point-set-topology

[umbral hamlet]

Ah ok

Im honestly unable to see why the current proof is have doesn't work tho

[rancid umbra]

lemme look

[umbral hamlet]

Am I wrong that image of union is union of image

[rancid umbra]

no, that is true. its the image of the intersection that you need to be careful about

[umbral hamlet]

I see

Ah I guess i implicitly assumed its true when it isnt necessarily

[rancid umbra]

here is a nudge in the right direction:
consider a continuous function f : X -> 2.
f is constant on the fibers of p since each fiber is connected, so f factors uniquely through p via a continuous function g : Y -> 2.

[umbral hamlet]

I think i may have gotten it

Suppose bwoc that p(U) \cap p(V) is nonempty

Pick a point y in the intersection

Then the preimage is open in X

[rancid umbra]

closed

[umbral hamlet]

Consider U and V restricted to the preimage of y, then its a separation

Sorry sorry

I meant connected my bad

So we have that the preimage of y is not in U WLOG

But thats a contradiction

[rancid umbra]

this is already a contradiction since p^-1(y) is connected

but sure, either way

do you want to continue?

[umbral hamlet]

I think the rest is straightforward

[rancid umbra]

i think this proof is pretty slick. you just need to know that a space X is connected if and only if every continuous map X -> 2 is constant, where 2 = {0,1} with the discrete topology

along with the universal property of the quotient space

[umbral hamlet]

We didnt quite talk about universal properties but I am familiar with them so it sounds cool

[rancid umbra]

if you have talked about quotient spaces, you have surely talked about like

if you have a quotient map p : X -> Y

when a function f : X -> Z

determines a function Y -> Z

right?

[umbral hamlet]

Maybe but I cant recall it

We honestly learned it quite poorly

[rancid umbra]

oh. this is like the whole point of quotient spaces

[umbral hamlet]

My prof isnt amazing

All I know is that quotient maps arise from defining equivalence relations on X

[rancid umbra]

yea. so if you have some equivalence relation ~ on a top space X, you can define a function p : X -> X/~ that takes a point x to its equivalence class under ~

but also, if you have a function f : X -> Z that also identifies equivalent points

then there is a unique function ~f : Y -> Z such that f = ~f o p

[umbral hamlet]

I see

I just found the theorem in munkres that establishes this

[rancid umbra]

yea!

this is the universal property of the quotient space

it essentially says that the cannonical projection is the smallest or best function identifying all equivalent points in the sense that any other continuous function that identifies equivalent points factors (uniquely) through it

[umbral hamlet]

I see

I recently learned about the universal property of the commutator subgroup in algebra and it reminds me a lot of that

I suppose thats the point of category theory

[rancid umbra]

yea! G/[G,G] is an abelian quotient, and in some sense, its like the best abelian group approximation of G

[umbral hamlet]

It was fascinating how smoothly the proofs involving it went

[rancid umbra]

maps out of G/[G,G] are determined by maps out of G that kill the commutator subgroup

[umbral hamlet]

Yeah

So essentially I just need a normal subgroup that contains [G,G] and I get an abelianization of G

[rancid umbra]

if N contains [G,G], then you will get an abelian approximation of G after quotienting

but it won't be universal

G/[G,G] is the smallest abelian approximation you could hope for

[umbral hamlet]

Oh oops, is abelianization specifically modding out the commutator?

I thought its just any abelian quotient

[rancid umbra]

the abelianization of a group G is specifically G/[G,G]

a quotient group G/N by a normal subgroup N is abelian if and only if N contains [G,G], that is, if and only if the quotient projection G -> G/N kills the commutator subgroup

[rancid umbra]

yea, universal properties are really nice

[umbral hamlet]

I wish I was more familiar with the one with the quotient topology but we went over them super quickly

[rancid umbra]

munkres and lee go over them well

there is another one, the categorical approach to topology book

[umbral hamlet]

The only thing I remember is defining the torus from quotients on R2

[rancid umbra]

yea, this is R^2/Z^2

[umbral hamlet]

We use munkres in class but I guess we either didnt cover that or i forgot it

I was having a pretty rough time in class around then

Luckily I feel much more confident with connectedness and compactness

But the middle section of pointset had me more lost than I expected

[rancid umbra]

the middle section meaning quotients?

[umbral hamlet]

Quotients, separability axioms, countability axioms

[rancid umbra]

feels so weird to put quotients there

[umbral hamlet]

Looking back i feel okay with all of these now

[umbral hamlet]

It was so quick compared to the other two

Feels like we took forever on problems like "show that R omega is first countable under the product topology" or things like that

[rancid umbra]

thats where all the good stuff is at lmao

[umbral hamlet]

Yeah who knows

Fortunately im an undergrad so ill go over all of this stuff again in just a few years

But this is my schools topology graduate course which is concerning

I've heard the professor is bad and I guess the weirdness of the curriculum is a big reason why

[rancid umbra]

hrm. yea, idk. i always found that when coming back to the material after the course or when using it in another course later is when the concepts from the previous course made the most sense

[umbral hamlet]

Thats what im hoping for with topology

[rancid umbra]

i think honestly yea, just come back to it and explore some more exciting areas like alg top or diff top

[umbral hamlet]

Im having a good time with algebra this semester bc its the second time im doing ig

[umbral hamlet]

Ill be taking intro algtop next semester with apparently a really cool professor which is exciting

[rancid umbra]

nice!

that should be fun. if you get to CW complexes, you will see more universal properties and quotients

[umbral hamlet]

The syllabus doesn't say anything about content so im not fully sure

But I've heard CW complexes are commonly covered so I bet we will

[umbral hamlet]

Let p: X to Y be a closed surjective map such that the preimage of every singleton in Y is compact. Show that if Y is compact then X is compact

Wanting to check the validity of my proof

Let C be a collection of closed subsets of X that satisfies the finite intersection property. Then p(C_i) is closed for all C_i in C, and so p(C) (bit of notation abuse) is a set of closed subsets of Y satisfying the FIP, thus the intersection of p(C) is nonempty as Y is compact.

Let y be an element of this intersection. Then its preimage is compact and every element of C restricted to the preimage of y is closed under the subspace topology, and so the intersection of these restrictions is nonempty, which implies that in X their intersection is nonempty, thus X is compact.

[rancid umbra]

yea, this looks good!

[fickle tendon]

Is the inverse limit of a system of irreducible spaces irreducible? If not, what are some conditions?

[warped helm]

<@&268886789983436800>

[rancid umbra]

do you know if the product of two irreducible spaces is irreducible?

i think it should be

https://math.stackexchange.com/a/2108133/803927

this gives a characterization which shows that the finite product of irreducibles is irreducible. it should generalize to arbitrary products i think.

now we just have to check equalizers…

ah but. from the same characterization, i don’t think that equalizers of irreducibles are boxes of irreducibles the product topology

yea like, if f : X —> Y is a continuous map between irreducibles, its inverse limit is {(x,y): fx = y} but this isn’t generally irreducible in the product topology. for example, if we take X = Y and f = id_Y, then the inverse limit is the diagonal in Y, and that isn’t a box of irreducibles in X x Y

[fickle tendon]

well the inverse limit of Y = Y is Y so that is still irreducible lol

[rancid umbra]

yea just realized :p

[fickle tendon]

if all the maps in the inverse system are surjective it seems that the resulting limit is indeed irreducible, the problem is that in the setting im working on (AG-related), the natural case is that of closed immersions

(Induced by a surjective direct system of algebraic structures, you take a spectrum-like contravariant functor to Top)

[rancid umbra]

hmm. do you have a concrete example i can think about? i’m not too familiar with this setting

[fickle tendon]

well, think of rings and their prime spectrum

[rancid umbra]

will have to go look at what a prime spectrum is ha! i’m sure somebody else will know the answer to this tho

[fickle tendon]

in this case, you can prove that a direct limit of integral domains is again an integral domain, so the result holds here (Spec as functor to Top anticommutes with direct limits)

[fickle tendon]

do you know of the right order topology on a poset?

[rancid umbra]

like open rays to the right form a subbasis?

[fickle tendon]

tl;dr you take the category of groups, define the "spectrum" of G to be its lattice of normal subgroups, and the right order topology on that. Along with sending f to the preimage function on normal subgroups is a spectrum-like functor from the category of groups

this functor anticommutes with direct limits, although here the result trivially holds too as every such "spectrum" is irreducible lol

[umbral hamlet]

Im actually not sure about this proof anymore

How do I know that the restriction on C satisfies FIP in the preimage

[rancid umbra]

p^-1(y) is closed. so for all the sets C that meet p^-1(y), you can just pull out p^-1(y) from the intersection

[umbral hamlet]

Which means I need their intersection region to meet in that preimage, correct?

