite subcover

[tiny ridge]

@unreal stratus I believe if you label the circles in T^2 x T^2 = S^1 x S^1 x S^1 x S^1 as a, b, c, d then the diagonal class is ab + cd + ad + bc. Here ab etc denotes the cycle given by the subtorus obtained by crossing the two circles

[unreal stratus]

Okay sure

[fathom steeple]

Oh. I read this as each Ball being subcovered by the union of all the balls

which made no sense

but i guess its saying the balls being covered by the open covering in the 2nd sentence

[tiny ridge]

You can derive this, as TTEG said, by using the canonical basis of the exterior algebra that is the cohomology of the torus.

[ebon galleon]

yeah idk why they don't label the covering

But yeah it's in reference to this open cover of S, which we assume for contradiction has no finite subcover

[tiny ridge]

A good sanity check is that the self-cup of the diagonal class is the Euler characteristic

[fathom steeple]

yep i get that

[tiny ridge]

This is why, unlike for product of two circles, the diagonal class of T^2_A x T^2_B is not A + B

The latter has cup product A^2 + AB + BA + B^2 = 2AB = 2

But Euler characteristic of T^2 is 0

This is a classic point of confusion when youre first introduced to the homology coalgebra

[fathom steeple]

Hm, he says "Reader should prove subset of totally bounded is totally bounded". Is this no trivial?

not*

oh, i guess he wants the centers to be in the subset

So then would the proof be to get a finite covering of balls with radius e/2 and then for each ball, if it intersects the subset then choose a point in the subset and create a ball centered at that point with radius e?

[ebon galleon]

Yeah something like that works iirc

[heady skiff]

would the union of all the open sets which make up the base work as the dense countable subset? i have a strong suspicion that that's the case and am going to try to prove it, just wanted to make sure (or have my hopes and dreams crushed)

bc im thinking about the set of all open intervals with rational endpoints as a topology on the real line

the union of all those is equal to Q and Q is dense in R

[gaunt linden]

No, the union of all of them is R itself, which is not countable.

[unreal stratus]

The union of sets in a base is the whole space by definition

[gaunt linden]

But you just need a small modification of your idea ...

[heady skiff]

oh wait you're right lol

[heady skiff]

very true

[heady skiff]

ok

will keep in mind

maybe the disjoint union of all the open sets....

[unreal stratus]

that isn't a subset of X really

And would have even bigger cardinality

[heady skiff]

fug

so i can't take intersections and unions cuz that would just be another open set

hmm

[queen prism]

i don't know if that's what you intended but https://en.wikipedia.org/wiki/Disjoint_union is a technical term and it's not the same as the other kind of disjoint union that people often talk about

[gaunt linden]

Perhaps think for a moment about what "dense" means for the set you're trying to construct.

[heady skiff]

ye i've been considering that, how for any x in X every neighborhood of x intersects this subset, so I have to base it off x somehow I suppose

i guess i can narrow it down to the question: what does every neighborhood of x intersect?

altho that's kinda stupid

idk

i'll try thinking some mroe

yea nvm that question's dumb

maybe if we can take a representative from each open set and form the union that'll work, and it'll also use the countability of the base

nvm that wouldn't even be well-defined lmfao

[gaunt linden]

You can assume the axiom of choice, I'm sure.

(The claim you're proving isn't necessarily true without).

[unreal stratus]

"representative" is sort of the wrong word though

idk if that's what led you to talk about well-definedness

[heady skiff]

i'm not so sure what's the right wrod

[unreal stratus]

i would just say "pick an element from each open set" lol

[heady skiff]

lol i guess that works but like i was tryna get it so that every neighborhood of x would intersect this set, and so the set would have to be dependent on x and you can't do that for every x in X, idk

but okay i hope this construction works

[gaunt linden]

"dense" means every (nonempty) open set intersects our set; it's not about neighborhoods of any particular x.

