Mathematics - 2024-09-06 (page 1 of 2)

nickbros123

03:28

@BenSteffan I can attest to that, but whether or not that is a good way to go about it is a different question :) Its been ~35 days into my semester, my first abstract algebra course, and we are already past lagrange, isomorphism theorems, actions and moving onto cayley and then sylow :) I am not gaining anything from this course except the well known fact that I dont know anything

Jakobian

@nickbros123 why not gaining anything?

those theorems are important, for group theory at least, combinatorics, and when you want to compute something

nickbros123

@Jakobian the course seems too fast for me, I have a tough time understanding the intuition behind these things / constructions. It does not help that I have 4 other courses of similar weight class.

Jakobian

courses are like that, they are fast and you can't understand much intuitively

leslie townes

dunno where you are, but 4 other courses seems like a lot

a lot of american universities would have departmental advisors telling you not to do that

Jakobian

focus on understanding Sylow's theorems, in fact, revise them right now

nickbros123

03:40

I got real analysis, linear algebra, groups & rings, numerical analysis, statistics, and quantum mechanics :)

Jakobian

group actions are also very important, in particular the theorems about sizes of orbits and so on

nickbros123

My main problem is that I dont have time to read the book. I love reading books and this puts me in a tough situation

Jakobian

revise that too, everything else is not that specific to group theory so you can just intuit it

leslie townes

it's not going to be easy if you don't have time to read the book

:)

Jakobian

just like, anything you could do, you could do right now

leslie townes

03:46

in systems where people regularly take on introductory classes in that quantity, students often depend on a kind of regularity and predictability in how they are going to be evaluated. e.g. there is a informal, student-maintained body of knowledge about what is tested and people direct all of their preparation to that

i'm probably not saying anything you don't already know

Jakobian

for example, there is this combinatorics exercise with braids, that you have some kind of necklace and you set up an action by cycling through those braids

I suspect this is important at least when trying to compute things, I had a roommate from computer science and seems like those type of theorems are important there. Of course for math itself this won't really matter that much

Sylow's theorems and theorems about group actions probably won't matter that much to you, so if you want to remember them I suggest you focus on revising them, everything else just repeats, be it in ring theory or other

but they do matter for combinatorics and finite group theory

nickbros123

@Jakobian I havent begun reading sylows, and cayleys etc (neither has the course). I am revising the 3 isomorphism theorems right now. But I havent been able to conjure up the time to complete the exercises in the book. What I am doing so far is just the assignment problems assigned to us, along with reading the relevant theory

@leslietownes there is something like that, a collection of problems that continually get asked. Theyre not hard really, but I would, for closure purposes (or to get a feeling that I have gained something in this 4 months) want to do the book properly

Jakobian

04:01

@nickbros123 Cayley's theorem isn't that important

The 3 isomorphism theorems are important, but not that much - you are going to be revising the same version of them, be it here or some place else

What's the structure of subgroups of $G/H$ and how do we quotient $G/H$ by a subgroup $N/H$ where $H\subseteq N\subseteq G$ is normal, that's important of course, but you'll have more than one opportunity to learn this

that $(G/H)/(N/H)\cong G/N$ by the "obvious" isomorphism (yes, it is sort of obvious)

leslie townes

yeah just cancel the H's

Jakobian

the fact that homomorphisms $f:G_1\to G_2$ can be factored $G_1\to G_1/\ker(f)\to \text{im}(f)\to G_2$

this is one of the other isomorphism theorems, you have a surjection followed by bijection and them injection

so this isomorphism theorem tells you that any map can be factored into simple components, sort of

leslie townes

@nickbros123 i'm of two minds about this, obviously you don't want the time to go completely to waste or cramming just for some set of exam problems, on the other hand, some of that intro stuff is relatively deep conceptually and takes a long time to sink in no matter how much time you have to approach it

obviously in the grandest scheme of things it's basically fine if you come out of it with only a vague memory of a lot of the details

Jakobian

yeah this ^

leslie townes

i think most of what i retained from my own first algebra course was just, attention to what definitions were, and reasoning in ways i could pin back to definitions

Jakobian

04:10

but it would be nice to remember as much as possible and you'd do that not by focusing on isomorphism theorems, but on Sylow's theorems and stuff like Burnside's theorem

they're the important ones and the ones that allow you to compute things about groups

leslie townes

my first course i think did try to at least state the sylow theorems but it was all a blur

a head thing and not a heart thing

Jakobian

I've described all the groups of order $\leq 60$ using Sylow's theorems but I don't even remember the statement and it's a blur

I mean maybe something is wrong with me at that point

leslie townes

yeah at a second pass level things end up breaking down to a point where, there are only so many basic moves with those tools, and if you use each of them a dozen times you kind of 'get it' and develop intuition for what they will give you and what they won't

i certainly didn't have that at the end of my first algebra course but probably did after the second

