point-set-topology

[heady skiff]

ye true

well

i guess i was thinking more about the point that munkres made

[gaunt linden]

(and even if it weren't path-connected, all of its path-connected components would be homeomorphic anyway).

[heady skiff]

pls pls let F be the given path homotopy between alpha and beta

🙏

[gaunt linden]

Not quite exactly -- you'll need to reparameterize the square to make that true.

[heady skiff]

agh

ok i'll try to sink some teeth into this problem

[gaunt linden]

Correction, you don't need a homotopy between alpha and beta. You need a homotopy between alpha+beta and beta+alpha.

[heady skiff]

oh right

that's what i meant

wait lemme check if it works

nvm

doesn't work

that would be too easy and we can't have easy things in math

oh

wait i think i got it

$F(s, t) = \bigl((\alpha * \beta)(s)\bigl)^{(1 - t)} \cdot \bigl((\beta * \alpha)(s)\bigl)^t$

[gentle osprey]

okeyokay

[heady skiff]

and that fulfills all the conditions of a path homotopy right?

$F(s, 0) = (\alpha * \beta)(s)$, $F(s, 1) = (\beta * \alpha)(s)$, $F(0, t) = e$ and $F(1, t) = e$

[gentle osprey]

okeyokay

[heady skiff]

and it's continuous since G is a topological group

[gaunt linden]

I think instead of trying to write down an explicit homotopy as a formula, it would be easier to say that "left and then top side" is homotopic in [0,1]² to "bottom and then right side", and the images of these two endpoints are exactly the concatenated loops in G you want to show are homotopic.

[gaunt linden]

I can't really make out what this means -- it looks like you're trying raise group elements to fractional powers?

[heady skiff]

oh yeaa

forgot only for integers

oops

[heady skiff]

wait sorry, do you mean showing that like the top left L of the square is homotopic to the bottom right L, and then show that the image of the top left L is \alpha * \beta and bottom right L is \beta * \alpha and deduce that they must be homotopic or smt?

or vice versa if the top left L is \beta * \alpha instead

[gaunt linden]

Yes.

(One of those -- I haven't bothered to read your problem statement closely enough to figure out which is which :-)

[heady skiff]

lol fair enough, i should figure that out anyways

okay, thank you!

[heady skiff]

i don't really understand the point of showing that \alpha is path homotopic to itself lol, or am i misinterpreting?

(this is the solution for the exercise i was tryna solve 😭)

[gaunt linden]

Hmm, that doesn't seem to make a lot of sense to me either, sorry.

[heady skiff]

it's alright i'll just ignore it

also when they say Let $(G, m, \tau)$ be a topological group, I'm assuming they're letting $m$ denote the binary operation? but like what does $\alpha \cdot \beta \simeq m\bigl(\alpha(t), \beta(t)\bigl)$ even mean if both of them are paths

[gentle osprey]

okeyokay

[unreal stratus]

Well like

you're multiplying two paths (pointwise) to get another path

i.e. the path t | -> m(alpha(t),beta(t))

[narrow cairn]

linear maps on real spaces are continuous right

surely

[limber wren]

finite dimensional yes, infinite dimensional not necessarily

[narrow cairn]

who cares about infinite dimensional vector spaces

unimportant

[limber wren]

lol then yes

ignorning infinite dimensional spaces, all linear maps are continuous

[unreal stratus]

Eh we care in functional analysis

And infinite dimensional spaces do appear in topology due to intersections w func anal

[high hill]

all the boring ones are finite dimensional...

[mossy rampart]

is the second part of b just intersection Talpha

where if the topology was any smaller it wouldn't be true that every basis element is a subset for Talpha

[ebon galleon]

For this exact reason, one typically doesn't care much about finite dimensional vector spaces when talking about continuity.

[narrow cairn]

lee?

[queen prism]

monkares

[narrow cairn]

ah

[west brook]

One man’s boring is another man’s well-behaved

[narrow cairn]

typesetting is similar to lee

[mossy rampart]

is this right

and then for the first part you would say that that if we take union of all Talpha and any other sets needed to create a topology any basis for this topology is unique

or no

any other topology containing union Taplha has a coarser basis

[narrow cairn]

yes you can just take the topology generated by UTa

[mossy rampart]

and I'm guessing I got all of b correct

[narrow cairn]

the intersection works yes

because nTa is a topology by (a) and is not only the largest topology but the largest set contained in all Ta

[mossy rampart]

is this logic correct

does it make any sense at all

[narrow cairn]

wait is that for the first or second part of (b)

[mossy rampart]

second

[narrow cairn]

well we dont care about if its any smaller we want the largest one

unless the intersection happens to be the trivial topology, there will always be a smaller topology

[mossy rampart]

I was trying to prove there is no smaller one

which makes nTa the largest

[narrow cairn]

uh

[mossy rampart]

nvm

[narrow cairn]

in order to prove its the largest

[mossy rampart]

i see it doesn't make sense

[narrow cairn]

you show there isnt a larger one

lol alr

[mossy rampart]

wait does that logic work for the first part

i think I'm mixing them up

The logic works for A because any smaller topology does not contain all collections of Ta

Again forgive me if I'm not making any sense

[soft zephyr]

If ur trying to prove uniqueness

Btw are u taking intersection of the unions for the first part of b?

U can’t just take intersections directly cuz u want the topology u create to contain all the Ta

[coral pawn]

What are minimal fibrations of simplicial sets useful for other than proving that the homotopy theories of CGHaus and SimpS are the same?

[distant lichen]

As far as I know this is the primary reason behind their existence

Geometric realisation being a Serre fibration does not seem like it'd be useless in other areas but I don't know of anything off the top of my head

[earnest ibex]

Does anyone know why Hatcher says that the projection $p$ of a covering space doesn't need to be surjective? For me it doesn't make a lot of sense that $p^{-1}(U_{\alpha})$ can be the empy set, because then how can it mapped homeomorphically onto $U_{\alpha}$ by $p$

[gentle osprey]

ImHackingXD

[tiny ridge]

is the geometric realization a genuine fiber bundle

[coarse night]

covering maps are always subjective

[earnest ibex]

As far as I understand, not according to Hatcher (chapter 1.3, in the definition of covering space)

[coarse night]

can you share?

your reasoning is correct, they are always surjective

[umbral panther]

p^-1(U) is not mapped homeomoephically onto U. It is mapped to U times a discrete set, which may be empty

[coarse night]

you saying \empty set is a covering?

[umbral panther]

Yes. It is not a connected covering, but if you allow disconnected covers, you should allow empty covers

[coarse night]

ok fine

[umbral panther]

Connected covers with a choice of lift of the base point correspond to subgroups. Without a lift they are conjugacy classes or something. General covers correspond to sets with an action of the fundamental group. A transitive action is a connected cover. If you allow multiple orbits, you should allow zero orbits

[earnest ibex]

ok, I guess that makes sense

thanks

[umbral panther]

You have to allow disconnected covering spaces so that the restriction of a covering space to an open subset is still a covering space. It’s harder to justify empty covering spaces, except through the formal definition, and because they make the category of all covering spaces have good formal properties

[heady skiff]

for 2a, would a suitable homotopy be F(x, t) = (1 - t)f(x)? (trying to show that given any map f: X --> I, it's homotopic to the constant map e0)

it's contained in the unit interval and it's continuous

[coarse night]

sounds right

(a good way to see why this should be true is to replace I by it's homotopy equivalent space, since we only care about upto homotopies. I upto homotopy is just a point.)

[heady skiff]

wdym

what would be it's homotopy equivalent space

[heady skiff]

For b), could I just define $F(x, t) = f_x(t)$, where $f_x$ is a path from $f(x)$ to $g(x)$? since $Y$ is path connected we have $F(x, 0) = f(x)$ and $F(x, 1) = g(x)$

[gentle osprey]

okeyokay

[heady skiff]

if I'm considering any two maps f and g from I to Y

[muted arrow]

I'm a little puzzled by what you mean here

@heady skiff

Is x a point in I?

