point-set-topology

[sharp frost]

Alright I’ll have another go when I get back to my computer, thanks for taking a look

[hollow harbor]

Actually I don't like that either. Wtf lol

[sharp frost]

Haha

[hollow harbor]

The inverse of the transformation just flips a sign

We're getting terms of different orders here, that wont help

[sharp frost]

Oh my god you found the one that works?

[flint cove]

Okay, so ignoring indices for a moment: eta „eats“ two vectors living in the s coordinates. So to get something that eats vectors in the t coordinates, we need to plug in the jacobian translating t to s coords in both erguments

And this jacobian is (del s / del t), because it tells us how s changes if we wiggle t

And then its just J^trans eta J

Does that make sense?

(sorry, corrected a letter flip)

[sharp frost]

Ah yep that makes sense

[hollow harbor]

Yeah, wait... Why are we getting a different answer here than from the usual formula (which should just be J^t eta J)? I completely agree with your computation, but we must have been doing something wrong ourselves.

[flint cove]

Oh, I think the trouble may lie in using alpha, beta for the indices of s

[hollow harbor]

Ok, it is different, and now the indices dont match up...

Yeah lol

Rip

[sharp frost]

If that matrix calculation works out then isn’t that all we need? What am I missing

[hollow harbor]

That is all you need, I'm just trying to figure out if some indices should be modified to upper / lower here

Like

[sharp frost]

Ohh right

The vector indices should really be upper I reckon

[hollow harbor]

Its been a while since riem geo but I feel like I recall a metric tensor in covariant coordinates having lower indices

And a metric tensor in contravariant coordinates having upper indices

Like eta^i,j

[sharp frost]

Oh really

[hollow harbor]

Yeah

[sharp frost]

Is that convention always used?

[hollow harbor]

This seems sensible to me, since it doesnt make any sense to compute eta_ijv_iw_j, but it also doesn't make any sense to compute eta_ijv^iw^j if eta contains dependence on contravariant coordinates. the dimensions don't line up.

It's not so much about it being a convention. It's about what rank of tensor the metric is.

[sharp frost]

There’s no dependence on the dual space, that’s all I know

[hollow harbor]

A metric is a bilinear form on each tangent space. So it operates on two covariant vectors to give a scalar, and therefore must be contravariant of rank 2.
You can also "raise the indices" to turn the metric into a form on the cotangent space, in which case it would operate on two contravariant vectors to give a scalar. But in that case it has to be covariant of rank 2.

In this case, our metric depends explicitly on the contravariant coordinate s_3. So it is written in contravariant coordinates, and therefore is covariant of rank 2 (i.e. we should write it eta^i,j for the purposes of matching indices in sums)

[sharp frost]

Ah okay I see

Thanks ryc and lux, you’ve been a big help

[hollow harbor]

This business of always keeping covariant indices up and contravariant indices down is why we were confused: there's a typo in the formula you're using, but it looks correct because the indices match up in pairs and if you change the formula the indices no longer match up. Well, that's because the indices on eta were also not in the right place.

Sure

This was a good refresher for me, thanks

[flint cove]

Since I started doing this anyway, that would be the equivalent calc. Without matrices:

[sharp frost]

Oh that’s pretty cool

And do you have any ideas on how the author found that change of basis in the first place?

[buoyant rivet]

https://lecture2go.uni-hamburg.de/l2go/-/get/v/27612
Great Differential geometry lectures

[pearl holly]

So this is the exercise that I am working with: Let $\rho$ be the uniform metric on $\mathbb{R}^\omega$ (the Cartesian product of $R$ with itself $|\mathbb{Z_+}|$ times). Given $X = (x_1, x_2, \ldots) \in \mathbb{R}^\omega$ and given $0 < \epsilon < 1$, let
$$U(X, \epsilon) = (x_1 - \epsilon, x_1 + \epsilon) \times \cdots \times (x_n - \epsilon, x_n + \epsilon) \times \cdots .$$
Show that $U(X, \epsilon)$ is not equal to the $\epsilon$-ball $B_\rho(X, \epsilon)$. So I think that $B_\rho(X, \epsilon)$ is not a subset of $U(X, \epsilon)$ but that the other inclusion works. That is because if $Y \in U(X, \epsilon)$ then the difference between $X$'s coordinates and $Y$'s must be less than $\epsilon$, so $Y$ must be in $B_\rho(X, \epsilon)$. The other reverse inclusion does not hold because if $Y \in B_\rho(X, \epsilon)$ then it does not necessarily follow that EVERY $|x_n - y_n| < \epsilon$. Is this right?

[gentle osprey]

older sister

[marble socket]

i think you got it backward

just to confirm by uniform metric you mean sup of the standard bounded metric right?

[pearl holly]

yes sir!

[gentle osprey]

det

[pearl holly]

But is if for every natural number n? Or is it only for a specific n?

[gentle osprey]

det

det

[marble socket]

(i shouldn't use made up notations)

[pearl holly]

Okay and this holds because the sumpremum is kind of the greatest element? (Not really the greatest element but you get the idea)

[marble socket]

so if rho(X, Y) < epislon < 1 then each d_b(x_n, y_n) = |x_n - y_n|

[marble socket]

yea supremum can actually equal epsilon

[pearl holly]

Oh okay... I mixed up sup with inf...

[marble socket]

for instance take y_n = x_n + epsilon (1 - 1/n)

[pearl holly]

Ohhh okay now I get it. I thought that supremum was the least element but I just mixed up inf with sup. Thank you so much!

[marble socket]

[pearl holly]

(Just realised that your colour name matches the colour of your pfp. That's nice)

[marble socket]

hehe uwu

[frigid river]

Is this right (please don't laugh at me if it isn't 0.0)?

[pearl holly]

I shouldn’t be the one answering this but it looks right. You could also look at the complement directly. Per definition, C is closed if R-C is open. So the complement of a singleton is a union of two basis elements, so it is open

[fathom cave]

One point set?

Set with a single element?

[frigid river]

yep

[fathom cave]

Such set doesn't have a limit point so it's closed trivially

[frigid river]

I could've also just written that R is Hausdorff, right?

All in vain

[fathom cave]

The think u did is helpful too

Thing

Nvm I have not read carefully

[empty grove]

It is all good

They're all the exact same proofs written differently

And good to have different ways to even phrase the same proof

[empty grove]

(except for Zero0's proof, that's an entirely different proof)

[frigid river]

Probably the simplest/best one though xD

[empty grove]

Yeah using past results is very elegant

[flint cove]

Well, but it's not correct in the generality it's given in
Just because something is a one-point set doesn't mean it doesn't have any other limit points.
Without further assumptions that would prove that every one-point set in any topological space would be closed, and that's definitely not true.

[fathom cave]

ofc it does not work all the time @flint cove

[flint cove]

well don't call it trivial then

[fathom cave]

we are working on R tho

[stray bane]

I have a topological space $X$ and two nonempty subsets $A,B$. If i assume $A\cup B$ and $A\cap B$ are both connected, then i want to show that firstly, if they are both open then $A$ and $B$ are both connected, and secondly if they are both closed then they are also both connected. This is my proof if they are both open

Assume wlog that $A$ is disconnected, then $A$ can be written as $A=A_1\cup A_2$, where $A_1$ and $A_2$ are open and disjoint. We assumed that $A\cap B$ is connected, and so $(A_1\cap B)\cup (A_2\cap B)$ is connected. This must mean that either $B\cap A_1=\varnothing$ or $B\cap A_2=\varnothing$. This however will lead to a contradiction since we assumed $A\cup B$ is connected. Assume wlog $B\cap A_1\neq \varnothing$ then $A\cup B=(A_1\cup B) \cup A_2$ is a way of writing the union of $A$ and $B$ as the union of two disjoint open sets. This shows that $A$ and $B$ are both connected.

I am however not sure what would change if they were both closed, if i am not mistaken, the proof would remain pretty much the same?

[gentle osprey]

Fro

[uncut surge]

I've seen this notation around a couple times now, what does it mean when I write, for a variety X and a field k the expression "X(k)"?

[wanton marsh]

it means the set of k-valued points of X ?