That was the thing i was worried about

[rancid umbra]

i mean, can’t we choose y such that y is in p(\bigcap C) since this is a subset of \bigcap p(C)? then y = p(x) is such that x is in \bigcap C

[umbral hamlet]

Oh of course

Im not very familiar with unions and intersections of functions clearly

[rancid umbra]

i hate proofs like this tbh lol

[umbral hamlet]

I know union of images and vice versa are equivalent

[umbral hamlet]

They just make me remember how little set theory i learned in discrete

Shouldnt have gone to a college for engineers my first semester but alas

[rancid umbra]

i always have to go back to this page to check

[umbral hamlet]

Wait doesn't this mean im done immediately

[rancid umbra]

yea

[umbral hamlet]

Holy am I saving this right now

[umbral hamlet]

But the compactness of preimages of points was unnecessary then

[rancid umbra]

no, since we need p^-1(y) to have FIP

right?

[umbral hamlet]

Idk

If i can find a point in \bigcap C

It follows its nonempty

Wait so yeah we cant choose a point in p(bigcap C)

[rancid umbra]

uhh. like, the collection p^-1(y) \cap C has the fip

so

by compactness of p^-1(y)

the entire intersection is non-empty

[umbral hamlet]

Wait wait yeah I need to step back

The preimage of y could be a bunch of scattered points

But by how we constructed y those points have to be scattered into Cs

[rancid umbra]

yea

[umbral hamlet]

So it still remains that they all intersect preimage of y

[rancid umbra]

right. we don’t know a priori that \bigcap C is non-empty

[umbral hamlet]

Im just still wondering if the restrictions of the Cs intersect in the preimage of y

The way we tried to show this just now was circular on accident

[rancid umbra]

oh shoot. right

okay. i have an idea

its a trick that has worked before

lemme try and find it

[umbral hamlet]

Wait no

The goal should be to show that any finite collection from C intersects in the preimage of y

So that we can apply compactness

I have an idea

[rancid umbra]

lmao

this is the same question

[umbral hamlet]

Oh nice lol

[rancid umbra]

i made the same error with u

[umbral hamlet]

But what i just came up with is instead looking at the images of finite collections of C

[rancid umbra]

i hate this proof

[umbral hamlet]

Then we can look at p(bigcap C_f)

Where C_f is some finite intersection from C

And we dont have any issues there

Hm or maybe not

Idk

[rancid umbra]

yea, that is exactly what i wrote in what i linked

[umbral hamlet]

I see

Seems you split C up into disjoint finite subsets

Maybe im reading wrong

[rancid umbra]

it’s exactly images of finite collections of C

[umbral hamlet]

I see

Man im kinda bummed I didnt get this one but whatever

[rancid umbra]

bleugh

this proof is bleugh

[umbral hamlet]

Yeah

[rancid umbra]

alr i gtg for a bit.

[rancid umbra]

will look at this after i cook some dinner

[fickle tendon]

oh of course take your time!

the problem isnt super interesting in this particular case but the idea is that a spectrum assigns to each group a set of normal subgroups that respect preimages and quotients (i.e. if N < G is a distinguished normal subgroup of G then f^-1(N) is of H and N/M is of G/M are for some homomorphism f : H → G and normal M < G containing N)

[gritty widget]

I’m reading a topology book and they’re denoting intersection by juxtaposition

Like $A_1’{}A_2’$ to mean $A_1’\cap{}A_2’$

[gentle osprey]

Eclipso

[gritty widget]

I was confused a bit upon reading it at first

Never seen this before

Is this common?

[unreal stratus]

Personally I dont think i have ever seen this, though I guess it doesn't sound like the worst thing in the world

One thing that is convenient and kinda common is writing e.g. U_ij = U_i cap U_j given an indexed family {U_i}, but normally you would indicate that that is what you are doing

[gritty widget]

Maybe it’s more common in older texts? It’s from 1948

[unreal stratus]

Ah yeah maybe - which book is this?

[gritty widget]

Dimension Theory by Hurewciz

[unreal stratus]

Oh wow lol

Pretty old then yeah lol

[alpine nest]

Flashback to the Hilbert-Ackermann logic book where they denote logical conjunction like this.

[fickle tendon]

wtf

[gaunt linden]

Early in the history of symbolic logic there seems to have been this never-quite-explicit vague assumption that it in order to be "allowed" to use symbolic expressions for logical claims, one had to use the same symbols that were already in use for arithmetic on numbers -- thus + for disjunction and juxtaposition for conjunction.

[queen prism]

def an older trend

[gaunt linden]

(Something similar was going on when Hamilton invented the quaternions. This was before R^n or abstract vector spaces, but it was becoming generally acknowledged that complex numbers could make a lot of plane geometry simpler by using algebra on single complex numbers instead of separate equations for each coordinate. Hamilton wanted to find a parallel to this in three dimensions, but he seems to have operated on an assumption that before one could use algebraic manipulation with values there were points or displacements in space, there had to be a notion of multiplying those things, like numbers ought to have. That's what he spent a lot of fruitless energy to make work in three dimension before he suddenly realized he could do it -- of sorts -- in four).

[hearty willow]

Something similar was going on when Hamilton invented the quaternions

invented or discovered

\throws self over cliff

[gaunt linden]

That's a metaphysical quagmire I'm not going to wade into now.

[wide kayak]

giggity

[umbral hamlet]

Invented

Theory is invented theorems are discovered

Thats a fat qed right there

[hearty willow]

telling people you did math in uni is like a calling card to linking shitty degrasse tyson videos about is math invented or discovered

its like telling family you are good at computers and now everyone needs help

[hearty willow]

sounds like a theory you discovered

gottem

[umbral hamlet]

Erm properly a theory is a collection of tools and perspectives

[normal swift]

what

[fickle tendon]

false, a theory is a category with finite products

[umbral hamlet]

Oops are u a logician

[normal swift]

emmm

am i too model pilled

[umbral hamlet]

Uhh given a set S and a formal language L over S, a theory is any subset of L

I was moreso trying to "define" theory as a sort of middle ground between colloquialism and rigor

[normal swift]

ok

"collection of tools and perspectives" sounds much cooler and mathy tho

[normal swift]

wait

thats not it right

[fickle tendon]

no lol

[normal swift]

yeah

[fickle tendon]

model theoretically a theory is a consistent set of formulas over a language right?

[normal swift]

but i hate model theory rn so i dont want to think about it

[umbral hamlet]

Just starting the model theory portion of my set theory course so idk much about it yet

[fickle tendon]

idk man im a universal algebraist and the trivial algebra satisfies any equational theory 🔥

[normal swift]

so firstly they have to be sentences, meaning every variable is a closed term.

wait

no they must be satisfiable

wait no

wait

[fickle tendon]

any formula can be turned into a sentence by just slapping on some universal quantifiers anyways

there must be a model of the set of sentences

[normal swift]

well yeah but this is model theory and were concerned about semantic technical details🤓 🤓 🤓 🤓 🤓 🤓

oh yeah...

[umbral hamlet]

Intuitively I just think the "educated guess" colloquial definition of theory is so wrong

Because neither in math nor science is that remotely right

[normal swift]

mhm.

[fickle tendon]

this just in: the general public has no fucking clue what theyre talking about 🔥 🔥

[umbral hamlet]

Yeah but i dont see them being that wrong about stuff super often

[fickle tendon]

that is true

until its about math

[umbral hamlet]

Like the colloquial definition of theory is almost like completely opposite what it actually is

[umbral hamlet]

Yeah fair math is almost immediately niche after calculus

[fickle tendon]

at least we dont get popsci'd

[umbral hamlet]

Yeah we definitely do

[normal swift]

evil and intimidating numberphile

especially knot theory and topology stuff

[fickle tendon]

not as much as, say, physics

[normal swift]

yeah, being a physicist must suck

[fickle tendon]

why is no popsci person talking about quandles 💔 💔 💔

[umbral hamlet]

People thinking mathematicians are dumb for saying 1 + 2 + 3 + ... = -1/12 not realizing thats not what analytic number theorists are claiming and not having the years worth of tools required to understand the truth

[normal swift]

"quantum thing very cool"

[fickle tendon]

and analytic number theorists are famously dumb, of course

[normal swift]

when you assume ive studied knot theory you immediately lose

[fickle tendon]

its an algebraic structure so algebraists can claim theyre doing knot theory

[normal swift]

damn

being an algebraist must suck

[umbral hamlet]

When you assume I've studied you immediately lose

[normal swift]

you have to memorise the quadratic foermula

[fickle tendon]

:despair:

i cant bro irs too hard

what are letters doing in math who invented this

[normal swift]

how do they even do that

[umbral hamlet]

Quadratic formula in the pile of things I only use when tutoring math

[normal swift]

my brother is a mathematician and he knows the solution of the 10th degree polynomial

[fickle tendon]

holy quintic formula

[umbral hamlet]

Along with integration methods and volumes of revolutions

[normal swift]

nah 10 is "biquintic" cause 2*5

[fickle tendon]

yes but that implies quintic

[normal swift]

ok

[fickle tendon]

S_10 is a solvable group omg

[umbral hamlet]

Actually its pentaquadratic so its solvable

[normal swift]

these are like the most applicable math thingys

[umbral hamlet]

Yeah but im a pure math student so applicability is in bizarro world

[normal swift]

k

[fickle tendon]

let me apply it to abstract algebra, famously the study of real numbers

[normal swift]

ok.

