point-set-topology

[tight agate]

the lemma tells you that the composition H^i(G, Z) ---> H^i(0, Z) ----> H^i(G, Z) is given by multiplication by |G|

but H^i(0, Z) is 0

so mult by |G| is the zero map

so all elements of H^i(G, Z) are torsion

[gritty widget]

and here

the first map is the map induced by the inclusion, and the second is the map induced by the inclusion

[tight agate]

the first map is restriction

[gritty widget]

sorry yes

[tight agate]

the second map is corestriction

it's in weibel

I think

[gritty widget]

okay thanks very much

i think i have enough to suss this out

[tight agate]

MSE moment

[pearl holly]

Okay so I think I got it but it seems kind of sketchy so I would appreciate if someone checked it. So let me change some variables because its unreadable now. So we have $A \to^\alpha B \to^\beta C \to 0$ and applying $\text{Hom}(-, G)$ gives $A^* \leftarrow B^* \leftarrow C \leftarrow 0$. Assume that $\varphi$ is in $\ker(\alpha^*)$, i.e $\varphi |{\ker(\beta)} = 0$ by exactness. Now we want to construct a $\psi: C \to G$ such that $\psi \beta(b) = \varphi(b)$ for all $b \in B$. We know that $C$ is iso to $B/\ker(\beta)$. So define $\psi(b + \ker(\beta)) = \varphi(b)$. This is well defined because $\varphi |{\ker(\beta)} = 0$. Now under the iso $C \cong B/\ker(\beta)$, the element $\beta(b)$ will correspond to $b + \ker(\beta)$. So now $\psi(\beta(b)) = \psi(b + \ker(\beta)) = \varphi(b)$ and we are done

[gentle osprey]

Tokidoki ✓

[gentle osprey]

Yeetus

[reef shore]

show both inclusions one by one

[empty grove]

Which inclusion are you having trouble with?

You'd take an element in one set and show that it's there in the other one

[pearl holly]

is this even supposed to belong here lmao?

[gritty widget]

Let T be S^1 \times ... \times S^1 (l times)

Connected subgroups of this are just S^1 \times ... \times S^1 (k times) with k<l

?

any disconnected subgroup must then just be S^1 \times ... \times S^1 (k times) and some finite group

but im not sure how to see where the finite sub groups come from

[long hornet]

@gritty widget The subgroups of T are in correspondence with those of R^l that contain Z^l. Maybe that helps.

[empty grove]

The diagonal should also be a connected subgroup right?

[reef shore]

@pearl holly looks correct for exactness at B*. Have you checked it at C*? Also all the work you did was just reproving the first isomorphism theorem/universal property of quotients

[pearl holly]

yo wtf I didn't even know that lmao

[reef shore]

can't tell if sarcastic or not

[pearl holly]

yee I have, the only thing I was stuck at was proving exactness at B* and I'm pretty sure my arguments hold for everything else. Thank you so much for verifying, I thought that my post was unreadible lmao

[reef shore]

It was unreadable because you were actually working with cosets 😵‍💫 disgusting

I didn't read the cosets part carefully but I am trusting you know how to compose functions

[pearl holly]

wait how did you do it then?

oh you edited lmao

[reef shore]

yeah all you've done is found a factoring of a map B → G through a quotient of B, and the first isomorphism theorem tells you exactly when such a factoring exists

[pearl holly]

ah yeah right

so in this picture I have constructed the \tilde{\varphi} right?

[reef shore]

yep

beta is exactly the quotient map

[pearl holly]

oh man lmao

I could've just done this

[reef shore]

in the sense that it satisfies the universal property of the quotient of B by ker beta

even though there are no cosets in C

but having cosets is not and should not be the definition of the quotient obviously

So in your solution you're also having to replace C with the quotient and beta with the quotient map because you're working explicitly with quotients

[reef shore]

we know this because first isomorphism theorem again

[pearl holly]

yeee I see. I should definitely use the diagram next time lmao

Okay I have to go lmao, thank you so much!

[gritty widget]

This does help thanks

@tight agate do you mind if I dm you?

[tight agate]

[charred root]

Dumb question: is there a way to write exterior derivative in terms of covariant derivative?

I know and understand the formulas for div and curl in terms of exterior derivative. I’ve also seen them written in terms of christoffel symbols, but I struggle with comprehending how to derive them this way.

[gritty widget]

you can (and it's not a dumb question, perfectly reasonable thing to wonder)

for one forms for example you have $$d\omega(X, Y) = (\nabla_X \omega)(Y) - (\nabla_Y\omega)(X)$$ (so long as $\nabla$ is torsion free)

[gentle osprey]

TTerra

[gritty widget]

and this generalizes to k-forms

[charred root]

@gritty widget Thanks a bunch! But… How?

How to derive the formula you gave me?

Just blow it up and check element by element?

That feels like an unenlightening thing to do 😢

[gritty widget]

yeah i guess that's one way to do it

i'm supposed to be watching an analysis lecture so i can't think of any more enlightening way to prove it

$$(\nabla_X\omega)(Y) = X(\omega(Y)) - \omega(\nabla_XY)$$ (like the product rule) now just expand the right hand side and move things around

[gentle osprey]

TTerra

[gritty widget]

it's also probably the quickest way to do it

[charred root]

🤔

Yeah, that is much better

Thanks!

[gritty widget]

no problem

it's an exercise somewhere in lee's book on RG if you want to find the generalization to k forms

i couldn't reproduce it off the top of my head

you could probably derive it by taking the "intrinsic formula" for dω and rewriting the lie brackets in terms of the covariant derivative

it probably simplifies to something nice

[fickle umbra]

Hey

How exactly does the notion of a topology formalise the notion of continuity?

[shy moss]

continuity of functions?

[fickle umbra]

I guess

[shy moss]

the topology of a space is for know what function are continuos

the topology gives you the open sets and a functions is continuos if the preimage of every open is also open

[charred root]

If I’m not mistaken, a mapping f of one topological space (X) to another (Y) is continuous iff preimage of every open set in Y is open in X

Ah

[fickle umbra]

Still I don’t see the intuition behind it

[shy moss]

do you know the definiton of continuity of functions in terms of epsilon-delta?

[marsh forge]

I think that continuity is not what topology formalizes

[lunar yoke]

maybe try verifying that this definition of continuity coincides with your concept of continuity of functions R -> R

[marsh forge]

Most things that already have an inherent notion of contintuity (e.g. metric spaces) don't need general topological definitions

Topology generalizes the idea of continuity and formalizes various ideas of connectivity

i.e. closeness and compactness and connectedness etc

as well as formalizing the study of "local" properties

[fickle umbra]

Yes

[shy moss]

you should check that definition in terms of epsilon-delta is the same as definition in terms of open sets for functions in euclidean space

[fickle umbra]

open sets are generalisations of open intervals right?

[marsh forge]

I would say no

open sets are generalizations of unions of open intervals

bases are generalizations of open intervals

but also a lot of more basic topological spaces were studied before the defn of a topolog was written down

[shy moss]

in R^n an set X is open if in every point x in X there is a open ball with center in x which is also in X

with usual topology

[shy halo]

Sus

[shy moss]

here, what exactly means to position the knot to lie almost flat on a table?

