point-set-topology

[solar turtle]

Then for the first example (first uncountable ordinal with the order topology) I'm still lost

Thanks for taking the time btw, I know I lack a fair bit of knowledge here

[opaque scroll]

So the first uncountable ordinal is exactly the set of countable ordinals, ordered by inclusion.

Now for any sequence of countable ordinals, their union is again a countable ordinal. So every sequence here has a supremum.

When you have a sequence with a limsup it's not hard to make a subsequence that converges to that limsup

For example just pick whatever element is larger than all already chosen elements that are smaller than the limsup

[solar turtle]

I get the subsequence part but it's not clear to me what "first uncountable ordinal with the order topology" means exactly

[gaunt linden]

(Just because the sequence has a supremum doesn't immediately mean it also has a limsup, though. It feels like there's a missing step in this argument saying something like the sequence cannot continue decreasing because ordinals).

[opaque scroll]

An ordinal is defined to be the set of all ordinals smaller than it.

So the first uncountable ordinal is exactly the set of countable ordinals.

[opaque scroll]

Yeah, you always have infimums of ordinals and limsup is defined as the infimum of the supremums

[gaunt linden]

OK, I'll buy that.

[solar turtle]

Alright sure but what does the "with the order topology" add to that?

[gaunt linden]

The order topology is a particular standard way to construct a topology on a totally ordered set.

[solar turtle]

So it's just that we consider an order defined by inclusion?

[opaque scroll]

In the order topology a sequence converges if limsup = liminf (assuming they exist) as you may know for real numbers

[gaunt linden]

Basically you take all subsets of the form {x | a < x } and { x | x < b } to be a subbasis for a topology, where a and b are arbitrary elements of the ordered set.

[gaunt linden]

Yes, that's the standard way to compare ordinals.

[opaque scroll]

(the usual topology for the real numbers is the order topology)

[solar turtle]

Ok I think I'm understanding

And it's not compact simply because it's infinite?

[polar storm]

the collection [0,x) forms an open cover but has no finite subcover

[gaunt linden]

This is generally true about the order topology on a total order with no largest element. (Or no smallest element).

The surprising thing about first-uncountable-ordinal is that it is sequentially compact.

[gaunt linden]

Just being infinite is not enough, though. omega+1 = {0,1,2,3,...} u {omega} is infinite but compact under the order topology.

[solar turtle]

Yeah I'm actually looking at https://en.wikipedia.org/wiki/First_uncountable_ordinal#Topological_properties which I believe says the same

[gaunt linden]

I was saying omega + 1 rather than omega_1 + 1, but both examples work.

[solar turtle]

So, just making sure:
- "omega", aka the first infinite ordinal, is just a shorthand for the set {0, 1, ...}, which is N the natural numbers
- then "omega+1" is {0, 1, ..., omega}
- "omega_1" is the first uncountable ordinal, and represents some set that includes all countable ordinals but not omega_1 itself, and so it's not compact
- then "omega_1 + 1" is that same set but with omega_1 included, and so it would be compact?

Alright thank you very much

[gaunt linden]

(The "1" subscript in omega_1 comes from a more general notation where omega_n is the smallest ordinal of cardinality aleph_n. As such, plain old omega can also be written omega_0.)

[lavish mountain]

wait, omega_1 has cardinality 2^(omega_0)?

nvm this is ch isn't it

[gaunt linden]

Yes, that is the continuum hypothesis. omega_1 always has cardinality aleph_1, but aleph_1 may or may not be the cardinality of P(omega).

(If you read a formal development, the definitions generally go in the other direction: First take every infinite ordinal that is not in bijection with any earlier ordinal. Put them all in a (transfinite) sequence according to the natural order of ordinals, and give each of them the name omega_n where n indicates its position in the sequence. Finally define aleph_n to be the cardinality of omega_n.)

[light phoenix]

What could topological analysis of a data reveal? Would it prove as exceptionally efficient than the traditional math we have been using in data science? In industries like healthcare for example?

[ruby delta]

It’s not a substitute for data science

But the idea is that sometimes there are topological features of the data space that can tell you certain properties of the thing you are getting data from that would be hard to measure otherwise

For a really informal introduction to what this might look like here’s a cool video illustrating how topology shows up when you do data analysis
https://youtu.be/YGLNyHd2w10?si=pPacWIHf8V4Fnpkg

[south halo]

A conference I went to presented a result where they found a 2d manifold in 20 dimensional space, so I guess it's useful like that

[iron bolt]

TIL that a topological space X is compact in the category-theoretical sense (Hom(X,-) : Top -> Set preserving filtered colimits) not iff it is compact in the usual sense, but iff it is discrete and finite

that's unfortunate, I'm trying to understand when e.g. a map of [0,1] into the space of germs of maps X -> Y around x could be understood as a germ of a map X x [0,1] -> Y around {x} x [0,1], but apparently if that even works it needs a far more specific argument than just [0,1] being compact

[rancid umbra]

yea, this was a bit weird when i first learned of it too. Top is not the most well-behaved category...

but um

if i recall correctly, this works nicely for more algebraic categories

[opaque scroll]

I wonder what the original etymology is.

You have that the compact objects of the category of open sets in X are the compact (open) sets, but this is not the typical category one would consider compact objects.

In a triangulated category you have that X is compact if any map to a coproduct factors through a finite coproduct, which mimics reducing to a finite cover, but this definition doesn't generalize.

I don't if there is some natural homotopy category where compact objects are exactly compact spaces...

[rancid umbra]

i believe it comes from compact objects in lattices

[opaque scroll]

Makes sense

[iron bolt]

by the way, I've since learned that you can actually characterise the compact objects of Top really simply, just not in the way you'd hope

turns out a topological space is categorically compact iff it is finite and discrete

[rancid umbra]

you can do something similar for non-empty connected spaces (a non-empty space X is connected iff Top(X,-) preserves coproducts)

[opaque scroll]

Sort of surprising that discrete is necessary

[iron bolt]

yep, I wouldn't have thought so either

this is where I got that bit of information from: https://golem.ph.utexas.edu/category/2009/05/journal_club_geometric_infinit_3.html#c023790

he didn't mention finiteness, but it's not hard to check that it is both also necessary and sufficient when combined with discreteness

nlab also has this lemma as a useful special case where things actually work out nicely:

but I don't know if it's possible to also apply a similar lemma for colimits along quotient maps, because e.g. [0,1] can be written as a colimit of a sequence of totally disconnected compact Hausdorff spaces along quotient maps, and the identity map obviously does not lift

[cosmic mirage]

The closest result I know of is that in the ∞-category of spaces, the compact objects are those with finite CW models (which will be the compact CW complexes) and retracts of those (which need not be compact)

by which I mean that the mapping space Hom(X, -) now commutes with colimits over filtered simplicial sets

[zealous berry]

is this proof right lol

feels too complicated

[kind egret]

I think it can be shorter

A base of the subspace topology on U x V (seen as a subspace of X x Y) is the set of all subsets S of U x V which can be written S = p_X^{-1}(Q) \cap p_Y^{-1}(R) \cap (U x V) where p_X, p_Y are the projections, and Q, R open subsets of X and Y. But clearly p_X^{-1}(Q) \cap p_Y^{-1}(R) \cap (U x V) = p_X^{-1}(Q \cap U) \cap p_Y^{-1}(R \cap V), which is the description of the elements of the basis of the product topology on U x V.

[lavish mountain]

this generalizes to any finite product since the product-intersection distributivity works

ah minor typo i meant to write (for some A x B sseq X x Y open) but apparently forgot half of that

[zealous berry]

fk

<-- noob

this is good, ty

[rancid umbra]

yea, there is a universal property way to look at this

where you don't have to get into the nitty-gritty point-set details

[zealous berry]

can we ban cat theorists from this channel

[midnight umbra]

that was my first thought, show the product topology has the universal property of the susbspace topology

or the other way around

[rancid umbra]

you can work it out using these two properties
the universal property of the subspace topology is that for S \subseteq X, a function f : Z -> S is continuous iff i_S o f : Z -> S -> X is continuous.
the universal property of the product topology is that maps into A x B correspond to maps into each factor.