[uncut surge]

https://en.wikipedia.org/wiki/Rational_point ah okay this

thanksie

[sleek thicket]

@tough imp

[tough imp]

Tfw

[sleek thicket]

So we can get that m is associated to B/A

So depth (B/A) = 0 and then by the SES inequality also depth B = 0

The homework problem from last week required that m wasn't an associated prime

[tough imp]

It says that (4) is satisfied right?

[sleek thicket]

How?

m is an associated prime of B, so 4 can't be satisfied

[tough imp]

No

B has depth 1 right?

It’s like A[some integral thing]

[sleek thicket]

oh yeah you can show there's an element which is regular on it

Well I still don't see how to use the hw problem from last week

But I can derive a contradiction from this

[tough imp]

Well now you have (4) yeah?

[sleek thicket]

Since I showed depth B = 0

[tough imp]

Oh sure I guess

[sleek thicket]

How do you get that a power of m annihilates B/A?

[tough imp]

Cuz

Support is V(Ann)

So since that’s just m

[sleek thicket]

Ah right okay that's what I'd forgotten

[tough imp]

You have m = r(Ann)

[sleek thicket]

I'm gonna do the depth >= 1 thing

(4) has so many conditions lmao

[tough imp]

Okay

[sleek thicket]

Just to write out

ty

[frigid river]

I want to prove that every order topology is hausdorff. Could you please give me a hint on how to start?

[empty grove]

Split it into 2 cases, depending on elements between x and y

[stone cipher]

,tex How do I show that, if $U$ is an open connected set of $\mathbb{R}^n$ ($n\ge 2$), then $U\setminus\left{x\right}$ is connected?

[gentle osprey]

DoJ

[gritty widget]

one way to look at it is that it's equivalent here to check path connectedness (because of local path connectedness), and if you have any two points in U, you can join them by a path which avoids the missing point by drawing a ball around it

[stone cipher]

,tex I know that in $\mathbb{R}^n$ path connected components and connected components are the same. How do we guarantee that in fact every connected is path-connected?

[gentle osprey]

DoJ

[gritty widget]

open and connected is the key here

[stone cipher]

Hummm

[gritty widget]

every open and connected set in R^n is path connected, but if you remove "open" it's clearly false

[stone cipher]

Got it

[gritty widget]

of course that's probably overkill, you can just use the definition of connectedness directly

[gritty widget]

I need help with some not terribly complicated geometry. I'm shit at this stuff and would like help.

if you have a question just ask

@gritty widget

Need help with this one

this belongs in #geometry-and-trigonometry

Oh. my bad.

[delicate hollow]

Question: Show each open set is a F-sigma set.
Attempt: I tried to intuition it and I cant. If F-sigma sets are a countable union of closed sets im pretty sure an F-sigma set is closed. Since its a real analysis book im just gonna assume open sets in R.
Let X be open in R, then X is a collection of open intervals, X = {(a_i,b_i)} for i in I
Set p_i = (a_i+b_i)/2 and let F_n=[p_i-(p_i - a_i) * d,p_i+(p_i-b_i) * d] where d = n/(n+1)
Then the countable union of F_n = X

[cerulean oriole]

Finite unions of closed sets are closed but countable unions of closed sets are not always closed.

[honest terrace]

(see for example U_[q € Q] {q} = Q, Q is definitely not closed but the union is countable and each element of the union is a singleton, so its closed)

[delicate hollow]

euros xd

i get the example though

does my construction work then?

[true robin]

Hint: try to show the statement for open intervals first

[delicate hollow]

i did that with F_n

{F_n} is supposed to be a countable union of closed intervals starting from the midpoint of an open interval and spanning outwards with its limit approaching the boundary of the open interval

[true robin]

Ah yes i see what you did, yeah that sounds right, now just justify why I can be taken to be countable and you’ll be good to ho

[delicate hollow]

I being the enumerating set?

[true robin]

Yeah

[delicate hollow]

my intention is that if u union all the \cup_{i=1}^{n}{F_n}=X

[true robin]

Wait first of all, are you only taking one n for each i?

[true robin]

Would it be possible for you to latex/send a photo of what exactly you are doing. I gind it hard to read

[delicate hollow]

oh i know you do

[delicate hollow]

We have that $p_i=(a_i+b_i)/2$ and $d_n=n/(n+1)$ then we define $F_{i_n}=[p_{i}-(p_{i}-a_{i})*d_{n},p_{i}+(p_{i}-b_{i})*d_{n}]$

[gentle osprey]

Marlin Flier

[delicate hollow]

$\bigcup_{i=1}^n F_{i_n} = (a_i,b_i)$ and $\bigcup_{i\in I} F_i = X$

[gentle osprey]

Marlin Flier

[delicate hollow]

sounds about correct to me :/

[true robin]

Yes

I guess it should be i,n not i_n, but i think that is just a typo. Also you should union (ai,bi)

Unless that is what you are defining as Fi

[delicate hollow]

i shouldve written that the union of (ai,bi)=union of Fi=X

also i never know when to use double subscript or just use commas

so i always default to double subscript

[true robin]

Cool

[rough obsidian]

Can anybody confirm if $\mathbb{R}^n$ is always a metric space? Thank you

[gentle osprey]

v2812ic

[true robin]

Do you know how distance is defined on R^n?

[rough obsidian]

No no no, I'm asking about a general case

[true robin]

What do you mean by general case?

[rough obsidian]

Nevermind nevermind

I was asking if I could always affirm that Rn is metric

But yeah, I understand what you are saying; if I don't define a distance Rn is not metric

[true robin]

I think what you might want to say is that R^n is homeomorphic to a topological space which has a metric (namely R^n with its usual metric)

[marble socket]

R^n is just a set. unless you tell me more about it... like if you tell me how to add two things, it becomes an abelian group, if you tell me how to multiply by elements of R, it becomes a vector space. If you tell me how to measure distance between any two points, its becomes a metric space.
So are you asking if every topology on R^n is metrizable?

[rough obsidian]

Yes yes

[marble socket]

Ah so answer to that is no.

if you have an infinite set X, then you can define the following topology. define X and every finite subset of X to be closed.

this is called the finite-complement topology.

notice that in this topology any two non-empty open sets must intersect!

because only finitely many points are outside each of them, but there are infinitely many elements in X.

(if you have taken a topology course, you might know that this topology isn't hausdorff and metrizable spaces are supposed to be hausdorff)

[marble socket]

anyway, this wouldn't be true if X was metrizable!

[empty grove]

the indiscrete topology on any set with more than 1 element is also non metrizable

for a simpler example

[rough obsidian]

Alright! Thank you very much!

I got it!!

[true robin]

My man has given himself the catking react, what a baller

[empty grove]

[rough obsidian]

Poor risitas

[true robin]

Oh yeah i heard he died recently, quite sad

[austere oxide]

Hello!
I don't know any topology, but I just wanted to get some intuition.
Would someone be able to explain to me on a high level, how the very formal definition of a set and a bunch of subsets satisfying some properties, relate to the cool shapes (donuts, mugs, and all that)?

Thanks!

[empty grove]

The subsets capture the idea of closeness. Given 2 points, if they are contained in a lot of common subsets, they are close to each other. So for example in the space that is the surface of a doughnut, these subsets can be thought of as disks (of any size) on the surface. If 2 points are closeby, there will be a lot of disks which contain both of them, in the sense that you will be able to use smaller disks to contain both, while if they are far apart, you can only use big disks. This is a generalisation of metric spaces where you have a distance function - given any pair of points you can say how far apart they are. In topological spaces you dont care about how far they are from each other in absolute terms, but only in relative terms, eg "are the points p and q closer to each other than x and y?". The reason we can think of topological spaces as stretchy is that stretching preserves relative closeness.

[austere oxide]

I see, thank you very much 🙂
How do these subsets differ for say a sphere vs a doughnut?

[empty grove]

In both cases we use disks, its just that the disks are somehow able to capture the shape of the space as a whole. (I should say that the actual topology has more things than just disks, but disks form a "basis" for the topologies in both cases, and thats why im using them to describe the toppology as a whole, similar to how you can describe metric spaces using just a basis)

[austere oxide]

As in, do they have the same disks, but there are some other properties that help differentiate them?

[empty grove]

same disks, in the sense that you just draw circles on the surfaces

so same to the extent that they are all the same shape, just on different surfaces and of different sizes

[austere oxide]

Alright

[empty grove]

Remember, topologies are just glorified semi-lattices.

