point-set-topology

[pastel thistle]

A2 is all points in S2 with z>1/2 and A1 all points in S2 with z<-1/2

the question is if such function could or could not exist.

I think not, but Im having trouble proving it

[haughty wave]

Can't you use urysohn's lemma?

[pastel thistle]

mmm

didnt learn it yet, but Im reading it

how would you use it ?

[haughty wave]

If you haven't learned it, then you can define a function like this

Oh wait, it should be flipped

[pastel thistle]

so you say that such function does exist ?

omg I was sure it didnt

[haughty wave]

But you can multiply that function (let's call it f) with (1, 1/2, 1/3, ...) and define g(x, y, z) = f(z) * (1, 1/2, 1/3, ...)

[pastel thistle]

oh okey

I think I got the idea thank you, so creative

[shy moss]

here why F(l,s)(t) is a homotopy?

boundaries

F(l,s)(0)=l(0) but l(0) is not id

[gentle osprey]

Moldilocks

Moldilocks

[shy moss]

thanks, i get it

[flint sinew]

I think I have some trouble trying to understand what standard topology means

[fading vale]

The standard topology on R^n?

[flint sinew]

I guess

not entirely sure, because the concept of std. topology is in itself conceptually foreign

[fading vale]

"standard topology" isnt a concept

It doesn't mean anything without the context of whatever set you're working on

[flint sinew]

I see

[fading vale]

if i have an arbitrary set X there is no "standard topology" on X in all cases

[flint sinew]

if that set is - say - all positive integers?

[fading vale]

"the standard topology on N" doesnt mean anything no

at least not anything formal

I guess it could mean subspace topology for subsets of R^n

which in this case yeah would be discrete

[flint sinew]

okay, let's ask it this way - if std. topology requires "context", what context is that exactly? are there any clear examples of a set and a context which establish whether something is or isn't a standard topology?

[fading vale]

"Standard topology" is a term used to refer to the usual metric topology on R^n and the subspace topology on subsets of R^n

In an arbitrary topological space it means nothing

[flint sinew]

R^n refers to space of n-tuples of real numbers?

[fading vale]

Yes

The metric topology being the one where open sets are unions of open balls

[flint sinew]

is open ball the same as soft ball?

[fading vale]

the open ball of radius r about a point x is the set B_x = {y | d(x, y) < r}

in dimension 1 this is an open interval

[flint sinew]

also I think I'll need to go even further to the fundamental vocabulary - what is a "metric topology" and what does the elaboration of "the usual metric topology on R^n" mean?

[fading vale]

If you are missing this much terminology you probably should just crack open a textbook

[flint sinew]

fair enough

[lyric quartz]

I'm trying to understand this proof from Weibel but to no avail. The part that I'm stuck at right now is that the intersection of D(A), the degenerate simplicial complex, and N(A), the nomralized simplicail complex, is trivial. I try to follow the proof but I don't see why ith face of y is x_i. Could anyone help? Thanks!

from here: https://people.math.rochester.edu/faculty/doug//otherpapers/weibel-hom.pdf

[plain raven]

I have gone through this proof not too long ago, let me see if I can recall

btw do you have a copy of Weibel's Errata on hand? It's substantial

you can find it on his website

Indeed consulting the errata it does seem that there is an error in the proof

the correct line should be

"If $y\in N_n(A)$ and $i$ is the smallest integer such that $\sigma_i(x_i)\neq 0$, then $y = \sigma_i\partial_iy = \sum_{j>y}\sigma_j(x'_j).$ By induction, then, $y=0$, and so $D_n\cap N_n=0$.

[gentle osprey]

diligentClerk

[obsidian socket]

Does anyone know what v represents in this?

[lyric quartz]

so I'm not getting this line either. Isn't y - \sigma_i \partial_i y = 0 right away because \sigma_i \partial_i = id?

[plain raven]

actually i don't think $\sigma_i \partial_i =id$ in general

[gentle osprey]

diligentClerk

[plain raven]

i don't think that's one of the simplicial identities

$\sigma_i\partial_i = \partial_i\sigma_{i+1}$

[gentle osprey]

diligentClerk

[tepid snow]

Hello! I am interested in understanding this paper on topology: https://arxiv.org/pdf/1912.11324.pdf Will an introductory topology book be sufficient? I already have familiarity with space groups, point groups, and Laue groups. What do I need to know to understand the terminology and develop a visceral intuition of the named spaces and groups. A group theory textbook as well?

[lyric quartz]

okay so I'm fine with everything except the induction. What's the induction on and how is it working?

[plain raven]

ok yeah let me take a look at this

Ok. Let $y$ be an element in $D_n(A)$. Now, assume that $y\in N_n(A)$, so $\partial_iy=0$.

[gentle osprey]

diligentClerk

[plain raven]

so $s_i\partial_iy=0$.

[gentle osprey]

diligentClerk

[plain raven]

(Recall that $i$ is the index of one of the degeneracies, so it's strictly less than $n+1$)

[gentle osprey]

diligentClerk

[plain raven]

So $y = y -s_i\partial_i y$.

[gentle osprey]

diligentClerk

[plain raven]

Since $s_i\partial_is_ix_i = s_i$ (because $\partial_is_i=id$), the $s_ix_i$ term disappears when we rewrite it this way

[gentle osprey]

diligentClerk

[plain raven]

and when $j>i$, we have that $sI_i\partial_is_jx_j=s_is_{j-1}d_ix_j = s_j s_i\partial_ix_j$

[gentle osprey]

diligentClerk

[plain raven]

Tehrefore, $y - s_i\partial_iy$ can be expressed as a sum of terms $\sum_{j>i}s_jx_j'$ for some new list of $x_j'$.

[gentle osprey]

diligentClerk

[plain raven]

We can view this as the induction step in an induction on $i$. We are proving, for each $i$, that $y$ can be expressed as the sum of degeneracies of the form $s_j(x_j)$ for $j>i$.

[gentle osprey]

diligentClerk

[plain raven]

But if this is true for all $i$, then in the end we can write it as a sum over no $j$, i.e., as an empty sum, and the sum must be zero.

[gentle osprey]

diligentClerk

[plain raven]

Each step in the induction kills off the lowest-index term in the sum and replaces it by higher-index terms. If you do this repeatedly you see that you can eventually eliminate all of the terms.

[lyric quartz]

oh okay, first given i, basically we move up indexing until you reach 0.

[plain raven]

yeah that's why he assumes that $i$ is the least index for which the term is nonzero

[gentle osprey]

diligentClerk

[flint cove]

Does anybody know the condition we need for a convergence to arise from a topology?

a centered convergence such that lim ℱ = \bigcap_{𝒰 ultrafilter containing ℱ} lim 𝒰 seems to be callled “pseudotopology”

Context: Sec2.3 in https://projecteuclid.org/download/pdf_1/euclid.rae/1403894903

lmao I actually wrote that down and found it again

See https://projecteuclid.org/download/pdf_1/euclid.rae/1403894903

As is tradition with nlab, the referenced page „relational β-modules” sounds way too scary for the problem at hand

Ah thank god, the relevant stuff appears to live in one section: https://ncatlab.org/nlab/show/relational+beta-module#bridge_to_a_concrete_description

Oh, the key difference from a pseudotop seems to be that we have a relation between ultrafilters

[verbal wraith]

but why?

[plain raven]

In 1970, Michael Barr gave an abstract definition of topological space based on a notion of convergence of ultrafilters (building on work by Ernest Manes on compact Hausdorff spaces). Succinctly, Barr defined topological spaces as ‘relational β\beta-modules’. It was subsequently realized that this was a special case of the notion of generalized multicategory. Here we unpack this definition and examine its properties.

michael barr is my god

but i don't know about this stuff, I mostly understand the relationship in the case of compact hausdorff spaces, it seems like barr is interested in trying to generalize this

For a CH space, every ultrafilter on the powerset converges to a unique point. conversely, you can characterize the CH spaces in terms of ultrafilter convergence as the category of CH spaces and continuous maps can be characterized as the category of algebras of the ultrafilter monad on Sets.

