point-set-topology

[ember maple]

@sleek thicket how do other people usually define affine variety? A tuple k^n for any alg closed field k?

Finitely generated mean (locally at least) it's polynomial with finite indeterminate?

[sleek thicket]

An irreducible closed subset of k^n in the zariski topology

Where the closed sets are the zero locii

Finitely generated means quotient of polynomial ring in finitely many variables

See Hartshorne or Gathmann's notes

This definition is weird because it's not classical AG but it's also less abstract than schemes

The structure sheaves are actual functions

[ember maple]

I'm not that big brain enough to imagine sheaf in general ring which is not polynomial /function

[sleek thicket]

Well the structure sheaf of an affine schemes is more complicated

It's a big tuple with some local compatibility stuff

[civic kraken]

this image must be preserved for all time

[bitter yoke]

aren't you a fucking geometer

[civic kraken]

fuck u

u always have to spoil my fun 😔

[honest narwhal]

@civic kraken wait what kinda geometry do you do? Your answer will either make us friends or enemies

[civic kraken]

mostly arithmetic

but im writing my thesis on teichmuller dynamics

i like geometry in general, but mostly i like number theory

[floral gust]

How are the definitions of T_3 and regular different?

[midnight jewel]

personally, I’ve learned them to be synonyms, but wikipedia says that T₃ is regular + hausdorff, our definition of regular included hausdorff as a condidion (or rather, that points are closed, which is equivalent in this case)

https://en.wikipedia.org/wiki/Regular_space just see here

[honest narwhal]

Okay so yeah that's fine we're friends. That meme I think makes fun of differential geometers anyway

[civic kraken]

that is correct

[floral gust]

@midnight jewel In Munkres, regular is defined as following Suppose that one point sets are closed in X. Then X is said to be **regular** if for each pair consisting of a point x and a closed set B disjoint from x, there exists disjoint open sets containing x and B respectively.

If I am correct, does that not imply Hausdorff? (just take the closed set to be another point

[midnight jewel]

it does

but not everyone takes the points are closed condition to be part of the definition

if you do then regular spaces are always hausdorff

otherwise they may not be

[floral gust]

In my notes however, I am provided that Regular = T_3 + T_1. Is it possible that the singletons are closed is provided by T_1?

[midnight jewel]

and T₃ being defined how?

literally none of these words have unambiguous definitions

[floral gust]

Closed and a point are separable by disjoint neighbourhoods

T_1 being two points can be seperated by neighbourhoods (not necessarily disjoint)

[midnight jewel]

apparently it’s equivalent (if I’m reading the wiki article correctly) but it’s not obvious to me

ah wait

let y in X and for each x define Oₓ to be an open neighborhood of x not containing y (exists by T₀). then the union of the Oₓ is precisely X\y, and thus {y} is closed

so T₀ already implies that points are closed

[floral gust]

For me T_0 means given two points atleast one of them has an open neighbourhood that doesnt contain the other.

So by T_0, we might as well have gotten a neighbourhood of y

[midnight jewel]

doesn’t T₀ mean that each point has a neighborhood not containing the other, but the neighborhoods don’t have to be disjoint?

[floral gust]

That's T_1 for me

[midnight jewel]

ah wait

I meant T₁ the whole time

fucking numbers

yes you’re right

just imagine I said 1 every time I said 0

[floral gust]

In my notes however, I am provided that Regular = T_3 + T_1. Is it possible that the singletons are closed is provided by T_1? Is this correct then

[midnight jewel]

yes

[floral gust]

Smh obviously...

[midnight jewel]

as I just proved

[floral gust]

Yes yes

Thanks!

[midnight jewel]

np

[finite turtle]

“how does this relate to upper undergraduate-level geometry”

John H Conway discovered most of these equilibrium honeycombs
The ones I showed are cubic honeycomb

A honeycomb is a space filling tessellation of polyhedra.
Cubic honeycombs, Archimedean and Platonic solids share great relation to subatomic structures. For example the Higgs boson is an E8 (quasicrystal composed only of platonic solids up to the eight dimension).

Everything except for the links was of my own work

I’m working on these honeycombs and making my own honeycomb structures because I’ve envisioned ways to build technology and advancement with these structures. Also for the interest and desire to find a cellular automaton capable of explaining all particle physics and occurrences via functions.

Technology

• Euclidian warping structures

advancements in fiber technology (using genuflecting bipyramids).

Impact safety features with an inverse metamorphose cuboctahedrille
Or other metamorphosis honeycombs

• Agriculture

Agricultural advancements in hydroponic automaton using space filling curves and AI data tracking systems.

Agriculture advancements using topology optimization with circular fields and the phi ratio.

• data compression with space filling curves

Etc.

[ivory dragon]

You didn't answer the question.

But either way, if this is as revolutionary as you say it is, I highly recommend you either:
a) get it written up as a paper and published
b) sell the patent to some big company

[midnight jewel]

quasicrystal composed only of platonic solids up to the eight dimension
that seems like a lot of fancy words without any meanings if you ask me

for one, platonic solids are by definition 3-dimensional

[ivory dragon]

rather than posting huge walls of links and technoblabble on discord without further elaboration

(also, iirc conway didnt construct the convex uniform polychora, but rather showed that they were the only ones that existed?)

(I believe he did discover the grand antiprism, but I think that's about it)

admittedly, it's uncited, but wikipedia corroborates my claim

though apparently the proof of completeness was due to Möller, not Conway-Guy

rather, Conway-Guy simply enumerated them and demonstrated some basic properties

and either way, I don't see that you're doing any actual mathematics

you seem to be mashing shapes together, which is nice exploration and all

but I haven't seen any, like, mathematical work

you threw out a bunch of buzzwordy terminology, but none that's actually used substantially in the fields involved (geometry, geometric topology, algebraic geometry, algebraic topology)

and I don't see any, like, theorems or claims or statements you've proven about your construction

just that "I made a thing"

and it's great to make things - concrete examples of geometry are quite interesting, and a great opportunity to explore mathematics in a physical manner - but I think you're greatly overestimating the impact and interesting qualities of your work, and it's bordering on mysticism, a la numerology.

but who knows? maybe you're the Cantor to my British Academia, and in a decade, graduate students will be studying your honeycomb constructions

somehow, though, I doubt it, considering I struggle to see why what you're doing is interesting or relevant.

I don't mean this to discourage you, but perhaps don't toot your own horn with such grandiose statements - or, if you do, do it to publishers/big pharma companies rather than to a random math discord.

Trust me, if this actually does have substance, it'll be better for you - and the world - if you communicate it through the conventional avenues for groundbreaking work.

[midnight jewel]

and work on your presentation cause like… I have no idea what you’re talking about you’re just saying words

[ivory dragon]

Yeah, perhaps if you explained it using the typical terminology and presentation of the field (or even, you know, mentioned what field your results belonged to), it'd be easier to understand your thesis.

(I'm not trying to gatekeep you or say "if you don't understand the fancy words used by mathematicians, you can't possibly make contributions"; but the fancy words are standardized for a reason: mathematicians understand them and there's no ambiguity.)

(Sadly, that's not the case for the terminology you're using. If you can find some alternate way to explain your findings, then go ahead! But clearly, what you've done above is not adequate; I can't parse any concrete claims, theorems, proofs, or hints at broad relevancy.)

[honest narwhal]

I'll ping you here because while you should read everything that's being said, this is administrative and thus you have to read it. @finite turtle if you're going to pursue your stuff further without following the recommendations of Namington and Sascha, from now on you are to do it in #chill

[finite turtle]

I explained the constructs using proper mathematical lexicon. And no, Platonic solids (regular polyhedra) are not restricted to just the 3rd dimension. You are correct, I am just putting shapes together, these shapes however do share relevance to subatomic structures and crystal growth patterns. The mathematics behind these constructs I’ve shared and explained with mathematical lexicon rather then equation.