[heady skiff]

ohhh, i probably confused the definition of dense then. my thought was that a subset S of a topological space X is considered dense if it's closure is equal to the whole space (that's the def), so to prove that S is dense you would take any x in X and consider two cases; either x is in S in which we're done, or x is a limit point of S. that is, every neighborhood of x intersects S at some point different than x

since the closure of S is the union of S w/ its limit points

[gaunt linden]

Well, but you get that if and only if your set intersects every nonempty open set.

[heady skiff]

hmm i'm not seeing that immediately but you're definitely right lol

let me think about that for a sec

[gaunt linden]

If there's an open set you don't intersect, then certainly everything in that open set will be outside the closure.

Conversely, once you intersect all the nonempty opens, then in particular you also intersect every neighborhood of every x.

[heady skiff]

oh ok i think that makes sense now, given that we're considering every single point x in X

at least i hope

[heady skiff]

ah ok there we go

yea that made me understand

okay thank you

what would the open sets the complex numbers be? just open balls right

[unreal stratus]

unions thereof

[heady skiff]

sorry wdym

oh

the open balls would be the base

and the unions would be the open sets is what ur saying?

[unreal stratus]

yes

just like with R basically

in fact like for any metric space X a basis is given by the open balls

[heady skiff]

oh right

yea that makes sense thanks

wait to show that a function is a homeomorphism you have to show that f and it's inverse are continuous and that it's a bijection right

nvm

i can literally look at the definition

[unreal stratus]

well that is slightly the wrong way round

well

okay maybe I am too pedantic lol

But in any case the way I'd describe it is that a map of spaces is a homeomorphism if it has an inverse

Where we only allow for continuous functions

[heady skiff]

ah is that bc like

in topology the only functions we consider are continuous

so it's kinda like a given

[eager vigil]

Anyone that could give a hint on where to begin to show S^inf is contractible?

[umbral panther]

There are many spaces called S^oo and many proofs of contraction

For example, there is the unit sphere in a Hilbert space, such as the space of functions on the interval. You can write down a deformation of the space of functions. Just slide the function to the right, replacing it with the constant function

S^n -> S^n+1 is easy to contract. So find a homotopy from the identity to the inclusion of a subspace
S^oo -> S^oo+1

[unreal stratus]

Another nice way is to apply Whitehead's theorem, which here says it's enough to show all homotopy groups of S^infty vanish

There is also a funny way using simplicial sets lol

[tawdry widget]

How? I want to learn this method by simplicial sets

[coarse night]

which is

[narrow cairn]

really having issues understanding what a "bump function" is

[coarse night]

Its a smooth function which is 1 in a ball and 0 outside a bigger ball

[narrow cairn]

ah i see

[feral copper]

But generically, loops in S^3 won't intersect. Surfaces in S^4 will. But yeah, I can believe this 🙂 however, there's no nice description of a basis for H_2, is it?

[unreal stratus]

It's described in a uh math overflow post

I'll try to find it again lol

But basically you take J, the category with 2 objects and each homset cardinality 1 (so two isomorphism objects) and show that the realisation of the nerve of J is S^infty

[tawdry widget]

Thank you.

I see. I need to show that disjoint union of (Δ^n times N(J)_n), under a equivalence relation, is S^infty… I will try

[tawdry widget]

ob(J)={-1, 1} I tried to define (( t_0,…, t_n), (X_0 -> X_1 ->… -> X_n )) to (… X_i sqrt(t_i) …)
Probably wrong, I can’t find inverse image of elements whose first few components are zero…

[feral sequoia]

I'm trying to prove that the interior of a convex set $A$ is convex. Is this a complete proof?

Let $x$ and $y$ be in $\text{int}(A)$, and let $z$ be in $[x, y]$.
For some $r_1$, $B(x, r_1)$ is in $A$, and for some $r_2$, $B(y, r_2)$ is in $A$.

We know that any point in $[x, y]$ is in $A$ by the convexity of $A$, so we consider $r = \min{d(z, x) - r_1, d(z, y) - r_2}$. Then the ball $B(z, r)$ is in $[x, y]$, thus in $A$.