Jakobian

do you know by any chance any quick proof that paracompact Hausdorff spaces are collectonwise normal by any chance?

leslie townes

nope

Jakobian

04:18

thought so. Like, you can't really generalize that paracompact Hausdorff spaces are normal proof

leslie townes

my advisor was sometimes a good source for stuff like that, because he ran into tons of technical issues in his early papers, he really knew the details of a bunch of things in topology and measure theory that other people had to look up

but now he's a ghost

Jakobian

I've seen some proof here that paracompact + normal is equivalent to fully normal, and fully normal implies collectionwise normal, but that requires me to know that numerable open covers and normal open covers are the same thing

where numerable means there is a subordinate partition of unity, and normal means there is an infinite sequence of star-refinements

but I feel like this itself is pretty advanced and I can't really prove it out of thin air

because its a statement that from covers you can create some continuous functions, it's like Urysohn theorem

leslie townes

burn incense to john kelley and see if he visits you

Jakobian

what can I say, I can see you're sad but I can't really do much about it, okay

leslie townes

haha kelley passed while i was at berkeley and he still got mail for years

it like piled up in the mail room

so much i wondered if he was somehow still corresponding with people

maybe send physical mail to him at the berkeley math department

who knows

Jakobian

04:28

I'm actually thinking of just going further into Normal topological spaces by Alo and Shapiro and figuring out the chapters about continuous pseudometrics and then, use that to prove that normal and numerable covers are the same thing.

I think it's just issue of me needing to absorb more theory

I was asking because in the same book there is this in chapter 1, but maybe the "in fact" was more of a statement that "Oh actually more is true" than "Prove that too"

leslie townes

if the former were intended, the nice thing to do would have been to put that remark in parenthesis

Jakobian

I'm still curious about fully normal spaces though, so I'll try to go with that goal in mind

@leslietownes or maybe it was this type of play like when you give some open problems on people's math test i.e. they wanted someone to come up with a short proof

leslie townes

oof, yes

there is evil in the world

one potato two potato

05:27

what would be a geometric interpretation of vanishing cocycle in first sheaf cohomology over a complex manifold where sheaf here is a sheaf of holomorphic vector fields @Thorgott

just vanishing one cocycle. the cohomology group itself does not have to vanish

ah it should just mean the element comes from a global holomorphic vector field

and I heard in certain dimensions, nonconstant global holomorphic vector field does not exist

Jakobian

09:36

but to be honest the exercises here are awful. They want you to show that totally ordered spaces are hereditarily normal (which is hard to do) and countably paracompact (which has dedicated article that takes two pages to prove it alone).
So one thing is confirmed at least - the authors are sadists

Alessandro Codenotti

10:14

@Jakobian There is a short proof in Engelking, 5.1.18

Jakobian

10:33

@AlessandroCodenotti do you know what bubble space on real numbers is?

Alessandro Codenotti

I've never heard this name before

Jakobian

Do you know if there exists $T_6$ non-collectionwise normal space in ZFC?

@AlessandroCodenotti oh thanks

Alessandro Codenotti

No clue. Are there non-ZFC examples?

Jakobian

Yes. There is Fleissner space under CH. I am not sure if it's perfectly normal under CH though

Thomas Finley

2

Q: Why am I getting two different answers when the countour integral is performed in two different ways?

If $C$ is the circle $|z|=2,$ taken with positive orientation then show that $\int_C\frac{2z}{2+z^2}dz=4\pi i.$ I tried solving the problem as follows: Let $I=\int\frac{2z}{2+z^2}dz.$ Then, $$I=\int_C \frac{1}{z-i\sqrt 2}dz+\int_C\frac{1}{z+i\sqrt 2}dz=4\pi i.$$ This is because, for any complex n...

Jakobian

10:42

ncbi.nlm.nih.gov/pmc/articles/PMC345971/pdf/pnas00443-0438.pdf

it's a perfectly normal non-metrizable Moore space according to my sources

Thomas Finley

None of the answers in here, seems to be getting to the point.

Jakobian

in particular, it's not collectionwise normal

although they might be saying that its perfectly normal under $\lozenge^\ast$

psie

I have a basic doubt. If we say $X$ is countable to mean $\mathrm{card}(X)\leq \mathrm{card}(\mathbb N)$, where this means there's an injection from $X$ to $\mathbb N$, is then "uncountable" simply defined by $\mathrm{card}(X)> \mathrm{card}(\mathbb N)$, i.e. we just flip the $\leq$ sign and where $>$ means there's a surjection $X\to\mathbb N$ but not a bijection?