[marsh forge]

Basically like

Try to flatten it without making it self intersect

It’s just a visualization thing

The important part is the next sentence

[shy moss]

so is like descompose a knot in arcs $\alpha_i$ and $\beta_i$ and then put all the $\alpha_i$ and $\beta_i$ in the table?

[gentle osprey]

Or x1

[cedar pebble]

sure

I mean think about how you typically draw knots on paper

it's a planar diagram, with the strands either overlapping or underlapping

since you can draw this in the plane with overlaps and underlaps, it means you can fit the picture into a disc x [0,epsilon]

[marsh forge]

The idea behind this exercise it to think of a knot as a bunch of arcs that meet at “crossings” and to use such a presentation to compute the fg group

[shy moss]

thanks

[long hornet]

If M, N are smooth manifolds, is there a smooth manifold which is homotopy equivalent to M v N (v for wedge sum)?

[marsh forge]

Not in general

This also doesn’t work if you drop smooth

[long hornet]

Too bad, why?

[marsh forge]

The easiest way to prove there is a counter example is poincare duality

I do not know of a good intuition other than, in general, there is no reason to assume one would exist

Being a manifold up to homotopy is very strong

[long hornet]

Aha

[marsh forge]

A good counterexample is to compute the homology of S^2 v S^2 over Z/2

[long hornet]

I will do that as soon as I can 😆

[fair idol]

Can someone explain to me what extra info is required to demonstrate that a manifold with boundary is smooth rather than just a manifold without boundary ?

[long hornet]

It makes sense though, they are of dimension 2 and intersect in a point, and it'd seem hard to bridge the gap.

[next spade]

I think im doing something wrong here

I want to show that Df maps TM to TN is smooth

But then i got stuck at the transition data of Df

f maps M to N is a smooth map then Df maps TM to TN is smooth

[fair idol]

Hmmm that sounds like a correct statement. I hope you get it lol

[abstract pagoda]

o

[abstract pagoda]

So prove that f smooth function between manifolds implies the differential of f is smooth between tangent spaces

[abstract pagoda]

What is Phi tilde?

[abstract pagoda]

Also the condition is that phi o f o psi^-1 is smooth for f being smooth

where psi is a parameterization of M and phi of N

ill do this exercise ok my own soon

[next spade]

Its a chart on TM that my professor had introduced

To make( TUα , phiα tilda ) chart on TM

[next spade]

I meant its a coordinate map

For TUα where (Uα,siα) is a Chart on M

[next spade]

Yes it (f) is between two smooth manifolds M and N

Similarly ,TM and TN are smooth

[abstract pagoda]

yes so you have a chart from one manifold to the other using that phi o f o psi-1 is smooth

[next spade]

And DF is a map btween them

[abstract pagoda]

ya

well

look at this right here

[next spade]

Ohh so this is what will come for the traansition data

[abstract pagoda]

yea

[next spade]

yeah makes sense

[abstract pagoda]

and look at this right here

the map taking M to TM is smooth

uh idk how thats useful nvm

but wait

i sorta get it

you want a smooth function between equivalence classes of curves

[next spade]

Hmm i remeber the natural projection map π: TM --M
Via (p,v)--p

[abstract pagoda]

and this can be given by the fact that phi o f o psi-1 is smooth

[next spade]

Yeah right

[next spade]

Yeah yeah i sorta get it

[abstract pagoda]

tangent spaces kinda complicated for me to think about

lmk the progress on problem

[next spade]

I think this would be the transition data

And this is smooth because

[next spade]

This is smooth

[abstract pagoda]

what is O_p

or D_p

Also the domain of $\tilde{\Phi}_\alpha$ is from U in M to Rn

[gentle osprey]

I Killed the Most

[next spade]

Its Dp

It is Df restricted to p

As we know that diferential of a manifold near point p is basically a linearization

[abstract pagoda]

oh

uh

like the point?

on the manifold?

[next spade]

Yeah p is any point in manifold M

Yeah

Like we know the tangent space TpM is a vector space , so

Dfp :TpM -- Tf(p)N is a linear and contnous map between vector spaces

[next spade]

Yeah right it is , but dunno why my profesor had taken it like that
I mean yes when we are defining homeomorphisms then it would be UxR^n

[uncut surge]

an emoticon from 2008

sorry that's the truth

q_q

[gritty widget]

[trail tiger]

any cute grad-level geometry book good for bedtime reading instead of intensive work

any topic

[plain raven]

this is not what you asked but i remember Pour-El and Richard's book on computability in analysis and physics being light reading

I only bring it up because you seem like you would enjoy a book on the practical side of analysis/more theoretical aspects of numerical computation

[trail tiger]

[abstract pagoda]

So I understand why CPn and RPn are compact. Because their construction using quotient maps has that the image of compact sets is compact since quotient is continuous

How do I show CP1 diffeomorphic to S2 though?

I am not sure I know how its true that RP1 diffeomorphic to S1

but from what Ive heard there is a homeomorphism from hemisphere to S1 but that isnt a diffeomorphism

[marsh forge]

The homeomorphisms are not hard to see when you think about the construction

I assume it’s not so hard to show they have to be smooth

Think about S^1 as the unit vectors of R2

And recall the construction of Rp2 using equivalence classes of vectors

[abstract pagoda]

So what would it mean to smoothly map a unit vector to an equivalence class of vectors?

Hmm

we only quickly went over why RPn is a manifold

So pretty much I need to construct a map from f:S1 to RP1 such that phi o f o psi^-1 is smooth

where psi is chart for S1

and phi for RP1

well S1 can be charted by e^ix

f takes e^ix as input and turns that into an equivalence class??

so like f(x)=[x]? lol idk

that isnt injective but meh?

does it need to be?

ok it definitely does

oh wait

it can be injective

if i look at hemisphere of S1

idk what differentiable is supposed to mean here though since i dont think you can differentiate a line

[bitter yoke]

For this to be smooth, you're asking that phi o f o psi^-1 is smooth?

[abstract pagoda]

Right?

Isnt that definition of f:M->N smooth?

[bitter yoke]

Yes thats right, so I'm wondering why you're asking how to differentiate a line

[abstract pagoda]

uh dont I have to define f so that the composition of maps is differentiable, You cant differentiate an equivalence class and if I define f:S1->RP1 by f(x)=[x] i run into a smol problem

[gritty widget]

This is done well in lee's smooth manifolds

[abstract pagoda]

what is done?

[gritty widget]

RP^n is a manifold

[bitter yoke]

yes, but you compose f with things?