[umbral hamlet]

Hey I mean lie groups

[normal swift]

good luck calculating volume of a cylinder in a group

[fickle tendon]

guh

not touching those, i do not know enough manifold theory

[dawn frigate]

I use the closure definition x in A bar iff there exists open neighborhood U in tau such that U intersect A is nonzero

[balmy nexus]

what is the question and the context please?

[balmy nexus]

What have you tried and where are you stuck?

[umbral hamlet]

Oh hey I remember getting exactly that problem horribly wrong

[twin swift]

Would like help on this

of course dim X >= dimUi for every i

but I am having trouble showing its the superemum

any hints?

[opaque scroll]

Take a sequence of irreducible closed sets in X, and think about the intersection with Ui.

There must be a Ui where it is non-empty

[quartz horizon]

What even is the dimension of a topological space

[opaque scroll]

Well there are a few different notions, but from the context I would assume supremum of length of irreducible closed subsets

This is the definition used in AG. Probably not that useful elsewhere as all hausdorff spaces are 0-dimensional

[twin swift]

yeah this is heartshorne

[opaque scroll]

Though interesting question: is the same property true for Lebesgue covering dimension? I would think so...

[radiant stone]

https://stacks.math.columbia.edu/tag/0054

[opaque scroll]

Okay, not true for "bad" spaces.

Like taking R and adjoining a point whose only open neighbourhood is everything.

But maybe for sufficiently nice spaces

[twin swift]

topology is weird

like its an amazing concept

but its still weird

[lucid ocean]

it's so distilled down and potent

[quartz horizon]

What do you find weird about it

[twin swift]

the exteremly weird examples

I rememeber something very long line

[quartz horizon]

I see I see

[crisp lintel]

the strange examples are an illustration of the generality and flexibility of topology

[gritty widget]

Separable+Metrizable is the usual assumption for topological dimension being “nice”, so maybe that would work?

[quick crane]

Hausdorff dimension maybe the most used one, but it's only for metric spaces

[gritty widget]

There’s lots! Brouwer defined two inductive ones, ind(X) and Ind(X), and, if X is separable and metrizable, ind(X)=dim(X)=Ind(X) where dim is the Lebesgue covering dimension

But as someone else already said that’s probably not the current dimension in that thing being discussed

[gritty widget]

Using the small inductive definition of dimension(empty set has dimension -1, a space has dimension <= n iff for every open set U and p in U, there is some open set V such that the closure of V is a subset of U, and the boundary of V has dim n-1), I think each non-empty open set has the same dimension as X? At least when the dimension of X is finite.

Suppose ind(X)=n

Letting U be some non-empty open set, and p in U, there is some open set p in V subset U such that the closure of V is a subset of U and the boundary of V has dimension n-1, because X is n-dimensional. Thus, ind(U)>=n

Nvm

Didn’t realize U could just have higher dimension than X I think?

Wait no I don’t think that’s possible for open U

I don’t think U necessarily has the same dimension then unless dimension at a point is constant across the space? For example the disjoint union of a line and a plane, line is open and has ind 1 but the space as a whole has ind 2

Darnit

Might still hold for open covers though

I’ll think more about it later

[tender halo]

subspaces dont increase in (lower inductive) dimension

because subspaces of zero-dimensional spaces are zero-dimensional, and boundary of an open set in a subspace is a subset of the boundary of its "parent" in the surrounding space

[gritty widget]

Ic

thanks!

[gritty widget]

I think I see the problem with my proof, then. I said the boundary of the subset V has dimension n-1 due to dimension of X being n, but the correct result is that the dimension is <= n-1 I think

[gritty widget]

What’s a good book for learning about (metrizable) Continuums in particular?

[smoky sigil]

In this Theorem from Munkres Topology why can we assume each A_n is nonempty?

[queen prism]

well if A_i is empty for a certain i what does that say about the union of the A_n?

[smoky sigil]

The union wouldn’t change if we kept A_i or left it out of the union.

[queen prism]

yea
in particular it wouldn't affect the countability of the union

because if you union a countable set with {}, it remains countable, and if you union an uncountable set with {}, it stays uncountable

[smoky sigil]

That makes sense to me. So for the proof what would happen if we didn’t assume all the sets were nonempty

[queen prism]

the bottom of the proof would have some issues as soon as you try to pick out an element from the empty A_i, so you'd have to work with just the nonempty sets within your family of sets
but it's not a big deal

in essence if you were to keep going, you'd look at the subfamily {nonempty sets B_m} c {all sets A_n} and prove the theorem for the subfamily, and then extend it to the superfamily

[smoky sigil]

Ohhhh ok ok I see

[lucid ocean]

use prime factorisation to produce an injection from the countable collections of elements into N, then hey presto

[rancid umbra]

fun-fact: this statement requires the axiom of countable choice in order for it to be true, and can fail in the absence of it.

[sly geyser]

a subbase of the whole topology?

[cosmic mirage]

sorry, i am confused about your question

what do you mean? S is a subbase of tau, i am not sure what other thing you are trying to rule out

maybe this clarifies it - a subbasis for a topology is not a basis of a subtopology

[sly geyser]

wtf is tau here

[cosmic mirage]

the topology on X

[sly geyser]

wait ok then what is a subbase?

I thought it was a base made from the subset of another base

[cosmic mirage]

do you know what a base of a topology is?

[sly geyser]

yes

[cosmic mirage]

okay cool

a subbasis S is a collection of open sets, such that the collection of finite intersections of open sets in S gives you a basis

[sly geyser]

......... I assumed that's what a basis was

do you have to be able to get every open set through only union from bases?

[cosmic mirage]

oh wait did i get the definitions mixed up lol

[young stone]

nope

[cosmic mirage]

lol nice i got the ping before the wikipedia link for basis loaded

[sly geyser]

ok I get it now

[cosmic mirage]

yeah a basis generates the topology under union, and a subbasis generates a basis under intersection

[sly geyser]

the way I view it a subbasis generates the topology under both union and intersection while a basis just does it under union

[cosmic mirage]

yes but both of us should be careful here - arbitrary union and finite intersections

[sly geyser]

same thing though of course

[sly geyser]

Ik

[fringe spade]

This feels like a very stupid question, but does there not exist an isometric embedding of Z into R (both with the Euclidean metric)?

I am learning about quasi-isometry and therefore trying to get a feel for the differences between isometry and quasi isometry

[quick delta]

There is an isometric embedding (namely, inclusion)
But there is no isometry (because there’s no non-constant map in the opposite direction)

[fringe spade]

Right because that would imply R is disconnected right?

So in general we require an isometry to be a bijective function?

[quick delta]

Yes and yes

[fringe spade]

Thank you!

[quartz horizon]

Funnily enough isometries are always injective

So being bijective is equivalent to being surjective

[fringe spade]

I did see that lol. Quite nice

[warm hedge]

when we have an equivalence relation E in a space X, if we say that E is borel or meager or open etc, does it means that the set $ {(x,y) | x \in X, y \in X and xEy } is borel or meager or etc in space X \times X$ ?

[hasty thicket]

Do we say tau is a discrete topology on a set X iff Tau = power set of X

[warped helm]

that is the discrete topology yes

[hasty thicket]

Okay so when we have a topological space, we know the intersection of any finite no. of open sets is an open set itself. What about the intersection of an infinite amount of open sets?

surely, even if the intersection is empty, then it’s still an open set as we know the empty set is open. Is there something else going on?

i’m using open and set too many times this is getting a bit awkward to read

[pulsar lagoon]

look at the sets of the form (-1/n, 1/n)

what is the intersection as n -> infty

[opaque scroll]

The empty set is open, but the intersection doesn't have to be empty

[hasty thicket]

oh that’s such a good example

{0} is not open so the intersection isn’t open aswell

thanks guys

[crisp lintel]

Spaces where arbitrary intersections of open sets are open are called Alexandrov spaces https://en.wikipedia.org/wiki/Alexandrov_topology

Which is kinda neat

[drifting plaza]

Send bro to the L hopital 😂

[cosmic mirage]

what?

[kind marlin]

is there a purely topological characterization of the complex numbers? inspired by this post that people were discussing earlier

https://mathoverflow.net/questions/76134/topological-characterisation-of-the-real-line?_gl=1*jls803*_ga*MTk4NTU3NDUwNS4xNzU2OTA2MDkw*_ga_S812YQPLT2*czE3NjEwMDI1MzgkbzcwJGcxJHQxNzYxMDAyNjg3JGo2MCRsMCRoMA..