[gentle osprey]

bordo99

[feral sequoia]

I just realized that B(z,r) is not in [x,y]

[tawdry widget]

I don’t think so. For example A= [0, 1 million] product [0,1] is a counterexample for this proof

[feral sequoia]

I was imagining things in R

How would I prove it then

[tawdry widget]

Well you can prove this:

U={tx+(1-t)y: 0<=t<=1, x,y from int(A)} is open

Since U is contained in A, prove this then U contained in int(A) contained in U, then done

(tx+(1-t)y. Any z near it , let h=z-(tx+(1-t)y), then z=(tx+(1-t)y)+h=t(x+h)+(1-t)(y+h))

[feral sequoia]

I see so all i have to do is prove U is open

[tawdry widget]

Yes, and the last message is a hint if you need it

[feral sequoia]

Alright thank you

[median sand]

Is the commutator of a topological group always a closed subgroup? I'm not very familiar with nets, but I feel like this should be true by using them.

[languid patrol]

Oh this is nice

[unreal stratus]

I learnt about it through here lol

#point-set-topology message

Really cool lol

[opaque scroll]

I mean you can endow any group with the trivial topology, so then the answer would be no.

My guess would be that the answer is still no if you require the group to be Hausdorff, but then an example might be harder to cook up

[median sand]

Hm, alright. For some reason I thought limits of nets would preserve commutators and their products.

[opaque scroll]

Appearantly the absolute Galois group of Q is an example of a (Hausdorff) topological group where the commutator subgroup is not closed
https://mathoverflow.net/a/363853/157483

[median sand]

@opaque scroll may I bend your ear on something #advanced-analysis message ?

[heady skiff]

can somebody give me a hint for the interior of c) ?

i don't know how to see this rigorously lol

[gritty widget]

what do you think the answer is?

[heady skiff]

i have no clue, i just put down R^2 - sin(1/x) for x positive

[gritty widget]

did you try out some random points to see if they were in the interior or not

like just test a couple of points and see what happens

try to come up with a reasonable guess

[heady skiff]

well pretending like the left hand side of the graph is nonexistent, we can find an open set for each of these points that's contained in R^2 - sin(1/x) for x > 0 which i'll just call A

like

intuitively very close to the origin it seems like you can't find an open set that contains a point there that's in A

i don't know how to show that rigorously tho because i don't know how to show "when" it becomes dense or whatever

anyways

this is what the solutions manual is for!

[heady skiff]

wait bruh

this is right according to the solutions manual

well then

[gritty widget]

what happens at the origin, or any point on {0} x [-1, 1] for that matter? can you find an open ball around such a point which doesn't intersect the graph?

i would encourage not looking at the solution manual until you have a decent idea of a solution

[heady skiff]

my intuition tells me no, but i don't know how to rigorously show that

i mean

the solution manual said my original solution was right

so is the solution manual wrong or.....

[gritty widget]

find out

draw an open ball around the origin. can it be disjoint from the graph? can you shrink it to make it disjoint from the graph?

[heady skiff]

no, i can't

[gritty widget]

so is the origin in the interior, or no?

[heady skiff]

no, but how do we know what points to "take" out of the interior? looking at a graph sure you have an idea but how can you be entirely precise?

ig that's my question

how do we know when, precisely, we can start having open balls about points contained in A?

at what point do the oscillations stop being "so frequent"?

idk how to make this rigorous lol

[tidal lynx]

Supplying a proof for that would suck lol

You would introduce like 5 magical constants minimum

[heady skiff]

yea, i mean if that's not the answer then what is if anyone knows

[tidal lynx]

What ttepa said probably

[heady skiff]

yea but what is it precisely

[tidal lynx]

like the region minus this set

sin(1/x) never touches that set (since it's not defined at 0)

but it approaches the y-axis, and in doing so oscillates an infinite number of times

if you think of sin(1/x) as the composition of sin(x) and 1/x, as x -> 0, we'll have 1/x -> inf, so sin will go through an infinite number of cycles