[rancid umbra]

yea, its not too bad

[zealous berry]

sorry im still like

very naive with topology and very very naive with categories

so im not particularly interested in categorifying this just yet

[midnight umbra]

i know 0 category theory

[rancid umbra]

yea, not trying to force anything on you

but um

[zealous berry]

very cool though

[rancid umbra]

idk you don't have to know category theory to use universal properties

[midnight umbra]

like if c squared pinged me rn and said "tell me the definition of a category or you're dead", im cooked

[zealous berry]

well universal properties are also not something i really know

[rancid umbra]

tell me the definition of a category or you're dead

[zealous berry]

um

something equipped with a class of objects and morphisms ?

[midnight umbra]

bro replied tot he wrong message

[rancid umbra]

they are just characterizations of familiar properties with morphisms instead of internal details

[rancid umbra]

yea. its a spicy graph

[zealous berry]

thankfully im not dead

[rancid umbra]

eff word

[zealous berry]

yeah i understand the concept of what a universal property seeks to do

just still very naive with it

[rancid umbra]

understandable

[zealous berry]

i know how the product and quotient are universal

and dats it

[zealous berry]

know is also a bad word

cat theory language is just hard to digest at times

[midnight umbra]

the universal property of the quotient space is pretty fun

[rancid umbra]

ye

[midnight umbra]

it gives a nice way to show different spaces are homeomorphic

[rancid umbra]

ik you have probably heard it a million times by this point, but it really is good to get used to thinking about things via universal characterizations.
if you want something concrete and accessible that doesn't mention category theory but still discusses universal characterizations/properties, check out Lee's ITM

he discusses universal properties of subspace, product, and quotient topologies without reference to category theory

[zealous berry]

hmm ok

[zealous berry]

doesnt that book do alg top as well?

[rancid umbra]

yes

[zealous berry]

ill add it to my list

[rancid umbra]

but yea, don't wanna keep beating a dead horse, i just like thinking about things categorically

[lavish mountain]

every universal property is like, every morphism from the diagram to somewhereland factors uniquely through the universal object

and everything on the page commutes

[rancid umbra]

universal arrows!

[zealous berry]

an object is initial/final in some accessory category

the quotient is such that maps from a set to the quotient to the image factors most optimally

if it were any other set, you'd need another arrow

[rancid umbra]

yea, so i think to rephrase, among maps which identify equivalent points, the quotient map is initial

[zealous berry]

yeah

there is a unique morphism between the quotient and every element of the image

that's true only of the quotient

[rancid umbra]

wdym by "every element of the image"?

[zealous berry]

mb i phrased it wrong

[rancid umbra]

there is a nice way to think about this. so, lets stay concrete and work with just topological spaces.
if you have a set X and an equivalence relation R on X, then R is a subset of X x X.

there are two projections p_1,p_2: R -> X x X -> X.

the quotient map q: X -> X/R is such that for any other cts map f: X -> Y satisfying f o p_1 = f o p_2 (that is, f identifies R-equivalent points), then there is a unique morphism g: X/R -> Y such that g o q = f.

this is the sense in which it is initial

[lavish mountain]

mfw colimit of parallel pair

[cosmic mirage]

0-category theory ... so preorders

[rancid umbra]

this is particularly cool because every congruence/equivalence relation in Top is the kernel pair of its quotient map

[cosmic mirage]

also adding on to the list of people who thought to prove the subspace topology on products thing using the universal properties

I almost sent a sketch earlier but decided I didn't want to be late to class lol

[cosmic mirage]

also ik this is a joke lol but worth noting how intertwined category theory and topology are historically, so good luck with that lol

[quick delta]

Tbf that’s more a #alg-top-geo-top thing than a #point-set-topology thing

[cosmic mirage]

It is but the algtop people also know pointset topology lmaoo

[quick delta]

given some homotopy theorists I know I wouldn’t be so certain /j

[cosmic mirage]

💀💀

[alpine nest]

pointless topology

[cosmic mirage]

well I don't think this is a contradiction because I'm one of those annoying homotopy theorists who says homotopy theory is not algtop 💀

[quick delta]

Correct opinion 🙃
As someone who’s much closer spiritually to low dim topology

[cosmic mirage]

I was going to say algtop people know and care about pointset topology but decided to toss out the last part since I think that's not true LOL

like the kinds of spaces in algtop are reasonable and don't have pointset problems

[stuck geyser]

Coequalizer jumpscare

[stuck geyser]

The tangent bundle is a coequalizer :3

[unreal stratus]

Would these peeps call themselves alg top peeps tho smh

[stuck geyser]

Potato has breeched the algebra channel containment barrier again

[quick delta]

mishu no know
mishu just want nibl them

[unreal stratus]

Oops

I am in the topology department at my uni

[stuck geyser]

I’m sorry to hear that hopefully you will recover soon

[cosmic mirage]

[novel acorn]

breeched

[quick delta]

Dw we will make you see the light of ggt :3

[urban zinc]

There's a topology department??

[covert nexus]

well presumably the topology research group

at very least there are a reasonable number of topologists at potato's institution

[prime elbow]

So when n = 3, is it projective plane?

[silver ridge]

n=2

RP^2 is the projective plane

[prime elbow]

Oh right

But there is a definition that the plane where each line intersects, right?

I mean in usual R^3 parallel lines doesn't meet, but we add the point 'infinity' where these all parallel lines meet

[ruby delta]

it's not a nice definition, you should think of it as just a nice property of projective space

[prime elbow]

Oh

[prime elbow]

But how is this definition implies all lines intersect

I mean any two lines intersect

[ruby delta]

you can go through the definitions and check it, but it's kind of a waste of time

[gaunt linden]

This definition of RP2 constructs it only up to homeomorphism -- it doesn't even say what "lines" are.

[opaque scroll]

A line on the sphere is a great circle. Ie. Given by a the intersection with a plane through the center.

Two planes through the center will intersect at a line through the center which then intersects the circle at antipodal points

[robust drum]

What does this mean

You could say any definition constructs only up to homeomorphism

[gaunt linden]

No, because RP2 also has a canonical projective structure.

In other words it is more than just a topological space.

[opaque scroll]

To be fair, you can inherit all the nice structure from S^2 using this construction. Even though the image only mentions the topology

[unreal stratus]

Yeah lol

oops

[unreal stratus]

What does this mean

[opaque scroll]

It should have a set of lines such that every pair of points is contained in a unique line and every pair of lines intersect in a unique point

[unreal stratus]

[opaque scroll]

I.e. you need to specify what the lines are

[zealous berry]

suppose $X = [0, 1) \cup {2}$. then $(0.5, 2]$ should be open in $X$ and refer to $(0.5, 1) \cup {2}$ right?

[zealous berry]

(with the order topology)

an interval $(a, b]_X = { x \in X : a < x \le b }$

[zealous berry]

wait what is (0, 2) here 🤔

its just (0, 1) in R right

[kind marlin]

that all seems correct

one way to think about this topology is that it's homeomorphic to [0, 1] with the order topology

so the open sets will look similar

[zealous berry]

havent gotten to homeomorphisms yet

im trying tos how the subspsace topology na dorder topology are not equal

but i am a bit scatterbrained rn cuz of other things

[kind marlin]

this is a good example yeah

[zealous berry]

oh i see

[0, b) is in the order topology

wait

hold on

[kind marlin]

a simple thing to look at is the set {2}

[zealous berry]

oh i see

{2} is open in the subspace topology

but obviously not in the order topology

wait

yeah

(1.5, 2.5) cap Y = {2}

so its open there

but you cant get {2} with the order topology

cuz 2 is the maximum, and all open sets that contain a maximum are rays

[zealous berry]

munkres needs more fun exercises.