Jakobian

@psie in ZFC this doesn't matter

You can simply define "uncountable" to mean "not countable"

outside of ZFC, I'm not sure how they define it, but they probably don't use surjections

I assume they either define it to be "not countable" or they define it to be "$\aleph_0 < \text{card}(X)$" with the meaning of there existing an injection $\mathbb{N}\to X$ but not a bijection

the two aren't necessarily equivalent if you don't assume AC

e.g. Schröder–Bernstein theorem relies on there existing injections. Having a definition that relies on surjections is not very useful, I would think

psie

11:00

hmm, ok

Thorgott

@onepotatotwopotato I'd interpret it in terms of Cech cohomology

@Jakobian it's consistent with ZF that you have something stupid like a set that is not countable but into which $\mathbb{N}$ does not inject, right?

Jakobian

11:13

@Thorgott Amorphous sets are not countable and incomparable with $\mathbb{N}$ in the sense that there is no surjection nor injection from $\mathbb{N}$ into them

So there is no injection nor surjection $\mathbb{N}\to A$ and there is no injection $A\to\mathbb{N}$

they can exist under absence of AC

tl;dr that's right

Thomas Finley

11:26

@Jakobian Can you please help me with this?

2

Q: Why am I getting two different answers when the countour integral is performed in two different ways?

If $C$ is the circle $|z|=2,$ taken with positive orientation then show that $\int_C\frac{2z}{2+z^2}dz=4\pi i.$ I tried solving the problem as follows: Let $I=\int\frac{2z}{2+z^2}dz.$ Then, $$I=\int_C \frac{1}{z-i\sqrt 2}dz+\int_C\frac{1}{z+i\sqrt 2}dz=4\pi i.$$ This is because, for any complex n...

complex-analysis solution-verification complex-numbers contour-integration complex-integration

None of the answers seem to be getting to the point.

Jakobian

@AlessandroCodenotti Bing's example H is an example of a $T_6$ but not collectionwise normal space!

psie

11:38

@Jakobian why do we require choice for $\mathrm{card}(X)> \mathrm{card}(\mathbb N)$ to mean uncountable or not countable?

Jakobian

did you miss the whole explanation I gave

psie

I don't see in which message you showed that AC is required.

Jakobian

Because amorphous sets can exist under absence of axiom of choice

psie

ok

Jakobian

if you want to, we can go into non-AC issues together, but most people just ignore them

psie

11:44

that'd be interesting one day...one day :D

Jakobian

Sure, it would be interesting for me too, as I am willfully ignorant of them

Although, there is some potential in doing topology under absence of axiom of choice that might be interesting

Alessandro Codenotti

@Jakobian which one is that?

Jakobian

11:59

@AlessandroCodenotti I'm not sure if you're aware, but Bing has an article where he has a bunch of examples of spaces relating to metrizability, the famous one is Bing's example G

Also, I found the theorem that totally ordered spaces are countably paracompact in Engelking as an exercise, which is nice since Engelking gives hints on how to prove his exercises

Bing, R. H., Metrization of Topological Spaces

Sine of the Time

@ThomasFinley with 4 answers and 15+ comments I doubt no one is getting to the point

Jakobian

@SineoftheTime you know how Thomas is

Gian's Pizzeria

hi

Alessandro Codenotti

@Jakobian Ah right I've skimmed this a long time ago

Gian's Pizzeria

@SineoftheTime are you free?

Sine of the Time

12:11

no I'm a slave

"Just ask; don't ask to ask."

Gian's Pizzeria

$(x-1)^2+y^2=1$ , Does $\theta$ vary between -π/2 and π/2 because near O(0,0) does the radius vector point to these two corners?

Sine of the Time

I don't understand the question

Gian's Pizzeria

I edited

Sine of the Time

do you want to express the area inside the circle in polar coordinates?

Gian's Pizzeria

Yes

Sine of the Time

12:18

centered in the origin?

Gian's Pizzeria

Yes

Sine of the Time

what are you fixing? A radius of an angle?

Gian's Pizzeria

Yes

Sine of the Time

12:41

?

Pizza

12:58

@SineoftheTime I discovered this thing that I used to do in another way: $\int \frac{2x+3}{(x-1)(x-4)} dx = \int \frac{A}{x-1} + \frac{B}{x-4} dx$

To find A and B I found this way

$x - 1 = 0 , x = 1$ Then I just need to insert $1$ in place of x in the numerator and denominator, but I ignore $(x-1)$ in the denominator. So $A = -5/3$

Alessandro Codenotti

@Jakobian do you want to look at example H and G together? The construction seems both interesting and pretty short

Pizza

I do the same thing for $B$ so $x=4$ and I find $B = 11/3$

Instead I did it like this: find the common denominator, collect common terms ecc

Instead it seems much faster to me this way

Can I apply this also when I have for example $\frac{Cx+D}{x^2+1}$?, Even though I don't know how

Gian's Pizzeria

@SineoftheTime staring at some rays I can see that near the origin it approaches 90°

one potato two potato

@Thorgott Ok so?