Most people don’t even know what a tetrahedron is
Sad
It can be a difficult task to estimate the educational background and retention rate of those that view your works. Especially given the retention rate of modern kind is less then that of a nat. speaking this in a simply stupid guide form would astronomically lengthen the explanation into something nobody would read.

I enjoy sharing my works on social media and chatting with others doing similar work.
This is a way for me to connect and meet new people interested in similar subjects, so I will continue to do so.

Sure I’ll do it in #chill

Tell me what’s wrong with my explanation please “This space-filling honeycomb, In Isometric Projection, produces the Rhombitrihexagonal Tiling Pattern.
The matrix is a truncated variant of the Cantic Cubic Honeycomb. It is a uniform matrix composed of tessellated Cuboctahedrons, Square Prisms & Semitruncated Rhombic Dodecahedrons.”

~ Discovery by Abyssal Dionysus ~

Info - Learn more
All the links I posted on words I used above
Showing each component of the matrix individual

[dim meadow]

This is terrible

Please share such drivel in #chill only, thanks

[finite turtle]

Why is there so much negativity towards this?

[umbral surge]

lol

[dim meadow]

Read Namington's post from earlier

Also listing stuff like "data compression with space filling curves" as an application of your work is an obvious sign of crankery

[floral gust]

He spread the same stuff on Kik as well and even accredited his work to the works of some crank physicist.

[honest narwhal]

@finite turtle this is an administrative order, from now on only chill

[tough hamlet]

k

[shrewd lagoon]

should i learn about topology before i start trying to learn about algebraic geometry? i just wanna get somewhat of an appreciation of it because it seems important

[frigid patrol]

I think so

[civic kraken]

100% learn topology

or u would make zariski upset

[sleek thicket]

AG doesn't actually involve any topology, the zariski topologies are fake

cofinite? more like costupid

wait fuck costupid would mean smart fuck fuck

[honest narwhal]

gg shamrock

[ivory dragon]

"AG doesn't actually involve any topology"

oh man i wish

[honest narwhal]

So I know the Zariski topology is there but how often do you do things with it?

[midnight jewel]

honestly getting to know an “ugly” topology in the wild (in this case: non-hausdorff) is like the only thing that’d interesting me about alggeo

[midnight jewel]

is this correct?

If $n$ is even, then $\delta c_n = \sum_{i=0}^n (-1)^i c_{n-1} = c_{n-1}$. Thus, for odd $n$, $c_n$ is a boundary, but for even $n$ it can’t be, as that would violate $\delta^2 = 0$

[gentle osprey]

Sascha Baer:

[dire warren]

Let X be a locally compact, noncompact Hausdorff topological space and X’ it’s one point compactification. Call such a space “compactification symmetric” if X’ minus {p} is homeomorphic to X for any point p in X’.

What are the compactification symmetric spaces?

[chrome dew]

spheres?

[versed pivot]

X' doesn't have to be a sphere
at the minimum it can be any compact (connected) manifold, just remove a point to get an X that compactifies to it

[dire warren]

Wait

Given any compact connected manifold, is M minus any point homeomorphic to M minus any other point?

[versed pivot]

https://topospaces.subwiki.org/wiki/Connected_manifold_implies_homogeneous

[dire warren]

Nice

[fiery hatch]

Hello

[errant rover]

I feel like there should be a proof that there exists no field structure on R^3 that respects R in the usual way as a subfield, starting by the fact that there is no topological group structure on S^2. I can't quite make it work, any idea? (I want to induce a group structure on S^2 via the multiplication, but actually making multiplication a map S^2 x S^2 -> S^2 doesn't always work, so I want to prove there is something homotopic to it.

[floral gust]

Is there some intuitive notion of "separability" hidden under the definition of a separable space as having a countable dense subset? I understand the notions of separability supposed to be captured by say like separation of a set or separability axioms. Not sure what sort of separability is being mentioned here.

[merry shale]

https://mathoverflow.net/questions/51494/why-the-name-separable-space

[floral gust]

Thanks!

[thorny flame]

I'm have trouble with parts of this problem from Munkres:
I understand the cases for the horizontal line and the vertical line but how can I figure out diagonal lines? Just working on the case for one R_L x R_L should be enough for me to understand the problem.

[sleek thicket]

Is R_l the topology generated by [a, b)?

[thorny flame]

Yes

[errant rover]

@thorny flame if L is a straight line in the plane and it is not vertical, then for every real number a, there is exactly one (x,y) in L such that x = a. This gives a bijection L -> R (think projecting down to the x-Axis). When you send open sets in L through this bijection, what happens to them and how do they look like?

(What you do on the back-end level is you give R a topology by sending open sets from L to R through this bijection. This makes this topology on R homeomorphic to L and this topology is easy to describe)

[thorny flame]

ah ha! ok that comment kind of helped thank you. I drew a picture out if it helps someone who is stuck

[floral gust]

Can we make a non-Hausdorff space Hausdorff by defining an equivalence relation such that x ~ y iff there exists a filter which converges to x and y

[errant rover]

Yep, a space is Hausdorff iff no filter has two limits

And I guess the space you get is actually the Hausdorff reflection, but I'm not quite sure how to prove that

[floral gust]

Would these definitions be correct?

[vocal wharf]

N_x are neighbourhoods of x?

[civic kraken]

it's easier if u just write out in english what ur trying to say with those symbols

[floral gust]

@vocal wharf Yes

@civic kraken I am aware but this helps in problem-solving much more (atleast for me). For me, the english definition seems to coincide with this, just double checking.

[vocal wharf]

the first definition is fine, if the index set i is from is countable

the second i'm not sure 🤔

[floral gust]

Secondly, is there any difference between metrizable and metric spaces?

[vocal wharf]

a metrizable topology is one that is induced from a metric

but you may not know that metric

[floral gust]

So I can use every result on general metric spaces on metrizable spaces?

and vice versa? (so long the result is not dependent on the choice of metric itself)

[vocal wharf]

yea

[floral gust]

So anyone has any idea about the second one?

[vocal wharf]

actually in both statements you would at least need $x \in B_{i}$

[gentle osprey]

Lochverstärker:

[floral gust]

Yes hence the x subscript in B

so B_x_i means for me the set that contains x and is indexed to be i

[vocal wharf]

ok, for the first line

that's not mentioned in second line?

because then i think it's fine, but it's a different definition of basis than what i'm used to and i cba to prove equivalence now

so someone else can comment

[floral gust]

Yes B_i merely means sets indexed as i and not necessarily neighbourhoods of x

[thorny flame]

Ok so I get that the lower limit topology is strictly finer than the standard topology so it must contain open intervals like (a,b). However, how can I construct an open interval (a,b) from intervals of the form [a,b) using infinite unions and finite intersections?

[dim meadow]

take the union of [x, b) over all x between a and b

@thorny flame

$\cup_{x\in (a, b)} [x, b) = (a, b)$

[gentle osprey]

Liquid:

[thorny flame]

ok so its like x is getting closer and closer to a but never actually touching it. So if we are on the R numbers then it should be an uncountable infinite union?