[unreal stratus]

Unique space whuch is a product of two spaces satisfying [topological characterisation of R]

[kind marlin]

Oh that’s fair C is basically just R2 in this context

that was boring, oh well

[cosmic mirage]

well it depends what you mean by C right

like if you want the topological field structure, R^2 is useless

[kind marlin]

yeah, but that’s just a separate question at that point, and I think I saw how C is characterized as a topological field somewhere

[cosmic mirage]

huh really? what was the characterization

[kind marlin]

C is the unique topological field that is algebraically closed, Hausdorff, locally compact and not discrete

[cosmic mirage]

ah sure

[kind marlin]

I forgot the important bit lol

[cosmic mirage]

lol yeah i was like what

i have wondered for a while if theres a completely non topological description for R as a field

[unreal stratus]

Special case of classification of local fields

[kind marlin]

wow there’s just a term for everything

[unreal stratus]

Local fields are very important in number theory, say

[kind marlin]

C is the only algebraically closed local field

[fickle tendon]

you can very nicely classify it up to elementary equivalence as a non alg-closed field with a finite degree algebraic closure iirc

(elementary equivalence means that all formulas behave the same)

[unreal stratus]

Degree 2 subextension of [algebraic characterisation of C] I guess?

[cosmic mirage]

but to even say local field you need it to be a topological field smh

[unreal stratus]

And then essentially for same reason as enpeace like

[cosmic mirage]

yeah these 2 makes sense

[unreal stratus]

Algebraically closed fields are determined up isomorphism by their characteristic and cardinality

So e.g. R is unique field of char 0 and cardinality 2^{\aleph_0} whose algebraic closure is a degree 2 extension

[fickle tendon]

https://en.wikipedia.org/wiki/Real_closed_field
the equivalent conditions are crazy

[cosmic mirage]

or as some may call it, \aleph_1

[fickle tendon]

char 0 isn't even a necessary assumption iirc

[unreal stratus]

Yes I believe this

[fickle tendon]

(implied by the elementary equivalence)

[unreal stratus]

Artin–Schreier theorem

[fickle tendon]

model theory is funn

y

[normal swift]

EXACTLY im so glad ppl are talking about this

[normal swift]

hey ive got a load of wiki articles i saved

all v cool articles

hmm wait, the reason i brought this up was i thought this included the definition for RCF but apparently i didnt save that

[fickle tendon]

Its so funny to me to see "closure operator" in the same list as forcing or representing a complete atomic boolean algebra

[normal swift]

what

[fickle tendon]

imo they should teach closure operators in an intro to math course

[normal swift]

no they dont

[fickle tendon]

they should

[normal swift]

exactly what benefit does it bring😭

[fickle tendon]

they pop up everywhere

as a mathematician the notion of a closure should be as natural as the notion of a function

[normal swift]

you have really uhh radical views

[fickle tendon]

I'm a universal algebraist, what can I say

[normal swift]

is that just category theory

[fickle tendon]

though reasoning using closure operators is super important in basically any branch of math that isn't like pure category theory

and even then closure operators are used as an intuition for monads

[fickle tendon]

no.

[normal swift]

im not goibng to ask what a monad is

[fickle tendon]

they're generalisations of the notion of "algebraic structure"

[normal swift]

[fickle tendon]

anyways, UA is the study of finitary monads over Set, i.e. the representation theory of small algebraic theories i.e. clones

[normal swift]

you really think i could understand that.

[fickle tendon]

I should've been sleeping 4 hours ago my brain isn't working properly

[normal swift]

please do.

[fickle tendon]

uuu I'm sure they're all the same notion categorically

waves hands

[normal swift]

even the nonstandard one

[fickle tendon]

maybe not

[normal swift]

we really have a knack for talking about stuff in unrelated channels fr

[fickle tendon]

I mean,,,

closure operators originated from topology

and like closure operators and lattices come up so much in UA that it might as well be the study of finitary closure operators 🫠

[normal swift]

ua must be really cool

i loved the general isomorphism theorems

[fickle tendon]

in a special case there is an analogue of the Jordan-Holder theorem lol

[normal swift]

yes, i know what that means

https://tenor.com/view/cat-bike-gif-9672975825298871464

[fickle tendon]

damn you don't know Jordan-Holder?

do you even know what a group is atp

[normal swift]

🧑‍🦱 🧑‍🦰 🤵 *100 = group

[quartz horizon]

I also don’t know Jordan-holder

[opaque scroll]

What it says it's that any composition series of a finite group has the same composition factors. Which loosely means that every finite group "factors" into a simple groups, and that what those factors are is unique.

[normal swift]

that.

thats why the classification of finite simple groups is so high and mighty right

[rancid umbra]

yea

from dummit and foote

[radiant stone]

me when:

[unreal stratus]

Idk Jordan–Holder either tbh

I know a similarly named theorem called Jordan–Hölder though

[fickle tendon]

grrrr

me when I'm too lazy

[unreal stratus]

Sorry

[fickle tendon]

lol it's okay dw

[prime elbow]

i am trying to show if X is Lindelof and Y is compact then X \times Y is Lindeolf.

so it is sufficient if we can prove for basis covering

say X\times Y covers by {U_i \times V_i }

now any hint?

because i don't see how X and Y's properties will help me here?

[lucid ocean]

I'm guessing it's gonna be like proving that X x Y is compact

you look at slices like {x} x Y

[prime elbow]

okay, thank you

[prime elbow]

i got it, thank you @lucid ocean

[lucid ocean]

[umbral hamlet]

what is the intuition for homotopy equivalence of spaces?

i understand that functions f,g are homotopic if you can basically have a "slider" that morphs f into g

but when im describing two spaces as homotopically equivalent, what should i be thinking about in my head?

[crisp lintel]

I think you can kinda think about it like a homeomorphism but you've relaxed the condition of being bijective in a loose sense

since you are allowed to not only mold the space but also compress it down

[umbral hamlet]

i see

[crisp lintel]

since for example the plane is homotopy equivalent to a point

[umbral hamlet]

i understand that i can shrink the plane down to a point

but i cant see how im supposed to properly understand how we can recover the plane from the point

[crisp lintel]

you can't

[umbral hamlet]

i mean i suppose it would just be the function from 0 to R^2 that gives us the homotopy equivalence

or wait no right

[crisp lintel]

yeah

but homotopy equivalence doesn't need to be a bijection

[umbral hamlet]

theres nothing about homotopy equivalences that says we have to be able to recover the spaces from each other

[crisp lintel]

ya

but you still have to do it continuously, so you can't compress the circle to a point obv

but you can compress an annulus down to a circle

[umbral hamlet]

i see

i dont really see how this is different from just looking at images of continuous functions though

i mean i suppose it will just turn out that homotopy equivalences preserve more properties than continuous functions do

[crisp lintel]

I guess another vague intuition you could use is that its sorta like a "homeomorphism that doesn't respect dimension"

in the sense that higher "dimension" spaces can be compressed down

perhaps too vague to be useful

but homotopy equivalence still captures a nice amount of structure (like the holes in the space)

which don't really have anything to do with the "dimensionality"

(which I'm being vague about it because its not really well defined)

[crisp lintel]

what do you mean by images of continuous functions

its very different than that

because it requires a genuine continuous function in the reverse direction

[umbral hamlet]

i suppose ill increase my confusion to

how is it different than just looking at a bicontinuous function

[crisp lintel]

well every two spaces you can find bicontinuous functions

just map everything to a point

in both directions

[umbral hamlet]

i see

but the requirement that the compositions are homotopic to the identity maps in each space is what allows us to come up with relevant info

[crisp lintel]

you need in particular that the composition is homotopic to the identity, meaning that the space can be "deformed down" into the image of the composition

in a continuous way

[umbral hamlet]

ahh

alright its registering to me why we really need to care about this now

i mean i knew homotopy was important i just wanted to know why it was

[crisp lintel]

homotopy plays nicely with paths and homotopy groups

[sturdy fox]

are transfinite topological objects legit? for example a circle with radius omega? which would appear to be completely flat for all finite amounts but would still be curved in the transfinite lens?

[opaque scroll]

I mean, curvature isn't really a topological property, so it's not clear how such a circle would be different to any other circle.

[sturdy fox]

i guess a transfinite manifold/geometry would be a better name

[opaque scroll]

I suppose, though in terms of Riemann curvature all circles are flat.

Maybe there is some other notion of curvature that would make sense...

[sturdy fox]

ig gaussian curvature

but well we can always think about spheres instead if we want riemann curvature

[iron bolt]

I think the closest actual thing to what you're asking are non-paracompact manifolds - i.e. locally Euclidean Hausdorff spaces, what you get by removing from the definition of manifolds the requirement that the space is metrisable / second-countable / paracompact

topologically there are some examples of these, for example the infamous long line (which is in fact defined directly in terms of a transfinite ordinal)

but it mostly servers as a counterexample afaik, with little to no actual use

I'm also not sure if it's possible to put a smooth structure on it, but what you definitely can't do is equip it with a Riemannian metric

because under such a metric, every point would have to be some finite distance away from the origin, which would allow you to construct an embedding into R

which is impossible because it is too long

[sturdy fox]

yeah i thought of the long line too and that wed lose the metric, what about using hyperreals or surreals?