[heady skiff]

i see

how do we know that {0.000000000001} x [-1, 1] is in the interior tho ig

actually nvm i'll just go to office hours

thanks for the help tho

[tidal lynx]

there's gonna be one sine cycle directly to the left of it and one to the right

[heady skiff]

yea i'm just confused as to why

sorry lol

[tawdry widget]

Have to bother you again… how do we prove the realization is homeomorphism to S^infty…? ((t0,…,tn) in interior of Δ^n, I tried mapping ((t0,…,tn),(X0->X1->…->Xn)) to (…εi sqrt(ti)…) where εi=1 or -1, depending on Xi is which object, this didn’t work)

[unreal stratus]

Well if you think of the construction of the realisation like

At each step you are basically adding two (non-degenerate) cells which form two hemispheres of the next biggest sphere

So like, we start off with two points

Add two non-degenerate simplices joining them together, so we get a circle [plus degeneracies but they're identified with stuff in the realisation]

Then we fill in that circle in two ways to get a sphere

etc etc

[tawdry widget]

Ahhh, thank you, got it

[heady skiff]

why can we assume that (s - \epsilon, s + \epsilon) is a subset of O if s is less than b?

[unreal stratus]

Think about what O must look like

[unreal stratus]

yeah I guess it's like

If you don't worry about why we get out S^infty too much, this is a simple proof

But yeah I guess it does need a formalisation of what I just said to make more rigorous

I just like how the fact an equivalence of categories leads to a homotopy equivalence of realisations is used rather explicitly here

[tawdry widget]

Yes, so fascinating.
No one dislike category

[ebon galleon]

You should try drawing pictures when you don't understand something

Instead of coming straight to discord lol

Like this is genuinely just a picture away from understanding

[unreal stratus]

ye

Pictures underrated

[unreal stratus]

lol

[heady skiff]

Nvm idgi

[ebon galleon]

Lmao

[heady skiff]

Any math wizards care to explain

[ebon galleon]

.

[heady skiff]

Big bro there was a picture right there

[ebon galleon]

Lil bro you can just picture it like wrapping the interval around the circle

So the two endpoints pretty much meet at (1,0)

And so if you have some neighborhood of 0, it would be like an open ball intersected with the circle. But then it must contain a bit of both ends of [0,1) in it's image on the circle

[heady skiff]

Ah that’s facts

Thank you!

[ebon galleon]

[tribal palm]

is bourbaki gaslighting me with their def of homeomorphisms here?

[void tapir]

no

[tribal palm]

but they don’t require the bujection to be continuos!? is that perhaps equivalent to them preserving open sets?

i still haven’t even gotten to how limits and continuty is defined in topological spaces

maybe i should just shut up and move on

[hoary breach]

no don't

make more noise

[gritty widget]

Can I ask something here?

[void tapir]

keep reading on and it will make sense

[gritty widget]

I'm a high school student (not good at maths) , and I wanna ask how many holes a t-shirt has

[void tapir]

a t shirt has 3 holes

[tawdry widget]

4

[hoary breach]

If (X,T) and (X',T') are topological spaces and f: X to Y is a bijection then f is a homeomorphism if and only the following holds: a subset U of X is open in X if and only if f(U) is open in X'.

think this is bourbaki's definition

[gritty widget]

So is it 3 or 4?

And..why?