[polar storm]

there is a fun exercise on Kuratotowski’s closure&complement problem in chapter 2.

[stuck geyser]

Algebra in my topology channel. Explode

[polar storm]

oh wdym?

[stuck geyser]

Kuratotowski’s closure axioms are algebraic :3

[polar storm]

ah

icic

[stuck geyser]

Not much fun in pointset unless you have something undiagnosed

[zealous berry]

idk i think it's fun

why does everyone think it's so boring

:(

[kind marlin]

i think its fun

[zealous berry]

da answer is finer yeah

its not strict cuz the additional elements in tau' when intersected with Y may collapse into an element of tau interssected with Y

[polar storm]

yee

[kind minnow]

I disagreed with the first half of this message, but, alas, the second half may be true

[fast dew]

creature spotted

[zealous berry]

hi gray

[polar storm]

👀

[opal coral]

oh i did this exercise today. super cool

the 14ness hinges on the fact that (X |-> closure of interior of X) is idempotent

[zealous berry]

do i need to justify more why i can move the pi inside the union

[trim spade]

a pretty basic thing about functions, I wouldn’t worry about it

[zealous berry]

yea thats what i thought

[rancid umbra]

-# left adjoints preserve colimits

[zealous berry]

category theory brainrot is real

sybau 😂 ✌️

[rancid umbra]

atp, it’s only you and holo who i pester with this non-sense

[unreal stratus]

where is this lol

[cosmic mirage]

lol yeah this was my thought too amazing

[zealous berry]

https://media.discordapp.net/attachments/743211245779419186/996996886806396938/caption.gif

[zealous berry]

i can invite you if you want ?

[unreal stratus]

ah dw lol i just wondered if it was some big server lol

[unborn hawk]

it's a private server

[warped helm]

this should be a subseteq instead of equality

er wait

its like that for intersections

[zealous berry]

yea

[zealous berry]

$A^\complement$, $A^c$, $A^C$, $\bar{A}$, $\complement A$, $\complement_U A$

[zealous berry]

which is best for complement of a set

well i guess 4th is out cuz it conflicts with closure

[queen prism]

(X \setminus A)

[gentle osprey]

nBladeoid

[polar storm]

I sometimes use A^c if the notation is dense. Such as this:

[zealous berry]

ugly n verbose

[zealous berry]

µ

gooooood i want to do measure theory already

but i cant do so many books at once

[polar storm]

I prefer using \setminus over A^c whenever possible. I personally think X \setminus A looks better than A^c.

tho I don't have a good reason for this lol

[zealous berry]

$\emptyset^C$

[zealous berry]

kinda cursed

[polar storm]

$\varnothing^c$

[gentle osprey]

Euiseok (Class of 3029)

[zealous berry]

hmm

$X - A$ or $X \setminus A$

[zealous berry]

i kinda like the -

easier to type

[polar storm]

I never use X - A cuz it can also mean a translate notation in topological vector spaces.

[gentle osprey]

Euiseok (Class of 3029)

[polar storm]

yeah, study point-set topology and multivariable real analysis before measure theory 🔥

[cosmic mirage]

paradox named russell:

i have literally never seen \complement used before lol. would definitely go with X\A or X-A or A^c (if the ambient set is very clear)

i personally use X\A most often but i think if i were more based i would do X-A

[zealous berry]

🫡

after i finish ch2 munkres ill go back to shurman

i want to finish both books by early july

well idk about the alg top part of munkres

[rugged escarp]

July? Ambitiouz

[zealous berry]

well idm if i go past it

[queen prism]

I write X \ A and read it as "X minus A"

[zealous berry]

lol noob

[delicate siren]

donuts and cups are homeomoprhic 😎

[keen cliff]

my cups don't have handles

[delicate siren]

so you cant morph them into donuts

nice

[still aspen]

oof

[small obsidian]

But they can be made into spheres. And as we all know, the hair on your cup cannot be combed.

[delicate siren]

anything closed is a sphere yes?

[small obsidian]

Anything without a "hole" can be made into a sphere of the same dimension

[delicate siren]

closed means without a hole right?

thats the poincare conjecturee

proven by that crazy guy rightg?

[nimble jolt]

closed just means compact and without boundary, these need not be spheres.

Also what is the rigorous version of the statement you are claiming kaynex?

@small obsidian

[raw nova]

Make sure you differentiate a closed manifold from a closed set though

[small obsidian]

Something something homology

[raw nova]

Just say it in terms of homotopy groups such as the first homotopy group of S^2 being trivial which establishes if something is homeomorphic to the sphere or not @small obsidian

[nimble jolt]

thats why I asked kaynex, the statement is kind of different in different dimensions

[small obsidian]

That's fair, I was thinking in lower dimensions. Which, come to think of it, isn't even true in two dimensions

[gritty widget]

Hello topology

[honest narwhal]

@delicate siren So, classification of surfaces is not too hard

The idea is that every surface is homeomorphic to either a sphere with handles or RP^2 with handles

Or, I mean every closed connected surface

The only one of those which is simply connected is S^2

Poincare says that even in 3 dimensions, a simply connected closed 3-manifold is homeomorphic to S^3

That's the hard result, and the main guy involved was Perelman. I might hesitate to say he's crazy, I know little about him, but he's... odd

I mean there was some controversy from I what I hear, supposedly Yau tried to downplay his role in the proof in favor of Hamilton (though I think he rejected the million dollar award because he felt Hamilton's work on Ricci flow/surgeries was more important). In any event some kinda mess happened, he ended up rejecting the Field's medal and left the public eye, I think even quit math

[warm rose]

Yup he turned down the Fields Medal because he said it would mean that he has to go 3days without doing Maths

[honest narwhal]

No, that's not quite true, he rejected it because he felt like the math spoke for itself and he didn't want to be idolized

[warm rose]

Hmm honorable man!

Can I trouble anyone with a question? How do I prove that the n-Torus embeds into R^(n+1)? I'm new to topology and the hint given just says to find a way to use induction

I know thats by writing T^n = S^1 x ... x S^1 and using the product of inclusion maps on S^1, gives me that T^n embeds into R^2n but how do I get the dimension it embeds into to decrease?

[delicate siren]

"Emptiness is everywhere and it can be calculated, which gives us a great opportunity. I know how to control the universe. So tell me, why should I run for a million?" Grigori Perelman

holy

ty wogo

[warm rose]

Can somebody help me with my question above?

[keen cliff]

@warm rose S¹ embeds into R²,
then S¹×R embeds into R³

then hmm

like somehow turn R into S¹ 🤔

[warm rose]

But we want n-copies S^1 x ... x S^1

[keen cliff]

ok like

assume Tⁿ embeds in R^{n+1}

then look at Tⁿ×R

using similar idea

[warm rose]

I also tried assuming T^n-1 embeds into R^n so there exists a continous, injective, open map call it f, that is homeomorphism from T^n-1 to the image of f

Now I want to consider T^(n-1) x S^1 and we need to show that there exists a new continuous, injective, open map from T^(n-1) x S^1 to R^(n+1) this will gaureente that map is a topological embedding.

the map I THINK we need is the product map something like (f x sigma) where sigma : S^1 to Reals, is the streographic projection map. But there is a problem because S^1 is not injective to R, but it is to R U {infinty}

[keen cliff]

oh I see, need to add a point

[warm rose]

So i was so close but I think this throws a wrench in the whole approach of considering T^(n-1) x S^1

Yeah but I'm not just allowed to add the point at infinty 😌

Do you know anyway if extending this to higher dimensions?

[honest apex]

🆙 | HyberCube leveled up!

[keen cliff]

woah donut surface

[warm rose]

Any way to extend this map F to higher dimensions?

[keen cliff]

probably

we can probably make a generating (n-1) torus

for example:
$$\mathbb{T}^1 = \mathbb{S}^1$$ given by $$x_1^2+x_2^2=r_1^2$$

[keen cliff]

then do

[warm rose]

Might this G^-1 map help us?