[abstract pagoda]

well yea but not my question , it might be helpful though

[bitter yoke]

in particular phi is going to send these equivalence classes of lines to R

[abstract pagoda]

yes

[bitter yoke]

and you can differentiate a map from R to R

[gritty widget]

I think if you look at the charts for RP^n this will help alot

[abstract pagoda]

But cant you apply chain rule to the composition and then you get one factor which ends up being an equivalence class? So I am wondering if defining f(x)=[x] the right thing to do? Im not sure what else honestly

[bitter yoke]

I mean, if you want to apply the chain rule, then you would get the differential of f is a map between the tangent spaces of S1 and RP1. You're still not "differentiating a line"

[gritty widget]

i think you are getting confused here

you are thinking about [x] as being an equivalence class

(which it is )

[empty grove]

"you can't differentiate an equivalence class" part is not really correct, the set of equivalence classes has a smooth structure on it so you can differentiate wrt that

[gritty widget]

but just think of it as in element in RPn

[empty grove]

What the elements are called is irrelevant

[gritty widget]

and use then read the part bout the charts on RPn

[abstract pagoda]

ok so smooth structures are the transistion maps right?

[gritty widget]

the charts for RPn will clear up how the equivalence class business works

[abstract pagoda]

we didnt go over that section in class yet

[gritty widget]

becuase even though [x] doesnt have a unique representative

in a given chart it does

[bitter yoke]

Smooth structures are the charts. You get the transition maps by composing one chart with the inverse of another chart

[abstract pagoda]

ok so i hope im not expected to write explicitly how f is continuosuly differentiable, but I can show its isomorphic. Im only a little intimidated by the technicalities in showing its differentiable, but maybe I can just say it is because phi is differentiable?

to me this is terrible reasoning for why it would be differentiable

maybe I can just write down the composition of charts with f like the phi o f o psi-1 one

[gritty widget]

a map of manifolds is smooth if it is smooth in the charts

if you look at the charts for RPn

you will see why this is obviously differentiable

there is essentially no point in trying to show that such a map is smooth without know what the charts for RPn look like

[abstract pagoda]

the charts cant be that bad right?

I have a few ideas myself

[gritty widget]

no they are very nice

[abstract pagoda]

so can they be given by just linear equations lol

like

y=mx+b

but ig id try to write that in different terms

actually im not sure what the chart would be, my first instinct would be to write it as slopes

Atleast in R2 because each line has a unique slope , but then i get trouble for the line x=0

but then ig you can write that in terms of x=my+b

wait no b lol

just solve for m

[bitter yoke]

How did you even define RP1 without talking about the charts

[abstract pagoda]

so x/y and y/x

[bitter yoke]

Like, how did you define the smooth structure on RP1?

[abstract pagoda]

prof talked about it handwavy just saying its set of all lines through origin

we didnt define smooth structure

[bitter yoke]

Uh

[abstract pagoda]

we didnt know what that was until two days ago

hw was assigned last week

but ig it isnt too complicated to read about

[bitter yoke]

You can't ask if two things are diffeomorphic unless they both have smooth structures on them

[obtuse meteor]

Your prof wack

[abstract pagoda]

yea thats why i struggle with this section lol

this problem set*

[bitter yoke]

Then shouldn't you start this question by figuring out what the smooth structure on RP1 is?

[abstract pagoda]

Ok so my question of how do I know f is smooth should have been what is smooth structure on RP1 so that I could know if phi o f o psi-1 is smooth

Well I think I mostly have the idea

[gritty widget]

o f o

[abstract pagoda]

Its getting a little bit late for me sadly, hopefully I can write up an explicit map tomorrow

I have to do same thing for CP1

but idk what a complex line is so im just assuming its whatever the charts are for RP1 but instead complex numbers

and they somehow come out as real numbers?

vector definition easiest for me to follow but idk any atlas for CPn or the proof that it is a manifold

[gritty widget]

[gritty widget]

it's a complex manifold and therefore a real smooth manifold, qed

are there nice conditions when a homotopy equivalence on two G-spaces is a G-homotopy equivalence

[long hornet]

According to @marsh forge, S^2 v S^2 is not homotopy equivalent to a manifold. Now if K is a simplicial complex such that the intersection of two faces "lowers the dimension" by 1, is |K| a manifold up to homotopy, or is there no reason to expect that?

[abstract pagoda]

well yea

the base point identification makes it not locally euclidean

[long hornet]

I am thinking up to homotopy

[abstract pagoda]

what classes you taking?

[long hornet]

None. I am vaguely familiar with what homology is about.

[gritty widget]

simplicial complexes are very in general not homotopy equivalent to manifolds

i don’t know what the inrswction of faces part is supposed to give

[long hornet]

My guess was that this happens because intersections don't behave well

[long hornet]

Well, apparently the counterexample doesn't need to be too high-dimensional, and I just couldn't come up with one

I didn't think really hard, though

[gritty widget]

Let $X$ be a space with a $G$ action. Suppose $i:Y \hookrightarrow X$ is equivariant with $Y$ and $X$ homotopy equivalent. Are $Y/G$ and $X/G$ homotopy equivalent

[gentle osprey]

lime_soup

[gritty widget]

damn they are not

[uncut surge]

saaanity check, if a space U has a CW-complex structure and is homeomorphic to V, then V also has a CW complex structure right

[tight agate]

what is your definition of a CW-complex structure

this should be a tautology if youre using the definition that I know

[uncut surge]

actually yeah it only really makes sense up to homeomorphism, doesn't it

[tight agate]

i.e. iso to a CW complex

[uncut surge]

yeah okay that's good lol

sometimes you do math in silence for so long, you just need a friendly soul to tell you "it's all okay"

thank you for fulfilling that role

[tight agate]

:petthecat:

[dim radish]

how does this work

if i choose X to be the torus, how is every map from I -> X homotopic

[true robin]

They mean non base point preserving homotopic

Sorry, end point preserving I mean

[dim radish]

oh yeah

thanks a lot!

[gritty widget]

Does anyone have now about excision in equivariant homology?

I'm not sure what the precise statement should be

do we need some notion of a G-subspace

[meager python]

The Hom(X, Spec R) -> Hom(R, O_X(X)) equivalence, what's your favorite way of "remembering" why it's in this "direction" of the terms? I think of it in the way of Hom(X, A^1) = O_X(X)

Also I guess it's about R-schemes so the direction is natural in that way also, but is there a more fundamental reason?

[tawdry widget]

？

It’s a bijection

Two directions not one

I like the way that it’s directly proved using a commutative diagram in algebraic geometry and arithmetic curves written by Q.Liu

[jolly creek]

Hey guys, really basic question, but I'm being asked to prove that Playfair's Axiom is a theorem of Euclidean Geometry.

I'm just wondering what criteria there is for this to be a theorem.

I proved a bunch of other things beforehand to get to this point, so I'm just trying to drive it home.

[gritty widget]

What's the prerequisite for topology?

barely anything, just some basic mathematical maturity

[unreal stratus]

Technically you could get by with just set theory (etc) initially, although some familiarity with analysis / metric spaces can be helpful ig?

[gritty widget]

5 seconds before this turns into a "you dont need analysis for topology" argument

[unreal stratus]

I didn't want that to occur dammit

[gritty widget]

No, I don't have familiarity with analysis and metric spaces.
I have Tao analysis with me. Complete it first?