If i remember correctly, a space needs to be Hausdorff in order to ensure that the point of convergence of a filter is unique

and compactness guarantees that if the filter is maximal, there is at least one point of convergence

[flint cove]

Because people introduce test functions and say „define sequences of test functions to be convergent if uniform convergence in every derivative and support is well-behaved“ and they never tell you why you are allowed to use topological concepts or e.g. why something other than the indiscrete topology should satisfy these conditions

That has bugged me for five years

or I'm just dumb and there's actually a way to characterize test functions in a polynormed way

[flint cove]

(samplesize N=2)

[verbal wraith]

Filters in general are convoluted ways to express simple things but I'm not sure that pseudo-topologies also express simple things. Maybe they're just convoluted?

[flint cove]

Nets are not better

and once you leave the realm of first countability you have to choose one of these generalizations

[verbal wraith]

I prefer semi-lattices to filters to nets

[flint cove]

You mean the Kuratowski axioms?

I'm not quite sure how one can describe a topology with abstract semilattices

Also, today I learned that locally convex TVSes are always induced by seminorms

And that finally explains why we care about local convexity, and now I'm kinda happy.

[verbal wraith]

A topology is a classification where the semi-lattice includes into P(X).

[flint cove]

I'm confusde, don't use the word semilattice in that paragraph
Furthermore, if we have a topology on X, what would Y be? just {0,1}?

Or would we have a classification in X×𝒯?

[verbal wraith]

The semi-lattice is the Y

it's just a poset with an intersection operation

the closed sets are the semi-lattice

[flint cove]

That wouldn't give you semilattices in general, but more specificallly meet-compllete semilattices

[verbal wraith]

I mean, I don't use that terminology.

[flint cove]

My inner Birkhoff is not strong enough for being really comfortable with this stuff tho

To be fair, you're the one using the word semilattices frequently just for bragging rights
You should be the one comfortable with the terminology lmao

[verbal wraith]

Well the problem is that its unwieldy.

[flint cove]

I'm gonna head off tho, it's >4am over here. gn

[empty grove]

Start of something big?

[verbal wraith]

I'm sort of getting distracted by foundations as I'm writing this because membership in a collection looks like a classification.

[woven juniper]

is it appropriate to ask about accessing papers in certain journals I dont have access to here?

it is in an alg/geo top journal

[gritty widget]

have you checked sci-h*b

[nimble jolt]

https://tenor.com/view/pirate-cat-cute-kitten-gif-15214073

[honest terrace]

took me some time to realize the cat isn't actually walking on two feets

[hollow harbor]

Finally, let me also say categorically that I myself never actually use SciHub, because it is illegal. I also specifically encourage my graduate students not to use it. In fact, just in case they might accidentally use it, I make a point to explain to them in detail all the steps that would be required in order for them to access articles via SciHub. I explain that it would be legally wrong for them ever to use that method to gain access to published articles for free, even though the authors of the articles (who are the true moral owners of the works) would want them to do so. I believe it is incumbent upon us all to instruct the future generation of researchers correctly in this matter.

[plain raven]

Finally, let me also say categorically that I myself never actually use SciHub, because it is illegal. I also specifically encourage my graduate students not to use it. In fact, just in case they might accidentally use it, I make a point to explain to them in detail all the steps that would be required in order for them to access articles via SciHub. I explain that it would be legally wrong for them ever to use that method to gain access to published articles for free, even though the authors of the articles (who are the true moral owners of the works) would want them to do so. I believe it is incumbent upon us all to instruct the future generation of researchers correctly in this matter.

[trail tiger]

Math narcs

[sweet wing]

Finally, let me also say categorically that I myself never actually use SciHub, because it is illegal. I also specifically encourage my graduate students not to use it. In fact, just in case they might accidentally use it, I make a point to explain to them in detail all the steps that would be required in order for them to access articles via SciHub. I explain that it would be legally wrong for them ever to use that method to gain access to published articles for free, even though the authors of the articles (who are the true moral owners of the works) would want them to do so. I believe it is incumbent upon us all to instruct the future generation of researchers correctly in this matter.

[honest terrace]

Finally, let me also say categorically that I myself never actually use SciHub, because it is illegal. I also specifically encourage my graduate students not to use it. In fact, just in case they might accidentally use it, I make a point to explain to them in detail all the steps that would be required in order for them to access articles via SciHub. I explain that it would be legally wrong for them ever to use that method to gain access to published articles for free, even though the authors of the articles (who are the true moral owners of the works) would want them to do so. I believe it is incumbent upon us all to instruct the future generation of researchers correctly in this matter.

[reef shore]

copylogy and pastametry 😋

[sleek thicket]

Finally, let me also say categorically that I myself never actually use SciHub, because it is illegal. I also specifically encourage my graduate students not to use it. In fact, just in case they might accidentally use it, I make a point to explain to them in detail all the steps that would be required in order for them to access articles via SciHub. I explain that it would be legally wrong for them ever to use that method to gain access to published articles for free, even though the authors of the articles (who are the true moral owners of the works) would want them to do so. I believe it is incumbent upon us all to instruct the future generation of researchers correctly in this matter.

[hallow swan]

ryc literally has all the good copypastas

[plain raven]

this one is originally due to set theorist Joel David Hamkins, I think, I saw it on twitter earlier

[hallow swan]

a comment from https://dailynous.com/2021/07/07/sci-hub-philosophy-grad-student-pirate-queen/

[gritty widget]

[uncut surge]

absolute king

[gritty widget]

Hi

I don't even know anything about topology but this isn't for me, it's for a friend

Do you know any topology books that are good for studying topology for the first time?

I'd really appreciate if you knew any book that is good for an introduction to topology

[reef shore]

Topology by Munkres

[gritty widget]

munkres good

Okay

there's also lee's introduction to topological manifolds, which (as the title suggests) focuses on a particularly nice class of topological spaces, but still serves as an introduction to the subject

munkres is the standard though

I'll send them both

Thanks @empty grove and @gritty widget

[reef shore]

[gritty widget]

[hazy nexus]

Finally, let me also say categorically that I myself never actually use SciHub, because it is illegal. I also specifically encourage my graduate students not to use it. In fact, just in case they might accidentally use it, I make a point to explain to them in detail all the steps that would be required in order for them to access articles via SciHub. I explain that it would be legally wrong for them ever to use that method to gain access to published articles for free, even though the authors of the articles (who are the true moral owners of the works) would want them to do so. I believe it is incumbent upon us all to instruct the future generation of researchers correctly in this matter.

[dim meadow]

I confess that I might be inclined to make fisticuffs with any individual person who claimed to my face that it was they rather than me who owned the articles that I have slaved over so carefully, polishing and editing and reediting to ensure that every word and symbol expressed exactly my intended meaning. It is entirely mine. How could someone else own this, just because of some established procedural publishing bureacracy? But alas, the fight will never come, because it is not a person but a corporate entity that is claiming to own to the work. I expect that many of us feel this way.

[bronze lake]

my summer research that's on knot theory somehow got to talking about the mayer-vietoris sequence

send help

[plain raven]

huh. that doesn't sound so surprising to me actually, I would have expected knot theory to use lots of algebraic invariants from algebraic topology

[hollow harbor]

Lol we were just talking about this in the context of derived functors the other day

The answer I got to "how do I understand mayer vietoris on any level beyond just knowing how to compute it" was "just know how to compute it"

[hollow harbor]

Thank you, that's nice

Yes, the diagram chase was only mildly useful to me when I learned it (and I completely forget how that goes)

[flint cove]

Is there a similar exposition that talks about manifolds or cw cohomology instead? That sounds pretty interesting, but I don't have the homotopy background to understand the paper yet.