[keen cliff]

ok wait

=tex r_1=\sqrt{x_1^2+x_2^2}

[keen cliff]

the donut surface does uhh

I found a stack exchange post:

https://math.stackexchange.com/q/307476/330017

the answer has an n-donut construction

but I do not fully parse it

[warm rose]

Thank you I'll check it out!

[keen cliff]

np

[keen cliff]

if a manifold M has the property (*): every point in M has a neighborhood diffeomorphic to an open subset of Rⁿ
does this imply M is a smooth manifold?

full disclosure: I am super not familiar with manifolds & smooth manifolds. (the only ones I have worked with are Rⁿ and matrix lie groups, that's about it)
I read a few stackexchange posts saying smooth manifold implies (*)

but nothing about the converse

[nimble jolt]

For you to talk about being diffeomorphic you need to put a smooth structure on M first. Otherwise its like asking if the set X is isomorphic to the group G, is X a group?

@keen cliff

[keen cliff]

oh I see

so I guess the question doesn't really make sense haha

[nimble jolt]

Nope, that's my point 😃

[keen cliff]

ok thanks (derp)

[nimble jolt]

np

Also the statement that a smooth manifold satisfies (*) is baaasically a definitional one.

[keen cliff]

oh

Ok

[still aspen]

= 🍺

[warm rose]

Technically that cup if full, it can contain beer but the torus cant. So the torus is not homeomorphic to a fulled cup 😋

Is**

[keen cliff]

property of being able to to contain beer is not necessarily preserved under homeomorphism

[wicked spruce]

Hi, I was wondering about a non-complete metric space that does satisfy the least upper bound property

I can't seem to think of one if someone could else me out?

[honest narwhal]

Wait so you want an ordered set satisfying the sup property, and you want to put some metric structure on it which is not complete

How does the metric interplay with the ordering?

[nimble jolt]

What is your definition of "complete"?

Cauchy complete?

[wicked spruce]

yes

sorry I'm not really sure how to respond

What do you mean by metric interplay with the ordering?

[nimble jolt]

LUB property implies Cauchy completeness.

Have to go to a seminar but try and prove this yourself.

The converse is close to being true, but needs an additional property, that the field is Archimedean.

[wicked spruce]

ahh

So I would have to find a cauchy complete set that does not obey the archimedean property

so maybe something that involves things like infinitessimals

[nimble jolt]

the example that comes to the top of my head does not involve infinitesimals, but see what you can cook up

[honest narwhal]

Wait a field?

I thought it was just an ordered set which was also a metric space, so my question was how the metric was defined exactly. If it's an ordered field with the sup property then it's isomorphic to R

[wicked spruce]

no just a set with a metric that is cauchy complete but does not have the LUB property

[nimble jolt]

You need to be talking about totally ordered fields for the Q to make sense.

[honest narwhal]

What does the metric have to do with the ordering?

You can consider R as an ordered set and put a stupid metric on it which isn't complete

A priori they have nothing to do with each other

[nimble jolt]

Yeah I will explain what I am getting at when out of this seminar.

[honest narwhal]

Oh I'm talking to mmmmy here

[nimble jolt]

🐱

[honest narwhal]

I'm just wondering if maybe he has a certain construction in mind for getting a metric out of an arbitrary (totally) ordered set

If it's an ordered field then yeah, it's R. If there's something else then maybe it's more interesting

[nimble jolt]

It's not a literal metric, in that it doesnt take real values.

non-neg elts of your field

[gritty widget]

wen i was littel ßoye my momma tol me yu no ned to type gud but talk gud so people felt heart warmong round yu

[wicked spruce]

hmm i'm just confused a little bit because my HW has a question that references the sup of a subset of a complete metric space

and then asks to find a counter example to show that the property does not hold with a non complete metric space

where completeness refers to cauchy complete

[honest narwhal]

Wait can you post the question verbatim? That'll help us understand

[wicked spruce]

Yes, it is for homework so I dont want to get any answers but the question is essentially 3.21 of rudin with the added question of finding a counter example for non complete sets

question 3.21 can be found on this website

https://minds.wisconsin.edu/handle/1793/67009

[nimble jolt]

Exercise 3.21 doesn't have anything to do with LUBs, it is in an abstract metric space. Which part of the question confuses you?

@wicked spruce

[wicked spruce]

@nimble jolt since it references the diam of the set E_n, then it must have the LUB property (other wise how would you take the supremum ) right?

perhaps im confused

[nimble jolt]

diam is the sup of the set of distances between points in your set.

the thing you are taking the sup of is a collection of real numbers

not a collection of points in your metric space

[wicked spruce]

ohh thats right

wow

my bad

[nimble jolt]

no problem

lesson learned

[wicked spruce]

yeah lol

[swift shuttle]

Hey i have a comprehension problem

Let E= ]1,2[ U [3,5]

Int (E) = ]1,2[U]3,5[

And Adh(E)=[1,2]U[3,5]

Does that mean that by example 2 is at the same time in Int(E) and Adh(E) ? <@&286206848099549185>

[still aspen]

Wait a minimum of 15 minutes after posting the problem before pinging helpers.

[gritty widget]

]1,2[???

And that.

[swift shuttle]

Oups

Sorry

@gritty widget sorry ?

[gritty widget]

What is ]1,2[

[swift shuttle]

Interval in R

[gritty widget]

First time seeing it

Ah open interval?

[swift shuttle]

Yes

[gritty widget]

Ah, then wdym by example 2

[swift shuttle]

wdym ?

[gritty widget]

What is Example 2

You referred to it

[swift shuttle]

Ok

Hem

i'm working on R

And i want to know that if E = ] 1, 2 [ U [3,5] is an open

or a closed set

maybe if i tell you that i'm working with the usual topology on R would help ?

@gritty widget

[gritty widget]

Oh, E is not open then ofc

[swift shuttle]

and not closed too

But how can i prove it rigourously

[gritty widget]

Yup

So do you know the property regarding

[swift shuttle]

My teacher is using adherence and interior notion

[gritty widget]

Interiors?

[swift shuttle]

hem

"The interior of a set is the largest open set contained in the original set. "

He said that

As a proof

3 is in adh(E)\E and 1 is in E\int(E)

But i know that int(E) = ]1,2{U]3,5[ and adh(E)=[1,2]U[3,5]

So 3 is in int(E) and 3 is in adh(E), as well...

no ?

I dont really understand

wtf ?

[gritty widget]

3 is not in Int(E)

Also if you read it again, you'd know if E and Int(E) are different E is not open.

[swift shuttle]

Ho ok ok ok

Ok i understand

[gritty widget]

Yup

[swift shuttle]

then

if you have E

And you want

[gritty widget]

Similar for closed I think

[swift shuttle]

to prove that

if E is not closed

You have to find that adh(E) \E is not empty

[gritty widget]

Yup

Or more easily

[swift shuttle]

is E\adh(E) the same ?

[gritty widget]

Adh(E) is not E

[swift shuttle]

Yeah because i saw that my teacher using open ball

(dont know if it is the same vocabulary in english)

[gritty widget]

That is.. more of a neighborhood in metric space

[ivory tusk]

Rudin writes open ball

so that's fine too

[merry shale]

Anyone familiar with the fundamental group of a topogical space?

[small obsidian]

The group of paths one can draw on a space, where two paths are equivalent if there exists a homotopy between them

I'm not very strong on the concept but maybe I can answer something @merry shale

[merry shale]

I'm just looking for what it would look like in simple cases. I know it's Z on a circle

I'm having trouble imagining it for a thing with two holes

like, if I poke one hole in the complex plane (that basically becomes a circle), I get the paths that can get back to a point (mapped to 0) then those that circle n times around the hole in one way (sense? direction? my english is bad) and m times the other way (mapped to n-m )

and these are homotopic because the hole fucks me if I want to get them back to anything else

[small obsidian]

More or less yeah

[merry shale]

now for two holes I don't have any idea

[small obsidian]

I'm thinking probably Z², but give me a sec to reason it out

[merry shale]

is there some canonical/simple way to choose thos paths?

inside their homotopy class?

