Mathematics - 2021-06-11

Thorgott

00:00

that's not well-defined yet

the key observation is that conjugation by $q$ fixes the purely imaginary quaternions, which are then identified with $\mathbb{R}^3$ via the standard basis $i,j,k$

leslie townes

copper isn't here so i'll say it: quaternions are overrated.

Thorgott

of course, one also has to check that conjugation by a unit quaternions is orthogonal and orientation-preserving (in fact, that the purely imaginary quaternions stay fixed is a corollary of the former observation as they're the orthogonal complement of the real numbers inside the quaternions, which form the center and hence are fixed by conjugation)

@leslietownes I'm not copper, but I will object

you use quaternions to construct exotic 7-spheres, can't beat that

leslie townes

thorgott, so what

Thorgott

grr

also, for the record, the fact that the map $\phi$ is surjective is a bit annoying to prove as well

Govind75

That's what I am doing now

Well, it's what I am following

Have you seen the 4D rotation homomorphism from $S^3 \times S^3 \rightarrow SO(4)$?

Thorgott

00:11

no, that's pretty wild

so Spin(4) is S^3xS^3, I didn't know

leslie townes

blegh

Ted Shifrin

00:40

You mean blech?

leslie townes

i do. i just mock-deposed a witness who will be deposed tomorrow.

i had to pretend to be a jerk for about five hours.

impossible task, i know.

Ted Shifrin

YeAh, anything less than 8 hrs is impossible for you.

leslie townes

the deconstruction of it was, here's what i was doing when i acted like that, here's what i was doing when i acted like this, here's what is going to happen tomorrow. he now thinks i'm a psychopath.

there's no way that tomorrow will be worse for him than the prep was. mission accomplished.

the worst thing was i could only bill 5 hours for it and not as you say the usual 8.

that reminds me of something, where do i send my invoices for my consulting fees. i mean in chat and also on main.

Andrew Micallef

Hey internet friends. Got my first math tutoring gig tonight :)

leslie townes

congrats andrew.

hyper-neutrino

00:47

Ooh, exciting. Congrats and have fun :)

leslie townes

don't yell at the person you're tutoring. tutoring is not deposition.

Andrew Micallef

Yeah, Im nervous. Im pretty sure the kid is gonna realise he knows more than I do. But my job is just to make sure he does homework ;)

I dont yell :)

leslie townes

when i began tutoring i started by not doing anything, not picking up a pencil even. just asking them about their homework.

Andrew Micallef

At least only in very special emergency circumstances

leslie townes

the witness we were preparing today is going to be deposed tomorrow by an attorney who yells. i needed to prepare him for it.

Andrew Micallef

00:50

@leslietownes so there is precedent !

leslie townes

i warned my wife in advance, this is all an act.

it might sound weird upstairs for a minute but i have to do this.

Andrew Micallef

That sounds unproffesional

Dont lawers have standards?

leslie townes

it definitely pushes the line.

i say 'yells,' it's not quite that, he acts like it's absurd that you can't answer him and then demands you answer him.

Andrew Micallef

Isnt there something wrong with giving evidence under duress?

leslie townes

the game that is played is you act like you can't believe what you're hearing, because you want to hear something else.

Andrew Micallef

00:52

@leslietownes sounds like my boss actually

Also sounds straighr out of the coercive control playbook

leslie townes

you leverage a lot of social norms, the weirdness of silence, the weirdness of disagreement, you push it so the person complies with what you want even if you haven't asked them to do it. it's a very sick game.

yes it is something people go to therapy to adjust.

i've read a lot of those books.

Ted Shifrin

@leslietownes Au contraire, you owe us.

@AndrewMicallef you can threaten him with Ted.

leslie townes

if they haven't done their reading, and if they haven't even put themselves in a position to ask questions, that's when you threaten them with ted.