[hollow harbor]

Yeah i mean if you don't know some basic stuff about open and closed sets in R then you're gonna have a rough time, but it's still totally possible

Last part added to shut tterra up

Not because i actually believe it

[gritty widget]

[unreal stratus]

I just think analysis serves one better initially than topology lol

[hollow harbor]

switch to pugh

No yeah looking at Tao 1 is fine im sure

Idk how thoroughly you need to know it

Just being comfy with set theory and reading a bit about open / closed / bounded sets in R, sequences, and maybe compact sets should be totally fine for you to start learning topology and know why they're doing all stuff that looks like nonsense at first

[gritty widget]

Can you suggest something simpler than Tao for analysis.
Quite difficult for me as I have no touch with advanced math for past couple of years.

[unreal stratus]

(for the record, i started topology before i knew analysis lol which had ups and downs)

[hollow harbor]

Well, idk about simpler or less simple, but the book "understanding analysis" by Abbott and the intro analysis book by Ross have different flavors

So you might find one of those good

[gritty widget]

Hmm, all these things are covered under regular set theory?
I mean I never heard about sequence in set theory.

[unreal stratus]

that was more an overview of intro set theory and intro analysis, can be helpful to look at those

[sweet wing]

yesh^ helps a lot to be somewhat comfortable with metrics and "topology of R" for topology

[hollow harbor]

No, those are basic analysis definitions

The point is, even though you CAN learn topology with just basic set theory, it could be painful to understand what's going on if you haven't seen the basic analysis definitions

[pearl holly]

This is what happens if you do topology before analysis lmfao

[unreal stratus]

lol

it's made me err towards avoiding sequences in analysis actually

[tough imp]

Oh sad

I thought this was gonna be when Nikita started talking about glorified semi-lattices lol

[unreal stratus]

Lol

[tough imp]

Also re: more natural to consider f instead of f^-1 tbh I don’t have a great reason to explain why intuitively f^-1 is more… topological

BUT I can say f^-1 is so much nicer than f in so many ways

[pearl holly]

it was actually lmao, scroll down a bit

[tough imp]

And eventually sometimes the thing that matters less is open sets and more about open covers and stuff, and what you really care about with continuous functions is that they pull back open covers

But this is like getting into Grothendieck topology shit so it’s not too relevant for a while lol

[gritty widget]

sequences are good for you

[pearl holly]

ah yeah okay I see

maybe it's also because intersections behave nicer with preimages?

I remember tterra talking about this

[tough imp]

Yes this is what I mean by f^-1 being good

[gritty widget]

what did i say

[pearl holly]

oh lmao

[tough imp]

Altho actually

Intersections commute with image of sets too

[gritty widget]

oh i remember saying a word or two on that

[tough imp]

Unions do not

[pearl holly]

a long time ago you explained why quotient maps are defined with preimages

[gritty widget]

why are they defined with preimages?

[pearl holly]

because f^-1(U \cap V) = f^-1(u) \cap f^-1(v) and there's no contradiction. THis is what you said

roughly what you said

[tough imp]

Uh

Should be \cap again

Not a u

[pearl holly]

oh crap lmao

[tough imp]

You mean oh \cap

LOO

I mean on the right side

You have f^-1(u) u f^-1(v)

[pearl holly]

oh lmao I didn't even notice

[gritty widget]

\intersect

[tough imp]

Me when I first had to latex the intersection of two sets

[pearl holly]

does it actually work tho?

[tough imp]

I don’t think it’s a command

[gritty widget]

$\intersect$

[tough imp]

$\intersect$

F you

[gritty widget]

lol

[pearl holly]

lmao

[tough imp]

Grrr

This is so sad

[gritty widget]

S

[gritty widget]

#point-set-topology

[tough imp]

This is where memes go right?

[gritty widget]

yes

[pearl holly]

yeeess

[gritty widget]

this is where i procrastinate on reviewing ant or doing my homework

[pearl holly]

#topology-and-memes

[tough imp]

#point-set-topology message

[gritty widget]

legendary message

[pearl holly]

lmfao

[tough imp]

If u bring topology to other channels that channel morphs into #point-set-topology for a while

This is why we try to keep each channel to only that content

[gritty widget]

lets do some geometry

[tough imp]

Nah

:swag:

[gritty widget]

yah

yUH

i regret not taking a geometry course this semester

[gritty widget]

Isn't it that Monkey can rewrite the entire work Shakespeare, given enough time.

S

[tough imp]

No monkey only type S and then break keyboard

[gritty widget]

So, this is against infinite monkey theorem logic.

[tough imp]

theorem

[sleek thicket]

S

.pin

Oh I already pinned this lol

[hidden crag]

Lmao

[marsh forge]

Rules don’t apply to regulars smh

This channel is a special zone anyway

Free of moderation

Namington and mniip have no power here

[fading vale]

But i do

[shy moss]

what are étale extensions?

[fading vale]

Like of a field?

That probably goes in #groups-rings-fields lol but theyre finite products of separable extensions

equivalently you can tensor them with a field extension E to get a free E-algebra

[stuck sleet]

How do you show that a basis is a base for a specific topology? I can do it for some topology, but I wanna show that the balls at rational points with rational distance form a basis for the Euclidean topology

[marsh forge]

It suffices to show that if you take a point and consider any open neighborhood then there is a basis element containing the point that is contained within the neighborhood

[tough imp]

No

This says the topology is finer

You need to know that you aren’t adding in a ton of open sets too

So you have to do something in the other direction too

[marsh forge]

It is impossible to do this if you know that all the sets in your potential basis are open

[tough imp]

What?

If your bigger topology is the indiscrete one

And you take the basis of “every subset”

Oh

I see

Yeah

I agree now

[little hemlock]

can anyone elaborate on the underlined part here?

P = [a, b] is saying that there is a straight path from a to b that doesn't leave G. How does one of the segments making up P having a point from A and a point from B allow us to assume this?

[lunar yoke]

well i think the point is that that segment is where things go wrong

so basically its just relabeling that specific segment to P again, since the rest of the original P probably wont matter for the coming argument

[little hemlock]

i mean i think i understand that much. I'm asking why does the sentence
"now a moment's thought will show that one of the segments making up P will have one point in A and another in B"
imply
"so we can assume that P = [a,b]"

[lunar yoke]

i dont think it actually formally implies this

just for this proof, you can restrict attention to that segment

[little hemlock]

but just because there is a polygon from a to b doesn't mean there is a line segment from a to b

[lunar yoke]

you also relabel a and b to the endpoints of that segment that crosses from A to B

[little hemlock]

ohhh

[lunar yoke]

it could certainly be worded better ^^

[little hemlock]

i thought something like this was going on so i was bewildered

where the red line wouldve been the new P

but, yea, now i see the orange line would be the new P

thanks

[limpid leaf]

this is more of a proof writing question but can someone sanity check me here that the union property is satisified by the topology (-infty, n) for n \in Z. like i know this is true but i am worried about bad faith grading and so i just want to make sure the jump from (-infty, m) subset (-infty, n) to union is equal isn't crazy .

(i am worried about bad faith grading ebcause the ta for this course is a notorious asshole)

maybe i should just include a picture.