[heady skiff]

ye

[muted arrow]

So you're letting f and g be two paths and your claim is that F is a homotopy between them?

[heady skiff]

yeah i think

[muted arrow]

Yeah this is fine, but its a little vague around what is this f_x

[heady skiff]

ah okay, yea i guess i should make it more explicit in my proof

thank you

[muted arrow]

It might be easier to just show everything is homotopic to the constant path

then you can explicitly contract along the path f

But whichever you prefer 👍

using basepoint invariance of pi_1 in a path connected space, once you have two constant paths in the same path component you know they lie in the same homotopy class

[heady skiff]

oh right.. i think that kind of makes sense

i'm learning about fundamental group stuff rn

to show that a contractible space X is path connected, could you just let any a and b in X and use the fact that iX is homotopic to e_b (constant map on b), so we have F(a, 0) = i_X(a) = a and F(a, 1) = e_b(a) = b which is then a path from a to b

where F is the given homotopy between the identity map and the constant map

[heady skiff]

wait so would it be something like transitivity

f homotopic to constant map

g homotopic to constant map

therefore f homotopic to g

[muted arrow]

yeah something like that

[heady skiff]

oh okay cool

would this work to show that if $Y$ is contractible then for any space $X$, the set $[X, Y]$ has a single element? let $f$ and $g$ be any two maps from $X$ into $Y$. Since $i_Y \simeq e_b$ for some $b \in Y$, it follows that $i_Y \circ f \simeq e_b \circ f$, so $f \simeq e_b$. a similar argument shows that $g \simeq e_b$, whence $f \simeq e_b \simeq g$

[gentle osprey]

okeyokay

[muted arrow]

what does e_b \circ f mean

What's e_b

constant map from I to b?

morally this is correct, yea, but your domains are just a little wonky

[heady skiff]

oh constant map on Y

e_b(y) = b for all y in Y

[muted arrow]

Then yea youre good

[heady skiff]

oh okay sick

thanks

[unique herald]

Statement : Give an example in the R² plane of a family of open parts whose intersection is not open.

Can someone explain it to me please?

[queen prism]

do you know how to do this in R

[unique herald]

no

[thorny agate]

I guess one thing to consider is if the family can even be finite?

[unique herald]

My teacher wrote this, but I don't understand very well, I'm having difficulty :

In R :

[thorny agate]

no hablo French >_>

[thorny agate]

lemme see if I can parse this

[unique herald]

In R : Posons ∀ n ∈ N*, O_n = ] - 1/n , 1 [

[thorny agate]

so is O_n open, or closed?

[unique herald]

The open balls of (R, I I) are the open segments

[thorny agate]

also answer this

[unique herald]

I don't know

the only statement I have is that

Statement : Give an example in the R² plane of a family of open parts whose intersection is not open.

[thorny agate]

Ok so consider properties of open sets

[unique herald]

Any union of open is an open

[thorny agate]

is the intersection of two open sets also open?

[unique herald]

Any finite intersection of open spaces is an open

my course say this

[thorny agate]

ok perfect

so if we want to have an intersection of open spaces that is not open

can it be a finite intersection?

or does it have to be infinite?

[soft zephyr]

Bruh isn’t that obvious

Ahaha

[unique herald]

finite ?

[thorny agate]

to you, maybe @soft zephyr

[thorny agate]

but didn't you just say a finite intersection of open sets is open?

[unique herald]

yes

[thorny agate]

so how can a finite intersection of open sets be not open?

we need an infinite intersection of open sets

if we want any hope of the intersection being not open

makes sense?

[unique herald]

ok

[thorny agate]

next question

for any n >= 1, is ] -1/n, 1 [ open or closed?

[unique herald]

I think open but can't tell you why

[thorny agate]

what's the definition of an open set?

in R

[unique herald]

r>0 with B(a,r)⊂U

[thorny agate]

yea, the open sets in R are the open intervals

[thorny agate]

ok so is this an open set or a closed set?

[unique herald]

open

[thorny agate]

cool

Ok so now consider the intersection over all n

[gentle osprey]

Spamakin🎷

[thorny agate]

can you compute explicitly what this intersection is?

(draw out the first few sets on the number line, maybe that'll help)

[unique herald]

argh I have difficulty doing that

[thorny agate]

I'll give you a hint that it'll be an interval of some sort (won't say if it's open or closed at some specific end)

so what do you think the upper end of the interval will be?

[soft zephyr]

It’s an empty set though u don’t want that either

[thorny agate]

hm

anyways how did you get empty set because I can immediately think of a few elements in every set of the intersection

[queen prism]

minus sign moment

[soft zephyr]

Something like (-1/n, 1/n) would work maybe

[thorny agate]

this also works

[unique herald]

I'm lost

[thorny agate]

have you drawn anything out

[unique herald]

like a line ?

[thorny agate]

I kid you not, draw out a number line, and draw out the interval ]-1/n, 1[ for n = 1, 2, 3, 4

and see if you can see the pattern

[soft zephyr]

I thought the 1 was a 0 lol

[thorny agate]

0 != 1 in most (but not all) circumstances

[soft zephyr]

Rip

[unique herald]

Is this what you asked me to draw?

[thorny agate]

yea

so does this give you a better guess as to what the full intersection is?

[unique herald]

yes

well maybe..

[thorny agate]

what is your guess?

[thorny agate]

I want an actual interval as a guess

[unique herald]

] - 1/4 , 1 [

[thorny agate]

] -1/4, 1[ is the intersection of this whole thing?

but what if I add another term to your drawing, ] -1/5, 1[

then -1/4 is no longer in the set, in fact nothing between -1/4 and -1/5 is

[unique herald]

] - 1 /2 , - 1 / 4 ] ?

[thorny agate]

no? are you now claiming that all the numbers between 0 and 1 are no longer in your intersection?

[unique herald]

]-1/2, 1 [

?

[thorny agate]

same issue

[thorny agate]

draw out a few more sets if you have to, you should notice the left sides of the intervals converging to a value

[unique herald]

converge to 0

[thorny agate]

yes!

will it contain or exclude 0?

[unique herald]

contain

[thorny agate]

so what is the interval?

[unique herald]

[0, 1[

?

[ebon galleon]

Yes.

[unique herald]

but I don't remember why we did all that

we were here

[thorny agate]

yes

So we have that

[unique herald]

My teacher concluded with "The open balls on R of (R, |.|) are the open segments

[thorny agate]

hold on

$\bigcap_{n \geq 1} \left] -\frac{1}{n} , 1\right[ = [0, 1[$

[gentle osprey]

Spamakin🎷

[thorny agate]

so the left is an infinite intersection of open balls in (R, |.|)

is [0, 1[ open? Why or why not?

[unique herald]

is close because 0 is included in the interval

[thorny agate]

it's not open or closed

closed is not the opposite of open

[unique herald]

it's not open

[thorny agate]

try to argue why [0, 1[ is not open

and then you are done!

we have constructed a family of open intervals whose intersection is not open

[unique herald]

but my teacher gave the interval ]-1/n, 1[ if he hadn't given it to me and I had to determine it alone, how can I choose the good interval?

[thorny agate]

play around with stuff

another good example is the one proposed earlier

[thorny agate]

this one

[unique herald]

And last question why (R, | . |) ?

[thorny agate]

real space is nice

lots of things work

we're used to thinking about stuff (hell you even drew some stuff out)

[unique herald]

ok thank you, well I need to think about R²

[ebon galleon]

Hint for R^2: ||Try the same idea, except make them squares instead of intervals||

[soft zephyr]

Makes me think if u can just literally take intersection of all open sets that contain a closed set and you would just get that closed set back for R

Feel like there’s a separability axiom where this is true

[thorny agate]

empty set is closed

it's also open

so that line of thinking doesn't really work

[ebon galleon]

Like C = intersection of {U open | C subset of U}?