Andrew Micallef

@leslietownes any good ones? I just had a difficult conversation with my partner about spending too much on a car...which I totally cant afford

@TedShifrin i'll stick with inviting him to ask his self questions. Socratic method doesnt require that much knowledge...right :P

leslie townes

i do have a particular book in mind, i can't think of the title at the moment but i will supply it when it comes up. it's broadly about leveraging social pressure.

i've read a whole lot of books about interrogation and i might actually be a sociopath.

Ted Shifrin

00:57

So that's my secret !

copper.hat

Influence by Cialdini is not a bad read.

Ted Shifrin

@leslietownes Might?

Andrew Micallef

@leslietownes lol. I recently read "Never split the difference" and "difficult conversations"... both incredibly enlightening

leslie townes

you slowly lean into the fact that the answerer isn't saying what you want. you do it slowly. you start with confusion and then depending on what is going on you ramp it up quickly or more slowly. it is a path to psychosis.

copper.hat

Ahh, interrogation.

leslie townes

00:59

cialdini is definitely on my bookshelf.

Andrew Micallef

That actually sounds pathalogical

copper.hat

closed pathalogical

Andrew Micallef

Maybe I dont want your advice anymore

leslie townes

it would be better if you did not adopt the leslie townes model.

Andrew Micallef

Gtg

copper.hat

01:00

probably antithetical in your context, but "getting to yes" is a good one.

more useful for getting vcs to part with money

leslie townes

i've got that one too. the sales context is different.

copper.hat

so much of negotiation is delaying at the most annoying moment.

leslie townes

negotiation is generally different from coercion.

i did frighten the witness today because i was turning it on and off.

copper.hat

no question.

@AndrewMicallef the 5th season of rake wasn't too bad. had a few belly laughs

leslie townes

some of those bell bottom people probably know a thing or two about talking to people

copper.hat

01:03

:-)

leslie townes

copper have you read patrick keefe's "say nothing". pretty routine investigation of something during the troubles.

but touches on a whole lot of cultural stuff.

copper.hat

i believe a pair of my old flares went into the dumpster in cork a few weeks ago.

no, looks interesting. at some point it is all a bit depressing.

when you realise its people around you.

not some tom clancy stuff

leslie townes

it's very much the usual, if you have any relatives who might have been into it

copper.hat

you can imagine

leslie townes

just depressing people selling drugs and storing weapons and disposing of things in fields.

otherwise a good book

Ted Shifrin

01:06

In fields, but not in groups …

copper.hat

my uncle was a veterinarian on the border.

the border was an interesting concept.

leslie townes

sometimes it's helpful to dispose of things in rings.

or ideals of rings.

you can quotient out by them

Thorgott

@Ted Say I have a Riemannian manifold $M$, $p\in M$ and a neighborhood $0\in U\subseteq T_pM$ that gets mapped diffeomorphically onto $M$ by $\exp$. Do you agree that if I take a constant vector field on $U$, $\exp$ pushes it forward to a vector field on a neighborhood of $p$ that is parallel with respect to any short geodesic passing through $p$?

Ted Shifrin

Good question to which I should know the answer immediately.

Thorgott

I feel like it should be true since $\exp$ is a "radial isometry", but I haven't been able to jot down something precise.

I've even drawn a picture of a sphere to convince myself

Ted Shifrin

01:22

Too special.

Need to do the computation in Gauss normal coordinates.

I will think later.

The constant vector field is parallel along any curves, so the diffeo is critical.

robjohn

02:29

@TedShifrin Just out of curiosity, how does one define parallel when the Gaussian curvature is not zero?

barista

If $R$ is a complete ring with $I$-adic topology then what is the multiplication of two elements? i.e $(x_i)(y_i) = ?$

More Anonymous

@TedShifrin I can't find any transformation which lets me take $ds^2 \to \lamda^2 ds^2$ and so I cant use the trick :(((

@TedShifrin Can you give me a metric which is scalable under coordinate transformations?