If you have two semi-lattices X and Y, and a monotone function f from X to Y then an element a of X is a sufficient factor for b in Y if for any refinement of X W, refinement of Y Z and monotone function f': W -> Z that extends f, for any element w of W, w subs a => f'(w) subs b. Likewise an element a of X is a necessary factor for b in Y if for any refinement of X W, refinement of Y Z and monotone function f': W -> Z that extends f, for any element w of W, w subs a <= f'(w) subs b. An element a of X is a determining factor for b in Y if it is a necessary and sufficient factor. The map f is factorable if every element of Y has a determining factor in X. This means that there exists a function f*: Y -> X.

What it means in topology for a map F: X to Y to be continuous is that the induced map f = cl o image_F, from the closed sets of X to the closed sets of Y is a factorable map.

[viral atlas]

I wonder if there's a good animation to show homeomorphism and how open sets are mapped to open sets.

[empty grove]

couldnt not

It will be pretty difficult given how many open sets there are

like I was googling open sets on sphere and couldnt get any good images

[viral atlas]

No like, a very rough representation would suffice

[empty grove]

[true robin]

Consider the null animation (nothing moves)

[empty grove]

how wrong I was

[viral atlas]

Smh internet is all of coffee cup and doughnut

One weirdass cow transforming into a ball GIF

[empty grove]

it is a good example

[viral atlas]

Sure

[true robin]

I would’ve expected a cow to have holes

[empty grove]

[true robin]

Yes i have stuck a rod straight through a cow, how did you know

[empty grove]

[empty grove]

I've always felt like this guy has cockroach hair

by this guy i mean the guy in the emote

not talking about myself

Im obviously bald

[true robin]

Well yeah, now that you mention it

[sweet wing]

animation would be more like homotopy/ambient isotopy instead of homeomorphism tbh

[viral atlas]

Hmmm

[raw sedge]

in the way that it's typically done, yes

but, here's what I have in mind

show a depiction of both spaces

and highlight a single point in each

and show them moving around continuously

then the highlighted points at each point in time correspond according to the homeomorphism

[sweet wing]

oooooo

that's smart

[flint cove]

ooh yes, that actually fits my mental image really well
I usually describe it as „if you wiggle a tiny bit on the left, it has to wiggle a tiny bit on the right“

[cerulean oriole]

I would be satisfied with a good animation explaining continuity of monotone functions between glorified semi-lattices.

[bleak helm]

[cerulean oriole]

Given a function f on (an open subset of) R^n to R^m, k times differentiable at 0, we can write a Taylor polynomial approximation
f(x) = f(0) + f'(0)x + ... + f_k (0)/k! x^k + E(x)
What kind of assumptions do we need to bound E(x) in various ways?

E.g. by this: https://math.stackexchange.com/a/63298 it seems that being k times differentiable at x (k >= 1) is enough to say that E(x) = o(|x|^k). What if I wanted to say that E(x) = O(|x|^(k+1))?

[cerulean oriole]

<@&286206848099549185>

[marsh forge]

i think the big issue is that for all "obvious" homeomorphisms this is easy to picture in your head and you can't animate the non-obvious examples well

[viral atlas]

That makes sense haha

[lean marten]

You might be able to represent it as the induced morphism between posets of open sets?

Or at least as that of an approximation of the poset of open sets

[gritty widget]

[marsh forge]

thats it all mentions of orders, posets, and lattices

are banned in this channel

[ivory dragon]

someone apply order theory to the zariski topology

[marsh forge]

namington please

lay down the law

[honest terrace]

Remember, topologies are just glorified semi-lattices !

[marsh forge]

outlaw these blasphemous orders

[verbal wraith]

Imaging saying 'f is a dominant map' rather than saying the induced map of f is surjective. 🤡

[marsh forge]

especially ordered topological spaces

ill live

[digital peak]

so am I reading this correctly that a TVS can be e.g. in the cofinite topology?

ah no, addition fails continuity doesn't it

[gritty widget]

Abstractly I see that for the root system of a lie algebra, they wely group should have an element of longest length

and this element will send the postive roots to the negative roots

but I'm not sure how I would actually construct such an element

[wanton marsh]

length ?

[gritty widget]

maybe height is more standard term?

From humphreys

[wanton marsh]

ah :o

I didn't know that

[gritty widget]

You didn't know this concept or the it being called length?

[wanton marsh]

it being called length

[gritty widget]

(just curious if you have another name)

ah okay

is there another name

[wanton marsh]

I suppose you can define it whenever you have a generating subset of a finite group

[pearl holly]

So a quick question. Munkres writes the following: We say that a subset C of X is saturated (with respect to the subjective map p:X -> Y) if C contains every set p^-1({y}) that it intersects. What does the "that it intersects" mean? If C contains p^-1({y}) then it certainly intersects C, right?

oh okay I see. Thank you!

[lunar yoke]

So I think it might help to start with $A_n$ and look at its explicit construction (see humphreys page 64). Essentially we look at the subspace orthogonal to $\sum_{i=1}^{n+1} e_i$ in $(n+1)$-dimensional euclidean space. The root system is $\Phi = {e_i - e_j \mid i \neq j}$, with simple roots $\alpha_i = e_i - e_{i+1}$ for $i = 1,..,n$. With this one can see that the weyl group of $A_n$ is actually isomorphic to $S_{n+1}$ with the reflection at $\alpha_i$ corresponding to the transposition of indices $(i,i+1)$. We know the length of an element is the number of positive roots which get sent to some negative root, and we want to maximize this. In our explicitly constructed vectorspace, the positive roots are $\Phi^+ = {e_i - e_j \mid i < j}$, and hence the negative ones are $\Phi^- = {e_i - e_j \mid i > j}$. So thinking of our weyl group elements in terms of transpositions on the indices of these $e_i$, we have to maximize the number of inversions. I'm pretty sure the permutation $(n+1,n,...,2,1)$ does the trick.

hmm the latex is not behaving like i though it would

[marsh forge]

yeah wait

wf

is your keyboard typing a nonstadard _

$a_b$

[gentle osprey]

weakly equivalent to max

[marsh forge]

/ are you using an international keyboard?

[lunar yoke]

I escaped all of the _ because discord tries to do some weird thing when multiple of them are in one msg

[marsh forge]

ohhh

okay

yeah it'll do that but the tex reader will read it properly

[lunar yoke]

oh ok

[gentle osprey]

Phil P

[lunar yoke]

actually writing this last permutation via the transpositions $(i,i+1)$ also gives you an idea of why this element is called the "longest" one. For $n=3$, if we let $\sigma_i = (i,i+1)$, it looks like $(\sigma_3\sigma_2\sigma_1)(\sigma_3\sigma_2)\sigma_3$

[gentle osprey]

Phil P

[marsh forge]

does anyone know how to derive the product in the homology theory of a ring spectrum

i cant find a reference

my friend gave some construction going through a spectral sequence but i cannot imagine thats the best way

[sweet wing]

i learnt some from https://etale.site/writing/an-introduction-to-spectra.pdf

[marsh forge]

huh

does that have my claim?

i know what spectra are haha

[sweet wing]

oop actually no

sed

are any of these of any interest to you

[marsh forge]

not quite

but they are neat

in theory what im thinking about exists as long as E is a ring spectrum and X is any spectrum

so like

$E_nX\otimes E_m X\to E_{n+m}X$

all the naive stuff doesnt quite work

[gentle osprey]

weakly equivalent to max

[marsh forge]

such a thing always exists when say X is an H space

but X shouldnt even need to be a (suspended) space

[sweet wing]

ahh i see

[marsh forge]

also possible my advisor is just wrong but thats never happened before

[pearl holly]

So I am currently looking at everything above the example 4 part and I am kind of confused with the very last paragraph before example 4. It says that p^-1(U) is just the union of the equivalence classes belonging to U, which I find kind of confusing because of the inverse function. It's hard to explain in words and that's why I have an example.

So let X = {a, b, c} and X* = {{a}, {b, c}} be a partition of X. Define p:X -> X* by the following:
p(a) = {a}, p(b) = {b, c}, p(c) = {b, c}.
Now p is surjective. Let U = {{b, c}} which is a subset of X*. Now p^-1(U) can be equal to two things. Either b or c. So is p^-1(U) = {b, c} by some sort of definition?