[thorny agate]

yea

[ebon galleon]

If so, the empty set doesn't break that.

[thorny agate]

I meant in the context of this question

like you aren't going to get something that isn't open

[ebon galleon]

Oh sure for the previous question that would not work for the emptyset

[soft zephyr]

Ya would have to artificially remove empty and whole space sets

It’s a strange condition

[ebon galleon]

This is true for T_1 spaces. (for A subset of X, take the intersection of { X - {x} | x in X - A }; don't even need A closed)

This is equivalent when we look at all subsets of X (note that this impliex X - {x} is open for each X). But not equivalent if this just holds for closed A (indiscrete spaces satisfy this for closed sets, but are not T_1)

[soft zephyr]

Hmm i think it would give back the set u started with but it won’t necessarily be closed

[ebon galleon]

If A is closed then you will get a closed set in the end, namely A.

[soft zephyr]

Oh ok lol I misunderstood don’t even need A closed

Ya I think it works then

[unique herald]

just why on R we directly define an interval ].., ..[ while R² we define a ball B(.....) ?

[ebon galleon]

Because for R, the open intervals ]a, b[ are precisely the open balls.

There is not as nice of a description for higher dimensions, without talking about balls.

[unique herald]

ok thank you

[pseudo ocean]

gosh what's a very elementary analogy of a topos?

Im trying to explain it to a friend and theyre just not getting it

(btw they dont know anything about topology or pure math)

[hidden crag]

Maybe topoi aren’t a good thing to explain then

[pseudo ocean]

theyre askinggg

I cant pass up this opportunity to teach pure math

[fickle elm]

Do they know something about topology or differential manifold?

[pseudo ocean]

a lil

though if youre gonna talk differentiable manifolds, #diff-geo-diff-top is your go to place

[fickle elm]

Sorry

I mean maybe you can try first talk about sheaves on a topological space

I just wrote a typo

[pseudo ocean]

well it's a collection of maps that takes open sets from your topological space to specific sets

[fickle elm]

Sorry again but what I mean is

You can start with this to your friends

[pseudo ocean]

OH WAIT I READ IT WRONG

[fickle elm]

First the idea why we need sheaves

[pseudo ocean]

I THOUGHT YOU WERE REFERRING TO ME 😭

[fickle elm]

Jump to topos may be a bit too quick

[pseudo ocean]

^

[fickle elm]

Start with sheaves is a good idea

[pseudo ocean]

oh god how the hell do i explain that in a very simple sense

[fickle elm]

Hmm, not very experienced in talking math to non math people.

Once talked about sheaves to a friend in physics

[hidden crag]

It won’t make a lot of sense

[fickle elm]

He is quite fond of the idea.

Though I doubt he cares about the details

[pseudo ocean]

it didnt 😭

[fickle elm]

He just like the idea that understand the space is to understand the maps to it

[pseudo ocean]

I had a very rough analogy with being able to mold something in a topos in clay

[hidden crag]

[fickle elm]

Maybe you should try another way

Told them about the history of topos

Not go into the details

[pseudo ocean]

howwww

[fickle elm]

Like some math article on quanta magazine

Talk about the invention of topos

Maybe try asking in #math-pedagogy . How to give a general idea of a modern math concept for non math people

[pseudo ocean]

will do

[pseudo ocean]

quanta magazine... ooof I dunno if he will even understand that article

now im not saying he's dumb or stupid

it's just his education level / interest isnt as high

[hidden crag]

Non math person doesn’t understand a very specific topic in mathematics

Shocking

[fickle elm]

I mean think about teaching graduate students topoi or even sheaves

It is not a very universal topic.

Naturally it is hard to convey it to non math people.

Wait a min.

I recall some philosopher is interested in topos

Maybe they have some better ideas on how to talk about it?

[pseudo ocean]

maybe

[pseudo ocean]

i dont have a formal education on pure math but i get it 😐

[merry geode]

What is this barycenter subdivision thing?

Professor vaguely mentioned some def of it but not in detail on how it works

I did not know intuition-based class was this hard to follow..

[hidden crag]

You get what?

[pseudo ocean]

the concept of a topos ig

though not an in depth one cuz i hadnt explored it that much yet

[hidden crag]

What is the concept of a topos?

[pseudo ocean]

my understanding of it is that its a Category that has similar properties to the category of sheaves on a topological space

[hidden crag]

Hmm

What are the relevant properties and why are they desirable

[pseudo ocean]

i suppose the objects ought to be sheafs of sets over a topological space

and there are sheaf morphisms between them that act as the morphisms of the category

and that, there's an identity sheaf morphism, and composition laws

that give commutative squares

apart from that, my knowledge of it is very limited haha

[hidden crag]

That is not what a topos is though, is it?

The category of set valued sheafs is one example of such and the rest of what you said is just parts of the definition of a category

[pseudo ocean]

i mean arent topoi categories?

[hidden crag]

They are

They’re a specific kind of category with certain properties

[pseudo ocean]

ack i havent even finished studying sheaf theory

[pseudo ocean]

whatre those?

[hidden crag]

Including finite limits and subobject classifiers but the important question is what do those mean/why are we demanding those

[pseudo ocean]

i uhh

[hidden crag]

I mean that’s in the definition of a topos

[pseudo ocean]

right.

oof yeah i am not ready to explain this

[hidden crag]

Neither am I, topoi aren’t simple

[pseudo ocean]

presh(D)? Grothendieck Sites??

utter nonsense

(I wont be calling it nonsense a couple years from now)

[fickle elm]

I only know that sheaves on a small site can be viewed as a Grothendieck topos.

IMO it is more concrete and it has some examples.

I guess you can start from here

[pseudo ocean]

i guess?

[fickle elm]

I am not an expert. That is all I know. Maybe for others topos has nicer generalization.

[pseudo ocean]

maybe

[rose vale]

the closure of a set S is defined as the union of S and its boundary

But doesn't this imply that S is taken to be an open set

Because otherwise the closure of S is just S

[queen prism]

no

you can define closure for sets that are open, closed, and neither

what do you get from closing up a closed set?

[rose vale]

nothing

[queen prism]

well you get nothing new, but you do get back the original set

[rose vale]

I mean the closure of a closed set is the same set, no?

[queen prism]

yes

[rose vale]

so how is it a new thing

[queen prism]

now think about what the closure of [0, 1) in the standard topology on R would be

[rose vale]

it would be [0,1] right

[queen prism]

why

[rose vale]

the boundary of the set is {0,1} if im not mistaken

so the union of the set and the boundary is the closed interval

[queen prism]

there you go

[rose vale]

i see now

[merry geode]

Do you guys define closure using boundaries?

[queen prism]

I like the intersection-of-closed-sets definition but I think this one is also fine

[merry geode]

I just thought of a terrible definition

A directed limit of closed sets containing the set

[rose vale]

wait are my lecture notes wrong?

E' is the set of limit (accumulation) points of E

[merry geode]

Nah, it is just a different definition.

[rose vale]

but a limit point of E can be outside of E

so by this definition, the closure of E is bigger than E itself

wait the example bladewood gave just confirmed this

no problem then

thanks

[queen prism]

yes it's bigger than E

(in general; it might be the same size)

[merry geode]

Was there an example where closure has bigger measure?

Well I brainfarted; Of course there is.

[queen prism]

bigger in terms of set containment, but in terms of measure yes that too

you can even raise the cardinality of a set this way

[merry geode]

Yea that is the example I thought of

[unique herald]

Statement : Give an example in the R² plane of a family of open parts whose intersection is not open.

I understand on R but not on R² ...