[small obsidian]

No wait, more than Z² because drawing over both holes is an extra path

Maybe Z³?

[merry shale]

you can also draw an eight around the holes

well not "around"

[small obsidian]

There's always homology that is a bit of a more "rigid" method

[merry shale]

what's that?

[small obsidian]

Reconstruct the space out of triangles and do the same thing, except an algebra finds the paths for you

[merry shale]

oh yeah by saying it got the same number of holes

I didn't see that result yet tho

[small obsidian]

No wait, going over both holes is a combination of going over each hole.

[merry shale]

how?

can you give me the homotopy?

(the idea, formulas would get shitty)

the way i see it you need to break the path

oh wait I started with a discontinuous path XD

seems to me like Z^3 yeah

oh yeah I see what you mean

[small obsidian]

Going over both holes is the same as going over the first, then going over the second

[merry shale]

like the eight is just two loops

there's two ways of going over both holes

[small obsidian]

So yeah, any path fits into Z²

Very easy way to show that the plane with one hole is not homeomorphic to the plane with two holes

[merry shale]

are you saying these are homotopic?

(I don't think it's that easy, the book I'm reading goes over a lot of trouble to even show that a circle is mapped to Z even tho the intuition is easy)

(but yeah you don't have to use the group to show that, just exibit that it can't be mapped to Z)

[small obsidian]

Where's your starting point? It's usually easier to assume one

[merry shale]

oh, you can take the junction in the second, and modify the first to have the top line bend over to join it

(btw have you seen the proof of d'alembert-gauss using these laces?)

[nimble jolt]

circle with two holes does not have fundamental group Z^2 or Z^3, its fundamental group is the free group on two generators.

In particular it is not abelian.

[merry shale]

oh cool

so its BIG

[nimble jolt]

If a denotes winding around one hole and b denotes winding around the other

aba^-1b^-1 is not the identity

[merry shale]

cool

[nimble jolt]

yeah its big.

I mean cardinality wise its the same as one hole, but whatever

[merry shale]

I think i saw a way to say the bigginess

using a metric on the words of a group

[nimble jolt]

yeah there is

[small obsidian]

Oop, yeah definitely not Z² if non-abelian

[merry shale]

oh i get the visualisation. basically you can knot your lace in shitty ways around the two holes that can't be unknotted

[nimble jolt]

yep

whereas around one hole you just get an integer, your winding number.

[merry shale]

is a sphere contractile?

[nimble jolt]

contractible? nope

[merry shale]

(identity homotopic to a constant function)

[nimble jolt]

I know

[merry shale]

(yeah, just to check if french contractile is english contractible)

[nimble jolt]

ah I see

[merry shale]

I don't think I visualize contractibleness right then

[nimble jolt]

😃

[merry shale]

My intuition gives me something like connected by arcs implies contractible and that's full of shit

[nimble jolt]

yeah, thinking directly in terms of homotopies and deformation retracts is more accurate imo.

can you continuously map your space into a family of subspaces which tend to a point.

sloppy wording, and vague, but I hope you get what I mean.

[merry shale]

oh yeah i do

didn't see it that way tho

that's a good way to put it

[nimble jolt]

If you punctured your sphere you could do it

just enlarge the hole, flatten out what remains of your space, it will be like a disk then

[merry shale]

like the diferential version of that would be something like explanable?

[nimble jolt]

and contractible in an obvious way.

explanable? have not encountered a differential version of this concept I don't think.

I mean you could phrase the concept in terms of smooth homotopies, and for spaces nice enough that that makes sense, I imagine it coincides with the purely topological notion.

[merry shale]

well it has not much to do with contractibleness but with the method you say. like this method would not work anymore to explane something. (explanable is "you can model it in paper")

like you can put a plane on it

[nimble jolt]

ahh I see

I thought you meant something to do with an "explanation"

[merry shale]

well with the roots of that word

[nimble jolt]

yep

But anyway, am no much of a topologist, so that deformation retract explanation is probably the best I can offer for what contractible means.

It's an easy property to detect once you have something like homology at your disposal, because homotopy equivalence preserves homology groups.

[merry shale]

i mean seeing the image of the function as a space you morph continuously seems good enough

[nimble jolt]

yep, and importantly this image is a subspace of your orginal space.

[merry shale]

yeah

cause nothing else exists in context anyway

[nimble jolt]

Yeah

Sounds like you are on top of the concept to me

[merry shale]

cool, I'm just studying this with a book so it's nice to chat about it a bit

that d'alembert-gauss proof tho

[nimble jolt]

fundamental theorem of algebra? oh yeah the most visually nice proofs of that are from AT.

I can explain them to a high school student, although heavy AT machinery is needed to make some statements rigorous.

[merry shale]

using laces and the fact that their group is Z on the sphere

yeah it's so ez to intuition

well here hs student don't even know d'alembert gauss tho

[nimble jolt]

The one I was thinking most of was:

Let P be a nonconstant poly. Then consider the image of the circle of radius r about 0 under P. For small r, this curve is near the constant coefficient of P which can be assumed nonzero, for large r, this is roughly the image of the circle of radius r under the leading term of the polynomial, which is some az^n. The winding number of P(C_r) transitions from 0 to n and so P(C_r) must intersect {0} for some r.

Not the easiest proof to make rigorous, but very visual.

[merry shale]

the one I saw was:
let P be a polynomial of degree k
let's also assume it has no roots on the sphere,
we can define a lace p of the sphere by normalizing P(z) for z on the sphere
if there are no roots inside the ball, we define an homotopy (that collapses the lace on a point) from p to a point : p is of degree 0
if there are none outside, we define another homotopy ( using t^kP(z) normalized) and show that p is of degree k

we conclude by unicity of the degree of a lace

[nimble jolt]

Ah yes I think we had that one on our AT exam actually.

Or maybe it just asked us to prove FTA, and that is the way I did it.

[merry shale]

I just started reading AT stuff today and it's already looking like a lot of cool stuff

[nimble jolt]

yeah its very powerful

am more of an analyst myself, but I did enjoy AT and use it from time to time.

[merry shale]

I'm doing measure theory in class rn, but we are still far from even defining lebesgue measure

I saw a lot of cool stuff in analysis but without measure theory it gets shitty

and I just get lost in the vocabulary when reading up on my own

[nimble jolt]

yeah you are a bit limited before you have measure theory.

[merry shale]

btw, what's dx in a lebesgue integral? (i mean it's the lebesgue measure, but why do we write it like that, is there an intuition for that?)

what would be d(x^2 or whatever) ?

[nimble jolt]

It is the measure induced by the differential form dx

which happens to be the Lebesgue measure

Have to leave to lecture, but the general chain of ideas is than on an orientable smooth manifold, if we have a top level form, we get a volume form, which by riesz-markov-kakutani gives us a measure.

[merry shale]

ty I'll look into it, good lecture

[nimble jolt]

ty

[frigid patrol]

In my class we write m for the lebesgue measure and so integrals are with respect to m, and there are dm's after the integrands

[nimble jolt]

\mu is common too.

[merry shale]

we use lambda for lebesgue, and mu when a measure is arbitrary

i think our teacher will write d\mu to not get the same question tho

[frigid patrol]

We use \mu for arbitrary measure

[frigid patrol]

Prove that if a connected set intersects the interior and exterior of a set it also intersects the boundary

<@&286206848099549185>

[gritty widget]

Can you clarify

Interior is included in exterior so..