[hoary breach]

but yea the book shouldn't write a simple definition in a confusing way like that lol

~~maybe try dugundji ~~

[tawdry widget]

Probably better in #chill or some other channels…

[void tapir]

if you imagine a transformation that tears the circle apart into a line segment [0,1) is this continuous?
well no, because for example [0,0.5) is open in the line segment but its pre-image in the circle is not open
this direction of preserving openness restricts 'tearing' and the other direction restricts 'gluing' if that makes sense

[hidden crag]

Why would you read bourbaki

[gritty widget]

Ah ok, someone in help told me to ask here

[tawdry widget]

If you have a sphere, with 4 disks on its surface removed, you can put it on, you know, for your waist, head, two arms

[tribal palm]

i’ve been looking for it :S

can’t get a hold of it tho

[void tapir]

the sphere with 4 disks removed has 3 holes

[tawdry widget]

What’s the definition of hole…

[tribal palm]

whoa

[tawdry widget]

While it’s same as a disk with three small disks on it removed

[tribal palm]

i am fascinated by the histories of these topics, and there’s no question they influental

[void tapir]

does a sphere with its top and bottom removed have 1 or 2 holes

[tribal palm]

this sounds amazing, i’ll keep it in the back of my mind, thank you

[void tapir]

it has one hole because it's a cylinder and then the circle

[tawdry widget]

I know it’s upper half of a genus-3 torus

[tribal palm]

and honestly their exposition and order in which they defined stuff and explained their relations is the clearest one i’ve seen yet

[tawdry widget]

Just not sure how you define holes, number of boundary components or what

[hoary breach]

#real-complex-analysis message watch

[void tapir]

im surprised they defined homeomorphism before continuous map

[tribal palm]

well, “things” being topological structures, neighbourhoods and bases

[feral copper]

The answer is: homology

[tawdry widget]

I know its first homology group is isomorphic to Z^3

So number of holes is defined to be rank of H_1 or something ?

[hidden crag]

I don't think it's a well defined term

[tawdry widget]

Okay

Yeah me neither…I only know genus, number of holes seems too daily

[feral copper]

For orientable surfaces it's half of the rank of H_1. For 3 manifolds, it's ambiguous. Except you could also relate it to the rank of H_1, because for closed oriented 3-manifolds, that's equal to the rank of H_2

Again, genus is only for surfaces

You can look at the special case of 1-handlebodies; then genus is the number of solid tori you glue together. But that's far from being a definitive answer for everything. Sadly, there's no "genus" for anything else but surfaces. Although homology is still interesting, and in fact even more than for surfaces!

[tawdry widget]

Half of rank? What if rank is odd. Like a genus-g surface having n boundary components, its H_1 is Z^(2g+n-1)…

[feral copper]

For orientable closed surfaces, it's always even!

For boundary surfaces, cap off with disks

Like the genus of a punctured torus is still one

But you have to glue disks

(Unless you know n)

[tawdry widget]

Okay, I see, then always n=0, yeah even, Z^(2g)

[feral copper]

As you said, 'a genus g surface with [...]'

[tawdry widget]

I thought you said when you have n-boundary components you cap off n many disks so it becomes closed

[feral copper]

You implicitly meant that you count the genus after capping the boundary components

That's what I said indeed

[tawdry widget]

Okay

[tribal palm]

fancy T being closed under arbitrary unions include uncountable unions right?

[gritty widget]

yes

[tribal palm]

yikes

thank

[ebon galleon]

Wdym yikes

This is very nice

If you only allowed countable unions, you wouldn't be able to show even simple statements like "a set that's a neighborhood of each of its points is open"

[tribal palm]

yeah exactly it’s wild how powerful it is

[unreal stratus]

i think you've probably just been corrupted by measure theory

[tribal palm]

i haven’t even touched it yet

…though i had a phase of constructive analysis

[unreal stratus]

Oh okay lol

Well it's just cause for sigma algebras ( = basically your set of measurable sets) you only require countable unions

[tawdry widget]

Can you guys recommend a reference of proof of cellular approximation theorem? Like some books , other than Hatcher…

[unreal stratus]

i think tom dieck and spanier have nice proofs

https://www.maths.ed.ac.uk/~v1ranick/papers/diecktop.pdf

tbh find it a shame there isn't a more elegant proof somehow lol

[tawdry widget]

Thank you☺️

[gentle osprey]

CoffeeMan

[formal tide]

Thoughts on using neighborhood systems? Coming from Munkres I just learned about them. Are they still used?