[dim meadow]

nah, you can do this with a countable union if you want, but in my case it is a uncountable one yeah

cause I was being lazy

[thorny flame]

i feel that

but ok yeah makes sense thank you

[sleek thicket]

@thorny flame unions of arbitrary cardinality are okay, but you could take x = a - 1/n if you wanted

[dim meadow]

^

[sleek thicket]

Or x being all rational points in (a, b)

[dim meadow]

that's not as nice

but yeah

[sleek thicket]

no but I like it better for some reason

[dim meadow]

haha

[sleek thicket]

It feels like a thing that could generalize to separable spaces somehow

But you need an order to talk about lower limit

So idk

[thorny flame]

ah i see because N is countable

ok

[dim meadow]

seperable metric spaces are at most the cardinality of the continuum

[thorny flame]

man infinity is freaking weird

[dim meadow]

👀

[thorny flame]

like everytime i think about it i'm like yeah but no but yeah ok i guess

[dim meadow]

and there are exactly continuum many of them

[sleek thicket]

Wait what? That's wild

[dim meadow]

nah, not so much

[sleek thicket]

Up to what notion of equivalence? Homeomorphism?

[dim meadow]

isometry

[sleek thicket]

Even weirder

Tbh

Can you sketch the proof?

[dim meadow]

basically completions are unique

so you need to find the number of possible metrics on a countable set

[sleek thicket]

And your completion is the completion of the dense subset, got it

[dim meadow]

and thats the same as the number of maps from N to R

and there are exactly continuum many such maps

that gives you an upper bound

[thorny flame]

what is continuum?

[sleek thicket]

Oh yeah, makes sense

Cardinality of the real line

Or the real line itself

Depending on context

[dim meadow]

yeah

[thorny flame]

is that just the Card of reals?

[sleek thicket]

Yee

[thorny flame]

oooo ok

[dim meadow]

all compact metric spaces are seperable

so there are exactly continuum many compact metric spaces

(there is an easy lower bound)

(at least for isometry)

haha this is a train of thought I've been talking about for a bit

I should probably stop sharing it lol

[sleek thicket]

It's neat

[dim meadow]

I guess something interesting would be showing a lower bound for homeomorphism

[sleek thicket]

I never really learned about metric spaces, just went straight to general topology

[dim meadow]

for isometry you just look at 2 point spaces

I'm taking a metric geometry class rn

so I've been thinking about metric spaces quite a bit

[sleek thicket]

I'm thinking about subsets of R^n

[dim meadow]

for a homeomorphism lower bound you can probably do some trivial stuff with connected components

hmm

[sleek thicket]

That seems plausible

I would be surprised if there were countably many homeomorphism classes of subsets of R^n

[dim meadow]

haha why not something in between

👀

yeah you can look at finite disjoint unions of spheres

n spheres for different n

[sleek thicket]

Concentric circles at the origin with irrational radii?

[dim meadow]

nah not even

just disjoint unions topologically

[sleek thicket]

Ah okay

I think my example gives homeomorphic spaces anyways lol

[dim meadow]

oh ok

[sleek thicket]

Where do you put the spheres?

[dim meadow]

you can make a bijection to binary expansions

[sleek thicket]

And stay seperable?

[dim meadow]

in a funny way

[sleek thicket]

Sure

[dim meadow]

you have either 0 or 1 of each n sphere

yeah

[sleek thicket]

Oh you mean like countable disjoint unions of S^n

[dim meadow]

no, thats finite

[sleek thicket]

I see it

Makes sense

What?

[dim meadow]

countable I mean

that's finite binary expansions

I need infinite binary expansions

[sleek thicket]

Why? You can keep adding spheres

And stay separable

[dim meadow]

I think youre idea of concentric spheres is good

[sleek thicket]

Like 101010... works, no?

For a point, a 2-sphere, a 4-sphere,...

[dim meadow]

S^\infty is compact, right?

[sleek thicket]

I'm not sure

But we're aren't looking at S^\infty

Oh I guess we are

[dim meadow]

no

[sleek thicket]

I don't think about the infinite sphere very much

[dim meadow]

it's not

[sleek thicket]

Oh okay, good

I was confused

[dim meadow]

haha

[sleek thicket]

Why doesn't the binary expansion trick work?

Convert 0 to 0 and 1 to Z in homology

Basically

[dim meadow]

cause each space only has a finite number of spheres

[sleek thicket]

Why?

[dim meadow]

cause they are disjoint

[sleek thicket]

Oh I mean you can't embed it in R^n

But it's still separable

And I think it's still a metric space

[dim meadow]

and a compact space has a finite number of connected components

[sleek thicket]

Sorry, were you were working on separable metric space homeomorphism classes or R^n subsets homeomorphism classes?

I was saying it would work for the first

[dim meadow]

compact metric space

sorry

[sleek thicket]

Ohhh

Sorry

Mb

This is probably dumb, but would it be useful to think about the cantor set? Doesn't every compact metric space show up there somehow? I'm talking about things I don't understand, sorry

[dim meadow]

Maybe you're thinking of the hilbert cube?

[sleek thicket]

Hmm maybe

Sorry

[dim meadow]

👌

[sleek thicket]

Ah I just checked Wikipedia™

"any compact metric space is a continuous image of the Cantor set"

That's what I was thinking of

But I don't see how to apply it here

It gives an upper bound but not a useful one I think

Like R^R

[dim meadow]

so if we can find a countable number of disjoint compact subsets of the plane, we can do something

not disjoint

i mean not homeomorphic

and actually there is an easy way to do this

I'll draw a picture

[sleek thicket]

Can't you do n circles in a line?

[dim meadow]

I was thinking you take a circle and put n chords in it

[sleek thicket]

Ah, nice

I think those are homotopy equivalent

[dim meadow]

oh cool

[sleek thicket]

If you put the chords in a certain way at least

[dim meadow]

and then you make a line of the circles with n chords in it

[sleek thicket]

Every chorded circle is homotopy equivalent to n circles touching at a point, for some n, which might be bigger than the number of chords

[dim meadow]

and have the circles get smaller

and do the trick with bijecting onto the binary expansions

[sleek thicket]

Oh gosh I might not be able to visualize this

It sounds good though

[dim meadow]

it's not too hard

[sleek thicket]

I think I get it

[dim meadow]

I'll draw a picture

if you need

[sleek thicket]

Please do, I thought I did but then realized I didn't

Again

Oh I think I have an idea

Put countably many touching circles in [0,1]

Getting smaller

Take a binary sequence

[dim meadow]

yeah

[sleek thicket]

Put a point in the middle of the nth circle if there's a 1 in nth spot

[dim meadow]

same idea

[sleek thicket]

Are those always not homeomorphic? I think so

I believe something like this works

Cool!

[dim meadow]

Here's my shitty picture

[sleek thicket]

Yee, I got it

[dim meadow]

But imagine there is space between the n and n+1st

[sleek thicket]

Okay so I don't think my example works

Because you can move around the points

They aren't ambiently homeomorphic

But if you have infinitely many, you can pull them all out into the plane

[dim meadow]

Yeah

Can you understand mine?

[sleek thicket]

Can you give it again?

[dim meadow]

I'm terrible at drawing pictures lol

So you have C_m, a circle with m chords in it

[sleek thicket]

Are the chords intersecting?

It shouldn't really matter, I just want to check

[dim meadow]

I don't mean chords

Actually

Look at my picture

Like this

[sleek thicket]

Ah sure

I think that's the same as chords though

[dim meadow]

Probably

[sleek thicket]

Just un-bend them and move the endpoints

Anyways

The endpoints of the chords/arcs line up

[dim meadow]

Yeah

So you put a bunch of them in a row

With a small amount of space seperating them

So that the infinite amount of them are contained in the unit box

[sleek thicket]

Yup

[dim meadow]

(you need to add in a single point to make them compact, but whatever)

[sleek thicket]

The nth one having n chords?

[dim meadow]

Yeah

And then you choose a subspace of that

Where a particular binary expansion tells you whether the nth one is there or not

I think that works

[sleek thicket]

Ah, I think I see it

I believe it now

[dim meadow]

Can you draw a picture?