I don't understand why my teacher say :

O_n = B(0, 1+1/n) the intersection is a not open ball B(0,1)

or

O_n = B(0, 1/n) the intersection is {0}

[queen prism]

what does it mean to be in the intersection of all the B(0, 1/n)s

[unique herald]

B(0, 1+1/n) you mean ? no ?

[queen prism]

I think this one will be easier to think about first

[unique herald]

Why B(0, 1/n) is a open balls?

[queen prism]

it's just an example

[unique herald]

I draw a circle ?

[queen prism]

B(0, 1), B(0, 1/2), B(0, 1/3), ... are all open
but is their intersection open?

what is their intersection?

[unique herald]

I don't see

[queen prism]

what does intersection mean?

[unique herald]

intersection: common values

between all this

[queen prism]

right
so what numbers are in B(0, 1/n) for every n?

[unique herald]

The number 0

[queen prism]

there you go

[unique herald]

and 0 is not open ?

[queen prism]

{x} is always closed in R^n

[unique herald]

And about B(0, 1+1/n)

[unique herald]

why?

r>0 such that (-r,r) C {x}

[ebon galleon]

That one also works.

[unique herald]

But the not open ball here is not {x}

it's B(0, 1)

[ebon galleon]

When you intersect those, you get a bit more than just the open ball of radius 1. For example, (1,0) will be in the intersection, even though it is not in B(0,1)

[dusty raptor]

The length of a metallic tube is one meter.Its thisckness is 1 cm and its internal diameter is 12 cm find the weightof the the tube if the density of the metal is 7.7 cubic cm

<@&286206848099549185>

[glacial frost]

morn

[dusty raptor]

?

[glacial frost]

nothing

[dusty raptor]

answer ?

[glacial frost]

https://tenor.com/Wfyc.gif

[unique herald]

Can you show me step by step why it's works please

[glacial frost]

idk for this one dude sorry

[ebon galleon]

Absolutely not topology. See #❓how-to-get-help and try a help channel

[dusty raptor]

Not a prob, its not even my class its like 8th mensuration i found it a lil tough idk why ?

[dusty raptor]

oh shet srry

[unreal stratus]

Is Serre's theory of classes of abelian groups still used much nowadays?

I seem mostly to find it in older books and things

[ebon galleon]

Well, what will the intersection look like. So if we take the intersection of B(0, 1+1/n) for all n, then this will be the set of everything whose absolute value is less than 1+1/n for any n.
Now, I recognize that we can't have anything with absolute value greater than 1 by the Archimedean property (if x has |x| > 1, then there is some N with |x| > 1 + 1/N > 1, so x is not in B(0, 1+1/N)). However, I also notice that everything with absolute value at most 1 will be in the intersection. That's because, if |x| ≤ 1, then for any n we have |x| ≤ 1 < 1+1/n.
Now, the set of everything with absolute value at most 1, {x in R^2 : |x| ≤ 1}, has a nice description, from which it will be clear that this it a closet set and not an open set. What is this set called?

[unique herald]

Still don't understand why is B(0, 1) :/

[viral atlas]

R Y X has pointed out above that the intersection above is not just B(0,1) but some more points (which is precisely why the intersection is not open)

Read their reasoning above carefully

[unique herald]

I didn't just read, I even took a sheet of paper and rewrote everything

in my native language

[queen prism]

B(0, 1) usually means an open ball
unless you’re weird like evans and use it to mean a closed ball

[unique herald]

here my teacher write "B(0,1) is a not open ball"

"The intersection is a not open ball B(0,1)"

or maybe yes the traduction it's closed ball

I don't know

[ebon galleon]

It's the closed ball, yes.

"boule fermée"

And it looks like there's some symbol above the B there to designate that it's closed.

[unique herald]

you trying to explain to me with your text how he deduced B(0,1)?

[ebon galleon]

Yes, that was explaining how he got the closed ball

[unique herald]

Well, what will the intersection look like. So if we take the intersection of B(0, 1+1/n) for all n, then this will be the set of everything whose absolute value is less than 1+1/n for any n.

I understand that

but here no

Now, I recognize that we can't have anything with absolute value greater than 1 by the Archimedean property (if x has |x| > 1, then there is some N with |x| > 1 + 1/N > 1, so x is not in B(0, 1+1/N)). However, I also notice that everything with absolute value at most 1 will be in the intersection. That's because, if |x| ≤ 1, then for any n we have |x| ≤ 1 < 1+1/n.

Why do you want absolute value greater than 1 ?

when

absolute value is less than 1+1/n for any n.

Eh bien, à quoi ressemblera l'intersection. Donc si nous prenons l’intersection de B(0, 1+1/n) pour tout n, alors ce sera l’ensemble de tout ce dont la valeur absolue est inférieure à 1+1/n pour tout n. | x | < 1 + 1 / n
Maintenant, je reconnais que nous ne pouvons rien avoir avec une valeur absolue supérieure à 1 par la propriété archimédienne (si x a |x| > 1, alors il y a du N avec |x| > 1 + 1/N > 1, donc x n'est pas dans B(0, 1+1/N)). Cependant, je remarque également que tout ce qui a une valeur absolue d'au plus 1 sera dans l'intersection. C'est parce que, si |x| ≤ 1, alors pour tout n on a |x| ≤ 1 < 1+1/n.
Maintenant, l'ensemble de tout ce qui a une valeur absolue au plus 1, {x dans R^2 : |x| ≤ 1}, a une belle description, d'où il ressort clairement qu'il s'agit d'un ensemble de placard et non d'un ensemble ouvert. Comment s'appelle cet ensemble ?
If |x| ≤ 1 then |x| ≤ 1 < 1+1/n.

__

We take the intersection of B(0, 1 + 1 /n) for all n

We say that it will be the set of everything whose absolute value is less than 1 + 1 / n

| x | < 1 + 1 / n

until here it's ok

If | x | > 1 we have 1 < | x | < 1 + 1/n

So x will be in B(0, 1 + 1/n)

because x ∈ [1, 1 + 1/n]

[ebon galleon]

So we can split it into two cases. If |x| > 1, then we can find some ball B(0, 1 + 1/N) which does not contain x. It might be in some of the balls, but not all of them.
Otherwise, we have that |x| ≤ 1, in which case this reasoning shows that x is in all of the balls.

[unique herald]

I do not understand sorry... 😦
Thanks for taking the time to try to explain,

doesn't matter I'm going to do another exercise

[umbral panther]

Serre invented his theory to prove that, eg, a simply connected space all of whose homology groups are finite has homotopy groups that are finite. It is the right tool for this theorem. But people today aren’t very formal about it and don’t bother to generalize to abstract Serre classes

He showed that the quotient of an abelian category by a Serre class is an abelian category. This does not seem useful. There is a special case that is popular, in which the Serre class is closed under infinite sums. I think this is called a localizing subcategory. Then the quotient category embeds in the original category. Not finite abelian groups, but torsion abelian groups

[unreal stratus]

Yeah sure

I've only really seen it for those original applications

Like showing homotopy groups of spheres are finitely and generated, and finite (except in the couple of special cases)

And then mod C Hurewicz, Whitehead

Not sure beyond that lol

[umbral panther]

There’s some competitor to rigid analytic geometry that phrases it as built from ring objects in a weird abelian category that’s a quotient

[unreal stratus]

Interesting

[umbral panther]

People use the mod C Hurewicz and Whitehead theorems all the time, for localizing subcategories C, such as 2-torsion. But this is equivalent to applying the usual theorems after localizing the spaces, which is how people these days phrase it

[chilly niche]

Can anybody tell me what this notation means:

So first we have:

But then we have to show the following

I thought it meant isomorph but im not sure

[unique herald]

1. Show that the union of a finite number of closed parts is a closed

2. Show that a point x belongs to the adhesion of E if and only if any open ball centered at x meets E

For the 1. we do that, but I don't understand the notation

and they say that we have to go through the complementary scheme?