[frigid patrol]

No its not

Interior and exterior are disjoint

Interior exterior and boundary from a partition of the whole space

[nimble jolt]

If it did not intersect the boundary, then the interior and exterior would give you a disconnection of your supposedly connected set. I.e. we have written a connected set as a disjoint union of nonempty clopen sets. @frigid patrol

[gritty widget]

Oof that kind of exterior

[visual grove]

[rugged swan]

Hey, if A satisfies Baire's property, is every closed set of A satisfies the Baire's property too ? I think not and does anyone has a counter example of it ?

[dreamy valley]

some dumb counterexample is like X = R, A = X, and B = {0} the closed subset of A, or i am being stupid

[merry shale]

I think you need to have that for every B part of A, the border of B is meager?

@rugged swan @dreamy valley

[dreamy valley]

i'm not exactly following what you're claiming: can you give a full statement? sorry

[merry shale]

baire property is "is called an almost open set, if it differs from an open set by a meager set"

[rugged swan]

my Baire's property is that A satisfies Baire's property iif for all sequence of open and dense sets of A, the intersection is dense in A

I've found one counter example

Take E = R^2

X = R+* x R U {0}xQ

X satisfies Baire's property

but his closed set {0}xQ doesn't

[dreamy valley]

ohh, is this "baire's property" what i would call "comeager"/"second category"?

i had the wrong definition in my head i guess (the one hazen was using)

[merry shale]

comeager = intersections of dense opens

(countable)

[rugged swan]

I think my definition is equivalent, but idk about meager set x)

[merry shale]

meager = countable unions of nowhere dense sets

[rugged swan]

my counter example is here : http://www.les-mathematiques.net/phorum/read.php?14,410001,411274

(french x))

[merry shale]

d'acc

[rugged swan]

:p

[dreamy valley]

oh ok it is maybe not quite the same, but i was confused about what definition you meant

[merry shale]

it is the same, the whole point of baire is to say that meagre parts are still nowhere dense i think

[dreamy valley]

it's not quite the same under the words i normally use, for stupid reasons, like, the disjoint union of R and Q is second category (it's not meagre) but you don't want to call it a baire space because the Q part is dumb

i like the $$R_+^* \times R \cup \lbrace 0 \rbrace \times Q$$ example, gives a good picture of what might go wrong

[merry shale]

(globalement, un fermé d'interieur vide c'est petit, une union denombrable de ces trucs ça reste petit mais quand même un peu plus gros : on les apelle maigres. un truc très gros ça deviens un truc comaigre (dont le complèmentaire est maigre). la propriété de baire revient a être presque ouvert a ce sens là)

I think "baire space" is wrong tho. the baire property applies to parts of a topological space, it is not intrinsic?

[dreamy valley]

the thing zak is calling "Baire's property" ("for all sequence of open and dense sets of A, the intersection is dense in A") is what people call "Baire space", right? Like, this is a definition of "A is a Baire space"

[merry shale]

yeah that makes sense

[dreamy valley]

and it's just not in general directly related to A being "almost open" in X? (though maybe a set with Baire property inside a Baire space is a Baire space or something)

[merry shale]

oh baire space != baire property, okay

baire property = almost open part / baire space = what you said

in a baire space, baire parts are dense

[dreamy valley]

what's a baire part?

[merry shale]

almost open

(parts that satisfy the baire property)

[dreamy valley]

open sets aren't dense in a baire space, so idk what you mean maybe?

[merry shale]

dense in themselves

wait

dense in the open set they differ from

[dreamy valley]

oh i see

is the claim that they're contained in an open set in which they are dense?

even the claim that they have a meagre difference with an open that contains them is nonobvious to me

[merry shale]

they are unions of dense opens of some open, so their complementary in this open is an intersection of no-interior closed sets

it's that comp(AuB) = comp(A)nComp(B) thing

their complementary in this open is an intersection of no-interior closed sets <=> "they have a meagre difference from an open set"

[dreamy valley]

i'm still a bit lost, maybe i should think a bit

[merry shale]

take a countable union of dense opens in the topology induced by A on X and prove it's complementary in A is a countable intersection of nowhere dense (no interior) closed parts of A with it's topology induced from X

that should shed some light

[west ocean]

Latex ?

[merry shale]

another intuition:
nowhere dense parts of an open set A of X are "small"
a countable union of small things is still small but bigger, we call it meager
a complementary of a small thing is big
a complementary of a meager thing is big but a bit smaller, we call it comeagre

this notion of bigness is tied to topology

and in a baire space it is nice enough to use

("nice enough" <=> almost opens form a sigma-algebra (tho I'm not sure you need a baire space for that) )

[dreamy valley]

the thing that i don't understand is, i think you claim that "baire property sets are dense in the open set they differ from", but baire property sets aren't a priori guaranteed to be contained in an open set whose difference with them is meagre, so i got confused

[merry shale]

everything is in induced topology in what I said

so you need a subspace of X

simplest way is to take an open set of X

it will guarantee the local topology looks like the topology of X

[dreamy valley]

you're saying that if a baire property set A differs from an open set U by a meagre set, then A \cap U is dense in U?

[merry shale]

A \cap U?

[dreamy valley]

intersection

[merry shale]

I'm saying AnU = U because it is implied in the definition of a baire property part that it lies in an open set

but yeah

[dreamy valley]

that isn't obvious to me

i am using wikipedia's definition atm, since i'm not particularly familiar with baire property

[merry shale]

what you said can be a definition too

there exists an U such that a defers from U by a lmeagre set

what I'm saying is that you can work directly in U

with the induced topology

[dreamy valley]

sorry what is the definition of baire property you are using

[merry shale]

there exists an U such that A defers from U by a meagre set

it's just how I interpret it

A is comeagre in U

with U seen as a topological space

[dreamy valley]

so certainly it's not the case that for any such U, that A is contained in U

[merry shale]

yeah you just have one

[dreamy valley]

i don't see at the moment why there even needs to exist one such open set

[merry shale]

to define almost open

[dreamy valley]

Like, okay, counterexample I think

$$X = \mathbb{R}$$, and $$A = \lbrace 0 \rbrace$$. Then $$A$$ has the Baire property (it differs from $$\emptyset$$ by a meager set), but it's not contained in any open that it differs from by a meager set.

[merry shale]

the purpose of defining an almost open set is to say formally that our notion of bigness (comeagreness) respects topology (if the notion of almost open is nice enough we will have that)

no set is meager in $$\emptyset$$

[merry shale]

exept shit like $$\emptyset$$

oh

latex fail

[dreamy valley]

Sorry, $$\emptyset$$ is open and $$\emptyset \triangle \lbrace 0 \rbrace$$ is meager, right?

[violet pendant]

Support the bot on Patreon!

Every little bit helps

[merry shale]

wow I fucked up

hmm

meager where?

[dreamy valley]

inside $$\mathbb{R}$$, that's a fair question

[merry shale]

oh yeah sorry

I had it wrong

yeah it's with the symetric diff it makes more sense

[dreamy valley]

right

[merry shale]

ohhh

comeagre = union of dense opens where?

like dense in what?

okay found it

dense interior***

[dreamy valley]

right, but for comeagre i think you need intersections

[merry shale]

A subset B of X is nowhere dense if for each neighbourhood U of X, the set B ∩ U {\displaystyle B\cap U} {\displaystyle B\cap U} is not dense in U.

[dreamy valley]

comeagre = countable intersection of sets with dense interiors?

that's the, uh, mindless switching of the words in the definition of meager but i didn't think about it hard

[merry shale]

yeah, wiki says : Given a topological space X, a subset A of X is meagre if it can be expressed as the union of countably many nowhere dense subsets of X. Dually, a comeagre set is one whose complement is meagre, or equivalently, the intersection of countably many sets with dense interiors.

and a complementary of a nowhere dense is a set with a dense interior

so it's intrinsical

baire space = unions of nowhere dense CLOSED sets are still nowhere dense

and the complementary of that statement

so intersections of dense-interior OPENS are still of dense interior

[dreamy valley]

that sounds believeable

[merry shale]

that should give you that almost opens are stable by countable intersection, making it a nice sigma-algebra to work with

ofc every "intersection" is to take as "countable intersection" if I forgot

[frigid patrol]

Is my classification of captial English widthless letters upto homeomorphism correct?