[bronze lake]

Loring Tu's "An Introduction to Manifolds" has a section on De Rham Cohomology and introduces the mayer-vietoris sequence in that context

[hollow harbor]

Thats essentially the context were I saw it

And I guess it didn't give me the right intuition lol

From Lee, not Tu

[flint cove]

Hm, skimming over it, it just says „By the zig-zag lemma, […] gives rise to a long exact sequencein cohomology“, so it leaves out precisely the interesting part for me
or am I overlooking something

Oh, the discussion of how to obtain the connecting homomorphism goes into more detail directly afterwards

thanks

I wouldn't say I have lots of discomfort with meyer-vietoris, but I still find it a bit mystic how cohomology of the intersection should somehow affect cohomology in higher degree

At the moment I'm just thinking about it in terms of chain complexes and reasonably happy with that, but still.

[woeful oasis]

Think of M-V as "How to put together two spaces and obtain the data of all of the corresponding dimensions of substructures". Connecting one dimension to the other is where you start looking at chains.

[marsh forge]

i think the split torus example is a nice one because it shows how the data of how two cynders intersect “creates” a generator in H2 @flint cove

basically the intersections are kinda like gluing instructions

and thats why they can affect how the higher dim stuff works

[flint cove]

Oh, that makes sense. Should even apply to the gluing of two open intervals to create S1, where a 1-cell appears

thinkking out loud And here the fact that stuff is glued together is witnessed by the nontriviality of H⁰(U cap V), i.e. the fact that there's actually stuff there in the intersection, and in fact enough so to cause higher-dim nontrivialities

I guess a good slogan that I will take away from this is that „gluing n-dim things things at a n-dim intersection together can introduce n+1-dim nontrivialities“

[marsh forge]

i would go further to say that it isn't just the nontriviality that matters but also the data given by the (co)homology maps induced by the inclusions

Like how it includes is clearly important

which is why those maps are involved as well

[flint cove]

Yeah, that makes sense

[plain raven]

i think the best way to understand mayer vietoris is in terms of mentally playing with the short exact sequence of complexes of differential forms. Take a space you understand like R^2 minus the origin, or S^2, give a cover of it by two opens, pick differential forms in an explicit coordinate system and just "play" - read the proof that the short exact sequence of complexes gives rise to the long exact sequence and spend some time thinking about the geometric content of each step (first few pages of Bott-Tu)

take differential forms and just move them left to right and up and down in the short exact sequence of complexes. I don't know if this makes sense lol

in the singular homology case I imagine a sphere S^2 covered by two open hemispheres U and V, and I have two simplices in the shape of closed hemispheres whose boundary is the equator in U\cap V that cover the sphere, the sum of the two simplices forming a generator for H_2(S^2).

think about how such a chain fits into H_2(U)\oplus H_2(V) -> H_2(U \cup V) -> H_1(U\cap V)

another way of thinking about this is that like, the most geometric understanding I have of singular homology is as being applied to integration problems for differential forms. and one of the cool things about simplices is you can subdivide them repeatedly into smaller and smaller triangles, so that given any chain in a manifold you can subdivide it until all your simplices live in a coordinate chart, and this makes analysis easier. you need some theorems that describe the relationship between a restricted homology that only deals with these small simplices subordinate to the open cover {U_i} by coordinate charts, and the full homology theory that deals with arbitrary simplices. what kind of information do you lose?

the answer is, nothing (up to homology) which is a beautiful result

the Mayer Vietoris is an immediate consequence of this when we take the case of an open cover with two elements

compare this to homotopy theory, where there's no such notion of subdivision of a map S^n -> X into a bunch of tiny little maps subordinate to an open cover

One last thing - the Mayer-Vietoris sequence for singular homology is equivalent to the excision property, which might be, if not easier to understand, at least give a different perspective. If you have nested subspaces A \subset U \subset X, and you're interested in the relative homology of X except that you mod out by all chains in U, then you're essentially trying to think about the homology of X where everything in U is just forgotten and quotiented out. It makes sense intuitively that if you just delete a subset A of U, then the relative homology of (X\A) modded out by chains in (U\A) should be exactly the same as the relative homology of X mod U, because the deletion happened in the area you were trying to suppress/forget. this result is true whenever A is closed and U is open

hmmm ok. neat. I have not heard of this technique

[bright acorn]

https://youtu.be/LelwRFs1hx0 Could someone tell me further upon what he is referring to at 59:30 ?

It seems interesting

Be he doesn't talk much about it besides some intuition

Thanks

I will give it a look

Yeah

Prolly a lot of category theory stuff that I don't know about lmao

It's pretty new stuff too

Seems to have been introduced around the early 2000's

Are there books on the subject or is it still just confined within articles? Maybe a book that covers more on homotopy theory, spectral sequences and such touches a little bit upon the subject.

Prolly with a little bit more of background on homotopy theory and spectral sequences I could give it a look.

But I am still reading on Hatcher's first book on AT

So I think it will take a while lmao

Anyways, I have asked this here before, but what books do you recommend on homotopy theory and spectral sequences?

[plain raven]

Spectral sequences are so great. This isn't a homotopy theory book but the book on algebraic topology and sheaf theory by godement covers sheaf cohomology and uses spectral sequences to prove lots of interesting results.

I don't know if there's an english translation 😦

[bright acorn]

What's the original language it's written on?

Ooh, too many people recommended me May's book :)

[plain raven]

It also is a good introduction to simplicial methods in homological algebra, although it doesn't introduce terminology of homotopy theory such as Kan complexes or homotopy groups
It's in french, Topologie Algebrique et Theorie des Faisceaux by roger godement. it's just mathematical french tho so it's not as bad, the grammar is very simple

[bright acorn]

So it must be really good.

[bright acorn]

I am currently trying to learn French lmao

So maybe this will be a way to get a little bit more used to "mathematical french".

I see

Hope the exercises are interesting then, if there are any.

[plain raven]

Yeah you can probably hack your way through it. I honestly love this book. I love the fact that it starts out with a focus on simplicial methods, I think it's a great introduction to spectral sequences in minimal levels of generality. But he doesn't include any exercises and it's not really that example driven, at least in the sense of explicit computations for a concrete choice of space

[bright acorn]

I still try to google it.

Prolly won't be able to find a physical point sadly.

That's really good

Thanks for the recommendations

[supple locust]

I have tried to write down detailed solution for this problem. Can someone take a look and tell me if it looks okay?

$X-V $ is closed so compact. For x, each $y\in X-V$, $U_y$ as in the problem cover $X-V$. So $U_{y_1},\dots, U_{y_n}$ cover $X-V$. Corresponding we have $U_{x_1},\dots ,U_{x_n}$. Each $U_{x_i}$ is open and closed so $\cap i U{x_i}$ is open and closed. As $U_{x_i}\cap U_{y_i}=\emptyset$, $\cap i U{x_i} \cap \cup i U{y_i}= \emptyset $ thus $x\in\cap i U{x_i}\subset V$.

[gentle osprey]

bert

[sleek thicket]

makes sense to me

[unborn lotus]

i don't understand why in (b) he says every neighborhood

so if i have a set E

and i choose say like a really big radius r

what if i find a q that's not in E?

i must be misunderstanding something

[frosty sundial]

you need to find one point in E

[unborn lotus]

oh One

a point

[frosty sundial]

who cares if there are points not in E

(not me)

[unborn lotus]

okay i see

yes ty for explaining

[frosty sundial]

also in this definition

[unborn lotus]

yeah i should have just reread

[frosty sundial]

you want to think of this as being really small neighborhoods

[unborn lotus]

so usually we're choosing r to be small enough

is what you're saying

[frosty sundial]

it's more like

no matter how small we choose r

we can always find some point

which is both in E and within r of p

[unborn lotus]

oh

right i can see this for R^n

but i don't know much about how R is constructed that gives it this ability other than stuff like [0,1] or the real number line has uncountably infinite elements

idk what got it there

[reef shore]

What ability are you talking about? This is just a definition

[frosty sundial]

what do you mean?