[sleek thicket]

Because you can tell the circles apart

[dim meadow]

Yes

Exactly

[sleek thicket]

Extremely believable

Cool!

[dim meadow]

But it's not a disjoint union

[sleek thicket]

No I suck at drawing lol

[dim meadow]

Because of that extra point you are adding in at infinity

Which is nice

[sleek thicket]

Yee

Next level of the question would be: can you do it on the line

But I need to get back to other things

Ty for the fun topology problem

[dim meadow]

Yeah, ttyl

[floral gust]

Are these two proofs correct?

[gritty widget]

Hi, I just started my topology course and I'm having troubles in proving that three topologies are the same generated by three metrics: 1st one being euclidean, 2nd one $d_s\left(a,b\right) = \sum_{i=1}^{n}|a_i-b_i|$ and third one: $ d_m\left(a,b\right) = max_i |a_i-b_i|$. The book says it's because three inequalities, which I don't understand why make the topologies equal: $d_e \leq \sqrt{n} d_m, d_m \leq d_s, d_s \leq \sqrt{n} d_e$. Could someone help me understand and write a proof sketch why those are equal?

[gentle osprey]

Godel:

[floral gust]

What is the criterion by which we understand that two topologies are equal?

[gritty widget]

Union of all open sets made by those metrics are the same?

[bitter yoke]

More that the open sets are exactly the same

[gritty widget]

I suppose we need to show inclusions like every ball d_s is in d_m and vice versa?

[bitter yoke]

If you know what finer/coarser topology means

I think that the inequalities show that d_e is coarser than d_m and d_m is coarser than d_s and d_s is coarser than d_e

So they must all be the exact same

[gentle osprey]

QuickMaffs:

[chrome dew]

I didn't think equivalent metrics meant they generated the exact same open sets, I thought it just meant they gave the same cauchy sequences

maybe I'm conflating open sets with open balls actually, since same radius open balls with different metrics are different open sets but doesn't necessarily mean they have to coincide in that sense I guess

[floral gust]

Open balls are the basis elements in a metric topology. They generate the metric topology. Open sets are elements in the topology. (taking the terminology from Munkres)

[chrome dew]

gotcha, yeah that makes sense

[gritty widget]

Not sure if that's what you meant, but $\lim_{n \to \infty} p_n = p $ is equivalent to any ball B centered at p contains all points $p_1, p_2, \dots $ besides countable many

[gentle osprey]

Godel:

[dim meadow]

@chrome dew (bilipshitz) equivalence implies homeomorphism trivially

[midnight jewel]

I didn't think equivalent metrics meant they generated the exact same open sets, I thought it just meant they gave the same cauchy sequences
equivalent metrics also generate the same topology, I’m pretty sure. but if you have two metrics that generate the same topology that doesn’t necessarily mean they are equivalent (e.g. you can equip ℝ with a bounded metric that generates the same topology as the standard one).

[nimble jolt]

Terminology varies. Sometimes authors use "strongly/weakly equivalent" to distinguish between homeomorphisms and bilipschitz equivalences.

and sometimes they just use "equivalent" to mean one of these, and they are consistent with that choice.

[midnight jewel]

not quite sure which channel to put this but since I (might) need it for diffgeo I’ll put it here
let’s say I have a closed injective curve in ℝ². by JCT I know that it has a well-defined interior (a bounded open set whose boundary is the curve). is the following claim true:
if for any two points on the curve, the line connecting them lies in the closure of the interior, then the interior is convex

I know the converse is true cause the closure of a convex set is convex

[west spindle]

should be true i think

pick any two points in the interior and connect them with a straight line

[rugged swan]

No, take ([0,1]^2\{(0,a), 0<a<1})\{(1,a), 0<a<1} in [0,1]^2

[midnight jewel]

hm that doesn’t have a closed curve as its boundary though

or, well,

it doesn’t fit into the description of what I have I’m failing to put it in words rn

plus it does actually have convex interior? it just doesn’t satisfy the assumptions

[rugged swan]

The interior in [0,1]^2 is itself

And isn't convex

[midnight jewel]

in ℝ²

and also where is my closed injective curve?

[rugged swan]

Ok

Idk, Idk what's tha interior of a curve

[midnight jewel]

I defined it

[rugged swan]

Oh you mean smth like jordan th ?

[midnight jewel]

JCT is the jordan curve theorem

[rugged swan]

Ok

[midnight jewel]

if you have a closed injective curve c in ℝ², then ℝ²\im(c) has a unique bounded connected component, which I termed the interior of the curve

anyway I’m not even sure if I need this

but I think what ann said is correct

you can take two points in the interior, take the line passing through them, and extend the lines in both directions until they hit the boundary. then the two points have a straight line connecting them which lies in the set+boundary, and since straight lines are uniquely defined by two points this is the same line as before

[rugged swan]

In R^2, I think it's true for bounded sets

[midnight jewel]

which was part of the condition

[patent escarp]

How could I prove that two continuous functions f,g; S^2 -> S^1 x S^1 are always homotopic?

nvm, finally did it

just need to show that any continuous function f: S^2 -> S^1 x S^1 is homotopic to a constant function

[dim meadow]

You have to use generalized lifting lemma

Although you probably know that

[floral gust]

For a locally finite cover, is it okay if there exists a such that all of its neighbourhoods have a non-empty intersection with no elements in the cover? This would not violate the locally finiteness of the cover right? Only when all neighbourhoods of x have a non-empty intersection with infinite number of elements in the cover?

[digital peak]

that violates the "cover" condition

because then a is not in the union of the open sets

[floral gust]

Thank you!

Does this seem correct? I am not particularly sure about how I arrived at the numbering

[midnight jewel]

you can’t start with “for each integer n”

because that assumes countability of U

but otherwise the argument works out

[floral gust]

How about saying. Take U. Find B_n. Index U to be n. The indexing is unique?

[midnight jewel]

yes

that’s how you should do it

deleted my wrong/confused messages

[floral gust]

Like this right?

[midnight jewel]

the indexing isn’t unique, actually. you can have multiple basis elements in the same U

however, the indexing is well-defined (in the sense that different sets will never get the same index)

[floral gust]

Oh yes true true. I should have proved the converse then. n = m \implies U_n = U_m

[gentle osprey]

QuickMaffs:

[floral gust]

Does the reasoning over here make sense? Particularly, I am confused about the injection I have built and whether the use of min lets me make an injection.

(The idea is to partition R^2 in 1 by 1 squares )

[burnt mirage]

I'm becoming aware that (pre)sheaves are a kind of natural pair to manifold structures, in the sense that "attach an algebraic structure to open sets" is similar in spirit to "attach homeomorphisms to R^n to open sets". But why do algebraic geometers care so much about the roots of polynomials? Is it just historical, or is there some deep geometric (or algebraic/categorical) reason why the roots of polynomials are things we should care about?

[midnight jewel]

I’m trying to understand what this question is exactly asking from me. In particular, I don’t understand what the function from $H_n(X; \Z/p\Z) \to \ker{\dots}$ is supposed to do. In particular, isn’t the function inside the ${ }$ injective?