The diagrams are partitions of a closed interval whose quotient space along those partitions give the associated letters

[dreamy valley]

I do not quite understand the diagrams yet, but it seems to me that Q and R, at least, have the same homomorphism type, at least the way you've drawn them?

[frigid patrol]

Oh dang that's true

[dreamy valley]

I think I agree with the rest

[delicate siren]

wtf dude

stay on topic!!

everything that is closed is a sphere babyy

woo

[steel needle]

what

[honest narwhal]

Uh

Yeah let's not post that kinda stuff here, eh?

[steel needle]

what are you gonna do, ban me?

(quote from banned man)

@everyone

fuck you

[honest narwhal]

Lol

[gritty widget]

😂

[merry shale]

There's two point of his R that are separated and have the property: if you erase them you cut R in two connected components, while there are none in Q

@dreamy valley

So i don't think they are homeomorphic

They are homotopic tho

[dreamy valley]

Oh I read the two legs of the R as coming from the same point

But either R is homomorphic to Q or A, right?

[merry shale]

A

But not his Q

Cause the line extends inwards

[dreamy valley]

I think the graph to the right of the R suggests they meant it to have the Q homomorphism type, but yes I see how it is not necessarily clear

[merry shale]

So removing the point gives 3 connected components. Yeah those exercices are just intuition builders anyway

[dreamy valley]

Tried to draw a clear picture

[merry shale]

Oh

[dreamy valley]

This R is clearly homomorphic to Q we agree

[merry shale]

Ye I think so

At least my arg doesn't hold

Yeah I have the homeomorphism

[dreamy valley]

(If you are picturing bending one into the other, picture doing the deformation in 3-space so that you can get inside the loop.)

[merry shale]

I just imagine moving in the R and drawing the Q as a function of that that I can reverse

So yeah, Homeomorph

The diagrams are weird

They seem ambiguous

[dreamy valley]

i think if read correctly they clearly define topological spaces? i was sort of picturing each as like, "directions for how to write the letter"

i don't think each letter gives a unique such diagram or anything, but that's no worry

[merry shale]

Oh okay there's a trick

They should define a partition of (0,1) which does define enough spaces for those letters

I just don't understand the partition in the diagram

And also how to quotient (0,1) to get A

[dreamy valley]

so imagine writing A on a sheet of paper with a single stroke

that defines a map from [0, 1] to "A"

and then just write down which points get sent to the same points

[merry shale]

Oh so it's not just finite partitions

Well points or finite

[dreamy valley]

i'm a little unsure why the word "partition" is getting used, but yeah, it's not just like, an equivalence on a few points

you have to identify intervals with other intervals

[merry shale]

An equivalence can be seen through the partition given by its classes

That's the partition

[dreamy valley]

partition of what? i'm confused

[merry shale]

Let R be a relation on E, you can build a partition of E such that xRy <=> x and y are in the same subset of the partition

If R is an equivalence rel

[dreamy valley]

ah i see ofc

ty

[merry shale]

Np

What leters are topological manifolds?

Which have borders as manifolds?

[dreamy valley]

D and O are the only topological manifolds I guess

[merry shale]

I is one

It has borders tho

[dreamy valley]

yeah if you add "with boundary" then all the ones that look like CGLMNSUVWZ are good

[merry shale]

Yeah

[raven garnet]

blinks

[frigid patrol]

The diagrams are like this: Identify marked points, before that identify marked line segments so that their arrows align

[merry shale]

Yeah I figured it out with the "draw the thing and look where you write twice and in what direction"

Thingy

[gilded hazel]

i like topology because you can bend the rules with shapes and sides and make things like "3 sided squares"

[humble cloud]

all 90 deg angles, but 3 sides...

on the surface of a sphere is easiest right?

[gilded hazel]

yep

you could also get a 5 sided square by using a pseudosphere

cause of its constant negative curvature

[humble cloud]

pseudosphere is an apple right 😉

[gilded hazel]

scree

[merry shale]

That ain't topology if you have angles tho

[gilded hazel]

I'm still learning about it im attempting to teach myself algebraic parts of it (sets, metric spaces, etc)

[nimble jolt]

not really accurate to talk about them as being algebraic parts of topology, especially with the subject of algebraic topology being a thing.

sets are omnipresent, a topological space is nothing more than a set equipped with a topology.

metric spaces are just a special kind of topological space that you would have encountered in calculus/analysis before trying to generalise appropriate notions to topological spaces.

[meager pagoda]

@nimble jolt deRham cohomology of some degree is a quotient algebra right?

then collectively it's like a diagram with all of them and there are certain morphisms induced by exterior derivative and pullbacks?

@calm shuttle am I totally off here

[calm shuttle]

I don't see that but I'm not very familar with de rham cohomology

[meager pagoda]

who is even

[calm shuttle]

huh

[meager pagoda]

sorry, you're not the first you tell me that today

[dreamy valley]

i am somewhat familiar if i can help

[meager pagoda]

pls

[calm shuttle]

you mean quotient algebra in the sense of universal algebra right?

[meager pagoda]

I thought it's literally an algebra

[dreamy valley]

what is the multiplication supposed to be?

[meager pagoda]

wedge

[dreamy valley]

that shifts degrees, so you don't get an algebra structure on the cohomology of a fixed degree

[calm shuttle]

the entire thing has the structure of a graded algebra IIRC

the complex

[dreamy valley]

ye

[meager pagoda]

so they're just modules for a fixed degree

but the entire thing's an algebra.

[dreamy valley]

yeah, modules over the ring of smooth functions, the whole thing is an algebra; on cohomology the exterior derivative is technically a morphism but it's not interesting (it's zero :P )

[calm shuttle]

when you say quotient algebra what do you mean?

do you mean the quotient of an algebra or a quotient algebra in the sense of universal algebra

[meager pagoda]

https://en.m.wikipedia.org/wiki/Quotient_algebra

[calm shuttle]

ok cool

what exatly are you trying to quotient by then?

[meager pagoda]

well I think for any given degree it'd be the quotient module of closed forms equivalent up to an exact form

when is a form not closed

[nimble jolt]

yeah other people have pretty much answered

[meager pagoda]

@nimble jolt analysis exam went by too fast

[nimble jolt]

in that it was so much fun you wish it could have lasted forever?

will assume so

[calm shuttle]

analysis sucks, just saying

[nimble jolt]

you suck, just saying.

[calm shuttle]

that's also true

[meager pagoda]

honestly just frustrating with short time exams

w/e

[nimble jolt]

yeah time crunch exams suck. was nice to be done with them

[meager pagoda]

why this true

middle part @nimble jolt

orz

[honest narwhal]

Where it integrates f' dx?

@meager pagoda

[meager pagoda]

si, senor @honest narwhal

[honest narwhal]

The point is that f is supported on [a,b]

So outside of [a,b], f = 0, so df/dx = 0

[merry shale]

F has compact support so it's not null exactly on a compact of R but compacts of R are bounded closed sets, so they must lie in an interval. Hence f is null except on an interval, then use the above

Don't forget that compacts of R are exactly the subsets of R which are bounded and closed

(This is true in every finite dimension normed vectorial space)

I'm not sure it answers you cause of the crop. (And after what i said it's just fondamental theorem of analysis)

[meager pagoda]

hold on

so the statement is that

$$H_c^1(\mathbf R) = \dfrac{\Omega_c^1(\mathbf R)}{\ker \int _{\mathbf R}}$$

[meager pagoda]

this is very much a statement about R i am realizing

was mostly not realizing why everything was a closed form

my mentor had a really nice geometric interpretation of closed & exact forms which I'm trying to convince myself is true

One thing depended only on the boundary, but the other one you needed to check homotopy stuff

gonna try to write things down properly

[merry shale]

Oh it's a way to reason modulo null measure sets

Like "this is almost everywhere true"

Would be the same as "it's true for pi(f) in H" where pi is the canonical projection in quotient sets

Pi(f) is the class of every g almost equal to f

Its a freaking equivalence relation i didn't know i love lebesgue i wanna have kids with its integral

[meager pagoda]

what

what's a null measure set

[kindred meadow]

I think you'd need to provide some more contexts. It's not completely clear what's your question.