[unborn lotus]

aghhh

this is horribly confusing for me to phrase

but

you're saying this is a definition?

hmm

[frosty sundial]

yeah

[reef shore]

It seems to be phrased as a definition in the image you sent

[frosty sundial]

this is defining "limit point"

after it defines "neighborhood"

also btw thsi isn't a standard definition of neighborhood but it will be equivalent for all the things you care about

(most people would just call it an "open ball")

[unborn lotus]

right

i guess im not totally convinced of being able to take r as small as we like

[frosty sundial]

what do you mean?

this is the definition of a limit point

the point 0 is a limit point of the interval (0,1)

because for any r > 0, the open ball around 0 with radius r

[reef shore]

Are you thinking about how this lines up with your intuitive idea of a limit point?

[frosty sundial]

contains a point of (0,1)

in this example p = 0, E = (0,1)

[unborn lotus]

hmm i suppose i can say every time if i wanted to find a point inside that open ball

just pick a point say

half of the radius away

or a third

[frosty sundial]

the point has to be in E as well

p = -1 is not a limit point of E = (0,1)

because if you take r = 1/2

[unborn lotus]

right that too

[frosty sundial]

the open ball around -1 of radius 1/2

contains no points of E

[unborn lotus]

when i saw the definition first i thought it was just referring to any point on the interior of a set

[frosty sundial]

you should think of p as not being a point of E

[unborn lotus]

so i can choose any p in the space that i want

but

[frosty sundial]

yes

[unborn lotus]

only some of those p have the property i want

[frosty sundial]

that's right

if your space is R and E = (0,1)

then p = 0 is a limit point of E

but p = -1 is not

[unborn lotus]

right

[frosty sundial]

you should think of limit point as like

"infinitely close to E"

[unborn lotus]

and this is where "every" comes in

[reef shore]

Another phrasing of the definition in English: A point is a limit point of E iff you can find points of E arbitrarily close to that point

[unborn lotus]

every open ball around p has to have at least one point that is in E for p to be a limit point

ok

and usually we fail if we pick points "outside"

since we can take a small enough radius to find that all of the points in this small ball are not in E

[frosty sundial]

that's right

example: E = Q = rational numbers

what are all the limit points of E?

(in the space R)

[unborn lotus]

hmm

so i don't want to think of p being points in E

[frosty sundial]

(yeah, we'll return to that in a second)

[unborn lotus]

maybe it's easier if i find points which fail

but

hmm

there are ... a lot of rational numbers

uhh

[frosty sundial]

that's right

[unborn lotus]

how am i supposed to find a ball which doesnt have one

[hallow swan]

between every two rationals there is an irrational

[frosty sundial]

that's a good question

[frosty sundial]

that's not really relevant here... also please let metal think this through.

[hallow swan]

yea i’m trying to solve this too

will solve privately

[frosty sundial]

no problem, as long as you're not spoiling it haha

[unborn lotus]

when i pick p as a rational number it seems fine, since if i pick a radius r i can just floor the radius and halve it

or

[frosty sundial]

floor?

[unborn lotus]

maybe im too attached to decimals when thinking of reals

[frosty sundial]

decimals aren't a bad way to go

[unborn lotus]

like im thinking of what happens in various cases

how to pick points inside the ball which are rational

when i pick p to be irrational i have to come up with a different solution

[frosty sundial]

let's try some specific cases

[hallow swan]

don’t look at this melia but would ||there is a rational between {rational, irrational} x {rational, irrational}|| be sufficient

[frosty sundial]

p = pi = 3.14159...

and r = 1/100

can you find a rational number within 1/100 of pi?

[frosty sundial]

yeah, though maybe it would be nice to justify that fact

[unborn lotus]

so my bounds are going to be around (3.131... , 3.151...)

or just a ball around in general ig

not just in the interval

[frosty sundial]

open balls in this case are intervals centered at pi

so now you have an interval. can you find a rational number in it?

(maybe you can write it down as a decimal)

[unborn lotus]

sure i can probably truncate something i wrote above or add half of a number and truncate

lets see 3.14 or similar

lots of answers

[frosty sundial]

great

3.14 is within 0.01 of 3.14159...

becuase their difference 0.00159... is less than 0.01

[unborn lotus]

right

[frosty sundial]

great, now let's try a smaller r

r = 1/1000

[unborn lotus]

3.141

[frosty sundial]

great

[unborn lotus]

but these seem convenient

you gave me r as

[frosty sundial]

let's try to do infinitely many r at once

[unborn lotus]

1/10^n

[frosty sundial]

r = 1/10^n

okay yes :P

[unborn lotus]

:3

[frosty sundial]

so we can do it for 1/10^n for any n

now what about any r?

for example

r = 1/20

[unborn lotus]

yeah that seems harder, cus to explicitly find one by hand my strategy is to express that radius as a decimal

and play the same game as above

[frosty sundial]

I claim that we have already done all the work we need to

[unborn lotus]

o

[frosty sundial]

I claim that we have already found a q which works

[unborn lotus]

oh

in expressing the radius as a decimal

we can use the previous result

i think

or at least in part

wait already...

🤔

lets see

and you mean already meaning like

we just

its like minimal effort

no fancy decimals stuff?

[frosty sundial]

that's right

[unborn lotus]

o

[frosty sundial]

you already figured out how to "win" if I pick r = 1/10^n

for example if I gave you r = 1/10000

and my claim now is that you dont have to do any more work

[unborn lotus]

i'd just truncate the expansion at the 5th or so decimal place

but

how do i know where to pick the place to truncate given r

is what i guess i need to find out

[frosty sundial]

well, if r = 1/20

where would you truncate

[unborn lotus]

well 1/10 is

wait

no 1/100

smaller

[frosty sundial]

great -- can you write that out as a complete sentence/thought?

[unborn lotus]

if you give me an r, i just need to work with a smaller r which may be more convenient to me

[frosty sundial]

yes!

[unborn lotus]

so in this sense no matter what r you give me

for any point

i can win

i can choose n large enough in 1/10^n

[hallow swan]

ok so ||i just convinced myself this was true without making a proper construction. i just figured you could perturb by the appropriate power of 10 in any case||

not the most precise

[unborn lotus]

even if r is irrational since i can just truncate r in any way which makes it smaller

[frosty sundial]

that's right

(also yes 123four that is fine. in some sense metal is making that more precise right now)

the way I might phrase it is "for every r > 0 there is some n such that 1/10^n < r"

[unborn lotus]

so all points in R are limit points

[frosty sundial]

that's right

[unborn lotus]

wow

that is an interesting result

[frosty sundial]

so the set of limit points of Q is all of R

[unborn lotus]

is this still true if you said irrationals instead of rationals for E

[frosty sundial]

yes, it's still true

[unborn lotus]

i feel like it is but it seems a little weirder

since the decimals thing is not the way i think about irrational numbers

[frosty sundial]

for any rational number p, and any positive real number r, there is some irrational number in (p-r, p+r)

[unborn lotus]

oh

[frosty sundial]

which is the open ball centered at p of radius r

[unborn lotus]

er how do we know this

[frosty sundial]

good question

[unborn lotus]

might be a weird question

[frosty sundial]

I claim that you can construct one directly

as a function of p and r

[unborn lotus]

alright lets exclude the case when p is chosen to be irrational

that case is easy

[frosty sundial]

(is it?)

can you write it out for me?

or at least give a sketch

[unborn lotus]

okay i suppose firstly

i need to know how to tell if numbers are irrational or not

what i was thinking initially for that case was to just halve r and add it to p

but

what

if

r is 2pi

p is -pi

that wont work

for halving

[frosty sundial]

what condition on r would make it so that you definitely still get an irrational?

so, p = some irrational

when can you guarantee that p + r/2 is still irrational?