[reef shore]

Not all subsets of Z have a greatest element, so you should be a bit careful with using max

[limpid leaf]

[reef shore]

And the union jump isn't crazy but if you're being extra asshole then that inclusion just proves that the union is contained in (-infty, n), not equality

[limpid leaf]

yeah thats what i figured

i guess proving the other direction of the inclusion is shrot

[limpid leaf]

i do not understand this

why do i not understand this.

[reef shore]

Z^op isn't well ordered

I mean you should write sup instead of max maybe

[limpid leaf]

yes

maybe

u know what. im going to take the L on the max thing just to test him

[reef shore]

no but ok heres an actual issue with that

what if n = infty

[limpid leaf]

😐

[reef shore]

or -infty

that's why max not good

[limpid leaf]

god

[reef shore]

not talking about the maxj

[limpid leaf]

ig sup works.

i kinda want to test him on this just to see how pedantic i need to be for the rest of the semester

[reef shore]

[limpid leaf]

is it worth the point

yes

these assignments are weekly

[reef shore]

tbh I would deduct marks if you missed the case of unbounded n_i's

[limpid leaf]

lmao

[reef shore]

especially if the problem says the set is {(-infty, n) : n in Z}

[limpid leaf]

okay ill right sup

[reef shore]

because then this is not union closed

because infty and -infty which are possible values for n once you union are not there in this

[limpid leaf]

this works i think

i cant possibly be more pedantic thna this

"We assume the positive integers are well-ordered"

[empty grove]

My point was exactly that n need not be in {n_i}

[empty grove]

If the problem explicitly says it like this, then it's not a topology for precisely this reason

[limpid leaf]

oh yes i see

my bad.

[empty grove]

Yeah so you gotta handle the unbounded n_i case separately

[limpid leaf]

u really think so

[empty grove]

Yes

[limpid leaf]

thats so fucked up

[empty grove]

So either max n_i exists or not

If it does then your proof works

[limpid leaf]

what do you even do if such an n_i doesn't exist

[empty grove]

If it doesn't, then n_i has to be unbounded and the sup will be infinity

And then union is R

[limpid leaf]

so its not a topology?

if none exists

[empty grove]

It is

[limpid leaf]

i am confused.

[empty grove]

Because they cheat by doing union {empty, R} at the end

Ok take the case Z = I

And n_i = i

Then there is not max n_i

[limpid leaf]

yes

[empty grove]

And then union is all of R

[limpid leaf]

Yes

but its still a topology because R is in the topology already?

[empty grove]

So I'm saying you need to handle this kind of case separately since your proof fails here

Yeah it's a topology

[limpid leaf]

okay

does this fuck up the intersection case

[empty grove]

Nope intersections will be finite anyway

[limpid leaf]

right

[empty grove]

And finite totally ordered sets always have min and max

[limpid leaf]

do i need to write sup still or does max work the same in this case

if I'm just going to assume such a max exists anyways

and then deal with it not existing seperately

[empty grove]

Max works then yeah

n = sup n_i. If n = infty then union is R. If n ≠ infty then it's a max so your argument finishes it off

[limpid leaf]

right okay

very cursed!

thank u though

[empty grove]

[coral pawn]

Can someone verify this?

The result seems too nice for a differential geometry class

[limpid leaf]

uh similar request for checking finite intersection for the topology

i feel gross for writing math like this! but i just want to make sure he has no excuses to dock marks xd

im sure there is a way to do the second containment directly

not sure what it is tho

[reef shore]

for each m<n, we have (-infty, n) sub (-infty, m)

this sounds wrong

maybe you just want the other inclusion

[limpid leaf]

oh yes i want n < m right

[reef shore]

right

[limpid leaf]

this is what i get for checking my assignment at 3am

[reef shore]

you should say what m is though lol

[limpid leaf]

who makes assignmetns due at 9am

Yeah

[reef shore]

maybe just say that m refers to some n_i everywhere

[limpid leaf]

okay for each i in I, n < n_i is prob better

[reef shore]

ye

like in the later arguments too

[reef shore]

you really shouldn't I would say these are important details

that you should be comfortable with

[true robin]

Moldi is a known asshole

[reef shore]

[true robin]

And operates on bad faith

[limpid leaf]

yeah

ur right

it just feels weird to write math defensively

but i suppose it is a good habit to work this stuff out lol

[reef shore]

yeah I mean if you feel like this argument has been done multiple times in class or some previous courses then sure

but given that you missed the unbounded n_i case for arbitrary unions I would say you should write this stuff

[limpid leaf]

yes

i wonder if i was supposed to notice that

i guess i should not assume that my very smart prof made a major oversight there

lol

[reef shore]

[limpid leaf]

okay yeah this looks correct

without telling me, is there a nicer way to do the second inclusion directly

i prob wont rewrite since I need to sleep but

[true robin]

Yes

[reef shore]

[true robin]

No need for contradiction

[reef shore]

ah

[limpid leaf]

guh

this is due in 6hrs (9am) so i wont work it out now but i'll try to figure it out tmrw afternoon since it bothers me that i couldnt immediately do it directly

[reef shore]

It is your proof pretty much

Like you assumed (not A) and proved A to get the contradiction

but you didnt use the assumption not A anywhere

[limpid leaf]

[reef shore]

so you could just remove that assumption and make it a normal proof

[true robin]

Well let’s just say that we can use the universal property of the minima

[limpid leaf]

go back to #category-theory

lmfao

[reef shore]

didn't he already do that

[true robin]

I mean directly

[limpid leaf]

guh okay before i am too tired to process this stuff, does this fail if X is a single element?

[reef shore]

oh right

[reef shore]

no?

[true robin]

No, I don’t think so

Wow, even this second problem has a categorical interpretation

[limpid leaf]

[reef shore]

your mom has a categorical interpretation

[limpid leaf]

wait i get it

[true robin]

Your mom is fat, ugly and has no nice categorical interpretation

[limpid leaf]

i was just worried that in the backwards direction the empty set would not be in the topologee

[reef shore]

[limpid leaf]

ok now i verify counter examples and sleep

what is the categorical interpretation of the first problem

[true robin]

Actually I can’t think of anything, I think I just said that because I shoehorned in the universal property. Apologies

[limpid leaf]

lmfao

my counter examples are correct

i am a god

oh might as well check that this is. valid

bnefore i hand it in

just asks if nested topology is smaller/larger/incomparable than usual topology

it does not request a proof

so i assume handwave is fine

[reef shore]

ball around x contained in U_n

but ye is good

[limpid leaf]

oh yeah i didnt know how to word that

what is the right way to say what i am trying to say there

because x is a point in Un but also it is a point in usual topology

[reef shore]

ball around x contained in U_n should be fine lol

if you really wanna be explicit say ball of the standard metric/top

[limpid leaf]

wait why is it contained in U_n

is the ball not in T_usual

[reef shore]

you can take epsilon smol enough

it is

[limpid leaf]

U_n \in T_nest

[reef shore]

but "ball" already kinda implies that it is in the metric

ball is an element of T_usual and subset of U_n

[limpid leaf]

yes

oh i see

lmao

[reef shore]

where the T_usual part is implied by the word ball

[true robin]

This may be a bit nit picky, but just saying that for any x in Un you cab choose an open ball feels like you are assuming that Un is open already

[limpid leaf]

oh

yeh

[true robin]

I feel like you should actually give the construction of epsilon

[reef shore]

saketh is a known asshole

but I agree

[limpid leaf]

no i swore off constructing epsilons after analysis 1 sorry

[reef shore]

~~probably because I am too ~~

[limpid leaf]

well the answer also just says to "explain"

[reef shore]

explain why U_n is open jesse

[limpid leaf]

because it's an element of T_nest

[reef shore]

[true robin]

Also maybe a cleaner way to do it rather than constructing the epsilon ball around each x you choose, you could probably just say that Un is the union over i<=n-1of B(i,1).