In fact, my whole question lies in my second section of my post. There's no need to read everything

[gritty widget]

So is p^-1(U) = {b, c} by some sort of definition?
by the definition of pre-image

[pearl holly]

oh okay!

One more thing real quick. Is a map different from a function?

[gritty widget]

map and function are usually synonymous

[pearl holly]

but p^-1 is not a function in my post above

so do I simply call it a map?

[gritty widget]

p^(-1)(U) doesn't mean the image of U under p^(-1)

it means the set of things in the domain of p which p sends into U

[pearl holly]

oh okay, I see

[gritty widget]

(if p is bijective, then it actually is the image under p^(-1))

[pearl holly]

yeah that makes sense. Thank you so much!

[marsh forge]

map usually means morphism

not function

fwiw

[gritty widget]

fine

personally there's no difference

i usually go with the more economical one

[marsh forge]

you can consider a preimage to be a function the codomain is just the powerset of the original domain

[marsh forge]

if you say “map of spaces” and don’t mean continuous ill stab you

[pearl holly]

oh okay I see now. Thank you!

[obtuse meteor]

Correct take

[marsh forge]

For the curious about my earlier question

I was misinterpreting a claim of my advisor

bonus points if anyone finds a way to make sense of this though

[sleek thicket]

I think tterra was saying function means continuous function

or map/function means smooth map/smooth function

[gritty widget]

i wasn't saying anything

[sleek thicket]

oh okay

[gritty widget]

just that i personally use map and function interchangeably

[sleek thicket]

I apologize for anthropomorphizing

[gritty widget]

and i thought that was common

now help me prove lee prop 7.34

[sleek thicket]

I think it is

No I have forgotten all of my riemannian geometry

[gritty widget]

what do you mean you don't want to compute $$-D^{\tilde{\nabla}}Rc + D^{\tilde{\nabla}}(\nabla^2 f) - D^{\tilde{\nabla}}(df \otimes df) + \frac{1}{2}\left(|df|_g^2+\frac{1}{2}S\right)D^{\tilde{\nabla}}g$$

[gentle osprey]

R2T2

[gritty widget]

[sleek thicket]

no, not particularly

Just look at an arbitrary index and expand everything out 4head

[gritty widget]

the MSE post i found says "after about 7 pages of calculation..."

[sleek thicket]

lmfaoooo

"Fuck you" - John M. Lee

[gritty widget]

but it does give something to aim for

i pray the rest of the book is not this computational

[marsh forge]

bad

altho i basically never use the word function

functions are structureless and i dont fw that

[gritty widget]

thanks very much!

[gritty widget]

@lunar yoke or someone else, i have a continuation question on this

any element in the wely group can be written in such a reduced form

is the longest element in the wely group not always going to just be

$\sigma_{\alpha_1}\sigma_{\alpha_2}...\sigma_{\alpha_n}$

[gentle osprey]

lime_soup

[lunar yoke]

No, this already does not work for $A_2$, where you get that $\sigma_1(\sigma_2(\alpha_1+\alpha_2)) = \alpha_1$

[gentle osprey]

Phil P

[lunar yoke]

if $\sigma_1\sigma_2$ were the longest element, it would need to send any positive root, in particular $\alpha_1+\alpha_2$, to a negative one

[gentle osprey]

Phil P

[gritty widget]

ah that helps alot

[lunar yoke]

wait i think i messed up the calculation

[gritty widget]

oh actually I am confused again

[gritty widget]

we should be able to write the longest element as a product of sigma_alpha, sigma_beta?

[lunar yoke]

yes

the longest element is $\sigma_1\sigma_2\sigma_1$

[gentle osprey]

Phil P

[lunar yoke]

or also $\sigma_2\sigma_1\sigma_2$

[gentle osprey]

Phil P

[gritty widget]

ah yes sorry

i kept on doing sigma_2 as flip in the yaxis

thanks that makes sense

[gritty widget]

and since the longest element is unique this is equall to the other

thank you very much

[lunar yoke]

So if we first reflect at $\beta$, then $\alpha+\beta$ is mapped to $\alpha$. If we then reflect at $\alpha$, we obtain $-\alpha$ as image of $\alpha+\beta$ under the reflection $\sigma_1\sigma_2$

[gentle osprey]

Phil P

[lunar yoke]

so it works out after all

[gritty widget]

thanks very much

[lunar yoke]

But it does not work for $\alpha$, which is first mapped to $\alpha+\beta$ and then to $\beta$ under $\sigma_1\sigma_2$, so this shows that $\sigma_1\sigma_2$ is not the longest element

[gentle osprey]

Phil P

[lunar yoke]

sorry for the confusion

[gritty widget]

no worries at all

i was actually trying to do this

https://math.stackexchange.com/questions/2995727/longest-element-of-weyl-group-for-g-2?noredirect=1&lq=1

but understanding the case for A_2 properly has made this clear now

[dim turret]

this is what i said:

d(fw) = z => dw wedge w + f.dw = z

=> dw wedge w wedge w + f.dw wedge w = z

since w wedge w is - in a 1 form

we see they are equal

does that make sense

[hollow harbor]

i have no idea what you are writing

are you not assuming the conclusion you're trying to prove in the first implication?

well, more so i just don't see where dw wedge w would come from

my strategy here would be to write this in coordinates and do it there. i.e. to pick a coordinate patch, write w = sum w_j dx^j there, and then use the local definition of the exterior derivative.

(since my assumption here is that we're proving all these nice coordinate free properties and can't use them. another way to define the exterior derivative is by picking several coordinate free properties which uniquely identify it, but this is basically one of those properties)

[dim turret]

wym write this in coordinates...

ok im not sure how you can just pick coordinates for a product? @gritty widget

figured it out

[gritty widget]

you're working on an open subset of R^n so your coordinate chart is just the identity map

omega is a 1-form, so it looks like sum a_i dx^i, for some sufficiently differentiable functions a_1, ..., a_n on your open set

f*omega is sum (f * a_i) dx^i

apply the definition of d

all you have to do is follow the definitions

[gritty widget]

Can anyone help me see why an inclusion of Dynkin diagrams induces an inclusion on the root systems

[gritty widget]

The only thing I’ve been able to come up with is

Taking a given Dynkin diagram and looking at the different cases for removing one node

[lunar yoke]

Say you have an isomorphism of Dynkin Diagrams $D_1 \cong D_2' \subseteq D_2$. Since the nodes correspond to basis vectors of your root system, this already gives you an isomorphism of vector spaces by sending basis to basis. You can check that the cartan numbers are uniquely determined by the Dynkin diagram, and from there deduce that the vector space isomorphism extends to an isomorphism of root systems.

[gentle osprey]

Phil P

[lunar yoke]

But I think maybe these Lie-Algebra questions fit better into the #groups-rings-fields channel?

[gritty widget]

Oh yeah probably sorry

[floral gust]

How does one calculate the boundary of something ? For example, why is the boundary of a cylinder just the two enclosing circles and not the entire topological boundary?

[gritty widget]

I’m not sure how to “calculate” this

But I think the the answer you are looking for is for manifolds by boundary we simply mean a different kind of boundary

Look up the definition of manifold with boundary

[empty grove]

There are many kinds of boundaries, you should clarify which one you mean

[gritty widget]

There is a nice section in guilleman and pollack on this

[floral gust]

Sorry I meant determine. But the confusing thing is that for any point that lies in the rectangular section of the cylinder, the neighborhood should be diffeomorphic to a neighborhood in the half space

[gritty widget]

Yes

The boundary

[floral gust]

So the boundary shouldn’t just be two enclosing circles right?

[gritty widget]

Okay so picture the half the plane

Pick a point in the cylinder away from the end caps

We can draw a neighborhood around this that is diffeomorphic to an open neighborhood of the half plan away from the line

By line I mean x-axis

[floral gust]

Oh I think I understand now. For hollow cylinder, the boundary is just two circles but for a filled cylinder the boundary is the disks along with the rectangle. Correct?