[opaque scroll]

So we have two rings, the first one consists of functions X -> F2 that are 1 on some subset A and 0 elsewhere. This is a ring under usual addition and multiplication.

Then we have another ring P(X) where addition is symmetric difference and multiplication is intersection. The statement is that these rings are isomorphic

[brave escarp]

I had an exercise where I have to compute the homology group $H_{k}(S^2\sqcup S^2,{p_1,p_2})$ where $S^2\sqcup S^2$ denotes the disjoint union of two 2-sphere, and where $p_1$ and $p_2$ denote the north pole of the first and the second 2-sphere. What I have is $H_0=H_2=\mathbb{Z}^2$ and $H_k=0$ for all $k\neq0,2$

[fickle elm]

This is relative homology?

[brave escarp]

yeah

[gentle osprey]

Gibzen

[fickle elm]

Try using the long exact sequence of relative homology

Since homology of spheres and points are easy

[brave escarp]

that's what I used

I just want to know if my answers are correct

[fickle elm]

For H0

Because for a connected space, homology relative to a point is the reduced homology.

So in your case H0 should be 0 as you have two point and two connected components

[brave escarp]

what is reduced homology ?

[unique herald]

someone understand this proof ?

[fickle elm]

OK maybe forget about that, think about the last part of the long exact sequence.

[brave escarp]

ok so my calculations are wrong ? I'l do it again

[fickle elm]

0=H_1(U,X)-->H_0(U)--->H_0(X)--->H_0(U,X). Here U is two points and X is two sphere.

[brave escarp]

why 0=H_1(U,X) ?

[fickle elm]

H_0(U)=Z+Z and H_0(X)=Z+Z and the map between them is induced by the inclusion of each north pole so both the ker and cokernel is 0.

[brave escarp]

ok and that's why also H_0(U,X) = 0

[fickle elm]

which part do you not understand?

[brave escarp]

so H_2(U,X) = Z^2 and H_k(U,X)=0 for all k =/= 2 ?

[fickle elm]

Yes.

[brave escarp]

ok thank you

[unique herald]

everything

[fickle elm]

OK.

1. think about how open sets are defined
2. In a set, what is the union of the complement.

[unique herald]

What X \ means ?

[feral copper]

Set difference

Complement of union is intersection of complements

That's all there is to this proof

[unique herald]

the statement is Show that the union of a finite number of closed parts is a closed

[feral copper]

Yes

[feral copper]

This is the main (and only) ingredient to this proof

[unique herald]

but I don't know why we look at the complements for the closed contrary to the proof "Show that the finite union of an open is an open"

and

don't know how to do that on a drawing, what should I understand about this notation?

[feral copper]

Something is closed iff the complement is open

[feral copper]

The open counterpart is that a finite intersection of opens is open

[limber umbra]

How much are (pointed unbased) fibrations relevant? (By that I mean that the lift of homotopy rel base point need not be rel base point.)
I'm reading about the Puppe sequence recently, and the hpty eq.of the homotopy fiber with the actual fiber for fibrations. And as far as I have understood, they are not necessarily pointed hpty eq if the lifts aren't rel basepoint.
Since homotopy groups are defined for pointed spaces, this seems to suggest that one can't make a long exact sequence of homotopy groups for unbased fibrations out of the Puppe sequence. Is this a problem?

[unique herald]

a finite intersection of closed is closed too

[feral copper]

Any intersection of closed is closed

[unique herald]

union and intersection is the same word ?

[feral copper]

No of course

Just... re-read my comments carefully, it's nothing involved at all

[unique herald]

Yes "Complement of union is intersection of complements"

[feral copper]

I'm not familiar with this at all, but wouldn't the answer involve groupoids?

[unique herald]

Why X \ is out in left and is inside in right ?

sorry

[feral copper]

Because again, the complement of a union is the intersection of complement... those things all live in X, so the complement of A is X\A

[unique herald]

Ok so X\A mean : The complement of A with respect to X

[feral copper]

Yes

[paper wedge]

there is an embedding phi: M --> V subset of R^m with M being phi(M)
so for that to be a chart do i just say that pi{1,....,n}(X) = pi{1,....,n)(phi(m)) for some m which is just (phi^1(m),phi^2(m),.....,phi^n(m)) ?
and ik phi is an immersion , so there exists a nbd such that its a local diffeo (hence a chart )by the inverse function theorem
so thats my choice of nbd
is that correct?

[limber umbra]

Some sources seem to suggest that the problem can be resolved if the domain of the maps to be lifted is well-pointed (the inclusion of base point being a cofibration). I'll think about it after I get some sleep,

[pearl trail]

What is the closure of ${(x,y): x^2+y^2 < 1}$? Is it ${(x,y): x^2+y^2 \leq 1}$

[gentle osprey]

GhostTheSavage

[paper wedge]

yea

[pearl trail]

Thanks. How should I think about the interior of this closure?

[paper wedge]

wdym

just like u woudl do

would do

with any set

[pearl trail]

I know know the interior of this. Is it the reverse of the closure here?

[paper wedge]

wdym by reverse of the closure

ifu mean the original set u started with then yeah

[pearl trail]

I mean is this true in this case Int(Cl(A)) = A

[paper wedge]

yeah

[pearl trail]

oh I was trying to come up with a counter example...to disprove this. Then this might be generally true?

[paper wedge]

no

[feral copper]

If A is closed this can't be true in general

[paper wedge]

yes

think of any connected top

if A is closed then that would imply A is open aswell

[pearl trail]

I’m doing this exercise:

[feral copper]

If A is open then yeah

[pearl trail]

oh so this is true if A is open, false if A is closed

[feral copper]

No, take a half open interval for instance

[pearl trail]

yeah part (a) it was closed, the counter example was easy, so it was false

[feral copper]

That's neither open nor closed

Actually no, if A is open this can fail too

A dense open subset is a counter example

[pearl trail]

So I need a different counter example, b/c it's true here for the open ball radius 1

[pearl trail]

So an open ball including the origin?

[feral copper]

Take the union of balls of radius 1/n centered around the n-th rational number

(Balls meaning intervals here)

Thats open and dense, so closure is R, and interior is R

Radius 1/n² sorry

You need a convergent series to say that it's not R in the first place

[pearl trail]

Why 1/n^2?

No simpler example?

[feral copper]

Because 1/n² ensures you have a finite measure

[feral copper]

If I got some sleep, probably xD

[pearl trail]

Need an explanation of this

[paper wedge]

maybe just

open up the rational numbers

[pearl trail]

I think there is a simpler example

[paper wedge]

epislon intervals around Q

this has measure 0

nope

mesaure epislon

[pearl trail]

You'll have to forgive me..I am a topology noob

[paper wedge]

measure*

[feral copper]

What does this mean?

[paper wedge]

r_n - epi/2^n

r_n+epi/2^n

r_n listing of rationals

union those up

[feral copper]

Yeah that's roughly my example

[paper wedge]

or u know what

R-{1}

[pearl trail]

these examples might be too advanced for my course

lol

[feral copper]

Except you're using 1/2^n instead of Basel

[paper wedge]

ur right

2^(n+2) ig

[feral copper]

Yeaaaaaah that's what a good night sleep example looks like

[pearl trail]

Oh! yes R -{(0,0} works right?

[paper wedge]

R-{a}

for any a

is open and dense

[pearl trail]

Doesn't have to be origin

any point

[paper wedge]

any point

[feral copper]

But less fun

[pearl trail]

But this is just a drive by topology question in my basic analysis class..can't get too fancy

[paper wedge]

yea i think R-{a} is cool

forget the measure theory stuff if u dont know about it

[pearl trail]

unless you want to explain topology to me for the next 30 mins, and measure

[paper wedge]

.

[pearl trail]

yeah I am no where near measure

[paper wedge]

just prove that R-{0} is both open and dense

and ur done

[pearl trail]

Yes

Closure(R^2 - {(0,0}) = R^2
Int(R^2) = R^2?? Which isn't R^2 - {(0,0)}?