Ted Shifrin

03:09

@MoreAnonymous: I need to think about this more, but I think the only such metrics are flat. In which case, your result — weird as it is — is not surprising.

@robjohn: This is not parallelism of lines. It's parallelism of a vector field along (a) curve(s). This is the notion of covariant derivative zero.

Bhavay

@TedShifrin Hi sir. Can you answer my doubt ?

Ted Shifrin

Huh?

If you have a question, don't ping me. Just post the question.

Bhavay

I posted it last night, got no response .

Ted Shifrin

Well, unless you link us to it, we aren't mind readers to just find it.

2

Bhavay

10 hours ago, by Bhavay

10 hours ago, by Bhavay

In A=QR , aren't we essentially taking dot product btw 3 vectors ($q_1,q_1^T , a$) ?

Hawk

03:12

A subset A \subset X is called "relatively" compact if its closure \bar{A} is compact, but relatively to what? This definition doesn't seem in parallel with other "relative" definitions in topology.

Ted Shifrin

@Bhavay: What do you mean by dot product "btw 3 vectors"? I only know how to take the dot product of two vectors.

Bhavay

Yes. That is the part I am confused about...I will explain what I mean by that.

Ted Shifrin

Your question makes absolutely no sense.

You know how Gram Schmidt gives the $QR$ decomposition?

Bhavay

To find matrix R we take dot product of $Q^t $ and $A$ , yes?

Ted Shifrin

No, you take the dot product of $q_1$ and $a$. Multiplying $q^\top a$ is the same as computing $q\cdot a$.

You don't take dot products of matrices. What does that even mean?

Bhavay

03:15

Okay , I will be need to more specific.

Ted Shifrin

No, you need to be more correct.

Because $Q$ is an orthogonal matrix, $Q^{-1} = Q^\top$, so, yes, in fact, $R = Q^{-1}A = Q^\top A$.

Bhavay

to compute the first element of R, we take dot product of $q_1^T$ and a , is it correct ?

Ted Shifrin

NO. I already corrected that.

Bhavay

@TedShifrin Okay. so the first element of R is $q\cdot a$ , is that correct ?

Ted Shifrin

Yes, $q_1\cdot a$.

So now what is your actual question?

Bhavay

03:20

Sir now to two get back to first element in the matrix A , do I need to take dot product btw first row Q and the first column of R . (which doesn't make sense to me , since I am taking dot product btw 3 vectors).

Ted Shifrin

you keep saying this garbage. There is NO dot product of three vectors.

Bhavay

you are missing the point ..

Ted Shifrin

You're multiplying the vector $q_1$ by the scalar $q_1\cdot a$.

NO, sir, you are not understanding.

Don't you dare tell me that I am missing the point.

Bhavay

You seem angry , did I offend you , if so I apologize..

ok.

David Choi

03:52

Hey guys I have a meta question:

I asked a question 2 days ago, and @TedShifrin gave me an answer that helped me understand the question better and answer my own question. I would like share what I worked out, but it’s going to pretty long. Should I add this by editing my original question, or should I add this as an answer?

robjohn

04:05

@TedShifrin Okay, along a non-closed curve I understand. It justsounded as if you were talking about something more global.

Ted Shifrin

04:31

@robjohn No, it was local. But no need for curves to be non-closed. Closed curves give the holonomy of the connection.

hyper-neutrino

05:00

@DavidChoi The standard for if you have an answer to your question is to post it as an answer to your question. Don't edit answers into your question, you should only edit fixes and clarifications in.

The core concept behind SE is to build a repository of questions and answers so beyond just getting the asker the help they need, it leaves a permanent resource up for anyone in the future.

Thus, adding an answer to your own question is actually good - in fact, that's why the "Answer your own question" checkbox exists when asking the question

on SE, it is perfectly normal to ask a question while already having the answer if you believe that some issue you ran into and found a solution for might be of help to other people in the future

Russian Bot Who Knows Your IP

nae

hello

{σ§h♂☼4é☺

Koro

05:24

Is it possible to find inverse of a derivative at a point using these hypotheses only?