[gentle osprey]

Sascha Baer:

[urban anvil]

@burnt mirage varieties are just the historical setting to work on, but in general schemes of finite type over a nice ring (which is what polynomials are) are nice objects to work with

I don't know if you think that's restrictive, but it's similar to asking that manifolds have charts onto R^n as opposed to R^infty

let me maybe be more explicit, I don't know how comfortable you are with these things.

assume our scheme is affine, Spec R -> Spec k

[burnt mirage]

I have no clue what a scheme is and wiki just makes me surer of that, but I'm trying to figure out the intuition here

[urban anvil]

that is, R is a k-algebra

finite type is just asking that R is finitely generated as a k-algebra

that is, that it's the quotient of the polynomial ring k[x1,...,xn]

the objects of study in schemes are algebras

its what you get when you map a ring to another ring

[burnt mirage]

Quotient of the polynomial ring in the same way as a clifford algebra is the quotient of a free algebra? modding out all but finitely many of the higher-degree elements

[urban anvil]

you don't need a further finiteness restriction like that

just k[x1, ..., xn]/I for some ideal I

by the isomorphism theorem

[burnt mirage]

you just go to town on the arbitrarily high-degree elements by subtracting elements of the ideal

[urban anvil]

no need

degree is fine

it's fine to be infinite dimensional as a k-module

the point is guess is that scheme maps are morally ring maps k->R, that is, rings R with the structure of k-algebras, and asking them to be finitely generated algebras is just asking them to be polynomial rings k[x1,..., xn]/I

[burnt mirage]

I think I'll need a bit [a lot] more time before I get comfortable with that idea, but one possible idea I had for the motivation behind varieties was that most geometric objects could be expressed as varieties

a circle is x^2-y^2-1 = 0, ditto for higher-dimensional spheres; you can take finite unions and arbitrary intersections, allowing you to build e.g. squares and cubes and whatnot, right?

[urban anvil]

geometric object usually means of finite type, sure

[burnt mirage]

and in a sense the theory of varieties studies how this construction attaches algebraic meaning to geometric ideas, and what can be done with these constructions

[urban anvil]

it's a natural condition

[burnt mirage]

I haven't been weaned off of finite-dimensional things yet ;~;

but is this a proper motivation for studying varieties? studying geometric objects like circles insofar as they're algebraic sets generated by polynomials

[urban anvil]

sure

they arent as restrictive as it seems

see GAGA

so that complex analytic spaces are actually really close to being complex varieties

[burnt mirage]

ok, this is great

my initial mental image of a variety was some set of points or a curve in C, which left me really confused as to why people liked doing this stuff...

[urban anvil]

C is dimension 1 as a complex variety

so smaller (closed) varieties are just finite sets of points, though you do have open sets

a better image is complex analytic spaces in C^n

or better CP^n

but even studying analytic open subsets inside C is quite rich

maybe you haven't seen much of riemann surfaces yet

[burnt mirage]

only riemannian manifolds

[urban anvil]

I guess the takeaway is that polynomials seem boring, but observe that vector spaces being geometrically boring doesn't stop differential geometry from existing

[honest narwhal]

@midnight jewel H_{n-1}(X) may have p-torsion, so your thing may have a kernel. They aren't telling you what the function is but the point of the problem is to find/prove existence of functions which make it work

[burnt mirage]

riemann surfaces are just the most general spaces in which you can do (single-variable) complex analysis?

[urban anvil]

well they're dimension 1

but yeah

it's general single variable complex analysis

[honest narwhal]

And yeah RS sound super dank

[burnt mirage]

yeah, i'm pretty interested in learning about this stuff now!

thanks @urban anvil !

[honest narwhal]

At one point in time I was considering going through Forster but never got around to it

Kinda still wanna do it though

Though atm I've got a lot lmao

[midnight jewel]

I tried to take a course on RS but it was above my level and I had enough other things to do so I dropped it

[midnight jewel]

also: today we learned about the christ awful symbols

the name is fitting, I agree

[dim meadow]

@midnight jewel for the problem you posted earlier, use universal coefficients for homology, then a long exact sequence from the Tor functor

And that does it

[midnight jewel]

I don’t know anything of what you just said

[dim meadow]

Do you know the universal coefficient theorem?

[midnight jewel]

nope

we know basically nothing about this

[dim meadow]

And you don't know about tor

Welp

[midnight jewel]

I only barely understand the notation

we pretty much only just started on this stuff

[dim meadow]

🤷

[midnight jewel]

he introduced exact sequences last week

[dim meadow]

I'm not sure how else to do this

[midnight jewel]

well I was gonna brute-force my way through explicitly finding the maps

[dim meadow]

Lol

That's pretty hard

[midnight jewel]

but the objects are fucking hard to understand

[dim meadow]

Good luck

[midnight jewel]

I mean the first map is pretty clear

[dim meadow]

The last thing is really Tor(H_{n-1}, Z/(p))

[midnight jewel]

I’m gonna wait till after next class

maybe we’re learning about these things now and the exercise was a bit premature

[dim meadow]

Hopefully

This is definitely a universal coefficients problem

I think this is the intended solution

[midnight jewel]

lemme check if the theorem came up in bredon before that question

we’re technically following that book but I’m kinda ignoring it

I don’t like it that much

[dim meadow]

[midnight jewel]

so we had the statement of this theorem here, but not the proof yet

and a bit after that is this, right before the exercise

[dim meadow]

That's not universal coefficients

[midnight jewel]

that last bit looks kinda similar

yea in that case it’s not here because that’s all there is in the “homological algebra” chapter before this exercise

[dim meadow]

Oh it is

In the second pic

[midnight jewel]

so is this in some way supposed to be obvious now or what?

[dim meadow]

Maybe not

[midnight jewel]

I can barely understand what it even says

[dim meadow]

One sec

I'll post a pic of universal coefficients

And I'll post the Tor stuff

[midnight jewel]

I swear to god this is all so much above me

[dim meadow]

And in Tor(A, B), if A is free abelian then Tor(A, B) is 0

[midnight jewel]

I guess one way to learn to swim is to get dropped into the ocean and be told there’s an island 50km to the west

except they forgot to tell me about the island

[dim meadow]

Notice $Z \otimes H_{n-1}$ is $H_{n-1}$

[midnight jewel]

yea I don’t really know anything about the tensor product either

[gentle osprey]

Liquid:

[midnight jewel]

but he just said don’t worry and just treat it as “the coefficients come from there instead”

[dim meadow]

And the induced map will be exactly multiplication by p

[midnight jewel]

we did so little in algebra I feel like

[dim meadow]

Oof I fucked up actually

[midnight jewel]

dw I didn’t even notice

[dim meadow]

I was computing Tor(Z/(p), H_{n-1})

[midnight jewel]

I’m not really following anyway

[dim meadow]

This is more for me tbh

[midnight jewel]

to be quite honest

[midnight jewel]

(also, the non-zero integer n is a prime p in the book)

[dim meadow]

Okay this actually isn't too bad @midnight jewel

I was wrong

You basically consider the chain complex C_n(X)

And take the tensor with the exact sequence

[midnight jewel]

the one from the top?

[dim meadow]

Yeah

And then you use the snake lemma

[midnight jewel]

…we haven’t had the snake lemma yet either we’ve had nothing

basically

[dim meadow]

You have

[midnight jewel]

not by name

is it the thing from the first image

[dim meadow]

The connecting homomorphism

Is the snake lemma

[midnight jewel]

aight

[dim meadow]

But yeah, this gives you a short exact sequence of chain complexes

[midnight jewel]

and C_n(X⊗ℤ) = C_n(X)?

[dim meadow]

Lmao

Tensoring is outside the parentheses

[midnight jewel]

look I don’t know what the tensor product actually does

[dim meadow]

Look it up lol

[midnight jewel]

I don’t understand it

[dim meadow]

So here's the construction

[midnight jewel]

the notation is all sorts of confusing me

[dim meadow]

You take the free abelian group on A and B

And quotient out by stuff

One sec

https://groupprops.subwiki.org/wiki/Tensor_product_of_abelian_groups

Here

[midnight jewel]

we were just said to treat Δ_n(X)⊗G as having coefficients in G instead of ℤ

and that was it

[dim meadow]

Sure

That's fair

And if you take coefficients in Z

Well that's what you're doing already

So it's the same group

You should learn this though

You should definitely know this much if you're learning Algebraic Topology

[midnight jewel]

yea it wasn’t prerequisite

tensor product is covered in the commutative algebra class

which is parallel and equally optional

and I’m not taking it

[dim meadow]

That's weird

It should be covered in a abstract Algebra class

Do you know what a module is?