Sine of the Time

@Gian'sPizzeria so you fix $r$ and then see the range of $\theta$?

jasmine

13:10

math.stackexchange.com/questions/4967919/….

Gian's Pizzeria

@SineoftheTime yes

Sine of the Time

@Pizza this is called cover up method if I recall correctly

Pizza

@SineoftheTime Can it be applied to the type of fraction I wrote above?

Sine of the Time

@Pizza do you want to factor $x^2+1$ in $\Bbb C$ or do you mean how to do it if one of the fraction has $x^2+1$ ?

@Gian'sPizzeria ok, but you have to find an expression as a function of $r$

this works near the origin

Pizza

@SineoftheTime the second one

Gian's Pizzeria

13:19

@SineoftheTime r=1

Sine of the Time

@Pizza it does not work because you have to find two constants

Jakobian

@AlessandroCodenotti sure, I don't mind, although I did search for it mainly so that there is an example of such space on pi-base

Alessandro Codenotti

Wanna go through it now?

Sine of the Time

@Gian'sPizzeria I'm not understanding what are you doing

Jakobian

From what I recall you need to first understand example G

Alessandro Codenotti

13:22

Yes

But example G appears to be self-contained

Sine of the Time

@Pizza can you type the whole problem?

Alessandro Codenotti

(I'm more interested in example G anyway)

Gian's Pizzeria

@SineoftheTime $r^2 (\sin^2(x)+\cos^2(x))=1$

Jakobian

There is another example of a space which is called Micheal's space, a subspace of Bing's example G

Sine of the Time

@Gian'sPizzeria where does this come from?

Pizza

13:24

@SineoftheTime Ah ok! So here we proceed normally by finding the common denominator, system etc.

@SineoftheTime Ok!

Gian's Pizzeria

I was wrong, I used another equation....

Lucky Chouhan

Hi @robjohn sir, do you remember you answered my question regarding branch cuts and singular points? I finally got it yesterday I read your answer again and your line "If one travels along the unit circle, the square root function comes back to being negative of what it was one revolution prior. If one travels along the unit circle counterclockwise, the log function comes back to being $2\pi i$ different from what it was one revolution earlier." changes my perspective :')

I was thinking what does it mean? Then I realize square root function reduces the argument of z by 1/2 that's why it gives one value in first revolution and second in second revolution then again first in third revolution and it continues...

And I have also got why there is not two branch cuts for cube root function. It simply because it returns same value in every third revolution, I have written something here please see mathb.in/79418 and drew its three possible values of argument on Desmos desmos.com/calculator/9hqch73lpx

Gian's Pizzeria

$r(r-2\cos(\theta))=0$

Alessandro Codenotti

Ok so example G is some topology on $2^{2^P}$ for an uncountable $P$. Most points of this space are isolated unless they are a principal ultrafilter on $P$ (disguised as functions $f_p\colon 2^P\to 2$)

Thomas Finley

@Jakobian Yeah, we all know him to be an idiot whose every response radiates a beam of stupidity which is enough to mesmerize someone who is only used to interacting with intelligent guys. Did I sum it up correctly? :D

Pizza

13:26

It was $\int \frac{x-4}{x^4-1}dx \Rightarrow \int \frac{x-4}{(x-1)(x+1)(x^2+1)}dx \Rightarrow \int \frac{A}{x-1} + \frac{B}{x+1} + \frac{Cx+D}{x^2+1}dx$

But I know how to continue here, I just wanted to know if that method worked here :D

Jakobian

@AlessandroCodenotti Yeah that's a good way to summarize it. Note that on pi-base $P = \mathbb{R}$

Sine of the Time

@Gian'sPizzeria if you want to descrive the inside, you have to put $\le$

Alessandro Codenotti

To give a nbhd of $f_p$ instead we fix some finite $r\subseteq 2^P$ and say that the $r$-nbhd of $f_p$ consists of all those $f$ such that $f$ agrees with $f_p$ on every element of $r$

Wait that's just the product topology isn't it?