Let there be a neighborhood $\displaystyle U$ of $\displaystyle x_{0} \in \mathbb{R}$ such that $\displaystyle f:U\rightarrow \mathbb{R}$ is one-one and differentiable at $\displaystyle x_{0}$ such that $\displaystyle f'( x_{0}) \neq 0$ and it is required to find derivative of $\displaystyle f^{-1}$ at $\displaystyle f( x_{0})$, if it exists.

So I proceeded like this: Based on given conditions, $\displaystyle f^{-1} :\ f( U)\rightarrow U$ exists. Let $\displaystyle p:=f( x_{0})$, for brevity. Now,

$\displaystyle \left( f^{-1}\right) '( p) =\lim _{t\rightarrow p}\frac{f^{-1}( t) -f^{-1}( p)}{t-p} =\lim _{t\rightarrow p}\frac{x-x_{0}}{f( x) -f( x_{0})} =\lim _{f( x)\rightarrow f( x_{0})}\frac{1}{\frac{f( x) -f( x_{0})}{x-x_{0}}} \ $

$ $Now, $\displaystyle f^{-1}( t) =x$ for some $\displaystyle x\in U$ and hence $\displaystyle t=f( x) \Longrightarrow f( x)\rightarrow f( x_{0})$

If $f(x)\to f(x_0)$ implies $x\to x_0$ then I am done! Also, I would like to add that denominator in the second limit can't be zero as $f$ is one -one.

That implication will require continuity of $f^{-1}$ but is it really required?

Hi @copper

copper.hat

Hi :-)

Koro

After taking vaccine, I had severe fever. It's fine now.

copper.hat

Bummer!

Where did $x$ some from all of a sudden?

Koro

Because f is one one from U onto f(U)

copper.hat

You need to more explicit, you can't just throw $x$ in there instead of $f^{-1}(t)$.

Koro

05:40

Will this work? : By definition of derivative $g'(a)=\lim_{x\to a}\frac{g(x)-g(a)}{x-a}$, where $x$ is in domain of $g$. Similarly, our $t$ is also in domain of $f^{-1}$ that is in $f(U)$, that is of the form $f(x)$ for some $x\in U$. @copper

copper.hat

It is too informal for me. You cannot write $\lim_{t \to p}$ with an expression that has $x$ in it. You need to have logical moves connecting the steps.

You need to define something like $x(t)$ or restrict the $t\to p$ or something.

Koro

Ahh @copper, you are right. Please ignore that limit.

Let's go to 3rd limit from the 1st limit directly. Is 3rd limit okay now?

If yes, then what follows (or should follow) 3rd limit is what confuses me.

copper.hat

Also, I think your hypotheses need to assume that $f$ is continuous near $x_0$.

Sorry, it's too sloppy for me to follow.

If you have not started with the appropriate hypotheses then there is something wrong somewhere.

Anyhow, I need to go to sleep now. Good luck, hopefully someone will help you :-)

Good night!

Koro

You are absolutely right copper. Continuity at $x_0$ is also a hyptheses.

Continuity of $f^{-1}$ at $f(x_0)$ and that of $f$ at $x_0$

I was trying to see how I can do this proof without saying "f is monotonic..."

Thank you! Good night :)

I missed that continuity hypotheses (of $f^{-1}$ at $f(x_0)$) somehow.

Govind75

10:56

What do we denote the set of pure quaternions?