[midnight jewel]

barely. we covered them but with basically no detail

[dim meadow]

Oof

[midnight jewel]

like we did the definitions, and then the theorem about finitely generated modules over PIDs and how to decompose them

and that was it

spent about … 4 hours on the topic?

[dim meadow]

Eh

[midnight jewel]

right at the end of algebra 1

[dim meadow]

That's not so good

[midnight jewel]

and algebra 2 was exclusively about galois theory

yea I passed algebra with perfect marks but I don’t feel like I learned anything there

the prof was a bit peculiar

I just checked a few years back and yea basically modules aren’t really part of that course

weren’t with that teacher either

and I know that prof, he’s good

so yea basically modules are covered in commutative algebra, which is explicitly not a prereq for algtop

they’re covered at the same time
and com algebra also only requires a basic course in ring theory as its prerqs

for alg top he writes:

You should know the basics of point-set topology.

Useful to have (though not absolutely necessary) basic knowledge of the fundamental group and covering spaces (at the level covered in the course "topology").

Some knowledge of differential geometry and differential topology is useful but not strictly necessary.

Some (elementary) group theory and algebra will also be needed.

[dim meadow]

That's gross

Tbh I learned all the alg top I know on my own

So probably us undergrad classes are just as bad

[midnight jewel]

I mean it’s a one year course I assume they’re gonna be able to squeeze in some catchup stuff

[dim meadow]

Hopefully

Do you know about tensor products of vector spaces?

[midnight jewel]

yea so about that
they were supposed to be part of the linalg course but we ran out of time and had to skip the topic, but it was only gonna be a short intro at the very end anyway.
I watched the lectures about them from the previous year on my own, but I didn’t do any of the exercises so my understanding is pretty rusty

but at least in the finite dim case they seemed pretty straightforward tbh

[dim meadow]

Yeah, vector spaces are the easy case

Cause you're just dealing with free shit

You should do problems from dummit and Foote

[sleek thicket]

I don't think the explicit construction is actually important, they just have a really useful universal property and satisfy a bunch of good rules

One useful way of thinking about tensor products is that they're a way to transfer operations on the ring to operations on the module. Like, R/I (×)_R M is iso to M/IM, and R[x] (×)_R S is iso to S[x]. Obviously this is incomplete because it doesn't cover the tensor products of two modules but still

[dim meadow]

@sleek thicket he's thinking about abelian groups, not modules

[sleek thicket]

Oh sorry lol

[dim meadow]

So I think the construction is alright if you're only doing the abelian group case

[sleek thicket]

I stand by what I said about the universal property though

Eh, the construction is still kind of gross

[dim meadow]

I kinda like free constructions tbh

[sleek thicket]

I like them because they let you get exactly what you need and nothing more

But once I've done that I don't care about the actual thing itself

[dim meadow]

That's fair

I do care about the actual thing though sometimes

[sleek thicket]

Fair

when I use tensor products I'm just applying abstract properties 90% of the time, but maybe that's because I haven't done enough with them

[dim meadow]

Apparently theres a k theory and characteristic classes class the same time as when I teach

Kill me

[sleek thicket]

Preemptively fail all the students?

[dim meadow]

Haha

I think its a seminar class

So it's even nicer

[sleek thicket]

I'm taking an AG course this year instead of an AT one (which gets to characteristic classes in the third quarter), and I really wish I was doing AT 😔

Schedule conflicts are the worst

[dim meadow]

Yeah

[dim meadow]

@sleek thicket oh shit I was wrong

[sleek thicket]

About the timing?

[dim meadow]

The k theory class is next semester

[sleek thicket]

Yesssss

[dim meadow]

🔥 🔥

[sleek thicket]

Have fun

I've been thinking about a topology thing and not making much progress

Give k^2 and k^2\{(0,0)} the zariski topology

Are they homeomorphic?

Obviously not if k is finite, but what about if k is algebraically closed? If k = the complex numbers?

They aren't isomorphic as varieties, but what if we just think about the topological structure e

And relatedly, is there a homeomorphism k^2{(0,0)< which sends the punctured coordinate axes to two lines whose closures in k^2 don't intersect?

[floral gust]

Should not the last be false? If uncountable then not second countable and hence not metrizable?

[wanton marsh]

are you saying that no uncountable space is metrizable ???

[urban anvil]

metrizable implies first countable

not second countable

[floral gust]

@wanton marsh Uncountable coupled with discrete topology

[wanton marsh]

i'm not very familiar with "second countable" and it looked like you were saying uncountable implies not second countable which then implies not metrizable

so it's the second implication that's wrong ?

[floral gust]

Oh yeah sorry I confused two notions. Lemme recorrect

[midnight jewel]

you can give any set with the discrete topology the discrete metric d(x,y) = 1 for x≠y

regardless of cardinality

[floral gust]

Oh nvm yeah true. I misinterpreted the claim If finite then metrizable <=> discrete as If metrizable then finite <=> discrete

[mortal shell]

Hi I do not quite understand the definition of partial derivates on manifolds

Why is the inverse of the coordinate mapping inside the derivative and not outside?

Like why is it D(f \circ x^{-1})(x(p)) and not D(f)(x^{-1} \circ x(p))

Nevermind figured it out was being silly

[fleet trench]

i'm trying to prove that given a Lie group $G$ with a bi-invariant metric, the geodesics starting at $e$ are the one-parameter subgroups. my plan is to show that one-param subgroups are geodesics so uniqueness gives us the result. i need to show that for left-invariant vector fields $X$ and $Y$, $\langle X, \nabla_YY\rangle = 0$, and i wanted to get at it by the definition of the levi-civita connection
\begin{equation*}
\begin{split}
2\langle X, \nabla_Z Y\rangle = Z\langle X, Y\rangle + Y\langle X, Z\rangle - X\langle Y, Z\rangle\
+\langle Z, [Y, X]\rangle + \langle Y, [Z, X]\rangle - \langle X, [Y, Z]\rangle
\end{split}
\end{equation*}
since if you set $Z = Y$ you get the $\langle X, \nabla_YY\rangle$ on the left. im not sure how to simplify the right though; i assume i need to use the left-invariance of the vector fields at some point, since without making assumptions about the vector fields i just get that $\langle X, \nabla_YY\rangle = \langle X, \nabla_YY\rangle$ which isn't helpful. do any of yall have any insight that could point me
in the right direction?

[civic kraken]

https://i.imgur.com/Bs6VDA0.png yo wtf is even going on here

well i got 3 lmao

i know the n(n+1)/2 is somehow encoding information about "we have 1 vector, then 2, then..." and terminates at n vectors

[fleet trench]

regarding 4, you can treat an element of GL(n) as a set of n basis vectors for R^n (since the elements have to be invertible). by changing them into an orthongal basis, you get an orthogonal matrix back, and i guess the R^n(n+1)/2 part has to do with the way you get to that matrix

[civic kraken]

yea that's excatly what i'm thinking

the O(n) part is clear

the idea is very similar to 3

[fleet trench]

yea

[midnight jewel]

is $\frac{\mathrm{D}}{\mathrm{d}t}$ standard notation for the covariant derivative casue this seems exceptionally ugly

[gentle osprey]

Sascha Baer:

[fleet trench]

yeah it's very standard

you get used to it haha

also if anyone is wondering, i solved my problem; you can use the left-invariance of the metric and vector fields to show that the inner product of left-invariant vector fields is constant. then differentiating along the vector fields gives you zero, so the first three terms in that koszul formula go to zero

[gritty widget]

Is it true that if d(x,y) is discrete metric on X and d' is an equivalent metric to d then there exists a>0 such that d'(x,y) >= a for all x=/=y in X?

should be, right?

but don't know to justify that

hmm, also I'm not about one thing which might be slowing me down: if we take any x, we can always find a ball such that its open, right? Not sure if my wording is correct, but if we take B(x,a), where a<1 then we have an open set?