Sine of the Time

@Pizza to find the linear terms, you can use the cover up method

i.e. you can find $A$ and $B$

once you've found $A,B$ you do the common denominator

otherwise, you can factor in $\Bbb C$

Gian's Pizzeria

$r(r-2\cos(\theta))\leq0$

Pizza

13:30

@SineoftheTime ah ok! Thanks, if I have to be honest I'm watching some videos by Blackredpen for the various strategies for solving the various types of integrals

Sine of the Time

I prefer solving the system after finding the antiderivatives, but this is up to your preference

Gian's Pizzeria

$r-2cos(\theta)\leq0 \to \frac{-2cos\theta}{r}\leq0$

$2cos(\theta)\geq0$

Jakobian

@AlessandroCodenotti yes its like product topology if $F = 2^{2^P}$ were given it

Pizza

For example here: $\int \frac{\cos x}{\sin^2 x}dx \Rightarrow \int \frac{1 \cdot \cos x}{\sin x \cdot \sin x }dx \Rightarrow \int \csc x \cot x dx = -\csc x + C$

Gian's Pizzeria

$\theta \in [-π/2,π/2]$

Jakobian

13:34

but we make everything that's not a principal ultrafilter on $P$ to be isolated

Alessandro Codenotti

Ok so checking normality is not bad, it boils down to checking that closed subsets of $F_p$ can be separated by disjoint open sets

Sine of the Time

@Gian'sPizzeria and what's the range of $r$?

Pizza

@SineoftheTime 👍

Sine of the Time

@Pizza what's your question here?

Jakobian

I find the notation $F_p$ somewhat unfortunate, I think $F_P$ would more appropriate since $f_p$ are functions indexed by $P$

Gian's Pizzeria

13:37

@SineoftheTime $r\in[0 , 2\cos(\theta)]$

Alessandro Codenotti

I agree, I also wrote $F_P$ initially and changed it after double checking the notation in the paper

Pizza

@SineoftheTime Is there a way to avoid using $\csc x , \cot x$?

maybe I can use $u = \sin(x)$

So -$\frac{1}{\sin(x)} = - \csc x$

Thorgott

@onepotatotwopotato so I don't have a better interpretation

Alessandro Codenotti

Alright the failure of collectionwise normality is a neat argument, I've seen similar arguments when dealing with forcing posets made up of functions

Sine of the Time

@Gian'sPizzeria ok

@Pizza yes you can

Alessandro Codenotti

13:50

Maybe a short proof of the failure of collectionwise normality can be given if the fact that arbitrary products of separable spaces have countable cellularity is already known

Sine of the Time

note that you fixed theta, not r @gian

Pizza

@SineoftheTime :)

Sine of the Time

@Pizza see example 3 here

Pizza

@SineoftheTime I'll look now, thanks :)

ah, I had one last question

Gian's Pizzeria

@SineoftheTime thanks

Pizza

13:57

Reviewing the exercises I had done

$\arcsin(x^2-y-1)$ , find the absolute extremes in the circle with center (0, −1) and unit radius. Here when I calculate partial derivatives, can I only consider the argument?

Sine of the Time

$x\mapsto \arcsin x$ is increasing

so maximizing the function is equivalent to maximizing the argument

Soumik Mukherjee

Hi

Pizza

Ah yes now I remember you told me this thing, but I didn't understand very well, reviewing it I had this doubt

Sine of the Time

@SoumikMukherjee sup

Pizza

@SineoftheTime But does this also apply to other types of functions?

Sine of the Time

14:07

to increasing functions in general

Pizza

@SineoftheTime So it can also be applied for example to $\log(x), \sqrt{x}$ etc.

Soumik Mukherjee

@SineoftheTime watching Hikaru vs Alireza

But more excited for Magnus vs Hans

Sine of the Time

@Pizza yes

will Magnus play in your opinion?

Pizza

Is the match at 8?

@SineoftheTime 👍

Jakobian

@AlessandroCodenotti because if you have a family of disjoint open sets in a product, you can restrict to a countable product which is separable, right

Sine of the Time

14:13

20:30

Soumik Mukherjee

@SineoftheTime yes, I think he will play

@Pizza which one? The Hikaru-Alireza match is currently happening

Alessandro Codenotti

Ok this is presented well in Engelking's (5.1.23)

Pizza

@SoumikMukherjee Magnus vs Hans

Alessandro Codenotti

He uses indeed that the product of separable spaces has countable cellularity to show that Bing's example is not collectionwise normal

Ryder Rude

hi

Alessandro Codenotti

14:18

Knowing that in a collectionwise normal space the cellularity must be at least as big as the maximum cardinality of a discrete subset (I forgot the name of the latter cardinal function) makes it clear that large products of separable spaces cannot be collectionwise normal. The issue is that they will also fail to be normal, but Bing's construction somehow fixes this

Soumik Mukherjee

@Pizza after 4 hours and 10 mins from now

Pizza

Oh okay!