Thorgott

11:29

I don't think there'S a standard notation

Govind75

11:45

$\mathbb{H}^*$

I've seen that somewhere

Thorgott

that should be the non-zero quaternions and any other use of that notation is evil

Jack Of Blades

I'm looking for information about something, but I don't know what the specific phenomenon/field is called(I know it's statistics/probability, but not the particular name of this exact thing).
For example if two people are flipping a fair coin 3 times in a row before it is the next person's turn, and #1 gets tails 3 times in a row. Is it that only #2's chances for heads is high, or is it that it is also high still when it is #1's turn again, because #1 ended with all tails?
I guess the question can sort of be formulated as "to whom does the next probability belong to?" or what scope the pro…

Thorgott

neither is the case

if you have a fair coin, the chance for heads is always the same as the chance for tails

what happened on previous throws does not impact the outcome of the next throw whatsoever

Jack Of Blades

So it's like the gambler's fallacy? I guess what I thought of was more like..

"The probability of getting 3 tails in a row is less likely than tail-head-tail for example. 4 tails in a row is much less likely than tail-tail-tail-head."

Since it would (if I'm not mistaken) be 0.5x0.5x0.5

Though then again they have the same probability hmm, maybe I'm confusing myself

tail-tail-tail-tail vs. the chance of that not happening is what I mean, I think

Sayan Chattopadhyay

The proof of hairy ball using the chern class is so damn cool

Thorgott

12:03

@JackOfBlades no, getting 3 tails in a row is just as probable as getting tail-head-teil

both half probability 0.5x0.5x0.5

however, it is more probable to get 2 tails and 1 head than it is to get 3 tails (not that I'm not specifying the order in which they are obtained)

@SayanChattopadhyay what proof? I guess just noting that non-triviality of the chern class implies non-triviality of hte bundle?

Sayan Chattopadhyay

Yeah but proving non triviality requires you to take the Kähler metric and then note that the chern class is calculated using the ricci curvature of the metric, which is different from the homological way of defining it

Jack Of Blades

That change in probability; who "gets to use it"?
So for example if I was to be at a casino and watch some guy not get some particular slot in roulette(45 for example), he gets 17,89,22,33,62 etc., so 10 times he does not get 45 but he gambled on 45 each time. If he then leaves the game, and I start playing, and I just keep guessing 45 each time for 10 rounds, do I have a higher chance of getting 45 than he did?

And if I play 10 rounds, and then he comes back and plays 10 rounds, does the probability "reset" if I was to get 45 in one of my 10 rounds? Or would he get the same probability I did?

Thorgott

ah, you're doing Chern-Weil theory, nice

wish I knew that

Sayan Chattopadhyay

Crazy stuff though the chern class. It can decide whether a line bundle is ample or not, hence give embedding for compact Kahler manifolds. I want to learn this proof sometime

Thorgott

hmm, is the Kähler metric on S^2 (I guess you take the Fubini-Study metric on CP^1) the same as the complexification of the standard round metric?

@JackOfBlades no, this is the Gambler's fallacy

assuming the casino is fair, of course

Jack Of Blades

12:14

Don't the tosses follow a uniform distribution? Don't all numbers eventually have to come up?

Sayan Chattopadhyay

I think it should be @Thorgott. The fubini study metric is defined as taking CP^n and looking at its tautological bundle, which sits naturally in C^n+1 and now you pullback the hermitian metric on C^n+1 (which is the natural complexification) using sections.

You can also do a calculation and show that the volume of CP^1 with FS metric is pi, which is the same as that of sphere with radius 1/2

love_sodam

12:40

$A$ is a noetherian local ring and $Q\subset A$ is a $m$-primary ideal where $m$ is the unique maximal ideal of $A$. Then why $A/Q$ is an artin local ring?

I know there are only two possibilities one is $m^n\supsetneq m^{n+1}$ for all $n\geq 1$ and other is $m^n =0$ for some $n\geq 1$ this is the case $A$ is artin local.