[paper wedge]

yeah

[paper wedge]

u are closer than u think

[pearl trail]

Hmm what's the intuition why Int(R^2) = R^2? We need the union of all open sets. That is all of R^2...because R^2 has no boundary?

[feral copper]

Especially if that's for analysis, you'll end up doing Lebesgue integrals sooner or later

[pearl trail]

It's like you, a stranger, were sent from somewhere to tell me this lol. Thanks

[feral copper]

The biggest open inside R² is R²

Cannot make bigger that the whole space

[paper wedge]

that the whole space

[pearl trail]

I've dropped the "notion" of "size" for interiors and closures. I just think of all intersections and unions, respectively. Is that okay?

[feral copper]

And the closure of R²{0} being R² is because you must have all limit points, and thus must contain 0

[pearl trail]

I noticed when I thought of largest and smallest, I got confused

[feral copper]

I meant this as the biggest for inclusion

[paper wedge]

he means that if u look at the definition u will see that the interior is the biggest open set that contains

the set

and closure the smallest closed set

isnt in the definition of a topological space

that the set itself is an element of the topology

hence open?

[pearl trail]

The "biggest" of course is the union of all open sets. Since the union of finite or infinite open sets is open, right?

[feral copper]

Yes

[pearl trail]

The "smallest" of course is the intersection of all closed sets. Okay. Those adjectives make sense now.

[paper wedge]

u have anything else?

cuz i wanna ask something too

it got buried up there 😦

[pearl trail]

Go ahead

apologies

[paper wedge]

ty

[paper wedge]

no need

[paper wedge]

.

any help

or not help ig just a proof correction

[gritty widget]

you can just assume that M is a subset of R^m, no need to work explicitly with the embedding

[paper wedge]

yea i think i went M ~ phi(M) subset of R^m

but yeah ur righ

t

is this correct?

[feral copper]

Does this help? https://math.stackexchange.com/a/3008369/259363

[gritty widget]

you just said "so that's my choice of neighbourhood," but i don't understand how you're concluding that one of these projections is a chart on it. there's no proof as far as i can tell

[feral copper]

It seems to me you're trying to Characterize submanifolds of R^n as 'slices' by some R^m with m<n

[gritty widget]

the fact that submanifolds are locally graphs of smooth mappings will probably be useful

[feral copper]

Aka local inversion theorem (?)

[gritty widget]

never heard the phrase before

[paper wedge]

the inverse function theorem gives a neighhourbood U such that phi:U-->phi(U) is a diffeomoprhism

so (U,phi) is a chart but pi is just images of phi ig

thats what i thought

[gritty widget]

i don't get it, sorry

[paper wedge]

i think he means constant rank theorem

[gritty widget]

maybe someone else will

[paper wedge]

then its probably wrong lmfao

[gritty widget]

please don't ping me so much

[feral copper]

Actually it's called inverse function theorem in en-speaking countries!

[paper wedge]

okay wouldnt pi just be image of phi?

when i consider it taking elements from the manifold?

6.27

[gritty widget]

i wouldn't call the fact i wrote down the inverse function theorem, but it's certainly in that box of multivariable calculus facts which are related to it

it's more implicit function theorem-y

[paper wedge]

yeah

[gritty widget]

(re: matplotlib)

[paper wedge]

yea it is

more implicit ig

[feral copper]

You're right

[paper wedge]

i sadly still dont know why am i wrong hahaha

if p is in M , p is phi(m) for some m correct?

[gritty widget]

yes, phi is an embedding. as i said before, do yourself a favour and forget about the embedding. just assume that M is a submanifold of R^m

[paper wedge]

yea i just wanna understand why what im saying does not make sense

pi would be just phi^1... phi^n

if this p is nonzero then there is some nbd around this p such that these phi's are coordinates

so this nbd that comes out of the inverse function theorem works?

thats what i was trying to say ig

[paper wedge]

cuz like immersion + topological embedding is local diffeo

no 1 😦

well ig i got what u meant now

its just a submanifold so its basically the vanishing of the other coordaintes so there exists some i_js of coordinates which do not vanish ig

[unique herald]

Statement : Show that a point x belongs to the membership of E if and only if any open ball centered at x meets E

Correction : Let us first show the reciprocal meaning. By contrast-
sition, it is a question of taking a point which is not in the adhesion, and of showing
that there exists an open ball centered at x which does not meet E. Suppose
that a point x is not in Adhe(E). The adhesion of E being a closed one, its
complementary is an open: there therefore exists a ball B(x, r) included in the complementary, which means that it is disjoint from Adhe(E), therefore also from E since Adhe(E) contains E.
Let's show the direct meaning. We still reason by contraposition: we suppose
that there exists an open ball B(x, r) which is disjoint from E, and we want to show
that x is not in Adhe(E). The set X \ B(x, r) is a closed one which contains
E. But Adhe(E) is included in all closed containing E,, so Adhe(E) is
included in X \ B(x, r). In particular Adhe(E) does not contain x

What I don't understand :

1.I don't understand why X \ B(x, r) mean here 2. I understand the reasoning, it's clear but the conclusion "In particular Adhe(E) does not contain x" no

[knotty vine]

Why is there no early university topology channel?

[unique herald]

sorry would bother you less the next few days, I'm just trying to understand a few things

[knotty vine]

I dont want to discourage you from asking questions!

[queen prism]

well which channel would be most suitable for metric spaces

[gaunt linden]

"Early university" is intended for things that are quite early in a typical undergraduate curriculum, and there's not really any topology there.

[knotty vine]

Our second year undergraduate had metric spaces and point set topology up to Brouwers theorem

I found ODEs and PDEs harder then!

[queen prism]

i think that's unusual, at least in 🇺🇸
in other countries maybe it's normal

[knotty vine]

Regardless, there's quite a range in this channel

[gaunt linden]

For better or worse, the categorization of channels follows US tradition. Supposedly "early university" is for topics that are generally taught in America without requiring the students to learn proofs.

[knotty vine]

I havent really looked at those channels but I guess that makes sense

Tho I dont know how you would do proofs-and-logic without proofs

[gritty widget]

that's the course where you learn how to write proofs

[gaunt linden]

That's a mystery for the ages. The explanation I got was that it's for the course that sits between "not proof based" and "proof based" mathematics, and therefore would feel equally out of place in both categories.

[gritty widget]

thank god my university didn't follow this US tradition 😵‍💫

[queen prism]

you could talk about it in terms of "mathematical maturity," whatever that means

[unique herald]

if someone know

[knotty vine]

Adhe is closure? And what is a truss?

[unique herald]

in all closed containing E not truss sorry , mistake of trad

Adhe is for Adherent

[knotty vine]

Could you define it?

[unique herald]

In mathematics, an adherent point (also closure point or point of closure or contact point) of a subset A of a topological space X is a point x in X such that every neighbourhood of x (or equivalently, every open neighborhood of x) contains at least one point of A. a point x ∈ X is an adherent point for A if and only if x is in the closure of A

[knotty vine]

Ah, I'd never seen that terminology before!

What is known about E?

[unique herald]

The statement is that : Show that a point x belongs to the adherent (or closure) of E if and only if any open ball centered at x meets E.

nothing more about E or something

[knotty vine]

But is the letter E used before? Is it a closed subset of a metric space? What is the context

[unique herald]

the context is the metric characterization of Closure

E a part of a metric space

[knotty vine]

How is Adhe defined precisely? The intersection of all closed sets that contain E?

X \ B(x, r) is just the complement of B(x, r). So it is closed.