Jakobian

@AlessandroCodenotti sorry, I didn't see this argument before and the talk about countability is a weird mistake. Still, the theorem uses that $\mathfrak{c}$ sized products of separable spaces are separable so I wasn't far off

@AlessandroCodenotti supremum because it doesn't have to be achieved, it's called spread I think

It's sometimes called sup = max problem in set theoretic topology, about those cardinal functions defined by supremum having it achieved

ams.org/journals/proc/1987-099-03/S0002-9939-1987-0875405-9/…

for example here

Alessandro Codenotti

14:35

@Jakobian Ah yes good point about supremum

Pizza

?

What was it?

Gian's Pizzeria

Aug 20 at 17:07, by Sine of the Time

$r\le 1$ and $r\le 2\cos \theta$ implies $r\le 1$ for $\theta \in [0,\pi/3]$ and $r\le 2\cos \theta $ for $\theta \in [\pi/3,\pi/2]$

Alessandro Codenotti

The way this is presented in Engelking is very nice. Given a space $X$ and a subspace $M$ let $X_M$ be the space obtained by making everything outside of $M$ discrete (more formally refine the topology of $X$ by throwing in $U\cup K$ for any open $U$ and $K\subseteq X\setminus M$). Then $X_M$ is normal as long as any two closed subsets of $M$ can be separated by open sets in $X$.

Gian's Pizzeria

@Pizza @SineoftheTime I was trying to do this exercise and I think you did it wrong

Or at least I don't understand where you found theta

Alessandro Codenotti

So Bing's big product fails to be collectionwise normal by the spread vs cellularity argument, while it is normal because it is of the form $X_M$ for a cleverly chosen $M$

Pizza

14:44

@Gian'sPizzeria let me see

Sine of the Time

14:57

where's the mistake?

Pizza

$\theta$ of the first circumference would be $\begin{cases} x^2+y^2 = 1 \\ (x-1)^2 + y^2 = 1 \end{cases}$, and find the point $P = \left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$ which corresponds to $\pi/3$

And from the graph you can see that it varies between $0 \leq \theta \leq \pi/3$

Gian's Pizzeria

ok so far

@SineoftheTime It is not clear to me how you found theta of the second Circumference

Sine of the Time

it's the same circle in the last ex you sent me

Pizza

@Gian'sPizzeria mm wait

I'm going to see

Gian's Pizzeria

15:14

@SineoftheTime I did it with brain, how did you do it mechanically?

I'm not clear on what you did

Ben Steffan

I guess the upshot about Atiyah duality is that nobody really cares since it's covered by ABG+

which makes sense

I guess now I have to go learn about $\infty$-topoi

Sine of the Time

15:35

Aug 20 at 17:05, by Sine of the Time

$(x-1)^2+y^2\le 1\implies r-2\cos \theta \le 0$ i.e $2\cos \theta\ge r$

Gian's Pizzeria

@SineoftheTime so $\cos(\theta) \geq r/2$?

Sine of the Time

yes, but you're fixing theta, why divide by $2$?

Gian's Pizzeria

But I have to find theta not r

Sine of the Time

why?

Obliv

How do I formally define the complement of a set $A$?

Gian's Pizzeria

15:42

@SineoftheTime Aren't they two different thetas?

There are 2 circumferences

Obliv

actually for example $(A\cap B)$ which is $\{a\mid a \in A \land a \in B\}$, how would I compute the complement?

Sine of the Time

I'm not following you

Obliv

intuitively it's the exclusive or, I think, where now $(A\cap B)^c = \{a \mid a \in A \lor a \in B$ but not both $\}$.

Gian's Pizzeria

@SineoftheTime But then $\theta_1 = \theta_2$?

I'm not understanding

Sine of the Time

me neither

Gian's Pizzeria

15:45

Then for the circle 1, r ≤ 1

Right?

Sine of the Time

yes

Gian's Pizzeria

For the circle 2 , $2 \cos(\theta) \geq r$

Sine of the Time

yes

Gian's Pizzeria

Now I also have to solve x ≥ 0 and y ≥ 0

Right ?

For theta

Sine of the Time

you find theta in the first quadrant

Gian's Pizzeria

15:48

Yes

Ben Steffan

@Obliv The complement of $A \subset B$ is $B \setminus A := \{b \in B \mid b \notin A\}$

@Obliv your intuition is correct :)

Gian's Pizzeria

For x ≥ 0, $r \cos(\theta) \geq 0$ so $\cos(\theta) \geq 0$ so 3π/2 ≤ theta ≤ π/2

For y ≥ 0 , 0 ≤ theta ≤ π

Now I only have to consider the first quadrant

3π/2 ≤ theta ≤ π/2 becomes 0 ≤ theta ≤ π/2

No wait

But the circle 2 has an extra part in the first quadrant

In the second circle theta varies between 0 and π

While for the circle one would vary between 0 and π/2

The intersection point is (1/2, √3/2) , so π/3

@SineoftheTime I don't understand what I find when solving x and y ≥ 0

Sine of the Time

you have both sine and cosine $\ge 0$

Gian's Pizzeria

ok so you can confirm that I find these

For x ≥ 0, $r \cos(\theta) \geq 0$ so $\cos(\theta) \geq 0$ so 3π/2 ≤ theta ≤ π/2
For y ≥ 0 , 0 ≤ theta ≤ π

Right?