[honest narwhal]

Don't worry about measure theory for now

The point is this

Your definition is

$$H^1_c(\mathbb{R}) = \frac{\ker d}{\text{im } d}$$

[honest narwhal]

Where you're restricting d to compactly supported forms

Now if you can show that every compactly supported form is closed, and a 1-form is exact iff its integral over R is 0, then you can show what you said above

Now the point is this. If you have a compactly supported exact form $$f' dx$$, then you take some $$[a,b]$$ such that $$f'$$ is $$0$$ at the endpoints and only non-zero inside. Then the integral over $$\mathbb{R}$$ is the integral over [a,b], now use FTC, gg

[honest narwhal]

So compactly supported exact forms are in the ker

[meager pagoda]

wait rudin says that for a "k-surface" $$\Phi: \in \hom(D (\subset \mathbf R^k) \to E (\subset \mathbf R^n))$$
$$\omega(\Phi) = \int_\Phi \omega$$

mathbot..

@nimble jolt what going on here

@honest narwhal

@merry shale since you mentioned integrals

[still aspen]

Hi @meager pagoda

[meager pagoda]

what @still aspen

[still aspen]

High

[meager pagoda]

what's new

[still aspen]

The other bot that does LaTeX

[meager pagoda]

you?

[still aspen]

@meager pagoda ya but woog kicked me

[gilded shell]

Stupid question here

How does he get the part above red?

How does he know that exists?

[small obsidian]

Have they already proven that every element is part of a basis element?

That's always true, btw. Try proving it, it's pretty natural

[gilded shell]

1stQ: Isn't that just (1)?

2nd. Sure, I assumed it was instant though.

[small obsidian]

What's (1)? I may use different notation

[gilded shell]

(1) on the definition of Basis

[small obsidian]

Oh, your basis is defined that way

Then yeah that's (1) lel

[gilded shell]

Uh... Ok, maybe I'm ahving issues because they defined "basis for a topology". But they haven't defined in my screesnhots "basis for a topology T".

[small obsidian]

I think it means ⊆

[gilded shell]

What is ⊆ of what?

[small obsidian]

Because you're right, there may not always be a basis element that is a proper subset of U

[gilded shell]

I'm using Munkres. I might be being pedantic but I don't see the Munkres using "basis for a topology T" anywhere before the Lemma. So...

[dreamy valley]

(It's the paragraph after the definition you posted)

[gilded shell]

Since I'm having these stupid issues can you tell me the proof you had in mind?

Language wise though he says "topology T generated by B".

Are you telling me he means to say that those two are exactly the same thing?

[small obsidian]

A basis B creates a topology on T. A topology T can be equipped with a basis B

[gilded shell]

Agreed

[small obsidian]

A basis is usually an easy way to write a topology, and they come with some nice properties

[gilded shell]

I've studied topology before but i noticed I suck at it now. I was trying to prove some basic theorems rigorously.

"rigorously". By that I mean very carefully

[small obsidian]

Lel yeah I could go through some of the stuff myself. Not an expert

[gilded shell]

Well you already helped though, I realized I'm not being 100% crazy. Thanks Kay

[dreamy valley]

Yeah, I see that munkres is kind of loose with the language. Saying "the basis B generates the topology T" and "B is a basis for the topology T" should mean the same thing, although the points of view are different (in the first case, you are imagining "starting with" the basis and "constructing" T, whereas in the second, you're imagining "starting with" T and then finding a "smaller" set that carries all the information inside). I am triple checking to make sure I am not crazy here

[gilded shell]

In the general situation that makes 114% sense.

But I didn't realize Munkres was doing that at the start

thanks

I wonder why they don't use the word generating set though. 🤔

Well... I guess that can ahve a different meaning

The set generated by a generating set is usually a set you obtain by repeated use of operations. And hopefully a finite number of times.
But a set generated by a generating set can be the smallest desired set closed under those operations. Which is kinda different I guess

but yeah

[dreamy valley]

Isn't that roughly the same? Here you want the coarsest topology (fewest open sets) such that every basis element is open

[gilded shell]

The problem is that if you obey the 1st definition you might still not obey the second.

Because a finite number of use of the operations is not enough. You might need an ordinal amount of operation uses.

Like the sigma algebra generated by all intervals isn't all possible counable unions of intervals, and all complement sets of intervals.
They're a basis according to my 2nd definition but not by the first.

Because actually you need to apply countable unions and complementation on intervals an ordinal amount of times.

Otherwise you'd be able to construct a non measurable set. Which you can't

You need AC for that jazz

[dreamy valley]

I don't think I know what "first" and "2nd" definitions refer to here

[gilded shell]

1st: The set generated by a generating set is usually a set you obtain by repeated use of operations. And hopefully a finite number of times.
2nd: But a set generated by a generating set can be the smallest desired set closed under those operations. Which is kinda different I guess

Also 5am here, sorry if I'm being incoherent. The time of the day is always a good excuse....

[dreamy valley]

ye

i can't say i really follow, but it is not as if i disagree with anything in particular so :p

[gilded shell]

Well have you learned sigma algebras?
I can't think of an interesting topology example.
But for example. If you start with the intervals, S. And then you obtain a set S1, by getting all coutnable unions of intervals and all complements of intervals. Then S2 by getting all countable unions of S1, and complements of S1.
And S_n etc.
Also S_omega. etc

When do you get to the Borel sigma algebra on R?

It's kinda interesting

[dreamy valley]

Yes, I know sigma algebras

well, I should say I know what a sigma algebra is and I understand the transfinite construction you are describing, more or less

[gilded shell]

Apparently it's ordinal corresponding to the cardinality of the reals. I'm using this language because I don't understand ordinals too well.

But yeah, I found that pretty cool

[dreamy valley]

Ah, I am not sure that's quite right -- I think you get to stop at the first uncountable ordinal (which is |c| iff the continuum hypothesis)

[gilded shell]

OH sorry

Hmm

So assuming the ContHyp

I'm right right?

[dreamy valley]

yeah :D

[gilded shell]

Yeah, I tend to always work in ContHyp tbh

God knows I'm right

You'll see when you die

[dreamy valley]

:P

[fleet trench]

Hey, I've got a problem that I'm not really sure how to go about proving (or if there's a counterexample, that works too---I'm pretty sure it's true though).

Consider an infinite nested sequence of connected subsets $$C_1 \subset C_2 \subset \dots$$ of a topological space $$X$$. Then the set $$C = \bigcap_{i = 1}^\infty U_i$$ is connected

It's trivially true for the finite case, but I'm not sure how to work with the infinite one. I tried working with it by contradiction, but $$U$$ being disconnected doesn't really give anything revealing about the sets in which it's contained as far as I can tell. Any hints on where to start?

[raven garnet]

I presume that's $C = \bigcap^\infty_{i=1} C_i$?

[fleet trench]

(we're just using the naïve definition of connectedness that has \emptyset be connected)

[gentle osprey]

Puerøsola:

[fleet trench]

oops, yeah sorry

[raven garnet]

Shouldn't that just be $C_1$ though, even in the infinite case ?

[gentle osprey]

Puerøsola:

[fleet trench]

oh, i meant \supset not \subset

[raven garnet]

Ohh right

[fleet trench]

with the typos fixed, haha:

Consider an infinite nested sequence of connected subsets $$C_1 \supset C_2 \supset \dots$$ of a topological space $$X$$. Then the set $$C = \bigcap_{i = 1}^\infty C_i$$ is connected

[raven garnet]

Sure :)

Hmm

Can't you take intersection?