Okay, so if d' is equivalent to d then md(x,y) <= d'(x,y) <= Md'(x,y). I had to show that d'(x,y) >= m for some m>0, so it's pretty much definition? Cause for x=/=y md(x,y) is just m?

Thanks, although it feels like cheating since this definition wasn't defined on my topology course, only in the analysis one haha

yup

Oh, also, there was this question that asked whether in Q \subset R every open set is closed. I'm not sure if my reasoning is correct - I think only empty set and Q are open so they also need to be closed. (We can't have a ball in any point in this space because its in R) Is that right?

[sleek thicket]

There are open sets other than those two

And in fact there are other clopen sets

Look more carefully at the definition of the subspace topology

[gritty widget]

I get how every point in closed, cause it's a point, and after your answer I feel like every point also might be open, but I have completely no clue how can that be possible.

[sleek thicket]

Try and prove or disprove it

What would it mean if {0} were open in Q?

[gritty widget]

it means there exists r such that B(0,r) \subset {0}

?

Im using definition taht $U \subset X$ is open if $\forall x \in U \exists r>0 B\left(x,r\right) \subset U$

[gentle osprey]

Godel:

[gritty widget]

in this case U = {0}, so there is only one element x, being 0

[sleek thicket]

Right

[gritty widget]

but no matter how small r is there will also be another rational

[sleek thicket]

Yup

Also, that definition only works if you have a metric. Do you know the definition of the subspace topology? If your class hasn't gotten to that nvm

[gritty widget]

Yes we have

I think

[sleek thicket]

So if we think of Q as a subspace of R, what would it mean for {0} to be open?

Your proof above does work, I just think this is a better way to go about it

[gritty widget]

it means that its either in the intersection of finite open sets or in the sum of any open sets?

[sleek thicket]

Q is a subspace of R, right?

When is a subset of a subspace open in that subspace?

[gritty widget]

if its complement is clsoed?

idk

[sleek thicket]

If X is a subspace of Y, and U is a subset of X, then U is open in X if there is an open set V of Y such that U = X intersect V

This is the definition of the subspace topology

Anyways, we proved that point sets aren't open in Q, but there could still be more complicated sets which are closed and open

Can you think of any possible examples?

[gritty widget]

Wait... we proved that point sets aren't open in Q?

you mean that arguemnt that you cant take a ball

because there is alwasya a rational near

[sleek thicket]

Exactly

[gritty widget]

I thought that if a set is closed and open then it implies its open

and implies its closed

[sleek thicket]

That's true

So point sets aren't both closed and open in Q

Because they aren't open

[gritty widget]

I'm thinking about a set in Q like (0,1]

its closed beacuse of 1

yeah should also be open near 0

[sleek thicket]

That's not really how open and closed sets work

[gritty widget]

if that even makes sense lol

yeah

[sleek thicket]

It's the corresponding set open in R?

*is

[gritty widget]

yes

[sleek thicket]

Nope

[gritty widget]

but only elements of Q

[sleek thicket]

It doesn't contain a ball around 1

[gritty widget]

nvm, read that wrong

Yes, that's right

(0,1) is open

[sleek thicket]

(0, 1] is a a good example of a set which is neither closed nor open

But we're looking for the opposite of that

[gritty widget]

any (a,b) should be open I think

[sleek thicket]

I agree

[gritty widget]

but how is that closed?

[sleek thicket]

How is it indeed

What do you mean by (a, b)?

That's not a subset of Q, right?

Really we should be looking at (a, b) intersect Q

[gritty widget]

Yes, my bad

[sleek thicket]

The set { r in Q : a < r < b }

You're good, I understood what you meant

[gritty widget]

wait

nvm I thought I got it

[sleek thicket]

Np

Let's try and find some examples of closed sets

(in Q)

[gritty widget]

[0,1] intersect Q

[sleek thicket]

Yup

And more generally?

[gritty widget]

yeah, any [a,b] for any a,b in Q

[sleek thicket]

Ah, but we can be even more general

[gritty widget]

and whole spcae

[sleek thicket]

Why are we restricting a and b to Q?

[gritty widget]

oh yeah, it can also be irrational

No, we can't - we can't take a ball in center a if its irrational, right?

the center has to be in Q

[sleek thicket]

That's true

But all we need is that the complement of [a, b] intersect Q is open in Q

And that means that for any rational x outside of [a, b], there is a ball centered at x contained in the complement of [a, b] intersect Q

Is that true?

[gritty widget]

yes

[sleek thicket]

So [a, b] intersect Q is closed in Q, for any a, b in R

And (a, b) intersect Q is open in Q, for any a, b in R

[gritty widget]

And you said there ARE closed open sets we haven't mentioned yet?

[sleek thicket]

There are!

Well I didn't say we haven't mentioned them yet

Just that we didn't notice they were both closed and open

Think about how we could use the thing I posted above about the intervals

[gritty widget]

Ohh I think I get it

[a,b] intersect Q if a,b irrational?

[sleek thicket]

Yes!

Can you prove it?

[gritty widget]

handwavy xD

we put a ball in any x in Q in interval above and we notice theres always a radius such taht the ball is still in the interval because there are still continuum rationals between x and a or b. And also complement of [a,b] intersection Q is open so the interval itself is closed

I gotta say, you helped me understand some concepts while doing that, thank you so much. So far I don't really get topology although I had only 2 lectures, it's for sure the hardest course I've taken so far.

[sleek thicket]

A little bit cleaner: [a, b] intersect Q = (a, b) intersect Q

You're welcome!

[gritty widget]

In this problem I have to determine if the set $\mathbb{R}^2$ without $\left{ \left(2^{-n}, 1\left) : n \in \mathbb{N}^*\right}$

[gentle osprey]

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

[gritty widget]

With this metric:

And I thought about one arguemtn for which I think it isnt open, I'll send a pic

wait, nvm, my thing doesn't work

[midnight jewel]

you haven’t actually said what that set is supposed to be

what do you want to show?

[gritty widget]

that R^2 without (2^{-n}, 1) for n =1,2,... is/isn't open with this metric

how do I type '' in latex?

\

[midnight jewel]

you can just use dollar signs to invoke math mode

if that’s the question

like this: $x^2 + 1 = 0$

[gritty widget]

hmm?

yeah

[midnight jewel]

I may have misunderstood the question

[gritty widget]

but my question is if that set is open or isn't

with the metric I showed above

[midnight jewel]

I meant the “how do I type in latex” one

[gritty widget]

oh

yeah I got it, but I remember there was a command for \ in math mode

doesnt matter

[midnight jewel]

what’s $\N^*$?

[gentle osprey]

Sascha Baer:

[midnight jewel]

oh, it’s \setminus

[gritty widget]

N without 0

yes!

[midnight jewel]

and like… \textbackslash in text mode I think

anyway lemme take a look

$d_e$ is the euclidean metric?