Is $Q = m^n$ for some $n$ ? then it make sense

Thorgott

it's probably just a calculation. CP^1 is the quotient of S^3 in C^2 by the scalar (isometric) action of S^1 in C, where S^3 carries the round metric. the map S^3 -> CP^1 -> S^2 is the Hopf fibration, I think, and can be obtained from the action of S^3=Spin(3), the double cover of SO(3), on the unit sphere S^2 in R^3, which is an action by isometries. somehow this all matches up, but I don't understand it well.

@JackOfBlades Yes, they're uniformly distributed. It's not like all numbers have to eventually come up, but they do so with probability 1. This does not contradict the fallacy.

leslie townes

13:09

if you were at an actual casino and 45 never came up over several days, you could safely assume that something was wrong with the equipment

but a statistical guarantee is not a guarantee of any given outcome in any run of tries and the wheel itself has no memory

one time i was illustrating a probability thing about the number of heads in a run of coin tosses in a class, and flipped four heads in a row. the universe might have a sense of humor

epsilon-emperor

5

Q: An integral inequality (Rudin $L^p$-spaces 3.12)

I've been learning about $L^p$-spaces and came across this problem in Rudin's Real and Complex Analysis: Suppose $\mu(\Omega)=1$ and $h:\Omega \to [0,\infty]$ is measurable. If $A= \int_\Omega {h d\mu}$, prove that: $$ \sqrt{1+A^2} \leq \int_\Omega \sqrt{1+h^2}d\mu \leq 1+A $$ I've alread...

Could someone help me understand why equality holds iff h = A a.e.?

like, if equality holds then h = A a.e. part of the implication

$\sqrt{1+A^2} \leq \int_\Omega \sqrt{1+h^2}d\mu$

I'm talking about the above (left) inequality

marmistrz

Question to American native speakers: is "thanks to" too informal to be used in mathematical writing? For example:
> both expressions are equal thanks to equivariance

leslie townes

it strikes me as something i might say in a talk, but not in writing

marmistrz

I've seen some authors use it, most notably John Lee (in Introduction to Smooth Manifolds), Ravi Vakil (in Foundations of Algebraic Geometry), L. Trefethen and D. Bau (in Numerical Linear Algebra)

leslie townes

although this is very much the realm of style and not substance

it isn't so informal that it would shock me in print

if ravi vakil does it it has my seal of approval :)

but i wouldn't myself do it

my mathematical style, to the extent i had one, was to write as if i had no personality and was not human. some people find this alienating or hostile to the reader but i prefer it.

this kept me from ever truly liking spivak's books, for example. it feels like he's right there in the room with you, which to my mind is too close.

Koro

13:21

@leslietownes Ahahaha

epsilon-emperor

lol

love_sodam

Ah I if $A$ is noetherian with $m$-primary $Q$, there is some $N$ such that $m^N\subset Q\subset m$

should be artin local then

aarbee

14:02

Hi. How come |w-w^2|<root3, where w is a cube root of unity?

I understand that the cube roots of unity are the vertices of an equilateral triangle

but does that mean the side length of that triangle is less than or equal to root3?

marmistrz

@leslietownes Oh, and I must admit I prefer a semi-formal style. I really like mathematical texts that are suitable for bedtime reading.

PM 2Ring

15:11

@aarbee Draw an Argand diagram. The length of the sides of that triangle is exactly $\sqrt 3$, and for the complex roots, $w^2-w=\pm\sqrt 3$

aarbee

15:39

@PM2Ring 1=(1,0), w=(cos60,sin60)=(1\2,root3/2). So, |1-w|=1

Is this incorrect?

PM 2Ring

16:03

@aarbee It's incorrect. The real part of both complex roots is -1/2. Their args are 120° and 240°.

Govind75

Is homomorphic a word?

If I were to say the map is homomorphic, is that correct?

Thorgott

homomorphic is a word, but not one that is used in mathematics

you would say the map is a homomorphism

Govind75

aww

so there's no such thing as a homomorphic map

Ted Shifrin

16:22

Nor a homeomorphic map.