[median sand]

Where can I find a thorough exposition of compactly generated spaces and how this interacts with uniformities and the compact-open topology? Entry mention of these topics I've seen so far has been in bits and pieces.

[tender halo]

engelking has like a chapter on k-spaces?

i haven't seen a super-coherent exposition

i imagine there should be a discussion in texts on convenient settings for topology

so maybe look at preuss' books

or something of the sort

the chapter on function spaces in "foundations of topology" seems to be talking about what you want

[hot bramble]

"Algebraic Topology" by Allen Hatcher, it is a nice starting point, introduces k-spaces, same e thing as compactly generated spaces

Lecture notes by charles rezk

a convenient category of topological spaces by NE Steenrod

A Concise Course in Algebraic Topology by J. P. May, I think maybe Chp.5, I forgor

[spice plinth]

just asked this in #real-complex-analysis, but realized this is probably a more appropriate channel: i was trying to prove X := [0,1]^[0,1] with the product topology is not metrizable, but confused myself with something. X should be path connected by interpolating, f + (g - f)t between any two points f, g in X, and so X is connected. but each set U_c = { f in X : f(0) = c } for fixed c in [0,1] is a basic open cylinder, and for c1 =/= c2, U_c1 and U_c2 are disjoint nonempty open sets, so over all U_c's, X can be expressed as the union of multiple disjoint nonempty open sets, contradicting connectedness. what have i overlooked?

(for the actual problem in question, i've so far got: X is compact by Tychonoff's, so is lindelöf as well, and a metric space is lindelöf if and only if it is separable, which X clearly is not by considering the restriction of any dense set to each of the aforementioned uncountably-many disjoint nonempty cylinder sets)

[kind marlin]

is { f in X : f(0) = c } open? let f = 0 uniformly. any open basis set containing f must correspond to R everywhere except finitely many entries, where we can instead use open intervals. but regardless, we will never be able to pick a basis set at 0 that only contains c

(not R sorry, just [0, 1] but you get the idea)

[spice plinth]

by 'correspond' you mean?

[kind marlin]

as in the entries of the product are [0,1] everywhere

except for finitely many entries

[spice plinth]

i mustve just mixed something up in the definition of the product topology then, i was thinkin the open cyliders of the form { f in X : f(x) = y } for fixed x,y in [0,1] were exactly the basic open sets

or at least a basis for the product topology

[kind marlin]

if you just write it as f(x) \in Y for an open interval y then that should work

[spice plinth]

yeah i think i was over-generalizing from baire space which im more used to, ofc the topology of each section [0,1] makes the basis more complicated

[kind marlin]

oh and specifically a subbasis

[spice plinth]

technically im rite if we give [0,1] the discrete metric lol 🤓🤓🤓☝️☝️☝️ ig i was doing that implicitly anyway

[kind marlin]

oh I was going to say that, I’ve seen that exact cylinder convention in measure theory when defining a topology on S^N for finite S in terms of the discrete topology on S

[gaunt linden]

S being finite there is a special case: the product will be a Cantor space.

[gaunt linden]

But then the interpolating path you used to argue for path-connectedness would not be continuous.

(Since the domain of a "path" is definitely [0,1] with the usual topology).

[spice plinth]

by c.c.c. then we can't prove X is non-separable this way, since we can only get countably-many disjoint open intervals {Yi} in [0,1] giving countably-many nonempty disjoint open sets in X of the form { f in X : f(0) in Yi } that we can restrict our dense set to. how else could we prove non-separability then?

[kind marlin]

Products of connected spaces are connected which gives us the result really easily

if that feels like a cheat you can try proving that instead

[spice plinth]

did not know this but also answers my question hehe

[kind marlin]

oh sorry I misunderstood what you were saying

I keep forgetting what separable means

[spice plinth]

tbf it is badly named

[gaunt linden]

My approach would be to say that in a metrizable space, {x} is a countable intersection of neighborhoods of x, thus also a countable intersection of basic neighborhoods of x, but each of those neighborhoods control only finitely many coordinates.

[spice plinth]

say i don't see why this would matter actually, wouldn't we just be working with [0,1] having the discrete topology for the construction of X = [0,1]^[0,1] but having the normal topology for the path? dont see why the first would necessitate applying the discrete topology in the latter case

[warped helm]

doesn't this depend on the topology used on the product?

[spice plinth]

(interpolation might still not be continuous tho i haven't checked to prove this)

[warped helm]

R^omega in the box topology (who care though its the box topology) is not connected even though the factors R are

[tender halo]

X is separable

[gaunt linden]

Yes, but assume that f != g such that f(a) != g(a) for some a. Let b be the arithmetic mean of f(a) and g(a) -- then the preimage of the cylinder { h | h(a) = b } under your interpolating map would be { 1/2 }, which is not open in [0,1] with the usual topology, showing that the interpolating map is not continuous..

[tender halo]

by hewitt marczewski pondiczery

the dense set is uhh points with rational coordinates with finite amount of distinct values

but idk whats the point of all this when its obviously not first countable

for many reasons, for example it contains an cantor space of uncountable weight as a subspace

T1 product topologies of more than omega spaces are not first countable

uhh with like factors not equal to 1 or 0 of whatever

because countable amount of basic sets at a point can only separate you from other points at countable amount of indices

alternatively, X is separable but not second countable

and therefore not metric

[spice plinth]

Just checking how to prove this. Let D = { f in [0,1]^[0,1] : f([0,1]) is a finite subset of Q }. Then for any basic open set U in [0,1]^[0,1] of the form U = { f in [0,1]^[0,1] : f(c) in W } for some open W in [0,1] and arbitrary fixed c in [0,1], we can pick f in D such that f(c) = q for some rational q in W, which must exist by density of Q in [0,1]. So f is in U and hence D is dense in [0,1]^[0,1]. This correct?

hmm is D countable tho

was that the set u meant, or something more precise?

[tender halo]

no i guess its not really true, you need to actually do the construction

i thought i get to cheat

no i mean the fact is true

the proof isn't

[spice plinth]

yea no it's def not countable the way i've written, the set of functions f_r(t) = 0 for t=/=r and f_r(t) = 1 for t=r over all r in [0,1] is uncountable but contained in D.

ill google hmp for the precise construction tho, thanks

[tender halo]

so, imagine c = 2^\omega (the index set of X) as a Cantor space. it is second countable, fix a base for it

lets look at the family of finite families of disjoint basic sets of c, it is also countable

our dense set will be functions from c to rationals in [0; 1] that are constant on some finite family of disjoint basic sets of c

[spice plinth]

(is there any reason to choose cantor space over [0,1] since they're both 2nd countable and have the same cardinality as index sets?)

[tender halo]

not really i guess, ur right

[tender halo]

oh and constant on their complement

(complement of their union, that is)

hewitt marczewski pondiczery is basically the statement that density \leq m is 2^m-multiplicative

which is why i tried to use the "natural" topology on 2^m

[umbral hamlet]

in class we have to answer questions like these after having known about homotopy only for 3 days and we have about 15-20 minutes to answer 8-10 such questions

is there a way that i can understand these types of questions without going through the whole process of checking for homotopy in my head

bc i dont have enough time on these quizzes to do so

i should say, every week we have to take a very short quiz with a lot of questions on what we learned that week

but often it feels like i havent had enough time to be comfortable with the ideas so i cant quickly apply them

[young stone]

Atleast for these questions, you don't have to check much because every function from a discrete space is continuous and every function to an indiscrete space is continuous

[young stone]

And there are "very few" continuous functions to a discrete space and "very few" continuous functions from an indiscrete space (what are they?)

[umbral hamlet]

Well for S0 to a singleton theres only one

Same for singleton to S0

S0 to itself there are four

[umbral hamlet]

I see

I guess i was just overcomplicating it

I usually do that on quizzes and exams

[orchid perch]

how do you show that if (X, T) is Hausdorff that any singleton is closed

pm if u can help

[crisp lintel]

try and show that the complement of a singleton is open

by writing it as a union of open sets

[orchid perch]

I don’t see how that could help

Or link it to the HD

[crisp lintel]

hausdorff allows you to obtain an open set around any point in {x}^c that doesn't contain x

[orchid perch]

but u know that X is HD not that X complement is HD??

[crisp lintel]

x is a point in X

then {x} is a singleton

[velvet salmon]

The point is that for a singleton {x} in X you can separate x from any y \neq x by opens since X is Hausdorff

[gaunt linden]

Shouldn't there be two?

[deep delta]

Good Evening, Im not sure if this is the right channel, but Im new to knot theory and Im curious why there is no Alexander polynomial for the sheet bend knot. It seems quite similar to the granny knot that has such a polynomial. Thanks for any answers.