[gritty widget]

Or just a neighborhood of regular old R^2

Yes I think so I’d need to double check that that is not a “manifold with corners”

But the boundary points

Like the points on one of the end caps

Are points that can only be diffeomorphic to an open neighbor hood that touches the x-axis

This is the picture to have in mind

[floral gust]

That makes sense. Thank you!

[gritty widget]

Nb

[lunar yoke]

Not sure if this helps but informally i like to think of the boundary as the points where, when moving inside the manifold, you may fall "off". So if you are moving on the folded rectangle part of the cylinder, you cannot really move into a direction where you would leave the manifold, but you can move towards the circles, and imagine that if you were to continue in that direction you would leave the cylinder

[floral gust]

So in this question dD should be the slanted cylinder thing from z=0 and z=2. And then dD cap {0<z<2 } should be the cylinder between z=0 and 2. And so dS1 should be empty set. No?

[hollow harbor]

well, for the purposes of say, stokes theorem, you would probably consider the top circle at z = 2 and the bottom circle at z = 0 to be the boundary of dS_1. stokes theorem asks you to provide a smooth oriented surface and a closed smooth boundary curve.
now, if you just want to say "what is the boundary of this surface in the sense of differential topology" and you don't plan on applying stokes (it really seems like you plan on that) then I guess you could just say it's the empty set, because the surface does not come with an explicit boundary.

[floral gust]

Can I not instead say the question isn’t well worded, because the surface they provide isn’t one on which stokes theorem is applicable, considering that the boundary is empty ?

[hollow harbor]

It is one on which stokes theorem is applicable

"boundary" is a word with a lot of definitions

The notion of "boundary" in stokes theorem is that of a smooth closed curve which (along with the surface itself) traveses the closure of an embedded surface in R^3. (i think that's somewhat accurate, it gets hairy if you start considering surfaces with fractal boundary).

Alternatively, one can think of it as the new stuff in the metric completion of the surface.

So you can see that this definition is extrinsic (depends on the embedding)

Or I guess it depends more on the metric you put on the surface

The point being, just because they wrote < instead of ≤ doesnt change how you use stokes theorem for this problem.

[low shell]

Can anyone help with classifying Euclidean isometries

[floral gust]

Ah okay okay. So instead of S1 I should rather apply stokes on the closure of S1 giving me the boundary of dD cap {0< =z<=2}

[hollow harbor]

Sort of, but note that d(D cap {0≤z≤2}) then includes the top disk and the bottom disk (which you dont want). This is why they use < instead of ≤.

[floral gust]

Ahhh nvm. I calculated the closure wrong. The closure of S1 should be the hollow cylinder whose boundary should be the two circles.

[hollow harbor]

Yep!

[severe stump]

I'm confused by case B on this page https://math.stackexchange.com/questions/79594/which-sequences-converge-in-a-cofinite-topology-and-what-is-their-limit

In the cofinite topology, the sequence 0,1,0,2,0,3,0,4,0,5,0,... doesn't converge to anything, right?

But the post and answers seem to imply that it converges to 0. What am I missing?

[obtuse meteor]

Yeah it converges to 0

The name of the game is take an open set U around 0

Now U^c is finite right?

In particular there’s an N so that if n >= N then n is not in U^c, aka n is in U

Bc U^c is bounded

Then we know that once we get to the point N, 0, N + 1, …

Every element of this will be in U

[severe stump]

@obtuse meteor I thought that it converges iff for all open sets S around 0, only finitely many members of the sequence is outside of S.

[obtuse meteor]

I took an arbitrary open set to start with so isn’t that what I just proved?

Make sure your quantifiers are in the right order

[severe stump]

I just realized I did a dumb

[obtuse meteor]

It happens ^^

[severe stump]

I confused cofinite with countable complement

🤦

[obtuse meteor]

Oop

[severe stump]

wait, i didn't even confuse it with countable complement

I have no idea what I was confused about lol

I think I added an extra "not" somewhere.

Oh wait. I thought that a set is open iff it's finite lmao

[cerulean oriole]

it's almost like it's a neighbourhood of infinity.

[gritty widget]

Do you people make lists of important theorems?

in my brain, yes

[obtuse meteor]

Nope

[hollow harbor]

oliver knill hours

[cloud owl]

yes, of course i do

i call them 'notes'

[obtuse meteor]

This is my strategy but it doesn’t work for everyone

One of my gfs really benefits from the writing lists

[glad nebula]

Stupid question but I'm trying to show that in an irreducible topological space that any nonempty open set is dense, but can't find the abstract definition of being dense. I think I remember that for a metric space $X$, $D$ is dense in $X$ if for any $x \in X$ and any $\epsilon > 0$ there is $b \in B$ such that $x \in B_\epsilon(b)$.

Does the definition extend from this so that for any $x$ there is $b$ with a neighborhood containing $x$? If this is the right idea, how is the definition properly stated?

[gentle osprey]

Kenya

[hollow harbor]

a set is dense if every open set contains a point in that set. equivalently, a set is dense if its closure is the whole space.

[honest terrace]

a definition of dense that holds in a general topological is that closure = whole space

crap too slow

[hollow harbor]

sniped

anyway the condition that every open set contains a point in the dense set is usually easiest to show

[glad nebula]

awesome thank y'all!

[frigid patrol]

Unironically based video idea

[marsh forge]

i would love it if ppl made videos like this w high production quality

that consistsntly got like 10 views

[pearl holly]

hmmm so this exercise seems kinda sus. It goes like this: Let p: X -> Y be a continuous map. Show that if there is a continuous map f: Y -> X such that p o f (the composite of p and f) equals the identity map of Y, then p is a quotient map. My solution goes like this: f is the inverse of p and therefore p is bijective and so is surjective. Since p is continuous, we know that if V is open then p^-1(V) is open. Now, since f is continous, given an open set U we know that f^-1(U) is open. But f^-1 = p so p(U) is open. Therefore, p is a surjective continuous map that is also open, hence a quotient map. It seems almost too easy so it feels like I've done something wrong, but I can't see it

[gritty widget]

p may not be bijective

only surjectivity is guaranteed (having a right inverse is equivalent, at least if you assume choice, to surjectivity)

[pearl holly]

oh yeah I forgot that

[marsh forge]

You are looking for something called a section

not an inverse

my advice

think about how to do this for sets

which is much easier

and then think about how to ensure continuity

oh wait i messed this up

sorry i realized what i was thinking is way too strong lmfao

its indeed false

[pearl holly]

hmm but even if it is surjective only, then my argument still holds right? Or is something else wrong?

[marsh forge]

i think up to modification this might go through let me think, the notation + the usage of inverse is throwing me off

[pearl holly]

yeah. My whole argument really lies in the fact that you can show that p is an open map from the fact that it has a continuous right inverse.

[marsh forge]

i think the idea works but i wont sign off until i see how you correct it hahaha

maybe a meta-comment: theorems this general will normally have very straightforward proofs

bc like

what else could you do

[pearl holly]

Okay so if I try to modify it then it would go something like this: We know that p o f = e (where e is the identity). This means that f^-1 = p. Now we know that f is continuous (it is given in the exercise) so given an open set U, f^-1(U) is open. But f^-1 = p so f^-1(U) = p(U) and p(U) must be open so therefore p is an open map. Now p is continuous (it is given in the exercise), surjective and an open map so it is a quotient map

[marsh forge]

f^-1 does not necessarily exist

[pearl holly]

but if f^-1 doesn't exist then p doesn't exist since they are equal, right?

[marsh forge]

well they arent equal

is my point

you are confusing set theoretic preimage with inverse function i think

you can make sense of the former even when the latter doesnt exist

[pearl holly]

but $p \circ f = e$ so $p \circ f \circ f^{-1} = f^{-1}$ which means that $p = f^{-1}$

[gentle osprey]

Tokidoki

[marsh forge]

again f^-1 doesnt exist

necessarily

[pearl holly]

oh yeah sorry lmao

[marsh forge]

if it does exist then thisnis correct

but you get a much stronger statement

(maybe figuring out what the stronger result is might help)

(hint if theres an continuously invertible map X->Y what is another name for that)

[pearl holly]

it's called a homeomorphism?

[marsh forge]

yeah

which is ofc an example of a quotient map

a very boring one

[pearl holly]

oh

[marsh forge]

but the thing youre trying to prove is stronger

[pearl holly]

so p is a homeomorphism?