Ted Shifrin

16:35

@aarbee This is wrong. Just compute directly.

robjohn

@aarbee If I'm not mistaken, it is equal to $\sqrt3$, not less than.

$\left|e^{2\pi i/3}-e^{4\pi i/3}\right|=\left|e^{\pi i}\right|\,\left|e^{-\pi i/3}-e^{\pi i/3}\right|$ do you recognize anything familiar about the difference on the right?

copper.hat

16:55

back to ace of base again with 'the $\sin$'.

aarbee

Thank you.

Hawk

In Topology, one defines a set A \subset X is relatively compact if the closure cl(A) is compact. Why is this called "relative"? It is not in parallel with other "relative" definitions in topology (e.g. subspace topology)

Thorgott

I much prefer the term precompact

Hawk

that's another name as well and makes a little bit more sense since precompact implies compact. But I just want to know if there was a motivation for using relative here that someone decided it was appropriate

Thorgott

I'll hand @leslietownes the responsibility of explaining this

I think the terminology comes from functional analysts

Simone

17:14

Hello.

Hawk

@leslietownes if you know, please @ me. Thank you.

Simone

I'm staking ALGO, ususally the function describing compounding with $a$ being the interests and $I$ the initial investment is $I( 1 + ax)^x$ after one year, however I pay $0.001$ ALGO each time I withdraw, so the new function should be $I( 1 + ax -0.001x)^x$ right?

copper.hat

@Hawk there is always hsm.stackexchange.com

Simone

17:31

sorry, the function is $I(1+ \frac{a}{100x} - 0.001x)^x$ where the interest is $a$percent

Hawk

@copper.hat oh history? I am seeking a mathematical reason, but history is secondary.

copper.hat

It is historical terminology, so both.

Alessandro Codenotti

I think it's "relatively" in the sense that it is not an intrisic property of A but rather a property of the embedding of A in X

Hawk

@AlessandroCodenotti hmm can you elaborate on that?

Alessandro Codenotti

Being relatively compact is not a topological invariant

It's not a property of the topological space A, it is a property of A embedded in a particular way into X

As a concrete example take $A=\Bbb N$, there are embeddings $A\to\Bbb R$ whose image is relatively compact, and embeddings whose image isn't

Ted Shifrin

17:46

Hint: $(0,1)$ is homeomorphic to $\Bbb R$.

That was a separate hint, but could apply …

Hawk

Relative to the embedding...thats a little hard for me to understand. Don't we just use inclusion for most simple cases instead of thinking about some other crazy embedding?

Alessandro Codenotti

Sure, but you can have two subspaces A and B of X which are homeomorphic while one is relatively compact in X and one isn't

Thorgott

yes, but that's not intrinsic either

Ted Shifrin

That's the point?

Thorgott

I was replying to Hawk

Alessandros message hadn't even loaded on my end when I sent mine

Hawk

18:04

@AlessandroCodenotti may I ask what embeddings did you have in mind in your example with $\Bbb N \hookrightarrow \Bbb R$?

Alessandro Codenotti

the usual one and the one sending n to 1/n

Ted Shifrin

@Hawk For my hint, embed $(0,1)$ in itself or in $\Bbb R$.

Hawk

@TedShifrin oh sorry i thought you were talking to someone else this whole time

okay i see what both examples mean now. But let me come to terms with if this really was original intention

Because this was quite the stretch in speculation initially.

hmm actually hang on. @TedShifrin this example embeds $(0,1) \hookrightarrow (0,1)$ and $(0,1) \backarrow R$, are you not changing the universal space

Thorgott

that's the point

Alessandro Codenotti

He meant embed into R with identity and embed into R with a surjective map

Hawk

18:17

hang on if u embed (0,1) into R with identity, we have relatively compactness, but if embed by a surjective map (0,1) \to X \subset R, he is saying there is no relative compactness? What is such a map?