[gaunt linden]

(I don't know the answer to your question, but knot theory is specifically mentioned in the channel description for #alg-top-geo-top, so you might try there if you don't get a response here).

((If you're comparing statements from two different sources, it may be because they differ in how they speak about knots with free ends of rope. The mathematical "knot theory" generally only treats "knots" without any free ends; but both a granny knot and a sheet bend as depicted in ropecraft books have such free ends. One source might implicitly imagine connecting those free ends together such that the existing math can be applied to it, whereas another says, the theory doesn't apply to a knot of this kind.))

[deep delta]

Thank you, I will try it there. And I thought about that but Im quite sure thats not the reason, but thank you for the answer.

[umbral hamlet]

Right idk why i said one

It was like 4:30 am when I said that so I was out of it i guess

[turbid mist]

Should I learn nets/filters/ultrafilters? I never read a book that used them and now I needed that for a proof. I found another proof that doesn't use ultrafilters, but should I learn it anyway?

[cosmic mirage]

I mean if you don't need it then that's not a reason to learn it right

but it is very cool

[crisp lintel]

yea they are quite useful and honestly not that hard to learn

[frigid sapphire]

wdym "there is no"?

just connect the ends in the canonical way, pick an orientation and find the writhe, and do the kauffman bracket

[dawn frigate]

Does U have to be open for 4

[opaque scroll]

It doesn't need to be, but you can choose it to be if you like since every neighborhood contains an open neighbourhood

[dawn frigate]

So i can say V is open in Y and B is closed and Y minus B is open in Y and X minus f inverse (Y minus B) = f inverse of B is closed in X and let U equals f inverse of B

[opaque scroll]

You're trying to prove the 3 implies 4?

And presumably you meant that B is equal to YminusV?

Then yes

[dawn frigate]

Yes

That is my intention

[dawn frigate]

So the sub space is the dictionary order topology do I have to show in each case it’s open in dictionary order topology

[opaque cloud]

holy handwriting, it's like Q in R

Embedding and Imbedding are the same things right? Just different ways to spell the same thing?

i.e an injective continuous map f : X --> Y that becomes a homeomorphism when you restrict the codomain to f(X)?

[crisp lintel]

I've never heard imbedding

[tender halo]

just a less common spelling

[iron bolt]

nope. this would be the case if every bijective continuous function was a homeomorphism, but counterexamples exist

for example, you can easily come up with a continuous map from ||[0,1)|| onto ||the circle S^1|| that's bijective but doesn't have a continuous inverse

what you suggest does hold for compact Hausdorff spaces though - every continuous map from a compact space to a Hausdorff space is closed, and closed continuous bijections are homeomorphisms, so in particular every continuous bijection between compact Hausdorff spaces is a homeomorphism

wait, my bad. I thought you were saying that every continuous injection was a homeomorphism onto its image, not that the maps you call imbeddings are

then that's indeed just the embeddings

[orchid perch]

you have to find a counterexample as to why this is wrong but the only thing i can imagine is trying to find a counter-domain where substraction isnt defined otherwise i wouldnt see why this wouldnt work

but like cant u always change ur codomain again to R for example

so im confused where the mistake lies

[prime elbow]

I think you are right, because it is not necessary that you have subtraction operation in co domain

[prime elbow]

How can I change my co domain to R?

[orchid perch]

i mean idk sometimes u can do it if u dont care abt it being surjective

but im just confused w the codomain properties

[opaque scroll]

What's the context here, like what is the codomain of f and g supposed to be.

You can't necessarily just change the codomain to be R if it isn't to begin with no

[orchid perch]

u dont know how f/g look

u just have to find a counterexample so that this doesnt work

[opaque scroll]

I mean if there's no restriction on what the symbols mean, then you can pick f to be "fish tacos" and g to be "square root of vegetables" and then nothing makes any sense.

But okay let's say f and g are continuous functions between topological spaces X and Y, and let's even say that the underlying set of Y is R (but possibly with a nonstandard topology), so that subtraction makes sense. Then the reasoning is still false, because it might be that h isn't continuous.

(I'm assuming you're supposed to conclude A is closed)

[orchid perch]

but if f & g are continuous wouldnt that make h automatically continuous

[orchid perch]

yea

the fault lied in these steps tho apparently

[opaque scroll]

It would not. Maybe you can think of why

Or perhaps try to argue why you think it would be true and then see where your logic fails

[orchid perch]

and this is probably a dumb question but like cant u always change ur codomain to something else like even if youd have f:{0,1}->{a,b} with f(0)=a en f(1)=b like if you know that a and b will lie in R cant u still say f:{0,1}->R

like cant u always expand the codomain

[prime elbow]

But what if a and b are not element of R?

[opaque scroll]

I mean sure, but why would a and b lie in R?

[orchid perch]

but like the thing is even if it isnt in R lets say it lies in C then

like cant u always have it be expanded

so that ur subtraction is still defined

[opaque scroll]

Say I define a topological space Y where the elements are {fish, noodle, cat}. How would you expand that to define subtraction?

Anyway, this is sort of a distraction since it doesn't work even when subtraction is defined

[orchid perch]

i cant find a continuous function f, g so that h isnt

[gaunt linden]

At the very least we'd need subtraction to be continuous rather than merely defined.

[gaunt linden]

Note that you probably shouldn't be looking for counterexamples to the conclusion (that A is closed if f and g are continuous), since that is true whenever the codomain topology is Hausdorff.
The interesting counterexample would be a topology with an example showing f-g is not necessarily continuous.

(If the topology doesn't even have to be Hausdorff, using the trivial topology on the codomain permits some boring counterexamples to the conclusion, too).

[opaque scroll]

I mean why not look for an example that isn't hausdorff?

[iron bolt]

this is probably not super related, but to give one naturally occuring example of a space with a subtraction that isn't continuous: on [-∞,∞] with its usual topology, subtraction is continuous everywhere but in (∞,∞) and (-∞, ∞)

with there being no way to assign values to ∞ - ∞ and (-∞) - (-∞) that makes subtraction continuous

[rotund prism]

so connected sets on wikipedia says that a formal definition of a space being disconnected is equal to the union of two disjoint open sets. And that a failure to be disconnected implies being connected.

But so um. why open sets?

Take a filled unit square in R^2. This is a closed set. Take the set difference of it with a line, eg a line that bisects the square. Call our cut square S. S is neither open nor closed. Intuitively, S is not connected. Intuitively, S seems disconnected in the important way.

But there is no open set that is equal to one of the halfs of S, right? So it fails Wikipedia's "formal definition" of being disconnected, which wikipedia says means S is connected? Is my line of reasoning good? Is wikipedia mistaken in some sense? Am I missing something?

[frigid sedge]

I am actually starting with Point set topology

and I will be following "topology without tears"

any sggestions or tips?

I am a second year undergraduate stud sem 3

[tiny obsidian]

the definition of connected on wikipedia you're looking at is connectedness of a full space, so you should be looking at the halves as subsets of S, and then they're open. Also S is open and closed in itself! (I assume you mentioned this because of the 2nd bullet point)

[rotund prism]

Huh? S is both closed and open? Its clopen? I thought that a point on the boundary of S not having any ball entirely contained in S makes it not open?

[tiny obsidian]

S is closed and open as a subset of S

[rotund prism]

ah

[tiny obsidian]

As a subset of the plane indeed it is not open because of the balls on the outer boundary

[rotund prism]

so is there just not much use in talking about the connectedness of a subset of an overall space, with respect to the overall space's topology?

[tiny obsidian]

There is, and I can even spend some time trying to remember a definition for it based off the topology on X, but it is equivalent to the subset being connected w.r.t the subspace topology

which I assume is why wikipedia doesn't include it explicitly

U', V' are open subsets in A a subset of X iff they are of the form U n A, V n A for U, V open in X (that's what the subspace topology is) and then you get some conditions on U, V based off of this

So it's something like "A subset X is disconnected if and only if there exist open sets U, V of X such that A is a subset of their union and their intersections with A are nonempty and disjoint"

[gaunt linden]

Disconnected.

[tiny obsidian]

yes..

[rotund prism]

So then yea, S here is disconnected, because we can seperate it into the two halves. These two halves have a union equal to S, and the two halves are both open with respect to S itself. Alright, yea thats fair, and I can see why we want the components to be open in that aspect

[gritty widget]

Connectedneds is also equivalent to the following property of a space X:

For every open cover U_a of X, and any two points p, q in X, there is a sequence U_1, …, U_n, of elements of the open cover, s.t. p in U_1, q in U_n, and U_i intersects U_(i+1) nontrivially. This might better fit your intuition for connectedness.

[lucid ocean]

a cake is connected if you cannot cut a slice out of it without leaving some on the knife

[opaque cloud]

Lol

[opaque cloud]

🫡

[opaque cloud]

Munkres seems to use this spelling

[lucid ocean]

weird variations in spelling everywhere

"embue" vs "imbue", "reflexions" vs "reflections", etc.