[marsh forge]

no sorry im confusing you

if f^-1 exists then p is a homeo and the statement is trivial

but its still true when f^-1 doesnt exist

[pearl holly]

well maybe f^-1(U) = p(U)? So f^-1 doesn't need to be p?

[marsh forge]

yes

but you have to prove that

[pearl holly]

yeah that's true

okay I will do it later, I realised that I need to go somewhere like right now. But thank you so much! I will figure it out on my way or when I get home!

[marsh forge]

from here it should just be (potentially tricky) set theory to prove those sets are the same

feel free to ping if u get stuck

[pearl holly]

okay great! Thank you!

[inner sluice]

Sort of noob question, but related to topology:

So T is a subset of the powerset of X. T is also a set?

[sacred quail]

yes, put simply T is a collection of subsets of X

[inner sluice]

Where does it imply it is a collection of subsets, because of the power set? A subset of a power set is a collection of sets?

[sacred quail]

The power set is the collection of all subsets of X
A subset of the power set is some of those subsets of X

[inner sluice]

How does this definition of topology imply connectivity?

So for (a), S1, S2 - each set defines a set of points potentially, and because they are in a set, it is implied that they have connectivity to one another?

FYI I'm trying to follow the book Topology for Computing

[sacred quail]

The points in each set imply some level of "closeness" yes. It's probably worth you reading the first few pages of http://pi.math.cornell.edu/~hatcher/Top/TopNotes.pdf for more precise motivation of topological spaces

[gritty widget]

what does connectivity mean here?

[sacred quail]

Also that Topology for Computing book looks interesting, what are you reading it for?

[gritty widget]

for computing

[inner sluice]

Am reading about bezier and bsplines - stumbled into this book because it seemed like a good intro to topology for computer science

Something I made yesterday

Am also doing CAD stuff

From reading the first few pages of this book, my hunch seems to be paying off in terms of understanding geometry decimation/connectivity reconstruction

@sacred quail That link was definitely much more precise than the book, thanks

It made it much simpler to understand this definition

Why does this definition seems nonsensical?

S is some arbritrary set? How does that mean its open? Does this mean a Topology is instrinsically contains open sets?

[wanton marsh]

elements of T are called open

a topology is the set of sets that it wants to be open

[river granite]

Does this mean a Topology is instrinsically contains open sets?
that's exactly the point

[inner sluice]

You can define what an open means though right?

[river granite]

you can in the context of e.g. metric spaces, where open sets are precisely those that contain an open ball at each point

[wanton marsh]

a topology decides who is open

when you define a topology you define who are the open sets

[river granite]

what's behind this though is simply the fact that these open balls are a basis of the metric topology. they generate the topology (i.e. the class of open sets) of that metric space in a precise sense

defining a topology through a basis is also common

[manic garden]

anyone have a good ref for chern simons?

[inner sluice]

Do you have any more information about this statement? I wonder if the Bernstein Basis defines a specific topology and whether this is a known form already

[river granite]

it's something you almost always see in a course on point-set topology, here's the definition from Munkres' book for instance

[river granite]

the Bernstein polynomials aren't a basis in that sense though; these uniformly approximate continuous functions by the Stone-Weierstrass theorem

and here we're talking about a basis for the open sets in a topology, defined as above

a basis in a topology has more or less the same role as the open balls in a metric space's topology

[inner sluice]

Hmm ok different definition

[river granite]

yep

in the case of these polys. you might be thinking of a https://en.wikipedia.org/wiki/Schauder_basis or similar

[inner sluice]

beautiful

[pearl holly]

Okay so I'm bakk and I think that I showed that f^-1(U) = p(U). So I did this using inclusion. So let x be in f^-1(U). This means by definition that f(x) is in U. This gives that p(f(x)) is in p(U). But p(f(x)) = x, so x is in p(U). The other inclusion is similar. Let x be in p(U). Then p(f(x)) is in p(U). This means that f(x) is in U. So x is in f^-1(U). This means that they are equal. Is this right?

[marsh forge]

yeah that looks correct

[pearl holly]

okay great! Thank you so much for the help and your patience, I really appreciate it!

[inner sluice]

So for a metric space for a set X and metric d, the topology is the power set of open balls in X?

[pearl holly]

I shouldn't be the one answering this but if I recall correctly then the metric topology on X has as basis all open balls in X

[river granite]

yeah

not every open set is an open ball (in most metric spaces anyway)

[marsh forge]

what

no

[river granite]

and the power set doesn't have anything to do with it

[marsh forge]

well

yeah no not quite at all

the topology is induced by the basis of open balls

[marsh forge]

but is in general much larger

yes

the transition basis->topology is unique and easy

[inner sluice]

So theres no explicit topology, but its induced by the basis

[marsh forge]

but its not what you described

well a basis induces a unique topology

so in some sense its explicit in that i can describe it for you

but there are more open sets than just the open balls

[pearl holly]

yeah every open set is a union of some open balls in the metric topology

[marsh forge]

for example in the real numbers the union of (0,1) with (2,3) is open but not an open ball

[inner sluice]

Yea I mean the r restriction for open-ball narrows it down

[obtuse meteor]

No

Lol someone already answered this problem

Lol

[inner sluice]

A = (0,1) and B = (2,3), A union B = (0,1,2,3) So the open ball with center 1.5, and radius 1.5 wouldn't be valid since the set doesn't contain such a value

[marsh forge]

huh

that notation doesnt make any sense

(0,1,2,3)

but yes that union is not the same as a ball centered at 1.5

it doesnt contain 1.5

[frigid river]

I have a brain dead question:

Is $\mathbb{R}_l$ coarser than $\mathbb{R}$?

[gentle osprey]

festina lente, cucurbita

[marsh forge]

can you define R_l for the class

[frigid river]

lower limit topology

[true robin]

It is finer

[marsh forge]

https://en.wikipedia.org/wiki/Lower_limit_topology

read the literal first sentence hahahaha

[gritty widget]

nice embed

[frigid river]

very nice

[marsh forge]

oh sorry first sentence under properties

fwiw I really dislike the terms finer and coarser for topologies bc ive seen them interchangeably (potentially on accident) too much

[true robin]

Notice that the union [a’,b) for all a’ strictly greater than a gives us (a,b)

[flint cove]

@frigid river if you wanna think about it in terms of convergence, note that 1/n converges to 0 in lower limit but -1/n does not, i.e. 0 is kinda „more disconnected“ from the points <0

and being finer usually means that the space becomes „more disconnected“

while being coarser means that the space is more „clumped together“ (think indistinguishable points or even the indiscrete top)

[frigid river]

Thanks.
And could you maybe help me understand why this is true?
$id:(X,\mathscr{T}_1) \to (X, \mathscr{T}_2)$ is continuous
$\Longleftrightarrow \mathscr{T}_1$ is finer than $\mathscr{T}_2$?

[gentle osprey]

festina lente, cucurbita

[marsh forge]

wait saketh

delete that message pls

@frigid river this is really an exercise you should do yourself

we can help you

but it is very instructive

❤️ thanks saketh

[true robin]

Np

[frigid river]

Sad part is that it isn't an exercise, but okay, thanks xD

[gritty widget]

everything is an exercise

[marsh forge]

well exercise is a generic term for "something you should but don't yet know how to prove"

like when you're reading a textbook, you should consider understanding the proofs in the book to be an exercise in and of itself

esp. if the author makes a lot of slick or concise arugments

[flint cove]

Most of the things you prove shouldn't be mandated exercises, it should be questions you ask yourself. Fight me.

[marsh forge]

esp. if they just claim stuff without proof or call it obvious

@frigid river i feel bad what i meant was "you should give it a serious try to prove it by yourself and let us know if we can help"

there is a really hard to explain difference in what you learn when you prove something by struggling yourself and when you just read a proof

[frigid river]

Okay, give me a few minutes then, imma try

[flint cove]

No haste. I sometimes took days for „easy“ exercises; it's the fight and thought process that makes it stick.

(and still do)

[frigid river]

Okay, it really doesn't seem that difficult.

Any set that's open in $(X, T_2)$ must also be open in $(X, T_1)$ by definition of continuity; therefore $T_2 \subseteq T_1$, and $T_1$ is finer than $T_2$.