[coarse night]

not really but it's a good perspective to have

also pointset topology is out dated now

[void tapir]

pointset topology is useful for analysis

[feral copper]

'outdated'
It's not like you can do any kind of topology without this basic knowledge anyways, can you?

[coarse night]

point is no one does point set topology so don't need to go very deep into it. Get the necessary things done and move on to something else

[quiet thorn]

*start doing something else then pick out whatever pointset you need along the way

[feral copper]

Sure, but also don't go to much to the other extreme of never doing any point-set and lose time interrupting anything you try to do elsewhere

[ebon galleon]

Certainly, defining things in terms of neighborhoods is outdated for 99% of topological purposes

Unless I suppose you're a psychopath that works with pretopologies

[broken nacelle]

@coarse night your name...

amen ✊

[gentle osprey]

CoffeeMan

[trail charm]

i also was very confused on that example

my go-to place for questions is mathse

because usually someone else has asked it and usually there is a really good answer

https://math.stackexchange.com/questions/1234070/mathbbr3-setminus-a-deformation-retracts-onto-s1-vee-s2

i really liked this one

idk which coffeeman to @ because there r several

[ebon galleon]

when there is no answer on MSE or MO

[trail charm]

the day that happens is the day i go and become a business major or smth

the day i get to a level of math where there is not a mathse post which is a duplicate of another is the day i quit math

[ebon galleon]

you gotta be strong when that day comes

cry a little maybe because you realize you have no way currently to figure out the answer anyways

but stay strong 💪

[trail charm]

yeah mhm yep yeah

[eager vigil]

Yeah, I also found that one, but I'd like to be able to do these sorts of reasoning myself if possible, so I was wondering if anyone had any references for texts with problems that cover these sorts of things

Or is it just something one must pick-up somehow?...

[heady skiff]

what is the best way to show that f is continuous? inverse of each open set in X is an open set in R?

or should i use the epsilon-delta analytical way

nvm

[ebon galleon]

well given that only one of them is a metric space, it'd be best to try the topological definition

[heady skiff]

fax

would an example of an open set under the finite complement topology be $\mathbb{R} - {0}$?

[gentle osprey]

okeyokay

[ebon galleon]

In fact, this is a special case of that even When you take the identity function id(x) = x as a function
id: (X, T) --> (X, S), this is continuous if and only if S is coarser than T. That is, if every set in S is open in T

Equivalently (for this, it might be best to think of closed sets), every set that is closed in (X, S) is also closed in (X, T)

[heady skiff]

i see

[heady skiff]

is this equivalent to showing that O is an interval essentially? since f is the identity function

because in order to show that $f$ is continuous, it suffices to show that if $O$ is an open interval in $X$, then $f^{-1}(O) = {x \in \mathbb{R} \mid f(x) \in O} = {x \in \mathbb{R} \mid x \in O}$ is an interval, which is the same as showing that $O$ is an interval in $\mathbb{R}$

[gentle osprey]

okeyokay

[ebon galleon]

no?

X is the set R with the finite-complement (/cofinite) topology. The open sets are those whose complements are finite. What does that make the closed sets under this topology?

[heady skiff]

wait can you explain why my reasoning is wrong

[heady skiff]

ah yea that's smart

[ebon galleon]

1. X as a set is just R, so the "intervals" would be the same. but
2. they aren't relevent at all to the topology on X

[heady skiff]

oh nah i meant if f is continuous and if O is open in X then f^{-1}(O) is open in R right, which is equivalent to saying that f^{-1{(O) is an open interval in R no?

cuz the open sets in R are the open intervals

[ebon galleon]

no

The open sets in R are countable unions of intervals

[heady skiff]

oh shit rlly

[ebon galleon]

For instance, R \ Z is open in the standard topology, since it's a union of intervals (n, n+1) for n in Z