Sine of the Time

so $\theta\in[0,\pi/2]$

Ben Steffan

16:02

love to spend a few hours working out a detailed answer to a question just for the author to decide to delete it

Sine of the Time

did the question have at least an answer?

Gian's Pizzeria

@SineoftheTime Yes exactly

Ben Steffan

@SineoftheTime no

Sine of the Time

you can try to ask in this room if you want

Gian's Pizzeria

so $\theta\in[0,\pi/2]$, (Circle 2) $2 \cos(\theta) \geq r$ , (Circle 1) 0 ≤ r ≤ 1

Sine of the Time

16:06

@gian since I've already written the answer in detail and I'm too lazy to type it a second time, look at my messages starting here

Ben Steffan

@SineoftheTime it would have been quite a general answer to a pretty specific question, so I think I won't pursue it further, but thanks

I might think about making it into a self-answered question

Sine of the Time

once asked, I think the question belongs to the community

Ben Steffan

let me put it like this: the question was roughly "how to show that Poincare duality holds for complex $K$-theory? can you give a reference?" The proof I offered shows is "here's how to show Poincare duality holds in a very general setting; the special case for $K$-theory follows as a one-liner"
so in a sense the answer is already not a great fit for the question

making it the answer to a more general question makes more sense, at least to me

Gian's Pizzeria

16:22

@SineoftheTime Ok so you saw that $2\cos(\theta) < 1$ when $\theta \in (π/3 , π/2]$ while $> 1$ when $\theta \in [0 , π/3)$

I understand, thanks

@SineoftheTime but so finding the intersection point was of no use?

Pizza

Yes for the first circumference

But I think you could also not find it.

Sine of the Time

16:40

@Gian'sPizzeria from a geometric point of view, you can find $\pi/3$ using the intersection point

otherwise, you can only use the relations that define the domain

Soumik Mukherjee

17:10

Are you watching @SineoftheTime?

Sine of the Time

yep

I want to watch the bullet

Soumik Mukherjee

Yeah, let's see if Hikaru can turn the tables in bullet

Sine of the Time

seems hard

Soumik Mukherjee

Will there be a third place match?

Sine of the Time

seems so

I'm not understanding this answer, if we don't consider p.v

30 min bullet won't suffice

Soumik Mukherjee

17:19

@SineoftheTime "I don't care chat, I literally don't care, no no, as I said, idc" incoming

Sine of the Time

yeah :D

Hikaru is the type of guy who does not care

psie

18:34

> Corollary 0.13: If $\mathrm{card}(X)\geq\mathfrak{c}$, then $X$ is uncountable.

Folland says the converse of this statement is the so-called continuum hypothesis. He writes it as "if $\mathrm{card}(X)<\mathfrak{c}$, then $X$ is countable." He is somewhat vague about the definition of $\mathrm{card}(X)$ as an object by itself (he doesn't really define it, only the relations $\geq$ and $<$).

I wonder, is $\mathrm{card}(X)<\mathfrak{c}$ really the negation of $\mathrm{card}(X)\geq\mathfrak{c}$? How would you show it is? I hope I am not asking something I've already asked about before.

(Recall, $\mathrm{card}(X)\geq\mathfrak{c}$ means there's a surjection from $X$ to a set with the cardinality of the continuum, and $\mathrm{card}(X)<\mathfrak{c}$ that there's an injection from $X$ to that set but no bijection.)

Thorgott

yes, there are two things here:
1. $card(X)\ge\mathfrak{c}$ is the same as $\mathfrak{c}\le card(X)$ (this, we discussed at some point before)
2. cardinals are totally ordered, i.e. given sets $X,Y$, there is either an injection $X\rightarrow Y$ or an injection $Y\rightarrow X$

additionally, it is true there is a thing such as "the smallest uncountable cardinal", i.e. there is an uncountable set that injects into any other uncountable set. the continuum hypothesis then translates into the statement that this smallest uncountable cardinal is $\mathfrak{c}$.

psie

ok. 1) and 2) are definitely in Folland 👍 for your additional statement, I guess I have to dig a little deeper into set theory :)

Thorgott

you don't need to worry about that, I just mentioned it for additional context

Sine of the Time

19:21

imagine if Hans does not win a single game

Soumik Mukherjee

19:32

He may win in bullet

@SineoftheTime tbh, Hans never won a single game against me either:)

Sine of the Time

:)

Gian's Pizzeria

Magnus Carlsen said it was an easy win

ILikeMathematics

Btw @BenSteffan Uni Bonn is very high up in topology and algebra, but what if you decide you want to specialize in e.g. analysis? Is Bonn strong on that end too or does e.g. Stuttgart 'beat' it there?