[queen prism]

just the author being somewhere between extremely old-fashioned, influenced by other languages, or just straight-up pretentious

👍

(they're all kinda the same)

[sonic crane]

🗿 is just a great emote

[sly geyser]

I don't understand (ok I'll say some more things later lemme think for now I guess)

hmmm

(I'm mainly thinking about the last sentence in the picture, about the projective limit in the category of topological spaces and all)

[unreal stratus]

i would say this 2nd definition is what one usually called "profinite space/set" btw

[sly geyser]

just need to ask

I know what an inverse system is of course

but I need to check

is it specifically an inverse system over N?

[quartz horizon]

i don't think so

it'd be over any cofiltered categoryy

[sly geyser]

so this but in reverse?

[quartz horizon]

yep!

filtered categories are categorifications of directed sets

a directed set is a preordered set where every finite subset has an upper bound

so a filtered category is a category where every finite diagram has an "upper bound", i.e. a cocone

[sly geyser]

btw when you say "an inverse system over a cofiltered category" do you mean "a normal system under a filtered category" or "a normal system under a cofiltered category"

[quartz horizon]

conversely, a cofiltered category is a category where every finite diagram has a "lower bound", i.e. a cone

[quartz horizon]

it just means you have a diagram whose shape is a cofiltered category

and you're taking the limit of that diagram

[sly geyser]

ok so "a normal system under a cofiltered category" seemingly

[sly geyser]

so in our inverse system there's a "lower bound" for every two objects?

[quartz horizon]

for any finite diagram, actually

which reduces to having a lower bound for any two objects, and a lower bound for any two morphisms

[sly geyser]

wtf was just sent here

[quartz horizon]

dw about it

[fickle tendon]

the better equivalent condition is as the prime spectrum of a Boolean ring of course

[quartz horizon]

The what

[fickle tendon]

Stone duality!

[quartz horizon]

wuzzat

[fickle tendon]

youve got a correspondence between Boolean algebras and Boolean rings (fully idempotent rings), and the Boolean algebra of clopen sets of Spec B is naturally isomorphic to B

where Spec B is given the usual Zariski topology

[quartz horizon]

how does that correspondence work

[fickle tendon]

formally its a correspondence of Lawvere / algebraic theories

[quartz horizon]

Ok hang on

What even is a Boolean algebra/boolean ring

[fickle tendon]

this is perhaps better suited to #foundations, lets move there

[sly geyser]

my man followed me from foundations 😭

[fickle tendon]

i am everywhere

though usually in the algebra channels

[sly geyser]

what is the topology in this case

[fickle tendon]

the closed sets are given by V(X) := { I ∈ Spec B | X ⊆ I } for X ⊆ B

the motivation for this is classical algebraic geometry

[sly geyser]

what is V(X)

[fickle tendon]

I just defined it

there

[sly geyser]

no I know it's a definition but what is it supposed to represent, is it a closed set or a set of closed sets (it looks like a closed set not a set of them from what I understand, the set of all ideals that are a superset of X being a closed set for each X I guess, but need to check)

[fickle tendon]

what

you let your train of thought spill into your sentences sometimes lol

not great for communication purposes

[sly geyser]

is V(X) a closed set for each X

[fickle tendon]

yes that's literally what this says 😭

[orchid perch]

$(x_0, y_0) R (x_1, y_1) \Longleftrightarrow x_0 + y_0^2 = x_1 + y_1^2$
and $X$ the corresponding quotient space. What is $X$ homeomorphic to?

[gentle osprey]

ana

[orchid perch]

i have the solutions (theyre in a diff language tho) but im having a hard time trying to see what were actually doing

[sly geyser]

@fickle tendon what this guy is asking is equivalent to R right?

[orchid perch]

and we use this diagram to create like that f bar so that f bar is a continuous bijection

[fickle tendon]

who knows, quotient spaces are weirddd ¯_(ツ)_/¯

[fickle tendon]

if you show that ker q = ker f, then fbar is bijective yes

then you still have to show the inverse is continuous I suppose

[quartz horizon]

This is why universal property is the way

[orchid perch]

yeah we then created this g so that later g = f bar ^-1

[fickle tendon]

yurrrr

[orchid perch]

but like how do we make that g

[quartz horizon]

Oh that’s more straightforward

You use the continuous map “z -> (z, 0)” from R to R^2

[sly geyser]

it's just that you can attach each positive and negative y together and get it to be the quotient on R x (R>=0), then it should be just a stretched and squeezed version of the equivalence of x+y on R x (R>=0) and that should just be equivalent to R

[quartz horizon]

Then you compose with the quotient map R^2 -> R^2/~

[fickle tendon]

using desmos, it seems that the equation x + y^2 = a sweeps out a family of parabolas as you vary a

and all those parabolas get collapsed to the same point

so it's some fucked up homotopy or wtv lol

[orchid perch]

and how am i supposed to see that without any tools

[quartz horizon]

No it’s just R

[quartz horizon]

Universal property helps for this

[sly geyser]

it's just R

I explained why

[fickle tendon]

well, you can see directly that \mathbbR^2 / R is in bijection with R by seeing that every equivalence class is represented by a single real number ig

[quartz horizon]

Specifically, you can think about what it means to define a continuous function out of your space

[fickle tendon]

sorry I meant a fucked up deformation retract of R^2 onto the x-axis

[sly geyser]

you attach (x,y) and (x,-y) together

[quartz horizon]

You just need to define a continuous f : R^2 -> Z such that if x_0^2 + y_0 = x_1^2 + y_1, then f(x_0, y_0) = f(x_1, y_1)

[fickle tendon]

like the projection onto R is given by [(x, y)]_R -> x + y^2

where [x]_R is the equivalence class of x under R

[quartz horizon]

This f is determined by its values on the x axis

And you can choose any continuous function R -> Z and build an f out of it

[sly geyser]

as (x,y) -> (x,y^2) is a automorphism on R x (R>=0), our x + y^2 quotient on R x (R>=0) is equivalent to the quotient under the automorphism, which is just the x+y quotient
it's kinda obvious from here that it's just R

[quartz horizon]

This isn’t quite the argument I’d use but sure

[sly geyser]

Z?

[quartz horizon]

Z is some arbitrary topological space

Essentially the idea is that $f(x, y) = f(x + y^2, 0)$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

So if you define $g : \mathbb{R} \to Z$ by $g(x) = f(x, 0)$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

Then $f(x, y) = g(x + y^2)$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

This sets up a natural correspondence between continuous functions R^2 -> Z respecting the relation, and continuous functions R -> Z

Hence the quotient is iso to R

[orchid perch]

and how do u show that f bar ^-1 is continuous

[quartz horizon]

I’m cheating a little here by using the yoneda lemma I guess

[orchid perch]

we havent seen that i think

im rlly new to quotientspaces so everythings a little too abstract rn and i cant follow

[fickle tendon]

I am like 100% sure you haven't

[fickle tendon]

well the yoneda lemma is pure category theory

[quartz horizon]

Right, sorry

I also found quotient spaces really tough to work with when I first learned about them

They’ve only started to make sense to me once I learned about the universal property

[orchid perch]

these r the solutions i have

im also not rlly sure why we make that g o f bar

[quartz horizon]

Perhaps - could I try to tell you about this universal property thing?

[orchid perch]

sure

[quartz horizon]

I think a simple example of a quotient space is a cylinder

It’s a quotient of a square

The universal property is about what the cylinder “does”, how it interacts with other spaces

Specifically, we think about embedding (or more generally, mapping) the cylinder into another topological space $Z$, which could be anything you like

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

So, we'll consider continuous functions $f : \text{Cylinder} \to Z$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

ok so far?

[orchid perch]

ye

[quartz horizon]

now, the cylinder is a square where two of the edges are identified

if we "unwrap" it, we'll get a continuous function $g : \text{Square} \to Z$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

defined by "take your point on the square, map it to the corresponding point on the cylinder, and then use f"

but we don't just get any ol' continuous function

the edges are identified, after all

so, the continuous function has to give the same values on the left and right edges of the square

$g(0, y) = g(1, y)$, if the square is $[0, 1] \times [0, 1]$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

does that make sense?