[unborn lotus]

when r is rational i feel very certain

its just weird if i add two irrational numbers together

ig

[frosty sundial]

great, I agree with both of those statements

so, if r is rational, we are set

[unborn lotus]

right

[frosty sundial]

for example, if r = 1/10^n we would be set

but what if r isn't? what would we say in that case?

[unborn lotus]

how do i know if when i add two irrationals together the sum is not rational

that seems like a difficult problem

[frosty sundial]

I agree

which is why

we are not going to answer it

[unborn lotus]

but i guess maybe we can circumvent it

try a different construction

involving both numbers

[frosty sundial]

so if r = 1/10^n

we are happy

[unborn lotus]

yes

[frosty sundial]

I claim that we are done

we dont have any more work to do

why is that?

[unborn lotus]

is this like before where i can just truncate the irrational radius into a rational radius which is smaller

and then i am done

[frosty sundial]

you tell me

[unborn lotus]

well.. it is!

i can find some way to express said irrational to a suitable number of decimal places and i can stop at any point

if i stop anywhere in the expansion i will have a rational number less than the irrational number

and then choose n large enough for r = 1/10^n to make this r smaller than the truncation i formed

[frosty sundial]

great

[sleek thicket]

pog!

[frosty sundial]

let's summarize this as "for any irrational number r, there is some n for which 1/10^n is smaller than r"

how does this compare to our last argument?

(when we were looking at limit poitns of Q)

[unborn lotus]

it is very similar

the idea seems to be the same except in dealing with r irrational in this case since addition of two irrational numbers is wonky

[frosty sundial]

can you maybe try to abstract a little bit and phrase it in a way which encompasses both this argument and the last one?

[unborn lotus]

okay

[frosty sundial]

(yes I know we havent yet dealt with the case where E = irrationals and p = rational, we'll get to that)

[unborn lotus]

ok wait

so in the previous argument for limit points p of Q in R we were saying that for any r given to me i can: find a smaller r in the form 1/10^n and then take the point p and truncate it at the n-th place when p is irrational or add on this smaller radius to p when p is rational to find a rational number in the ball

then

for this one to find an irrational number in (p-r,p+r) i was considering cases

and this case we just did is when r is irrational and p is irrational

since if p irrational, r rational we are done

[frosty sundial]

yeah

so more what i'm interested in is our strategy for handling both cases

in each argument we had a situation like "for some values of r we are happy... how do we deal with other values of r?"

[unborn lotus]

when we want rational numbers we tried to make rational + rational

when we want irrational numbers we tried to make irrational + rational

in some way or another

in this most recent one we find it easier to mess with r being irrational since it's easier to make a smaller r which is rational

and so when p is irrational we can then go into the irrational+ rational case by finding the smaller r via our r = 1/10^n, suitable n strategy

usually seem to be happy with rational r

but then what happens when we want to find an irrational in (p-r,p+r) and p is rational and r is rational

now i am not happy

agh

it's very easy to turn irrationals into rationals

er

not turn but

to find smaller numbers

[frosty sundial]

so everything you're saying is right

[unborn lotus]

which are rational

[frosty sundial]

yes!!

that last thing is what I wanted you to get to

[unborn lotus]

oh

i see haha

[frosty sundial]

it's not about rational or irrational

it's that you can always make r smaller if you want to

or the converse of that is

if some particular value of r works

then all larger values of r will automatically work

so as long as you can find a q for sufficiently small r, you're golden

[unborn lotus]

ah i see

[frosty sundial]

you don't have to do it for every r, as long as you can do it for small enough r

in some cases, it's convenient to do r = 1/10^n

[unborn lotus]

yes this q would work for all r greater than the one we computed with our game

[frosty sundial]

if you want to do some decimal stuff

that's right!

that is the connection between the two cases we've done so far

[unborn lotus]

ah okay

[frosty sundial]

that idea

[hallow swan]

fuck now I want to read this convo

[unborn lotus]

this is me getting my hands dirty

this experience is very good i think

[frosty sundial]

dirty hands

[hallow swan]

filthy hands

[unborn lotus]

yes

but now im trying to figure out how to find an irrational between two rationals

cus before it was easier to deal with two irrationals and to make one smaller AND rational

but uh

[frosty sundial]

yeah so this part is kind of a different argument

[unborn lotus]

how do i make a rational number smaller and irrational

yeah i assume this is some weird construction

er

if there is a construction

at all

[frosty sundial]

just so you can move on i'll kind of give you the idea, also like, this idea isn't really useful later

[unborn lotus]

doing this has made me realize how little i know about irrational numbers

[frosty sundial]

as opposed to the "you can always make r smaller at no cost to you"

which is an important idea

[unborn lotus]

right

[frosty sundial]

basically you can do something like

[hallow swan]

add k*pi to the smaller one?

[frosty sundial]

q = p + pi/10^n

for large enough n

just make n gigantic and you're okay

[unborn lotus]

ah okay

yeah using a known irrational is clever

cool cool

[hallow swan]

the /10^n is just for intuition right ?

[frosty sundial]

another thing you can do is something like, consider sqrt(2) in the interval (0,2)

then scale by r

(assuming r is rational here)

then add p

[unborn lotus]

right

[hallow swan]

like we can just say there exists s.t. p + k*pi

[frosty sundial]

so like, p + r*sqrt(2)

or maybe divide by 2 or sometyhing

[unborn lotus]

right the 10 is not too important

just

whatever makes it smaller and we're not dividing by an irrational

[frosty sundial]

yeah, it's just a convenient way to get a really small irratioal number

[unborn lotus]

alright

well this was very good

er

we worked out the example u gave me

[frosty sundial]

next example

[unborn lotus]

ok

[frosty sundial]

what are the limit points of Z

the set of integers

(and: why is this question maybe a little subtle)

[unborn lotus]

well if i pick p to be a non integer, what if my radius is pretty small and i don't have any integers enclosed

then that won't work

[frosty sundial]

that's right

[hallow swan]

Z isn’t dense is the main problem?

[unborn lotus]

but then if i choose an integer.. same issue right? i make my radius smaller than 1/2 and i get problems

[frosty sundial]

density isn't relevant here

[hallow swan]

oh

[frosty sundial]

saying "E is dense" just menas that the set of limit points of E is R

(the term you want to refer to Z is that it is nowhere dense)

[hallow swan]

ok I thought it meant something else

[unborn lotus]

if my radius is smaller than 1/2 when i choose p to be integer, there are no integers in the ball

so that's not good either

[frosty sundial]

really? i can think of an integer in the ball (1/2, 3/2)

(p = 1, r = 1/2)

[unborn lotus]

wh

hmm

[hallow swan]

i can’t ?

don’t you need another integer besides p

[unborn lotus]

no integers other than 1?

[frosty sundial]

yes both of you are right

[unborn lotus]

i thought u can't pick p as q

[frosty sundial]

your'e right, you can't

that's what's slightly subtle -- it seems like a really random part of the definition

but this is exactly why

in order to be a limit point, you need other points in E to be close to p

[hallow swan]

haha

trying to trick us

[unborn lotus]

hmm yeah it is kind of weird that we have to choose other points

[frosty sundial]

so I will say that this is a matter of convention

some authors allow any point of E to be a limit point

but not adding the q \neq p condition

the one you see here is the more common one though

[unborn lotus]

ah

[hallow swan]

i feel like we did more nt than analysis

let alone topology

[frosty sundial]

i mean, we did futz around with decimal expansions a little bit

but i wouldn't really call that NT haha

[unborn lotus]

i thought that was just part of the shrinking r argument

doesn't really have to be 10

[frosty sundial]

yeah

[unborn lotus]

10 is nice cus we work in decimals and im used to truncating decimal expansions

[frosty sundial]

I will say that the shrinking r argument is actually pretty useful

for example, there are uncountably many r, but this shows that you can replace that with just a countable sequence of r's

like 1/10^n

and that can be useful in some cases

[unborn lotus]

ah

[frosty sundial]

sometimes you might use 1/2^n instead

[unborn lotus]

right

[frosty sundial]

or other things

[unborn lotus]

cool cool

[hallow swan]

buncho goat teacher

literally gives 1 hour lecture for free

[unborn lotus]

yeah this was really good

[sleek thicket]