[limpid leaf]

what is B(i, 1)?

oh just radius 1 ball

[true robin]

Ball of radius 1 around i

Oh and i ranges over integers

Just to clarify is this the lower order topology on Z or on R

Because the details of what I said will change depending

[limpid leaf]

on R

[true robin]

I see, hopefully what I wrote is correct then

[limpid leaf]

i see no issues

much nicer

okay thank u moldi and saketh you are both very good people. i will now get four hours of sleep.

[true robin]

Good night

[coral pawn]

Is there any way to do this without having a bunch of different cases?

[plain raven]

left adjoints preserve colimits. every set is the coproduct of copies of the singleton, so the discrete topology on a space should be the coproduct of singleton spaces

that's my categorical interpretation of the second question

[shy moss]

why the (p,q)-torus knot only exist when p and q are coprime?

[bitter yoke]

You can look at it when they share a common factor, but then it'll decompose into disjoint knots

[empty grove]

They won't be disjoint right? You would just go over the same knot gcd(m,n) times?

[shy moss]

thanks

if K is a (p,q) torus knot then why $\pi_1(R^3-K)=\pi_1(S^3-K)$?

[gentle osprey]

Or x1

[gritty widget]

is there some nice way to show that ab and ba have the same characterstic polynomial

where a and b are square matrices over a general field

i thought there was something slick like using the irreducibility of affine n space

[true robin]

Here is a cool method: prove that det(I+AB)=det(I+BA). If A is invertible this is obvious, and then use the zariski denseness of invertible matrices to conclude generally

det(x+AB)=det(x+BA) can be done in the exact same way

[gritty widget]

yes this what I was looking for!

so what we are using here is that

if V is an algebraic set of affine n space

and p is a polynomial

if p is zero outside of V p is zero everywhere

else we would contradict the irreducibility

[true robin]

If you don't like zariski denseness, there is an even crazier argument which goes as follows, consider both these sides as polynomials in variables x,a_11,..a_nn (basically treating entries of A as a formal polynomial. we can divide by det(A) on both sides as it is a non zero polynomial and conclude that as formal polynomials these expressions are the same, hence they are equal when evaluating at whatever values you want to put for the a_ij's

[gritty widget]

i like that

can you explain a bit more to me the zariski dense arguement

so we know that

invertible matrices are dense

[true robin]

What we actually did was create a zariski dense open set, and proved that a zero polynomial had a zero set containing it. since zero sets are closed (algebraic set) we see that the zero set is the entire thing

oops replied to the wrong thing

now invertible matrices are dense because they are the set of matrices A st det(A)/=0, that is some xariski open set

and all nonzero open sets are dense

because of irreducibility

[gritty widget]

so the set of invertible matrices is dense

[true robin]

yes

[gritty widget]

and our statement

char(ab)=char(ba) now holds for a dense set of matrices

[true robin]

Yes. you can think of it as det(xI-AB)-det(xI-BA)=0 as a polynomial on a dense set of matrices

we should probably move this to abstract algebra

[tough imp]

For Zariski dense ness are you considering the zero set of the ludicrous polynomial you get by subtracting these two determinants?

[true robin]

yes

[tough imp]

[gritty widget]

ah yes thank you i see this now

oh i think maybe we actually need to consider

det(I-XA)-det(I-AX)=0

[tough imp]

Also fucking Hurb

[gritty widget]

thank you very much though

[tough imp]

As I type my question you’re typing the answer

[gritty widget]

whats a Hurb

[tough imp]

Hurb

Bruh backwards = Hurb

Asking what Hurb is is a Hurb

[true robin]

A is just what I'm putting in for the n^2 variables

[gritty widget]

oh okay

i think what i said only gives

for a given A

char(AB)=char(BA)

but

its arbitary A

[true robin]

yeah that is what my argument gives as well

(well, i suppose it is the same argument)

[gritty widget]

sorry

I am just confused about the x in your argument

[true robin]

just another formal variable

[gritty widget]

okay okay i see now

[gritty widget]

affine space is only irreducible when the field is algebraically closed

or

atleast infinite

?

[tough imp]

Nopeeeeeeeeeeeeeeeeee

It’s always gonna be irreducible for a field

I thinks

[true robin]

Infinite field

[tough imp]

At least as a scheme it is

[true robin]

Finite field is the discrete topology

[tough imp]

🤣 🔫

That doesn’t sit right with me

Why is Spec F_p[x] irreducible but the varsity for A^1_F_p reducible

[true robin]

I mean, the topologies are different

I guess the moral of the story is that having (0) as a point makes all the difference

[tough imp]

TRUE

This meme brought to you by generic points

[prisma seal]

I've proven all the norm properties and that omega is dense -> nu is a norm, but I haven't done nu is a norm -> omega is dense. I see how I could do it in a metric space, but I don't see how to do it in the general topological setting.

I've proven that compact, Hausdorff spaces are T_4, so if x is a point not in the closure of omega, there are disjoint neighborhoods U and V containing x and the closure of omega respectively, and that feels very helpful, I just don't see exactly what to do from there. Any hints appreciated

and I'm wondering if there's an easy topology theorem I'm just missing that makes it obvious

the thing I want to do is for contradiction assume that there's some x not in the closure of omega. Then let f(x) = 0 on the closure of omega, and let f(x) = (something nonzero) at a point outside the closure of omega, and then use the disjoint neighborhoods to show that such a function can be extended to something continuous on the space, but no clue how to do that last part, or if it is even possible

[true robin]

Every compact hausdorff space is normal

So use urysohn’s lemma

[prisma seal]

I do not know Urysohn's lemma

and yeah I proved that compact, hasdorff spaces are T4 (so obviously normal) so I got that part

[prisma seal]

ok a quick google and that's literally what I needed ty lol

[true robin]

It says that for any two disjoint closed sets A and B in a topological space there is a function f that is 0 on A and 1 on B

[prisma seal]

Thank you so much

My school does not have an undergrad course in point-set topology for stupid reasons so I've only seen topology in metric spaces, and this is for a functional analysis class. The prof treats us as if we know point-set topology. I've had to pick a lot up very quickly so there are some gaps

[true robin]

that seems really unreasonable, urysohn's lemma is a very hard theorem

[gritty widget]

someone shared here a nice exposition on torsors the other day

does anyone still have that

[shy moss]

if i have an open set X which is contractible and compact, then X minus one point is simply connected?