[gentle osprey]

Sascha Baer:

[gritty widget]

yes

[midnight jewel]

aight so that’s the SCNF/paris metric then

[gritty widget]

so if they lay on different lines going through (0,0) we add euclidean distances form each of the points

[midnight jewel]

do you know how open balls look in this metric?

[gritty widget]

its like lines ending with 'normal' balls?

if the radius is biugger than dthe distance to (0,0)

[midnight jewel]

if I’m understanding you correctly, not quite
so for example let’s say $x = (2,0)$, which points have distance 1 from it?

[gentle osprey]

Sascha Baer:

[gritty widget]

its just going to be from (1,0) to (3,0)

[midnight jewel]

yea

[gritty widget]

if the radius was, lets say, 4

i'll draw

[midnight jewel]

it would be an euclidean ball of radius 3 around the origin, and the line from (1,0) to (5,0)

I’ll spare you the drawing

[gritty widget]

nvm hope you didnt see that

[midnight jewel]

don’t worry I did

[gritty widget]

but I get it, issa ball and a line to the right

[midnight jewel]

yea okay, so now take a point in your set

[gritty widget]

I guess it breaks for (1,0) right?

no wait, swrong metric

seems like it is open

[midnight jewel]

it is yea. the interesting points are those which are collinear with one of the (1,2^-n)

for all others you can just take like, a ball of radius (so that it reaches halfway to the origin)

or sth like that

and then you can argue about those that are collinear

and you should be able to give a description of an open set surrounding each point that way

[gritty widget]

yeah co if they are colinearm only one point that was excluded from our set R^2 will be on the line, so we can calculate the distance to it and tkae that as raidus right?

[midnight jewel]

yea

or half that, but it doesn’t particularly matter

I like giving myself a bit of “room”

[gritty widget]

lol

[midnight jewel]

even if it’s not necessary

[gritty widget]

ok thanks for the help:)

[midnight jewel]

np

[midnight jewel]

@dim meadow I’ve been able to solve the exercise btw
I saw through the matrix after we did the proof of the snake lemma in class (we had seen the statement but I didn’t really grasp it before seing the proof)

[dim meadow]

Nice

Let me recall the problem

[midnight jewel]

it’s kinda interesting, I think these (algtopo and diffgeo) are the first subjects I’ve had that I both

enjoy and struggle with

it was usually one or the other

struggle with ⇔ not interested in it

but hey, I’d be in the wrong degree if I didn’t want a challenge eh

[dim meadow]

That's true

Did you tensor in the end and take the induced long exact sequence?

And mod out by the kernel for the first one, and restrict to the image for the last one?

[midnight jewel]

yea

I didn’t really understand what you meant with the tensoring when you said it but it made sense once I realized what I actually needed to apply the snake lemma

I gotta work through that proof again tho

gotta get used to diagram chasing

[signal venture]

How would I approach this: "show that any nonempty open subset of an irreducible topological space is dense and irreducible"?

Say the space is X and the open subset is Y. A proof for irreducibility using the definition would suppose Y is written as a union of proper subsets Y_1 and Y_2, each of which is closed in Y

Being closed in Y means each $Y_i$ is of the form $Y \cap C_i$ for some subsets $C_i$ closed in X

[gentle osprey]

Auvera:

[signal venture]

So I get $Y= Y_1 \cup Y_2 = (Y \cap C_1) \cup (Y \cap C_2) = Y \cap (C_1 \cup C_2)$, which says Y is contained in $C_1 \cup C_2$

[gentle osprey]

Auvera:

[signal venture]

But now I'm not sure what to do

Not sure how to use irreducibility of X here

[urban anvil]

if your open set wasnt dense, then the closure isn't everything, so closure plus complement is a finite union

so set isnt irreducible

[tawdry knoll]

How would one go about proving that the closure of a topological group subgroup is itself a subgroup?

[signal venture]

@urban anvil that makes sense. How do I prove irreducibility though?

[sleek thicket]

@signal venture what's the closure of the union of two sets?

[signal venture]

The union is already closed

[sleek thicket]

Oh lol that's a better way to see it

Actually I'm being dumb

Ignore me

[signal venture]

All good 😂

[sleek thicket]

So suppose we have U = Z_1 union... union Z_n

All closed in U

Or just U = Z union Z'

What do we know about Z and Z' as subsets of X?

I guess you already said this above

We get closed subsets C and C' of X such that Z = C intersect U and Z' = C' intersect U

Right?

[signal venture]

Yup

[sleek thicket]

So then U = (C intersect U) union (C' intersect U)

Yeah?

What can we do with that?

[signal venture]

Well I factored out U but I'm not sure if there's an alternative

[sleek thicket]

That's what I was getting at

So U = U intersect (C union C')

What's the relationship between U and C union C'?

[signal venture]

Lol U is contained in the union. I did all this up there in my attempt at the problem 😂

[sleek thicket]

Omg I'm sorry

I didn't see that

Well you were almost there

Is the bright side

What does X look like?

Draw a venn diagram bunch of blobs representing the sets on paper

You have U, and C, and C'

How do we cover the bits outside the union of C and C' by a closed set?

[signal venture]

Hmm. Take compliment of the interior of C union C'?

[sleek thicket]

How do you know that contains it?

We know X\Int(C union C') = Cl(X \ (C union C')) = Cl(X \ C intersect X \ C')

Wait maybe I'm being dumb, give me a sec

[signal venture]

Yeah I think this works

[sleek thicket]

So that will contain X \ C intersect X \ C'

Nvm

You're right

This feels wrong to me

Oh it's because I made a typo

I wrote union instead of intersect

So this might be too small

[signal venture]

You write X as (C union C') union (X\Int(C union C')). Both sets in the outer union are nonempty and closed

[sleek thicket]

How do you know that's all of X?

Can you prove it to me?

[signal venture]

If an element isn't in C union C' then it's in the compliment of it, which is contained in X\Int(C union C')

[sleek thicket]

You're saying X \ (C union C') is contained in X \ Int(C union C')?

[signal venture]

Yes

[sleek thicket]

Oh yeah, duh

This still feels wrong to me

Oh well

[signal venture]

Haha how come

[sleek thicket]

It feels too general

We forgot about U

Like, pick and two closed C and C'

You can write X = C union C' union X\Int(C union C')

I think I see the issue

They need to be proper subsets

How do you know X\Int(C union C') is a proper subset of X?

[signal venture]

Well if it were X then we get Int(C union C') is empty. Not sure what that implies

[sleek thicket]

Well, the interior is the largest open subset of C union C'

So what would that imply about open subsets of C union C'?

[signal venture]

The empty set is the only open subset

[sleek thicket]

Yup

Can we figure out why that's false?

[signal venture]

Oh wait U was contained in it, and it's open. I think it's assumed nonempty also

[sleek thicket]

Oh yeah it should be

[signal venture]

Yeah it is

[sleek thicket]

There we go

[signal venture]

And C union C' is also not X...?

[sleek thicket]

What if it is?

X is irreducible, remember?

[signal venture]

Ah right lol

I'm too tired to be doing this haha. I think that's everything though

[sleek thicket]

Well there's one more step

Since X is irreducible, either C or C' is all of X

And then C intersect U or C' intersect U is all of U

But besides that, you're done

[signal venture]

Right, that makes sense

Thanks for the help! I'll go through this again from the beginning in the morning

[sleek thicket]

Np. The core idea is pretty simple: if U is covered by closed sets of U, that lifts to a cover of U by closed sets of X, and then you add in the complement of U

[urban anvil]

U is dense already

so if you use that you get a cleaner proof

cover U by U-closed sets

their closure is a covering of X

[sleek thicket]

Oh yeah, that's why I originally said "what's the closure of the union of two sets?"

I shouldn't try and help people with math while getting dinner

[urban anvil]

hmmm