🐐

[unborn lotus]

🐐

[sleek thicket]

buncho is an animal

(pejorative)

[unborn lotus]

ty very much buncho

i learned a lot i think

[frosty sundial]

i- uhhh

well

i also think you learned a lot

:)

[unborn lotus]

:3

i should ask more questions in here

i remember in spring when i was learning beginning algebra shamrock helped me a lot

[frosty sundial]

i feel like i'm more likely to help people who i already know

who i think will actually appreciate it and take it seriously

[hallow swan]

yea so

you were helping melia

and then i sort of joined in unannounced

👍

[frosty sundial]

lucky you I suppose ;P

thanks for not being an asshole about it

some people take that as an opportunity to like

show off how much they know

[unborn lotus]

yeah a lot of people like to snipe the problem

[frosty sundial]

and it's like... this convo isn't about you... please leave

[unborn lotus]

which is annoying indeed

[frosty sundial]

but you were nice :)

[unborn lotus]

yes

[hallow swan]

cool 😎

[unborn lotus]

is there a reason they're called "limit points"

part of me feels like it has to do with being able to get arbitrarily close to said points

like limits

[frosty sundial]

that's right

have you defined limits in general yet

[unborn lotus]

i don't think so

[hallow swan]

epsilon delta stuff

[unborn lotus]

i've seen the epsilon delta definition before but i think we never got into it too deeply

[frosty sundial]

you can phrase it in terms of open neighborhoods as well

in any case, I'll tell you the lemma without telling you the definition

and let you try to fill it in if you'd like

a point p is a limit point of E if and only if there is a sequence of points in E whose limit is p

[unborn lotus]

okay

ill keep this in mind when i get to limits

[bleak helm]

Note the sequence of points should consist of points other than p if you use the definition you guys seemed to be using above

[frosty sundial]

good point

I forgot thatd etail

[empty grove]

That'd etale

[fathom cave]

🐐

[golden gust]

by the way, if you drop this condition then you have the definition of an adherent point

so the set of limit points of Z is empty, but the set of adherent points of Z is Z

[forest rover]

@marsh forge do you have notes from your talk?

[marsh forge]

i will upload it

i just am lazy

[unborn lotus]

got it, ty 😌

[gritty widget]

yw

[viral atlas]

Sideurk moment

[quartz loom]

I have a quick question, relatively non-formal, but I couldn't think of a better channel than this one to ask it in

[gritty widget]

[quartz loom]

This question comes from the first unit of exercises from M. A. Armstrong's "Basic Topology". It reads "Imagine all the spaces shown in Fig. 1.23 to be made of rubber. For each pair of Spaces X, Y, convince yourself that X can be continuously deformed into Y. There are three examples: 1.) X= cylinder with a puncture, and Y= Disc with two punctures, which makes sense. 2.) X= punctured torus and Y= Two cylinders glued together over a square patch, which took a bit but I see now, but the third I still struggle with

I found this

And I can somewhat convince myself that this is true, through a little push, but I have no idea how one would see this process, finding this process intuitively

[bright acorn]

Ok so, I have heard before that there were some similarities between Galois Theory and the theory of Covering Space and the Fundamental Group.

I mean, I can see why some results can be sort of seem as similar. For instance, the isomorphism between the group of deck transformations of the universal cover of a suitable topological space, or the fact that if you have a path connected topological space there's a correspondence between the subgroups of its fundamental group and its covering spaces (Which can be seem as a sort of analogous to the Galois Correspondence in some way)

Even though, these similarities are sort of just vague intuition, but how could you make such a thing more formal?

It seems a really interesting topic

[coral pivot]

oh yeah this is cool. I dont know the full formalization but yesterday we were doing the inverse galois problem for C(t) and it came up. it was p cool

basically if you have a covering map to riemann surface X, you can make this a holomorphic map and give a complex structure to the covering space, and similarly if you have a holo function from Y to X where Y is compact then you can remove a few points and make this a covering map

this map now induces a homomorphism from M(X) to M(Y) where Y is the covering space and M(X) is the field of meromorphic functions on X

so you can now think of M(Y) as a field extension of M(X)

[bright acorn]

Oh, I see it.

Too bad that I couldn't really find anything that goes on this topic a little bit further

But yeah, I intend to take a class on Riemann Surfaces next semester and I am doing a little bit of revision on my AT.

[coral pivot]

This turns out to be really nice, you can now say acontravariant functor from the compact riemann surfaces with maps into X to the field extensions of X is sending Y -> M(Y)

[bright acorn]

This seems to come up a lot

[coral pivot]

and turns out the categories are anti-equivalent under this

yeah riemann surfaces and AT are cool!

[bright acorn]

Seems to be a pretty tough class tho.

[coral pivot]

yeah I am pretty weak at this stuff myself, but honestly this type of relations show up in so many places that im motivated to learn riemann surfaces a lot

[bright acorn]

You do Number Theory iirc, right?

[coral pivot]

ye

[bright acorn]

It seems that the last grading on this class

Is going to be about

Giving a talk

On a certain subject of your choice

Are there some interesting algebraic number theory problems where Riemann Surfaces come up?

I was thinking that maybe giving a little introduction to the Weil Conjectures and the proof of some of them would be a nice talk.

But really hard to do

[coral pivot]

one of the easy ones is this

[bright acorn]

So I was thinking about something simpler

[coral pivot]

eliptical curves are riemann surfaces that are isomorphic to some complex structure on the torus

(solutions in C to elliptical curves that is)

and you also inherit the group structure from a group structure on the torus

[bright acorn]

Oh, I have heard of this classification theorem before. Just didn't go any further.

I think Arithmetic of Elliptic Curves goes over this.

Might be a good topic to talk about for sure

[coral pivot]

mhm, you can look into the wierstrass p function which gives you this isomorphism, the way you figure this out is very cool

[bright acorn]

Thanks for the suggestions :)

[gritty widget]

👀

[gritty widget]

this is so cool

[coral pivot]

yep

[gritty widget]

my complex analysis course built up to this

very pleasing

[coral pivot]

oh nice, i heard it mentioned in silvermans book and then researched this with a few ppl to figure it out

[bright acorn]

"Galois Theories"

That's intimidating

https://ncatlab.org/nlab/show/Galois+Theories

Found also this article that goes over a bit of the motivation of the book

It also mentions function spaces on Riemann Surfaces too

[plain raven]

That book also treats monadic descent, which is nice

i think i had a general question about Grothendieck fibrations at some point and someone pointed me to this book

and i was like, ooh very cool

I haven't read much of it because I don't know very much ordinary Galois theory

[unreal bay]

p adic numbers are homeomorphic to Cantor sets minus a point. But Cantor set is a compact hausdorff space. Then why is p adic numbers not a compact Hausdorff space?

i see it is because subspace of compact may not be compact?

right

we need closed

[marsh forge]

honestly

this still is kinda wack to me

i have a terrible mental image of the cantor set though lmao

[unreal bay]

subspace of compact is compact if and only if subspace is closed?

[tight agate]

the p-adics arent compact?

do you mean Z_p

?

or Q_p?

[unreal bay]

p adic integers are compact but not numbers

Q_p

[tight agate]

👍

[unreal bay]

But how we loose just a point from Cantor set when finding its homeomorphic

it is unclear

How Q_p homeomorphic to Cantor minus a point

I am studying this topic , thats why asked all this

[verbal wraith]

This seems unintuitive because you can also interpret these pictures as 2D linked shapes and those are not equivalent.

[bright acorn]

Cantor set is weird

One of the most weird striking facts for me is that any compact metric space is a continuous image of the Cantor Set.

[sweet wing]

oh wha

[gritty widget]

cursed

[bright acorn]

This seems weird because for me compact metric spaces are somehow really well behaved topological spaces, while the Cantor set is weird and strange.