[orchid perch]

yh

[quartz horizon]

the point of the quotient topology definition is that we can go in reverse

if we have a continuous function $g : \text{Square} \to Z$ such that $g(0, y) = g(1, y)$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

then we automatically get a continuous function $f : \text{Cylinder} \to Z$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

in other words, continuous functions out of the cylinder correspond to continuous functions out of the square that satisfy g(0, y) = g(1, y)

[orchid perch]

okay cool

[quartz horizon]

this turns out to be true for any quotient space!

if you have a topological space $X$, and an equivalence relation $\sim$ on it

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

then continuous functions $f : X / \sim \to Z$ correspond to continuous functions $g : X \to Z$ such that $x \sim y \implies g(x) = g(y)$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

the reason this is nice is that checking continuity of f is often hard to think about, because you have to directly work with what open sets in "X/~" are

on the other hand, checking continuity of g is significantly easier, because you only have to work with the open sets of X

then you just need to check that $x \sim y \implies g(x) = g(y)$, which doesn't actually involve any topology

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

do you see what i mean? this trick lets you define continuous maps out of a quotient without having to work with the quotient topology directly

instead, you define a map out of the original space that satisfies some (non-topological) conditions

[orchid perch]

Yeah

[quartz horizon]

@orchid perch

oh, sorry

[orchid perch]

sorry mb

Haha

[quartz horizon]

so, we can use this to help with your problem!

take our original space $\mathbb{R}^2 / \sim$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

then continuous functions $\mathbb{R}^2 / \sim \to Z$ correspond to continuous functions $f : \mathbb{R}^2 \to Z$ such that $f(x_0, y_0) = f(x_1, y_1)$ whenever $x_0 + y_0^2 = x_1 + y_1^2$, right?

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[orchid perch]

Yea

[quartz horizon]

Then, any such function has to satisfy $f(x, y) = f(x + y^2, 0)$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

because $x + y^2 = (x + y^2) + (0)^2$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[orchid perch]

mhm

[quartz horizon]

so if we get $g(t) = f(t, 0)$, then $f(x, y) = g(x + y^2)$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[tender halo]

||here we see category theorists in their natural habitat, stalking the chats and jumping at unsuspecting prey, making them learn what a quotient object is||

[quartz horizon]

note that $g : \mathbb{R} \to Z$ is guaranteed to be continuous

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[fickle tendon]

you say this but eventually you will come crawling back to them when you need their help

[quartz horizon]

make sense so far?

[tender halo]

no! no! i refuse to learn what a filtered colimit is!

[orchid perch]

yeah

[quartz horizon]

we can also go the other way!

if $h : \mathbb{R} \to Z$ is any continuous function

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

then we can define a continuous function $f : \mathbb{R}^2 \to Z$ by $f(x, y) = h(x + y^2)$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

and that satisfies our requirement of f(x_0, y_0) = f(x_1, y_1) whenever x_0 + y_0^2 = x_1 + y_1^2, just because h is a function

[orchid perch]

Okay

Tyyy

I’ll review my notes again and write your commentary but ty for the extra information I think this will help a lot

[quartz horizon]

mhm mhm

it's quite easy to finish off from here, actually

what we've shown is that you can take a continuous map $\mathbb{R}^2 / \sim \to Z$, and convert it to a continuous map $\mathbb{R} \to Z$, and also do the reverse

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

if you set $Z = \mathbb{R}^2 / \sim$ and $Z = \mathbb{R}$, then you get continuous maps both ways which are inverses of each other

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

one of which is $\bar f$ actually

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

that's how you get the homeomorphism!

[orchid perch]

Okay ty

[quartz horizon]

to be fair i think cat theory was really helpful for me wrt working with quotient objects in topology

[tender halo]

i mean makes sense, given the original aims of cat theory

algtop bookkeeping etc

[quartz horizon]

the quotient topology is just really awkward to work with directly a lot of the time

quotients in general honestly

[tender halo]

yea its like

the worst way to make a new topological space frankly

unless its proper and then its cool

[quartz horizon]

proper?

[tender halo]

closed with compact fibers

(alternatively, just compact fibers, which is the same if the codomain is locally compact)

[quartz horizon]

hm i see

[tender halo]

maybe you've heard them called perfect maps?

[quartz horizon]

nope

[tender halo]

(the difference is that perfect ones are surjective)

basically its the nicest type of quotient map you can realistically have

preserves metrizability for one

and a lot of other properties

[mortal narwhal]

Hello, I'm new to this server and still an undergraduate from Spain (I'll begin my third year in college). Due to (reasons) my university doesn't give a fuck about me and won't let me enroll in the thesis this year although I've enrolled in everything else I need for the degree and even more.
So if I don't want to waste time on beaurocracy on the long run I'd better start studying on my own in order to eventualy do resesearch on my own since my university doesn't seem to be bothered to help
My university books also pretty much made me hate most branches, but I came to like point-set topology and now I'll start with Munkres and see if I find algebraic topology interesting later down the line (I better do).

Having said that, I'll be hanging around the topology channels, probably asking for help when absolutely necessary or guidance towards what to study.
This would be my introduction, nice to meet you

[crisp lintel]

ok month late but I just thought of something in the shower about this:

I'm going to stick to discrete topological spaces, but I think the intuition carries over in the general case. Recall that a space X is compact iff every ultrafilter on X converges. Also recall that each point x in X generates a principal ultrafilter that always converges to x itself. For a discrete space X, one construction of the Stone-Cech compactification is as the set of all ultrafilters on X together with a suitable topology. Then the space X embeds within the compactification as the set of principal ultrafilters. We can think of this as taking every ultrafilter on X that doesn't converge, and giving each one a unique limit point. It is in this sense that it is the largest compactification

conversely the one point compactification is as if you invent a single extra point and then force every non-convergent ultrafilter to converge to that point

[solemn willow]

is continuity a required assumption here?

if $\sigma$ covers X, then it also covers $p^{-1}(y)$ for each $y\in{Y}$. since each $p^{-1}(y)$ is compact, there's a finite subcover $U_y\subset\sigma$ for each $p^{-1}(y)$. Applying the hint to $V_y=\cup{U_y}$ we get $W_y$ neighborhood of $y$ such that $p^{-1}(W_y)\subset{V_y}$. doing this for all $y\in{Y}$ we get a cover of $Y$ containing all such $W_y$. since $Y$ is compact, theres a finite subcover $W_{y_1},\dots,W_{y_n}$ that covers $Y$. Hence, $p^{-1}(W_{y_1}),\dots,p^{-1}(W_{y_n})$ covers $X$. Since $p^{-1}(W_y)\subset{V_y}$ we can take $V_{y_i}$ for all $1\leq{i}\leq{n}$ and this covers X. since each $V_{y_i}$ is a finite union of elements in $\sigma$, the claim is proven.

the fact that p is surjective and closed is used to prove the hint, but i cant seem to find where continuity is used. i assume my argument is wrong because of this but not sure where

[gentle osprey]

coal

[crisp lintel]

p sure the hint requires continuity

or maybe just closed implies that

yeah you're right

or does it

ah ok you just take U^c and map it by p

[solemn willow]

yeah

[crisp lintel]

yea I think maybe you don't need continuity for this specific result

but perfect maps in general ofc you want continuity

[solemn willow]

ah ok so he probably just did that for the perfect map remark

alright

[rancid umbra]

can't believe i never noticed that

v cool

[orchid perch]

Take $\gamma$ an index collection, then pick $\beta \in \gamma$ and set $A = \gamma \setminus {\beta}$. Show that
$ \Bigl(\prod_{\alpha \in A} X_\alpha\Bigr) \times X_\beta \cong \prod_{v \in \gamma} X_v $.

[gentle osprey]

ana

[orchid perch]

can i get a hint or something

im having a hard time to even visualize this

[velvet salmon]

This is the generalized version of $(A \times B) \times C \cong A \times B \times C$ so try to see if you can translate this iso to the general case

[gentle osprey]

Moorts

[queen prism]

"\beta \in \gamma" disturbs me mentally

[velvet salmon]

Yeah that's really weird naming

[orchid perch]

its not actually a gamma i just didnt know what the symbol in my book was called

[velvet salmon]

I guess this is topology, so maybe replace iso with homeomorphism

[orchid perch]

😭

It’s whatever that is

[velvet salmon]

Idk either tbh

[queen prism]

https://detexify.kirelabs.org/classify.html seemingly \aries

prob not right

oh well

[orchid perch]

upsilon

analysis is killing m

e

[queen prism]

true

[novel acorn]

greek letter upsilon

[hasty thicket]

am I being silly?? how can it be that a+c/2 is in the open interval but not in the closed interval that contains the open one in the first place?

[queen prism]

well that's the contradiction yea
but you know (a + c)/2 is in (a, c) because a < c (can you justify this?)
and c < b so (a + c)/2 is in (a, b)
on the other hand (a + c)/2 < c means (a + c)/2 can't be in [c, d]

[hasty thicket]

omds jesus christ i shouldn’t do maths with 4h of sleep

thank u i was being very silly

[queen prism]

dw I was in #real-complex-analysis on 1 hour

[halcyon bone]

Why is paracompactness useful?

I’ve been told “partitions of unity are useful for manifolds”

But like I don’t intuitively see WHY it’s useful for this

[tender halo]

it is a sort of "local niceness" condition

[crisp lintel]

well you really want the local finiteness to build partitions of unity

since you don't want to be having infinite sums at any point

partitions of unity are of course vital to define integration and are just useful in general

it is a bit of a technical condition though

there are a decent number of conditions in topology that seem a bit more technical and less natural, but are really just chosen because they are the weakest conditions that allow us to do nice things