With that in mind, you should draw a diagram involving x, B(x,r), E, Adhe(E) and X

[unique herald]

Adhe(E) = Closure of E

is like Closure(E)

[knotty vine]

I understand

[unique herald]

Let us first show the reciprocal meaning. By contrast-
sition, it is a question of taking a point which is not in the closure, and of showing
that there exists an open ball centered at x which does not meet E. Suppose
that a point x is not in Closure(E). The closure of E being a closed one, its
complementary is an open: there therefore exists a ball B(x, r) included in the complementary, which means that it is disjoint from Closure(E), therefore also from E since Closure(E) contains E.
Let's show the direct meaning. We still reason by contraposition: we suppose
that there exists an open ball B(x, r) which is disjoint from E, and we want to show
that x is not in Closure(E). The set X \ B(x, r) is a closed one which contains
E. But Closure(E) is included in all closed containing E,, so Closure(E) is
included in X \ B(x, r). In particular Closure(E) does not contain x

for correction

[knotty vine]

The notation X \ B(x,r) is the set difference: https://en.wikipedia.org/wiki/Complement_(set_theory)#Relative_complement

[unique herald]

but why Closure(E) does not contain x ? that I don't understand, it's not clear for me 😦

[knotty vine]

You said you understood the reasoning, right? Just the conclusion is the problem?

[unique herald]

It's strange, I don't know how to explain it. I understand what they say, I follow their reasoning, but I do not understand the link between the reasoning and the conclusion

[knotty vine]

If $\operatorname{Cl}(E) \subseteq X \setminus B(x,r)$ then that means that $\operatorname{Cl}(E)$ and $B(x,r)$ are disjoint (they share no points). But obviously $x \in B(x,r)$. Hence $x$ is not in $\operatorname{Cl}(E)$

[gentle osprey]

Semer

[knotty vine]

If I'm honest, the proof that is given is rather indirect. I think a proof without contrapositives is possible

[unique herald]

Are you sure a can we just say that?

it's a bit weird

Cl = closure ?

[knotty vine]

Yeah

Adhe in your source

If you draw a Venn diagram with all these sets it should become clear

[unique herald]

I'm not confortable with Venn diagram

[knotty vine]

It's just circles representing some random X and E and x showing how they are included in eachother

It's not a proof method, but just a way to make it clear to yourself

Like this

Since B is open, its complement X \ B is closed. We knew already that B was disjoint from E, hence E is in X\B. But since Cl(E) is included in every closed set containing E, Cl(E) is included in X\B. But then Cl(E) is disjoint from B. Since x is in B, x is not in Cl(E).

[unique herald]

Ok Thank you very much !

I'm going to read it all again in a few minutes.

[heady skiff]

here, would p^{-1}(W) be one of the V_\alpha, since it's open? so the partition would just have to be the trivial one, p^{-1}(W) itself

[ebon galleon]

No. Consider an countable stack of real lines, Z x R (given the usual topologies), with p : ZxR ---> R being the projection. For any open set U, the inverse image looks like p^{-1}U = Z x U, so p evenly coveres any open subset of R [the V_alpha here being {n} x U for any n in Z]. But notice: If we have an open subset W of U, then p^{-1}W = Z x W, which is clearly not just one of the V_alpha

Since each V_alpha is homeomorphic to U, you would just restrict each V_alpha to the part that corresponds to W through the homeo.

[heady skiff]

ohh i see

got it thanks

i'm not quite following why V_\alpha intersects the preimage of b in a single point?

because if V_\alpha all have this point in common, wouldn't that contradict them being disjoint?

oh wait

wait never mind it's not the same point

[lime sable]

the stack of pancakes picture is very important

[heady skiff]

wait no it is

ye

but i'm trying to rigorize

[lime sable]

wikipedia has some good pictures

[heady skiff]

no munkres has one

which is very helpful intuitively

but like i require RIGOR!!!

wait wtf why am i not seeing this

why every V_\alpha intersects p^{-1}({b}) in a single point

moreover how does this show that p^{-1}(b) has the discrete topology

[lime sable]

the preimage of an evenly covered set U looks like n copies of U

if you restrict to a copy, it is just a homeomorphism

so everything appearing in U should appear identically n times in the stack of n pancakes

[heady skiff]

so each of the pancakes is a V_\alpha right

[lime sable]

yeah

and also you can replace "n copies of U" with "product of discrete space and U"

but it is easy to visualize when there are finitely many pancakes

[heady skiff]

so it's kinda like the elements of p^{-1}(b) break up amongst the pancakes

i'm confused

[lime sable]

so if you have your pancakes, the projection map pushes them all down onto U

[heady skiff]

right

[lime sable]

so the preimage of a point consists of the points floating above it

[heady skiff]

ah

i kind of see...

wait but l ike what if the preimage of a point is two points in a pancake

[trail charm]

another nice one is the helix as a covering map for S1

[heady skiff]

so then would we have two points floating over those two points

in each of the pancakes

[lime sable]

it can't be because the projection map is a homeomorphism when restricted to a pancake

[heady skiff]

ahh right

i see

ohhh okay that makes a lot of sense now

i think that was the missing puzzle piece

thanks! appreciate it

[lime sable]

np

[pearl trail]

I’m pretty sure I have numerous errors in this proof. Could anyone correct or provide cleaner writing for this proof?

[fervent root]

Would i be right in saying that R^2 with the lexicographic toplogy is not 2nd countable or hausdorff

[muted arrow]

Why would it not be hausdorff

[ebon galleon]

I agree it's not second countable though, but yeah it should be Hausdorff.

[merry geode]

Maybe they meant something like T4?

Or is it normal

[muted arrow]

Why wouldn't it be normal

[merry geode]

Yea it is normal. Hmmmm

[muted arrow]

Maybe it isn't perfect?

[ebon galleon]

I think it should be perfectly normal. I think this should look like a disjoint sum of |R| copies of R with the usual topology.

[muted arrow]

yeah, its homeo

Restriction to [-1,1] \times [-1,1] isn't perfect though

pretty neat

[ebon galleon]

Yup, difference here is the behavior around the edge, since [-1,1] has min/max elements

But you don't get that weird behavior with R^2. So since it looks like a sum of perfectly normal spaces, should be easy to check it's perfectly normla.

[muted arrow]

You can just feed the function through the homeo yeah?

[muted arrow]

In any event, this is super duper hausdorff

There are two types of open sets in this topology and you can always use the easier one to separate your points

[ebon galleon]

Yeah T6 is invariant under homeo. So for disjoint sum it should just be using the universal property of coproducts

(i have not checked this but it sounds right)

[fervent root]

hm okay, we havnt worked with the topology very much so i guess i’m not super comfortable with the open sets in it

i thought it was locally euclidean too

[muted arrow]

What are the open sets in the lexicographic topology in R^2

I feel all your questions will be answered once you answer this

[fervent root]

aren’t they just like {a}x(b,c)

like they’re vertical strips

[merry geode]

Huh, is it sum?, I.e. coproduct?

Maybe I am confused because I thought only finitely many should be smaller in coproducts.

[tidal lynx]

I remember a while back I brought up how in R^n it is true that interior points are automatically limit points.

Today I figured out that this holds because R^n is perfect, and it turns out that perfect spaces are the only spaces for which this holds. (i.e. that a space is perfect iff for every subset, among its limit points are its interior points)

[merry geode]

How come I never learned about perfect space?

Even tho I took multiple topology classes

[tidal lynx]

FWIW I haven't real seen them in the hypotheses of big theorems

[merry geode]

Hm

[muted arrow]

so I'm not sure how this encodes "vertical strips"

but there are two types

The lexicographic order is as follows:
(a,b) < (c,d) if (a<c) OR (a=c AND b <d)

Your open sets in the topology are the set of points (x,y) such that (a,b) < (x,y) < (b,d)

[muted arrow]

Using this, you'll have two "types" of open set:

• (x,y) s/t (a,b) < (x,y) < (a,d) for some b < d
• (x,y) s/t (a,b) < (x,y) < (c,d) for some a < c

Draw both of these to get an idea for what they look like

Once you've done that, determining if it is locally euclidean will be very easy.