But just shows how messed up continuous functions can be

[gritty widget]

the cantor set is the continuous image of the cantor set

[bright acorn]

[empty grove]

It makes sense if you think of it as Cantor set being the largest cardinality compact metric space (you can prove this separately) and from it being totally disconnected, meaning most functions from it will be continuous

"most" used very loosely

That thing that you can prove separately is a really nice exercise btw, there like 5 different proofs (one of them from ryc if I recall correctly )

[gritty widget]

oh yeah this one time some dude tried to describe the cantor set using category theory language in my analysis class and the prof told him to go on MSE

[bright acorn]

Lmao

[pearl holly]

I think that Munkres had an exercise about this but I skipped it lmao. Let me check

[bright acorn]

I mean, you could somehow thing of the Cantor set as being the initial object in the category of compact metric spaces because of this??? Like, very loosely because there's not necessarily only one continuous function from the Cantor set to a compact metric space such that this general compact metric space is going to be the continuous image of.

How tf did he try to describe the Cantor set using category theory tho?

[empty grove]

Initial also has a uniqueness requirement

[bright acorn]

Yup

Yeah

[gritty widget]

like that, mistersystem.

[bright acorn]

Yeah

I thought so

But it can be

[gritty widget]

he was just asking if his interpretation was correct

prof told him "this isn't the place" despite it being a reasonable question

funny, but also kind of annoying if you're the student

[empty grove]

UGCTs are so oppressed 😔

[bright acorn]

Analysis professors must hate category theory

Is that correct?

I mean

I would understand tho

Jk

[pearl holly]

yeah, the exercise is about proving that it is totally disconnected, compact blabla bla and then to conclude that it is uncountable

[gritty widget]

Do you like Algebraic Geometry?

Me too

Bruh

I’ll enter in IMPA’s website for take some pdf’s of geometry and abstract algebra

@gritty widget Would you like to receive some pdf’s about them?

Alright

[sleek thicket]

Cantor set...

[plain raven]

here's something fun

personally I usually think of the Cantor set by means of the theorem that every point in it can be expressed in a unique infinite binary expansion 0.0101011011000... , with this correspondence it's not hard to see that the Cantor set is homeomorphic to 2^\omega with the discrete topology on {0,1}

anyway these sets of the form 2^A are interesting because, if X is a T_0 space, and T is the set of open subsets of X, then X embeds into 2^|T| by the function which sends each x to the set { U | x\in U}

actually this inclusion has a continuous retraction as well

furthermore, sets of the form 2^A (in particular, the Cantor set) are injective objects in the category of T0 spaces

so this shows that every T0 space embeds into an injective space

[swift fjord]

I think using ternary expansion might be easier to see by virtue of the construction of the cantor set.

Easier to see the function

Axtually nvm

We did it like that in my class but we showed that its cardinality Is the same as all numbers without 1 in their ternary expansion

So it ends but being the same

Maybe a bit easier to understand that construction tho

[lean marten]

The correspondence between binary sequences and decimal expansions is given by a map which sends a binary sequence to the number with that ternary expansion but 2s instead of 1s

Because we want to remove all middle thirds which correspond to a 1 at some point in the ternary expansion

If ternary is the right word lol

[supple locust]

I wrote solution to this problem. Can someone take a look and tell me if its okay. Also ig there must be better way to do it.

(a) $f(x_1,x_2)=(\min (x_1,x_2),|x_2-x_1|,1-\max (x_1,x_2))$. So as min and max and abs value functions are continuous f is continuous. \
(b) Observe that $f(x,y)=f(y,x)$. Let $X={ (x,y) \in [0,1] \times [0,1] \ | \ y \leq x }$ and $Y= { (x,y) \in [0,1] \times [0,1] \ | \ x \leq y }$. Observe that $f$ is injective on $X$ and $X $ is compact. Thus as continuous function from compact space to hausdorff space is homeomorphism, $X$ is homeomorphic to its image under f. Define $C=A\times B \cap X \cup A\times B \cap {(y,x) \ | \ (x,y) \in A\times B \cap Y }$. C is closed. $C \subset X$. Then $f(A\times B)=f(C)$ is closed.

[gentle osprey]

bert

[swift fjord]

Generally if a space is not T1 then normal doesn't imply regular right? You need T1 to ensure singletons are closed

[supple locust]

here is another problem whose solution I have written.

(a) Suppose B (WLOG) is not connected. Let $A=U\cup V$ where $U$ and $V$ clopen in A. and $U\cap V=\emptyset$. As U and V are closed in closed set A they are closed in the topological space X. $A\cap B= (U\cap B) \cup (V\cap B)$. Note that $U\cap B$ and $V\cap B$ are closed ssubsets of $A\cap B$ so $A \cap B $ is not connected - a contradiction. \
(b) Consider $A=(-1,0)\cup (0,1/2)$ and $B=[0,1]$.

[gentle osprey]

bert

[supple locust]

I should add that if $V\cap B=\emptyset $ then $A\cup B$ is not connected.

[gentle osprey]

bert

[supple locust]

actually part b just follows by continuous image of compact is compact

[pearl holly]

Hmm I am kind of confused. I am re-reading Hatcher from the very beginning because I just read small portions of it before when I wanted to do exercises with veryhappyperson and now when I look at it in detail, I don't get this part: If a space X deformation retracts onto a subspace A via $f_t: X \rightarrow X$, then if $r:X \rightarrow A$ denotes the resulting retraction and $i: A \rightarrow X$ the inclusion, we have $ri = 1$ and $ir \cong 1$, the latter homotopy being given by $f_t$. I understand that $ri = 1$, but how is $ir \cong 1$? Isn't $r \cong 1$ by the homotopy given by $f_t$?

[gentle osprey]

Tokidoki ✓

[pearl holly]

Is this true: if a \cong b, then is fa \cong fb?

That is true, right?

You can just compose the family of functions that induce that homotopy to get fa \cong fb, right? Or am I missing something?

[true robin]

Well the homotopy f is actually from the identity to ir, reason being that at any time t f_t is a function from X to X. It is just that ir is “pretty much the same” as r

[pearl holly]

Hmm well this is what I am thinking now: the induced retraction is f_1 since f_1(X) = A and the restriction to A is the identity function. We also know that f_0 is the identity. So therefore, r and the identity are homotopic. Is this wrong?

[pearl holly]

and all these fancy facts that I am stating come from the definition of a deformation retraction just to be clear

[true robin]

The reason why r and Id can’t be homotopic are purely formal, they are maps between different spaces. It only makes sense to ask wether ir and Id are homotopic, and indeed they are, it is given by f

[pearl holly]

but r: X -> X and so does Id? I don't really understand

[true robin]

r:X->A here

[pearl holly]

oh yeah lmao sorry

[true robin]

Np

[pearl holly]

But how would you prove that ir and Id are homotopic? I don't see how f induces that homotopy

Because f_1 is not ir, right?

[true robin]

It is

So it is pretty standard to use r and ir interchangeably (for obvious reasons) so hatcher may have said f1=r but formally it is ir

[pearl holly]

But the induced retraction is literally f_1, isn't it?

[true robin]

Well r is not f1, r is f1 once you have “reduced the codomain” of f1 to be A. In other words f1:X->X, but r:X->A takes the same values as f1 on all inputs. so we see f1=ir

[pearl holly]

ooohhhh okay now I see lmao

Hatcher uses r: X -> X and r:X -> A at the same time lmao

Anyway, thank you so much! I was confused about this but now you cleared it all up!