All the stuff that's open in $(X,T_2)$ is open in $(X, T_1)$ by definition. Therefore, the preimage of any set open in $(X, T_2)$ is also open in $(X, T_1)$, and the function is continuous.

I can't find the latex error 0.0

[gentle osprey]

ok-hi

festina lente, cucurbita

[frigid river]

Ahh, thanks.

But why is that necessary? 0.0

[opaque totem]

[gentle osprey]

ok-hi

[frigid river]

Ahh. The first never posed any problem though, it just looked different. I suppose the underscores confused the bot

thanks again

[opaque totem]

no problem

[marsh forge]

https://math.stackexchange.com/questions/4164435/signs-for-products-defined-homotopy-theoretically someone answer my stack exchange question my family is starving

[frigid river]

How would I go about proving/disproving that $\mathbb{R}$ and $\mathbb{R}_l$ are homeomorphic?

[gentle osprey]

festina lente, cucurbita

[frigid river]

(Sorry, Max, I hope you are fine with me posting)

[verbal wraith]

You can try to find a property that R_l has that is preserved by homeomorphisms that R doesn't have.

[gritty widget]

i cannot answer but i have given you an upvote

[frigid river]

Thanks, I found someone using the argument that R has a countable basis, and R_l doesn't. But this would require me to prove that having countable bases is preserved by homeomorphism, which I don't know how to do either :/

[verbal wraith]

well homeomorphisms distribute over unions so f of a basis is a basis

[marsh forge]

Dont just answer questions for people

[true robin]

Union? Oh you mean the meet of the poset. You should really strive to be more clear

[marsh forge]

lmfao

[frigid river]

don't worry, that didn't answer anything for me, sadly xD

[marsh forge]

hahaha

all right

heres my hint

basically every topological property (and in a formal sense, every topological property) is preserved by homeomorphism

so basically if I were you id just like

write down the formal statement

and then try to prove each axiom

homeos are so nice that it should be smoothish once you get your bearings

(youd be surprised how often explicitly writing down the thing you are trying to prove makes it easier)

[frigid river]

I really don't know how I should connect those 3 "axioms" (can you call them axioms?) of homeomorphisms to bases :/

[pearl holly]

what an interesting problem!

but why are you doing problems at like 01:30 lmao???

[frigid river]

xD That's when I work most efficiently.

[pearl holly]

lmao but where do you get these exercises? I need the sauce

[frigid river]

From YouTube lectures, actually. The problem is he just assumes that this is the case, without much talking about it.

[pearl holly]

ouch...

[marsh forge]

okay so

what are you trying to prove @frigid river

i know what youre trying to prove i just wanna make you do it hahah

[frigid river]

I formulated it this way:
X is homeomorphic to Y
$\Rightarrow$ if X has a countable basis, Y has a countable basis.

[gentle osprey]

festina lente, cucurbita

[frigid river]

I really don't know if this is a smart way to phrase this problem.

[true robin]

Yes this will work

[frigid river]

I assume I also only care about the continuity part, because bijectiveness doesn't really have anything to do with bases.

[honest terrace]

lmao

[marsh forge]

okay

[true robin]

Look at the injection of {1} into Rl, the lhs has a countable basis, but the rhs doesn’t continuity is not enough

[marsh forge]

thats true

lets work on it

what should the basis be

let {B_i}

be a basis for X

and let f:X->Y be a homemorphism

what should the basis for Y be

[frigid river]

{B_j}, maybe? xD

[marsh forge]

haha does that use f?

we want to take a basis in X and make a basis on Y

how do we do it

[true robin]

Try to find something that is inside of Y using Bj’s

[marsh forge]

pls

[frigid river]

I really don't know. I know that it must "cover" (also don't know if I used that right) all the stuff in Y, but that's about it.

[marsh forge]

well

all you have is B_i

and f

can you think of a way to combine them?

[frigid river]

Since B_i is open in X, it also is open in Y?

[marsh forge]

is B_i a subset of Y?

[frigid river]

should be, else it couldn't be open, right?

[marsh forge]

well the second half is correct

[frigid river]

-.-

[marsh forge]

you still havent used f 🙂

actually this is a spoiler but i want to make sure

[frigid river]

I don't know how, I don't know anything about the function :/

[marsh forge]

do you know that you can apply a function to a subset

[frigid river]

like with the definition of continuity? Yes?

[marsh forge]

so let A be a subset of X

do you know what f(A) means?

[frigid river]

f(A) is a subset of Y?

[marsh forge]

well yes do you know which subset?

[frigid river]

no

[marsh forge]

okay

so A subset X

then f(A) is the set

{f(x) | x in A}

[frigid river]

yes

xD

[marsh forge]

okay

so do you have an idea how to turn B_i into a subset of Y

?

[frigid river]

f(B_i)

[marsh forge]

Okay yes that is a subset of Y

So if I have a basis {B_i} then I have a collection of subsets of Y {f(B_i)}

with me so far?

[frigid river]

Yes.

[marsh forge]

Okay step one you already hinted at

Can you tell me why f(B_i) is open?

[frigid river]

Basis elements of a set are always open in that set. And since the function and its inverse are continuous, f(open set) is open.

[marsh forge]

Yes!

Okay so what tests do you know for bases

there are some slick ones but I'm not sure whether you know them. If not we can just use the definition of a basis

or if you want we can just use the definition of a basis anyway

[frigid river]

I honestly only remember the two "axioms" of a bases xD

definition

whatever

[marsh forge]

Okay what are they

[frigid river]

every element of the set must be contained by some basis. And if two bases contain one point, the intersection of those bases must be a basis as well.

[marsh forge]

Okay we can do it with this, let's start with the first part

let y be an element of Y. Do you know why y is in f(B_i) for some i?

[true robin]

I feel like you should clarify that this definition is a bit wrong

[marsh forge]

This seems identical to the wikipedia defn unless im misreading something

[pearl holly]

It's the definition given in Munkres

[true robin]

Intersection should contain a basis, but need not be a basis element itself

[marsh forge]

I think that is a convention

Oh no

I see your point

yes

This is stronger even though it doesn't make a practical difference

[pearl holly]

I thought that you said that the first axiom is wrong put now I get it lmao

[pearl holly]

THIS is the definition given in Munkres with the first axiom that happy person stated

[marsh forge]

@frigid river There's a slight difference. The second axiom is too strong. Compare with: https://en.wikipedia.org/wiki/Base_(topology)#Definition_and_basic_properties

[true robin]

Yeah I think for a basis for R^2 would be very weird with this definition instead of the usual open balls

[marsh forge]

Well you can just take a basis as you are defining saketh and throw in some extra stuff without issue since the topology generated by a basis is unique and opens are closed under finite intersection

so theres no effective difference

[frigid river]

I am still confused about this, just because B_i "covers" (still don't know if that's the right word) X, why does this mean that B_i covers Y? We only know that B_i is open in Y.

[marsh forge]

(thanks for pointing it out though since the definition you are pointing out is way more common)

[marsh forge]

You have to prove it!

It's not entirely obvious

[frigid river]

how should I know xD

[marsh forge]

Covers is the right word, by the way

again, don't worry about like

not seeing the full answer right away

its okay to take small steps and follow your nose

I'll start with a hint

Do you agree that f(X) covers Y?

(and do you know why?)

[frigid river]

because bijective

[marsh forge]

Great

So if all of the B_i cover X, and f(X) covers Y, do you see why all of the f(B_i) should cover Y?

[frigid river]

Okay, yes. It must, because it's bijective.

[marsh forge]

Yes, now let's actually prove it

[frigid river]

huh

isn't that the definition of bijective? xD

[marsh forge]

it depends hahah

if you're full convinced

then this is fine

but for newer ppl its better to be explicit

up to you if ur not getting graded

[frigid river]

I'd leave it like that, then, please. I feel like the rest is difficult enough 0.0

[marsh forge]

hahaha okay

because I can't resist and you can read this later: ||Let y be an element of Y. Then there is some x such that f(x)=y. Let B_i contain x. Then f(B_i) contains y||

Okay next step is the intersection thing

do you understand the wikipedia statement (and why yours is too strong?)