I heard Bonn is rather average in analysis?

Sine of the Time

19:52

Hans creating drama

Pizza

What happened to Hans's PC?

Sine of the Time

he's complaining about something (probably clock) and now they're reviewing the footage

psie

20:16

@Thorgott am I right in observing that $\mathrm{card}(X)\not\geq\mathfrak{c}\iff \mathrm{card}(X)<\mathfrak{c}$ only holds under the axiom of choice? Somehow that seems odd to me, because then the statement of the continuum hypothesis seems to depend on the axiom of choice, i.e. taking the converse of the Corollary I posted above results in a certain statement under the axiom of choice and...I don't know what if we don't assume choice.

Thorgott

yes, both of the two bullet points in my previous message rely on choice

psie

ok, I see

Thorgott

you can state the CH just fine in ZF, but the converse of the Corollary might simply not be equivalent to CH if we don't assume choice (I do not know whether it is or isn't)

psie

ok

Jakobian

20:39

@psie to introduce some notation, the smallest uncountable cardinal is called $\aleph_1$ while the smallest infinite cardinal is called $\aleph_0$

that is $\text{card}(\mathbb{N}) = \aleph_0$

continuum hypothesis says there is no cardinal strictly between $\aleph_0$ and $\mathfrak{c} = 2^{\aleph_0}$

the strong continuum hypothesis says there is no cardinal strictly between $\kappa$ and $2^\kappa$ for any infinite cardinal $\kappa$

The cardinals defined by exponentiation are the so called beth hierarchy

$\beth_0 = \aleph_0$ while $\beth_1 = 2^{\beth_0}$

the continuum hypothesis says there is no cardinal between $\beth_0$ and $\beth_1$

the strong continuum hypothesis says that every infinite cardinal is equal to $\beth_\alpha$ for some ordinal $\alpha$

in other words, the continuum hypothesis says that $\beth_1 = \aleph_1$

Alessandro Codenotti

Just a quibble on notation, it is usually called the generalized continuum hypothesis or GCH rather than the strong continuum hypothesis (I complain about this only because the acronym SCH is also used in set theory, but that's the singular cardinal hypothesis, a weakening of GCH)

Jakobian

Yes, I only wrote "strong" instead of "generalized" because I forgot

According to my sources, $\mathfrak{c}$ doesn't need to be a cardinal in ZF, so beth numbers seem to only make sense in ZFC

35

A: How to formulate continuum hypothesis without the axiom of choice?

You are correct that without the axiom of choice $2^{\aleph_0}\newcommand{\CH}{\mathsf{CH}}$ may not be an $\aleph$. Therefore the continuum hypothesis split into two inequivalent statements: $(\CH_1)$ $\aleph_0<\mathfrak p\leq2^{\aleph_0}\rightarrow2^{\aleph_0}=\frak p$. $(\CH_2)$ $\aleph_1=2^...

there seems to be at least four different continuum hypotheses without AC

psie

20:59

ah nice! :) that's what I was looking for

Jakobian

the thing to take away from this is that without axiom of choice, math as we know it breaks and everything is sad

psie

haha, yeah it'd be a tragic day if we stopped using it

Soumik Mukherjee

AC makes math cool

Sine of the Time

ice what you did there

Thorgott

@Jakobian I don't know what a cardinal in ZF is or isn't, but the statement "there is no set with cardinality between N and R" still makes perfect sense without choice, that's what I had in mind

Jakobian

21:52

@Thorgott a cardinal is simply an ordinal of smallest cardinality

Thorgott

is it true without choice that every set has a cardinality?

no, it isn't

Jakobian

yes, we went over this

were you part of that discussion, I don't remember

Thorgott

so I guess the issue is I don't know what $\mathfrak{c}$ is without choice

Jakobian

Scott's trick allows you to define cardinality as simply all sets in bijection to $X$ of the smallest rank

this is a set, consisting of all "simplest" sets in bijection with $X$

from what I recall it wasn't you, but Joe and Alessandro

okay, I typoed

a cardinal of $X$ is simply the smallest ordinal with the same cardinality as $X$

having a corresponding cardinal is equivalent to being well-ordered

$\mathbb{R}$ can't always be well-ordered

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