@fervent root

[merry geode]

..wait. R^2 with lexico ordering is locally Euclidean

… huh

[fervent root]

i understand how the first one works, it is like a vertical line

but the second one changes x coordinates so i’m not sure how that works

[muted arrow]

Consider (0,0) < (x,y) < (1,0)

[muted arrow]

then, fix your x-coordinate and vary y. plot all the points that satisfy this relation

then fix y and vary x

this should give you an idea

[fervent root]

so it’s everything in the band between them

but only including the part of the band above the first point and below the second point

right?

[muted arrow]

Yep!

One of these is obviously homeo to R^n for some n. Do you see it?

[fervent root]

the ones where the horizontal component is fixed and the y varies is my guess

i wanna send it to R^2 but i wouldn’t be able to construct a map to prove it i guess

assuming that’s even the right open set that is obviously locally euclidean

[muted arrow]

Well those aren't open sets

[muted arrow]

The open sets are this

[muted arrow]

And this

Could you draw a picture of both and send so we know we're on the same page?

[muted arrow]

Great. Now which of these is clearly R^n?

Like no thought required it's immediate

Brain empty

[fervent root]

this is scaring me

[muted arrow]

It's the first one!

Literally R!

It's just an interval

[fervent root]

that’s the one that fixes the horizontal component yeah

[muted arrow]

Oh that's what you meant, I thought you were referring to this, sorry!

[fervent root]

wait so it has dimension 1

[muted arrow]

Yeah. Locally R. Super weird right

[fervent root]

i was wrong about the dimension so i’m glad we’re going through this

[muted arrow]

This is where you definition of locally euclidean really matters

If you define (X,\tau) to be locally euclidean if for all x\in X there exists U\in \tau such that x\in U and \exists n such that U is homeo to R^n, then we are done because all (x,y) in (R^2, lexicographic) are contained in one of those vertical lines which are homeo to R

But this is the standard definition so we should be happy :)

[muted arrow]

This might give you some insight into this comment as well

It looks like a bunch of disjoint copies of R because it is locally R1

[fervent root]

so like if i wanted to construct a map

it would be something like f(a,b) = (a,y-b) maybe

[muted arrow]

Well that doesn't land in R

Looks like you're landing in R^2 to me

If I is your vertical line, neighborhood containing an arbitrary point (x,y), your homeomorphism will be a map f: I -> R

[fervent root]

i think i need to think about this some

i’m trying to fill in the details for why this is inverse continuous and a bijection

[muted arrow]

"This"?

Nobody has written the map down yet

Here is a useful Lemma:

All nonempty open intervals in R are homeomorphic to R.

Proof hint: ||translate and scale to an appropriate interval. Then apply a certain trig function which has range R.||

[rapid olive]

that's a pretty long line you got there

[unreal stratus]

||trig is overly complicated smh||

[muted arrow]

||listen here u lil root vegetable||

||I'm currently teaching trig so it was on my mind ||

[fervent root]

tangent!

[opaque cloud]

what about closed/clopen intervals

[viral atlas]

there are no nonempty clopen intervals (intervals like [a,b) for instance are neither closed nor open)

For bounded closed intervals the reason why it cannot be homeomorphic to R should be easy to see (||compactness breaks for R||)

[opaque cloud]

hm right they are neither closed nor open

[queen prism]

R

[unique herald]

Statement: We place ourselves on X = R² provided with the Euclidean distance. Let A and B be two subsets of R². We define the whole:

A + B = {a + b ∈ R², a ∈ A, b ∈B}

Let A and B be two open ones. Is A+B open?

how to show this please ?

[knotty vine]

How far did you get? First step is: how do you show that anything is open?

[unique herald]

I have the correction but I don't understand

An open of E is a part of E which satisfies ∀x ∈O , ∃ε > 0, B(x, ε) C O

[knotty vine]

Right, so given some $p \in A+B$ we need to construct an open ball around it inside of $A+B$

[gentle osprey]

Semer

[knotty vine]

But for any $p \in A+B$ there exist $a \in A$ and $b \in B$ such that $p = a + b$. We know that $A$ and $B$ are open, so we get open balls around $a$ and $b$.

[gentle osprey]

Semer

[knotty vine]

Lets call those open balls B(a) and B(b). Now, we wanted to construct an open ball around a+b. Our first guess ought to be something like B(a) + B(b)

What should we check now?

[ebon galleon]

Well I mean that's by definition though, no?

[unique herald]

B(a+b, ε) C A and C B ?

or triangular inequality

[knotty vine]

We have $B(a,\varepsilon_a) \subseteq A$ and $B(b,\varepsilon_b) \subseteq B$. My first guess was to look at $B(a,\varepsilon_a) + B(b,\varepsilon_b)$. But another good guess could be $B(a+b, \varepsilon_a + \varepsilon_b)$. Can you draw both of these? Are they equal?

[gentle osprey]

Semer

[knotty vine]

We are trying to create an open ball around p in A+B, so we should check that whatever we have constructed is

1. A subset of A+B and
2. Contains (or in fact is equal to) and open ball around p

[ebon galleon]

Or no, I thought perfect was a closed set such that every point is a limit point? So something like the cantor set is perfect even though it has empty interior in R

[queen prism]

every point is a limit point and every limit point is a point

[muted arrow]

Yep!

[karmic smelt]

Do you know if topology and differential calculus are linked? Or are they two independent domains?

[quick bough]

they are very much linked, look up differential topology

[tidal lynx]

Yeah, I just never made the connection lol

Like I had noticed before that “interior point => limit point” in R

And I was like “hmm this seems pretty important why have I never seen it written down anywhere”

And then yesterday I learned what a perfect space is

And in particular they’re exactly the spaces for which my property holds (and this is shown quite easily from the definition of perfect space, like you say)

[tidal lynx]

A + B is equal to the union of sets of the form a + B over all a in A. Now try to show these sets are open, and then you’ll be done since the union of open sets is open.

[tulip bluff]

simply connectedness of S^2 implies that any map S^1 -> S^2 is nullhomotopic. Does this generalize to maps S^{n-1} -> S^n for n > 2?

[umbral panther]

Yes

[tulip bluff]

do you need to consider the extension of the map S^{n-1} -> S^n to D^n -> S^n or is there another way to prove this?

[viral atlas]

Good catch, I reread my statement thrice to avoid pitfalls and still missed this

[umbral panther]

There is an analogy to covering spaces, but it’s very fancy. The usual way is to argue that a differentiable maps are dense in all maps, so a small homotopy gets you to a smooth map, which is not dense. So it misses a point, so it’s image is in a copy of R^n, so it is homophotopic to a constant map

[heady skiff]

is the identity map i: X --> X a covering map, because for each point x in X, we can let U be an open set containing x, then the preimage is just gonna be U which is open, and we just take the trivial partition (i.e. just U) which is obviously homemorphic to U

[knotty vine]

Clearly yes

[heady skiff]

just wanted a sanity check lol

[knotty vine]

Can't you just use the suspension theorem?

[umbral panther]

How do you prove the suspension theorem?

[knotty vine]

I dont think I've ever seen it proved using differentiable maps at least

[umbral panther]

You could use piecewise linear maps, but that’s hardly different

[heady skiff]

why does the preimage consist of an interval of the form (0, eps)? for example, if i take eps = 1/2, then wouldn't the image of that be what i've outlined in lavender

which is not a neighborhood of b0 since it doesn't even include b0

oh is it because it's contained in one of the V_ns?

[knotty vine]

(0,epsilon) is not a neighbourhood of b0, its in the inverse image of a neighbourhood of b0

It's just one of the pancakes (a half a pancake here I guess) above U

[knotty vine]

I mean I've never seen the use of some class of maps being dense for this

[umbral panther]

Look at the proof again