[true robin]

Do you mean topological space X instead of open set, or is your X some open set in euclidean space or something

Because in euclidean space, open sets are never compact

[obtuse meteor]

if you mean a topological space, then no

[gritty widget]

have to find tangent space of $\mathbb C[\frac{1}{x_1 x_2^2}, \frac{1}{x_2}, x_1 x_3, x_1^2 x_2 x_3, \frac{1}{x_1^2 x_2^3 x_3}, \frac{x_3}{x_2}]$ at point $I = (\frac{1}{x_1 x_2^2}, \frac{1}{x_2}, x_1 x_3, x_1^2 x_2 x_3, \frac{1}{x_1^2 x_2^3 x_3}, \frac{x_3}{x_2})$ how to do this?

[gentle osprey]

Empty2

[shy moss]

if i have an open neighbourhood contratible $B=B' \cup \lbrace \infty \rbrace$of $\infty$ in the compactification of R^3, where $\infty$ is the added point, then $B'$ is simply connected?

[gentle osprey]

Or x1
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[marsh forge]

Uh

You have to phrase it more formally

But like

A disk minus a point

Sounds like a counterexample

[true robin]

I think in the more specific question about the compactification of R^3, indeed B’ is connected

[marsh forge]

It says simply connected

If B is a small disk around infinity then I still think what I said is a counterexample

[true robin]

Equivalent in euclidean spaces right?

[marsh forge]

Do you mean path connected, saketh

[true robin]

But small disks are not open in R^3

[marsh forge]

Simply connected means trivial fundamental group and path connected

[true robin]

Ah yes, I meant simply connected, apologies

[marsh forge]

Path connected

[true robin]

Ah yes

[true robin]

Wait no, I did mean simply connected

Basically, my intuition is that a simply connected open neighbourhood around infinity should look like a sphere

And spheres minus a point are simply connected

Does that seem right to you?

[shy moss]

i know that spheres of dimension at least 2 are simply connected but why spheres minus a point are simply connected?

[true robin]

Apologies, I meant ball and not sphere

Ie: It has the sphere and all of the points in the sphere

[shy moss]

balls minus one point are simply connected because R^n minus a finite number of points is simply connected and there is a deformation of R^n onto a ball?

[true robin]

Yes, but n must be greater than or equal to 3 for R^n minus some points to be simply connected

[plain raven]

i also recommend learning the tietze extension theorem. a good summary of the topology you need for analysis can be found in chapter 4 of follan'ds book on real analysis

[shy moss]

o yes, thanks

[marsh forge]

Ah I misread as R2 somehow

[gritty widget]

Hello I started my topology course and I have to prove that and I have no ideas how to start :/

I understand the two basis have a different nature

I try figure how to explain that certain sets of R can be described with 1 element of R_l and not by 1 element from R_k and vice versa

Sorry for my english

[tidal cedar]

So you have to show that neither topologies contain each other

As a collection of subsets

[gritty widget]

Something like that ?

[plain raven]

anyone want to dm me a link to the algebraic geometry syndicate discord

[plain raven]

nvm.

[shy moss]

how exactly can i define a deformation retraction of $S^3-T_{p,q}=(S^1 \times D^2)\cup (D^2 \times S^1)-T_{p,q}$ onto $S^1 \times I/ \sim$ where $\sim$ is the relation generated by $(z,0)\sim (e^{2 \pi i/p}z,0) $ and $(z,1) \sim (e^{2 \pi i /q}z,1)$ and $T_{p,q}$ is the $(p,q)$-torus knot?

[gentle osprey]

Or x1

[rancid umbra]

is there a characterization of compact subsets of Q?
there is the result that a subset of Q is compact in Q if and only if it is compact in R, but i mean like a heine borel for subsets of Q. since for example, [0,1] as a subset of Q is not compact, then closed and bounded isn’t enough to imply compactness.

ye, was just about to say that

[hollow harbor]

it's basically going to be like

unions of sequences which converge in Q (including their limits)

such that the limits form a discrete set

something like that

the limits might not need to form a discrete set

[rancid umbra]

hmm. so something like {1/n} U {0}

[hollow harbor]

sure

and like

you can have unions of things like that

but

probably not big unions

finite unions of things like that def work

my question is, can you get infinitely many limit points in by having them all converge to another limit point?

[rancid umbra]

yea countable doesn’t work i don’t think. u can probably get Q from unions of sets of that form

[rancid umbra]

yea i think

[hollow harbor]

like e.g. you pick the sequences {1/n + 1/(100n)^k : k in N} for each fixed n

and then the limits are {1/n}

[empty grove]

There are sets whose limit points are exactly the 1/n

[hollow harbor]

and then you throw in {0} too

[empty grove]

Nice lol

[rancid umbra]

so take the set 1/N = {1/n: n in N} and then squish it between 1/2 and 1. then squish 1/N between 1/3 and 1/2 and so on

[hollow harbor]

however

idk if this thing is

well

i think it's closed

[empty grove]

It's compact yes

[hollow harbor]

it should be, i don't see how it wouldnt be

ok

and you can do it forever too

you can keep fitting in smaller things convering to the last set of stuff

so it's very weird

[empty grove]

It's bounded because of your choice of elements

[rancid umbra]

which one are we talking about, ryc’s example?

[empty grove]

Ye

[hollow harbor]

thanks real chmonkey

[rancid umbra]

i think mine does the same thing

[empty grove]

I'm even verified

I guess you could just say sets which are bounded, closed in Q, and no irrational limits lmao

[rancid umbra]

bruh that’s just sequentially compact

[empty grove]

Equivalent to compact for all metric spaces

[rancid umbra]

yea but like, this is just as good as saying that the compact subsets of Q are exactly the sequentially compact subsets lol

not saying it’s bad

just kinda funny

[empty grove]

No, that would be if you said "compact subsets are exactly the subsets in which all sequences have convergent subsequences"

[drifting sundial]

for this, why can't we use (C) to get a finite collection of cells which cover the closure of e, then let the collection of closures be the desired subcomplex?

[long hornet]

It's not obvious that it will be a finite subcomplex

[drifting sundial]

is there a counterexample?

or is it true but just hard to prove

[long hornet]

It is true, but it is the statement of the lemma. If I understand you correctly, your proof is circular.

[drifting sundial]

I'm using one of the conditions of a CW complex, where the closure of each cell is contained in a union of finitely many cells

if you then take the closures of those cells it should form a subcomplex right?

[long hornet]

Yes, finitely many open cells

A priori, we don't know how many open cells are in cl(e) - e

I mean even if you say cl(e) is contained in the union of e_i, the subcomplex composed of the cells {cl(e_i)} is not proven to be finite yet, because we don't know whether taking the closure will introduce infinitely many open cells or not.

[drifting sundial]

yeah I was thinking something along those lines but couldn't really come up with an example

[long hornet]

Well, there is none! The statement is that taking the closure only introduces finitely many cells. It's what we are trying to prove.

[drifting sundial]

ah I see

so it's sort of a refinement of closure finiteness?