[frigid river]

yes

[marsh forge]

Okay, so suppose $y\in f(B_i)\cap f(B_j)$. We want to prove that there is some $f(B_k)\subset f(B_i)\cap f(B_j)$ containing $y$

[gentle osprey]

MaxJ (Cali Surfer Arc)

[marsh forge]

Maybe reading the thing I blacked out above will give you a hint on how to do this

its the same idea

[frigid river]

I know there exists some open set containing y, but no, I have no idea why there should be a basis doing so.

[marsh forge]

Okay here's a hint

Let y be an element of Y. Then there is some x such that f(x)=y.

^ thats the first step

well let y be the same element from before

the first step is picking an x in X so that f(x)=y

do you have an idea of what to do next?

[frigid river]

Wait, if I have x in U, and U in X, does this mean f(x) is in f(U)?

[marsh forge]

yes!

[frigid river]

then it's easy. But why is this true again? xD

[marsh forge]

(recall my definition of f(U))

[marsh forge]

.

A=U here

[frigid river]

Okay, then, if $x$ is in $B_1 \cap B_2$, then $f(x)$ is also in $f(B_1) \cap f(B_2)$.

[marsh forge]

(almost)

[gentle osprey]

festina lente, cucurbita

[marsh forge]

Okay so what you've written is true

but

only because f is a homeomorphism

The statement thats equivalent to what we agreed on is

$f(x)\in f(B_1\cap B_2)$

[gentle osprey]

MaxJ (Cali Surfer Arc)

[marsh forge]

You want $f(B_1\cap B_2)=f(B_1)\cap f(B_2)$

[gentle osprey]

MaxJ (Cali Surfer Arc)

[frigid river]

Wait, if I have x in U, and U in X, does this mean f(x) is in f(U)?
So is this also only true for homeomorphisms?

[marsh forge]

no thats always true

see what I wrote

Here's an example. Let $X={a,b}$ and $Y={c}$. Let $A={a},B={b}$ then $A,B\subset X$ and $A\cap B=\emptyset$. However if we define a function $f:X\to Y$ by $f(a)=f(b)=c$ then $f(A)\cap f(B)\neq \emptyset$

you need f to be a bijection

not necessarily a homeo

sorry

oh shoot

my {} dissapeared

[gentle osprey]

MaxJ (Cali Surfer Arc)

[frigid river]

Thanks for your great effort, but I can't think at all anymore, sorry xD I will just go to bed now and try this tomorrow again.

,ti

[gentle osprey]

The current time for veryhappyperson is 02:34 AM (CEST) on Sun, 06/06/2021.

[frigid river]

0.0

[marsh forge]

hahah

feel free to ping altho ur timezone is kinda different

[cerulean oriole]

[cerulean oriole]

It makes a practical difference if you want open balls in R^n to form a basis and not "finite intersections of open balls which are non-empty".
Which I hope everybody wants.

[true robin]

Damn, did I mix up join and meet? I can never keep them straight

[empty grove]

You were correct, the poset is under reverse inclusion

[marsh forge]

my point was that the open balls in R^n generate this more awkward definition anyway

So like

practically speaking the difference is linguistic

obviously the weaker defn is better though

[empty grove]

What why tho

Can't you just dualize everything

[marsh forge]

this is why you should randomly choose between limit and colimit

[winged viper]

How do you show that there cannot exist a pairwise disjoint sequence ${E_n} \subset \mathbb{R}$ of closed, nowhere dense sets such that $\bigcup_{n = 1}^\infty E_n$ is complete?

It supposedly follows from the Baire category theorem, but I'm not able to see it

[gentle osprey]

Frank

[empty grove]

R is complete, so cant be a countable union of nowhere dense sets

By Baire's theorem

[winged viper]

right

but the union only needs to be any closed subset of R right

[empty grove]

I didnt get that

[winged viper]

If $\bigcup_{n = 1}^\infty E_n$ is any closed subset of $\mathbb{R}$, then it is complete

[gentle osprey]

Frank

[empty grove]

right

[winged viper]

so why does this lead to the contradiction

[empty grove]

oh

I didnt read the question properly, sorry

[winged viper]

oh np

[dim meadow]

wouldn't the complement of your union be dense in R by BCT

[winged viper]

right

[dim meadow]

idk if this is enough

one sec

[empty grove]

is it not?

oh wait yeah the complement is dense

[winged viper]

also sorry, the problem should be this:

How do you show that there cannot exist a pairwise disjoint sequence ${En} \subset [0, 1]$ of closed, nowhere dense sets such that $\bigcup{n = 1}^\infty E_n$ is complete?

[gentle osprey]

Frank

[winged viper]

idk if the fact the En's are compact helps

[dim meadow]

it does actually

[winged viper]

ah my bad

[dim meadow]

I think

why can't you have each E_n be a different element of a convergent sequence and have E_0 be the limit

this seems weird

[winged viper]

hmm

ok lemme just send the thing im confused on haha i probably missed some condition

[dim meadow]

like the classic E_0={0}, E_n={1/n}

[winged viper]

answer to part b

i don't get the last line

yeah but the set E is only the endpoints

[dim meadow]

oh so that's a very particular set

[winged viper]

yeah

[empty grove]

Each E_n is nowhere dense in the subspace E

So BCT is being appied directly to E

[empty grove]

This is what that part with the open balls is proving

[winged viper]

oh wait

so E has the baire property because it's complete

[dim meadow]

yeah

[empty grove]

yep

[winged viper]

Each En is nowhere dense in E

[empty grove]

yes

[winged viper]

yet the union is dense in E

[empty grove]

yeah

[winged viper]

contradiction?

[empty grove]

[winged viper]

oh man ok i completely misunderstood that oops

[empty grove]

wait are you asking what the contradicition is?

lol

[winged viper]

is the contradiction what i wrote

[empty grove]

yep

[winged viper]

just union of nowhere dense sets is dense

ok whew got it

[empty grove]

yeah

[winged viper]

makes sense, thanks guys

[empty grove]

[winged viper]

wait so is this true?

could you apply BCT to the union?

[empty grove]

nope, liquid gave the counterexample

[winged viper]

oh right duh

[empty grove]

You cant because there each E_n is only given to be NWD in R

but it may not be NWD in E

for example no set is NWD in itself

other than the empty set

[winged viper]

oh so like if E = {1 / n : n = 1, 2, ...}, then the set {1/2} is not nowhere dense in E

because {1/2} is open w.r.t. E

[empty grove]

yep

[winged viper]

ah ok that clears things up, thank you

[empty grove]

[frigid river]

Could anyone help me here, checking the second condition for a basis?

Also, is it "biijectivness", or what's the noun for that? xD

[empty grove]

Bijectivity

Also, just proving what you are trying to prove is not enough, that just showed that it's a basis for some topology

A more direct way is to show that a union of basis elements maps to a union of images of basis elements

[frigid river]

[true robin]

What do you mean by second condition? That the intersection of two basis elements contains a basis? Try the same maneuver of just applying f or removing f from everything

[empty grove]

Doing this for both f and f inverse will do

[true robin]

Also show that what you are getting is actually open too

[frigid river]

That's all great, but what's the "image of basis elements" again? xD

[empty grove]

Lol image under the homeomorphism

[frigid river]

and what's an image? 0.0

[empty grove]

So you're showing that if {B_i} generates topology of X then {f(B_i)} generates topology of Y

[frigid river]

Yes.

[empty grove]

Image of B under f = f(B)

[frigid river]

ah, okay

[frigid river]

So, now I know what this means. That doesn't mean I know how to do it though 0.0 Isn't it the same problem I am facing with my (incorrect proof) all over again?

[empty grove]

Say U is the union of B_j's

You need to show that f(U) is a union of f(B_j)s

So you see how to do this?

[frigid river]

of course not

[empty grove]

Try to prove both inclusions

And see which one you get stuck on

[true robin]

Your mom is the union of Bj’s

[empty grove]

[frigid river]

So basically, I need to show this, right?
$$f(\bigcup B_n)=f(B_1)\cup f(B_2) \cup ...$$

[gentle osprey]

festina lente, cucurbita

[empty grove]

Yes

[frigid river]

beautiful formatting btw xD

[empty grove]

Use single dollars for better formatting lol

[opaque totem]

and \cdots

instead of ...