[heady skiff]

ah facts

[heady skiff]

the finite sets right

[ebon galleon]

correct. A subset if closed in X iff it is finite. So the statement that this f is continuous is just that finite sets are also closed in the standard topology on R

[heady skiff]

oh yea and then the last part is easy to show i think

ok thanks! ur the goat

wait no i'm stupid why is that closed in R, so we would be considering this set C under the standard topology of R which we know must be finite, how do we know that it's complement is open? bc for instance what if C is not a closed interval, in other words it skips some real number if that makes any sense

o

nvm

[heady skiff]

here to show that f is not continuous, i could just take an open interval (a, b) in R and then the image is obv (a, b) but the complement of that is infinite

[ebon galleon]

I assume you mean for f not a homeomorphism

but yes that should work for the reason you said, provided (a, b) is not all of R

[narrow cairn]

given a topological space X, define a topology on P(X) by:
U is open if all x ∈ U are open in X or if U = P(X)

what properties does this inherit from X? (this is a topology right)

i just made this problem up to be clear no idea if this is a thing

if X is countable, is P(X) second countable?

if X is second countable, is P(X) second countable? if X is hausdorff, is P(X)? connected? path connected? im not asking for like hints im legit just curious

[umbral panther]

This is very degenerate. It is a bunch of open singletons and a bunch of points whose only neighborhood is the whole space. It hardly remembers anything from original space

[narrow cairn]

well the first 2 i asked arent true if you take X to be Q with the discrete topo P(X) dies

[narrow cairn]

thats true

so it definitely isnt hausdorff unless X is discrete

is there any reasonable way to define a topology on P(X)?

[umbral panther]

This was recently asked on MO. I think the answer was no. But you can look at the compact open topology of the subset of P(X) consisting of continuous functions to various two point spaces

[ebon galleon]

I believe (at least in certain circumstances) one topology you can take is called the "Scott topology"

It is actually quite useful in characterizing the exponentiable topological spaces (in Top)

https://www.cs.bham.ac.uk//~mhe/papers/newyork.pdf

I suppose this is a bit different in that it's a topology on the original topology itself

rather than one defined on all of P(X)

But still a related idea

[trail charm]

reading about homology and i dont quite get the intuition of 0-chains

eg if we have a graph, its vertices generate a free ab grp

but what does it mean to add/subtract vertices?

like 1-chains make sense — youre moving along the edge

[unreal stratus]

Well I mean I'd just say it doesnt mean too much, but the image of the preceding map does have good meaning

Since it corresponds to endpoints of paths

So when you quotient out we are just imposing the relations given by paths connecting points

[trail charm]

oh okay

that kinda makes sense

[tidal cedar]

the way I often think of it is using H_1 as the motivation

extending that to other dimensions gives you all the higher homology groups

as well as the connected components, because of the 0-simplices

[trail charm]

mmm okay

in the case of S^1 as a Δ-complex formed by a point and an edge, is the simplical H_0(S^1) ≈ Z because u can think of each point as a point on the spiral? the “spiral”, ie, the covering space of S^1

i drew a picture somewhere

[unreal stratus]

The covering space is n really relevant

[unreal stratus]

From my description hopefully it is clear H0 corresponds to path components

What you say is more relevant to H1 ig

Though using the covering space for that is sorta overkill

[trail charm]

yeah ig im just trying to draw the connection between π1 and H1

since the latter is the abelianization of the former?

[unreal stratus]

Yes

[trail charm]

or did i remember smth wrong

yea ok

[unreal stratus]

Was saying H0 here a mistake

[trail charm]

hmm ok

i’m not sure i understand why H0 ≈ Z then :(

i get that ker of boundary hom d_0 is every v, since v-v=0

heres context for what im trying yo understand

oh wait

maybe because elements of H_0 are cosets of im \partial_1, but im \partial_1 is cyclic <v>, so H_0 ≈ Z?

[tidal cedar]

so for H_0

it's $\frac{Ker \partial_0}{Im \partial_1}$

[gentle osprey]

the pavement won't answer me