[true robin]

Np

[gritty widget]

homo toe pee

[bronze lake]

I went to a seminar where they pronounced ha-MAH-tuh-pee

I hated it

they also said "beta" like "bee-tuh"

and "corollary" like co-RHO-luh-ry, second syllable like the greek letter

[hollow harbor]

i had a russian prof who used these pronunciations

[plain raven]

it's also common in Australia

[gritty widget]

#geometry-and-trigonometry

wrong channel

[pearl holly]

Imaging that being all of topology

[gritty widget]

i would hate topology more than i already do

[rancid umbra]

lol why do u hate topology

[gritty widget]

general loathe of mathematics

[rancid umbra]

bruh💀

[bronze lake]

I'm the same way, I hated geometry in elementary/middle school but here I am now

triangles can frick off

[bronze lake]

yeah but at least I can pretend that they're balls instead

[sweet wing]

simplices moment

[tidal cedar]

all is triangle

[sharp frost]

I might be having a brain fart but what do they mean by "unit map" in this definition?

[gritty widget]

ohmygod can't they just say bilinear map

[sharp frost]

lol

[frosty sundial]

the unit map is eta

they are defining "unit map"

[empty grove]

Those are just the names for those maps, when you are talking about monoids in a monoidal category. The intuition is that it seems to be acting like the identity wrt that multiplication

[sharp frost]

oh ok so I shouldn't necessarily recognise the unit map like I recognise mu (as multiplication of the algebra)

its a new thing

[frosty sundial]

mu is the multiplication map and eta is the unit map

[gritty widget]

"a k-algebra is precisely a monoid in the monoidal category (Vect_k, \otimes, k)"

agony

love it

[empty grove]

meme math

[plain raven]

A is a ring, and eta is an embedding of the field k into A such that the 1 of k coincides with the 1 of A. in this sense eta tells you what the unit of A is, as you can extract it as eta(1)

monoidal category theory is pretty interesting tbh. I put off learning it for a long time as it seemed dry and contentless but imo there are a lot of interesting results

[sharp frost]

Ohhh thank you

[hoary drift]

why set of points of continuity for am G-delta set?
I have the answer but can anyone provide me intuition please. Please tag me when responding

[pearl holly]

I’m reading the proof that the group operation in the fundamental group is associative and I’m confused. How do I know that the f(gh) is a reparamentrization of (fg)h? What does the graph on the right represent?

I guess that it changes the "amount of time" it takes to traverse along those paths or something?

[marsh forge]

the idea is that (fg)h does f at 4x speed and g at 4x speed and then h at 2x speed

and then f(gh) does f at 2x speed and g at 4x speed and h at 4x speed

and up to homotopy this is just a reparamerization

@pearl holly

[pearl holly]

yeah okay that makes sense I guess. But how do I know that it is a reparametrization? Why does composing with a map change the amount of "time" it takes to traverse along those paths?

[marsh forge]

thats the definition of path composition

you always reparameterize so everything is a function from [0,1] instead of like [0,2]

Every loop takes 1 second so doing two loops should take 2 seconds

You speed it up so it takes 1

[pearl holly]

Hatcher defines a reparametrization of a path to be a composition f \phi where \phi: I -> I is any continuous function such that \phi(0) = \phi(1) = 1. Here, it's normal composition and not path composition. I don't see how this changes the "time"

[marsh forge]

Do you understand what I’m saying?

Ignore hatcher for a second

[pearl holly]

okay let me re-read

I don't even think that I understand reparametrization lmao

What does it do to the path?

[marsh forge]

Think a function out of the interval as happening over time

Then a reparameterization is just the same function done faster / slower / whatever

[pearl holly]

yeah okay I get that, but how can I see that from the formal definition?

[marsh forge]

Ignore the formal defn lol

Or just think about it harder

One of the two

[pearl holly]

yeah okay lmao, I will think about it harder. Thank you so much!

[marsh forge]

Honestly like

I mean it when I say ignore the formal defn

If doesn’t matter if you understand what I said

And eventually you’ll look at this

And it will make complete sense

[pearl holly]

No okay lmao I get it now

I'm just tired

You're changing the time according to the graph

smh

[gritty widget]

[marsh forge]

Its worth thinking about associTivity in the version of pi1 I taught u

[pearl holly]

Oh yeah that works too! Thanks! (Sorry for not responding lmao, I took a nap)

[marsh forge]

Lmao dont sweat it

I have absolutely no idea what timezone you and very are in tho

[pearl holly]

We are both in Europe so it's 20:29 for us now

(It was a bad idea taking a nap now, I know)

[gritty widget]

how dare you nap

you must constantly be doing mathematics

[pearl holly]

This is what happens when you are doing math while being tired

"constantly be doing mathematics" for me is like "constantly being dumb" if I'm tired

[abstract pagoda]

Source of the proof is from Allen Hatcher Algebraic Topology
I understood most of the proof until the implication that

$$h(s+\frac{1}{2})=-h(s)\implies \tilde{h}(s+\frac{1}{2}) = \tilde{h}(s) + \frac{q}{2}$$

[gentle osprey]

celina baeza

[abstract pagoda]

Why is this true?

[abstract pagoda]

Having an exact definition of the lift would help me understand easier, but is it that
h-tilde(s)+q/2 gets mapped by p to p(h-tilde(s)+q/2)=exp(2ipih-tilde(s)+ipiq) and the ipiq part corresponds to -h(s)

Thank you very much

[abstract pagoda]

Oh ok I see exactly why the implication holds after doing scrap work. A work in progress for me is being able to know what “h represents q times a generator of pi1(S1)” means.
One thing I wish I knew more about are the purpose of lifts in these proofs. I am thinking that their main use is to tell us something about the homotopy class of whatever is being lifted.

[gritty widget]

that's a great book. you might wish to supplement it with e.g.
"advanced calculus" by folland (he does all of the topology spivak does but in a much more "modern' and friendly way)

"analysis on manifolds" by munkres (expands on a lot of things spivak does)

hubbard and hubbard's book (don't know much here)

spivak's approach to topology is a little strange and i think folland does a better job there

[gritty widget]

make sure to do all of the exercises, including the incorrect ones

yeah the book has a few typos

luckily they're all well documented

just the two you found

plus various MSE threads

there's no one list, sadly

i can tell you off the top of my head that the stuff that's not in the problems that's bugged is:
-statement, proof of the rank theorem in R^n (last theorem in differentiation chapter)
-proof of "integrable iff null set of continuities"
-statement of partition of unity theorem
-proof of sard's theorem

those are the big ones

oh also the proof at the start of chapter 5 of the equivalence of various definitions of manifolds

id also consider the topology section to be bugged just because he uses open rectangles instead of balls

this is a book i have read too much of

there's actually no loss of generality in doing this

any hausdorff, second-countable "abstract smooth manifold" can be embedded into euclidean space of some dimension by whitney's embedding theorem

no spivak is a good book

as long as you keep the errata open

that's a joke

it's a great book

many good problems

[gritty widget]

not a bad idea

but the book is already terse enough

exercise 2: rewrite the book with no errors

[empty grove]

Must have been a spelling error

[frigid river]

If I want to show that the fundamental group is a (covariant) functor (between Top_* and Grp) using the axioms in Riehl, how would I do that?

For any composable pair f,g in Top_, Fg \cdot Ff = F(g \cdot f).
I suppose f and g are morphisms, hence continuous functions in Top_. Then I'd have:
\pi_1(g) \cdot \pi_1(f) = \pi_1(g \cdot f)

What's that even supposed to mean and what's the fundamental group of a function? Also, what does it mean to \cdot 2 groups?

(I'm very sorry for the formatting)

[gentle osprey]

TTerra

[gritty widget]

that's what the "fundamental group of a function" is (this wording is misleading, since you're not assigning to each morphism a group)

remember that a functor has to take object to object and morphism to morphism

so there are two things going on here

one's the fundamental group you know and love (object to object) and one's the thing i just defined (morphism to morphism)

the \cdot here probably just means composition of functions

max is not online so i must make do

[frigid river]

thanks

[gritty widget]

[gritty widget]

what server is this emote from

personal

invite

no

personal

and im a person

therefore invite

no you're clearly a dog

person all

[frigid river]

Tinkiwinki is typing...