[long hornet]

I think so

[loud scarab]

I have a question about this theorem, after showing that one-point sets are closed in a Hausdorff space, doesn't this imply that an arbitrary union of one-point sets is closed, which means that we could have a set with infinite points which is closed? Then why is the theorem stating that every 'finite' point set is closed when we could do an arbitrary union?

[reef shore]

arbitrary unions of closed sets aren't closed

That is for open sets

[loud scarab]

dammit, thank you!

[long hornet]

Is the functor H_k (X ; --) exact? H stands for simplicial topology

Homology*

[tawdry widget]

I think not. You will have a long exact sequence instead I think

[long hornet]

I see, thanks

I think my questions are a bit premature and dumb, but anyway: If we know all H_k (X ; Z/n) for all n, do we know H_k (X ; Z)?

[quartz edge]

i should like to see a quick example of the computation of homology groups for some simple manifold

where can i find this

[true garden]

Hello. Can anyone give a simple example of two limit point compact Hausdorff spaces whose product is not limit point compact?

[long hornet]

If they are compact, then their product is also compact, and therefore limit point compact. So if X x Y has this property, at least one of them, say Y, is lp compact, but not compact. Do you know such a simple Y?

The only Hausdorff example I know is the minimal uncountable well-ordered set in the order topology.

[reef shore]

omega1 x omega1 also seems to be limit point compact though

countable product of sequentially compact is sequentially compact iirc, so omega1 won't work

You will need a limit point compact space which is not compact and also not sequentially compact

wait does sequential compact imply limit point compact? It should I think, since for any infinite set you can take a sequence with distinct terms in it and that should have a subsequentiall limit and that should be a limit point of the original set

[long hornet]

Yeah. Maybe we take X to be different than Y, but I still can't see how that will make things easier.

[long hornet]

That makes sense

[reef shore]

https://math.stackexchange.com/questions/538674/the-product-of-limit-point-compact-hausdorff-spaces-is-not-limit-point-compact

I doubt there are any simple examples

The linked answer also seems very complicated lol

[long hornet]

Yeah

But I think it proves a strictly stronger result

How about the Sorgenfrei plane? It's a product, and has an uncountable discrete subset (so not limit point compact). Is R_l limit point compact?

[reef shore]

The Sorgenfrey line is mentioned in the answer linked in the answer on the page that I linked, is that related?

[long hornet]

Actually we should pick some bounded subspace

[reef shore]

I don't know what it is

[long hornet]

I think that was a counterexample to something else

The Sorgenfrei line is R_l (R in the lower limit topology). The Sorgenfrey plane is R_l x R_l.

[reef shore]

ah

I see why you want a bounded subspace

Don't think that works though, because you can take the sequence 1 - 1/n, and the limit "should be 1" but it's not because lower limit shenanigans

I mean even viewing the sequence as just a set

[long hornet]

I just checked and found that 1st ctbl + lpcompact implies seq. compact, so it doesn't work, as you said

[reef shore]

I see

[long hornet]

Ahaa, right

[reef shore]

right yeah since first countable spaces are sequential

[long hornet]

That's nice!

In finer spaces, compactness is harder, right?

I mean if you have X and two topologies on it, one of which is finer than the other

[reef shore]

Yeah seems legit

open cover in coarser is also one in finer, so take finite in the finer topology and return

[long hornet]

Yeah

I think one can often check whether a property is "increasing or decreasing"by looking at the discrete and indiscrete topologies

[reef shore]

yeah true, if you know that the property is one of the two

[long hornet]

It'd be amusing to formalize this

[reef shore]

R with the topology generated by the basis {(r, infty) : r in R} is limit compact, not sequentially compact

but I think the square of this is limit compact too

yeah it is

[long hornet]

Isn't that the upper limit topology?

[reef shore]

the basis for that would be {(a,b] : a,b in R}

[long hornet]

We can intersect

[reef shore]

like these are going all the way to the right

[long hornet]

Oh, my bad. That's not a subbasis

[reef shore]

ye

[long hornet]

I doubt this cheap fix will work

But what if we take X = [0, 1] with lower and upper limits?

I mean X with the union of these topologies

[long hornet]

But uh, it still suffers from this

So the space should not be first countable

[reef shore]

if you take both lower and upper then I think you just get discrete

intersection of (a,b] and [b,c)

[long hornet]

lol

[reef shore]

fuck this problem

[long hornet]

Agreed

I saw this question in MSE once

About whether there is a topology T on C, such that analytic = continuous in T

It wasn't answered

[reef shore]

does seem more interesting though

[long hornet]

More precisely, we want f : (C, T) --> C to be continuous iff it's analytic

[long hornet]

Yeah

Is being normal increasing or decreasing?

[reef shore]

don't think it would be either

Yeah R_l x R_l is finer than R^2

[long hornet]

Yeah. On the one hand, we don't want too many closed sets. But on the other hand, we want many open sets!

[reef shore]

and for another example take indiscrete topology

Yeah exactly

[long hornet]

Aha

[reef shore]

wait indiscrete top is always normal

wait no

being T_1 is also included in normal

[long hornet]

Yes, and so is discrete.

[reef shore]

I guess it might be a matter of definition

[long hornet]

Minimal T1 = cofinite

So maybe we should just consider that

[reef shore]

Yeah that should work, doesn't seem normal

any 2 open sets intersect

non empty open sets

assuming infinite space

or actually lol

R^2 is coarser than R_l^2 is coarser than discrete on R^2

normal < not normal < normal

[long hornet]

Great

I was thinking of invoking Tietze's theorem

For no really good reason, just that extensions are restricted in complex analysis

[reef shore]

makes sense

[long hornet]

I think C* is not paracompact

But I don't think paracompactness is increasing, so

[plain raven]

Compactly generated Hausdorff spaces are used in homotopy theory.

Does anyone have a good idea of like, how much stuff about locally compact Hausdorff spaces goes through for CG spaces?

like

lots of analysis relies on results about LCH spaces.

A lot of sheaf theory, sheaf cohomology etc works well over LCH spaces

Just trying to get an idea of how practical it is to think of compactly-generated as a slight weakening of locally compact.

[dusk heron]

If we define the Riemann curvature tensor, in abstract index notation by $R_{\alpha\beta\gamma}^\delta V^\gamma=(\nabla_\alpha\nabla_\beta-\nabla_\beta\nabla_\alpha)V^\delta$, on a Riemannian manifold, then the right definition of the Ricci tensor $R_{\alpha\beta}$ should be $R_{\alpha\beta}=R_{\gamma\alpha\beta}^\gamma$, right? The book "Spinors and Space-Time" by Penrose and Rindler instead defines $R_{\alpha\beta}=R_{\alpha\gamma\beta}^\gamma$, which gives the opposite sign.

[gentle osprey]

gustavn64

[gritty widget]

want to find intersection of ideal $I = (4 - \frac{x_2}{x_3} - \frac{x_3^2}{x_1x_2}, 4 - \frac{1}{x_1 x_2} - \frac{1}{x_3 x_3})$ with $\mathbb C[x_1,x_2,x_3]$ any idea how to do it?

[gentle osprey]

Empty2

[long hornet]

Like what?