Alessandro Codenotti

You can write down one explicitely with a tangent and some rescaling of the interval

Hawk

tangent? By scaling, like x \to 2x? how does that give no relative compactness?

Alessandro Codenotti

Because R is not compact

Like x\to tan(2\pi x) or something close

Any homeomorphism (0,1)\to R works here

Hawk

no but the closure of (0,1) under 2x is [0,2] that's compact

oh that's what u meant by tangent

hang ong, isn't the closure still [0.1]?

under tan(2pi*x)

Ted Shifrin

18:37

@AlessandroCodenotti no I didn't.

Include into $\Bbb R$ and you get compact closure.

Oh, compared to just embedding in itself by the identity.

I'm granting that the interval and $\Bbb R$ are homeomorphic.

Astyx

18:55

@Hawk What's the range of $\tan(2\pi x-\pi)$ when $x\in]0,1[$ ?

Hawk

oh the range, it would be R then.

i was computing the domain for no reason

Xander Henderson

20:33

I am writing an exam in which I introduce my students to the notion of a symmetric difference, and ask them to find the symmetric difference of two small sets. However, I don't want to use the term "symmetric difference", as this is very easily Googled. Anyone have any suggestions for a made-up term which means the same thing as the symmetric difference?

Ted Shifrin

From Dr Dolittle: the pushmi-pullyu :)

Xander Henderson

Hah!

I loved those books as kids.

Ted Shifrin

It sorta fits!

Xander Henderson

It does.

It sounds a little too non-mathy for my taste, but if I can't come up with anything else, it may work.

Ted Shifrin

I like literary allusions

hyper-neutrino

20:40

remind me; symmetric difference of A and B is {a in A | a not in B} union {b in B | b not in A} ?

Xander Henderson

$(A\cup B) \setminus (A \cap B)$.

So... yes.... what you wrote.

Ted Shifrin

$A-B \cup B-A$

Xander Henderson

It is all of the stuff in either $A$ or $B$, but not both.

Ted Shifrin

Whence the term.

Xander Henderson

@TedShifrin That, too.

hyper-neutrino

20:42

well basically just something is in the symmetric diff IFF it's in A xor in B ¯\_(ツ)_/¯

my math notation is not so great; i use coding notation sometimes and that's not always accurate

Xander Henderson

@hyper-neutrino Yes, that is exactly what it is, if you want to be computer sciency about it.

hyper-neutrino

:P

Ted Shifrin

@hyper That's the interpretation of what Xander wrote.

hyper-neutrino

i mean i could think of more roundabout ways to write it probably :P but hmm, lemme see if i can think of any other wording that'd make sense but not be so googlable

Xander Henderson

The "distinguishing set" (as is it is the collection of stuff which makes the two sets $A$ and $B$ different)?

Ted Shifrin

20:45

That's not bad.

hyper-neutrino

hm, I think that's quite fitting

I might've also suggested something like "the distinct items" or smth, but I think yours makes more sense

copper.hat

21:25

Elements that are exclusively in one set or the other?

Xander Henderson

21:42

@copper.hat I was looking for something short and pithy and not super duper Google-able.

I settled on the "distinguishing set", and decided on the notation $A \ddagger B$.

Just to be annoying.

Ted Shifrin

May the pushmi-pullyu live on …

Xander Henderson

(One of the first assignments that I give to students in calculus is a worksheet where I ask them to read a bunch of definitions, and give them some exercises about understanding definitions; e.g. as planar set is convex if... give an example of a convex set; give an example of set which is not convex; is [this set] convex?, etc).

PM 2Ring

It's the complement of the intersection (relative to the union), so you could call it the outersection. ;) FWIW, Python overloads its exclusive-or operator ^ for the symmetric difference.

hyper-neutrino

I mean, a set is just as much a collection of elements as a boolean function from the universal set, so it makes sense to overload boolean/bitwise operators to mean similar things when applied to the membership property :P

it is quite nice - i really like that syntax

very good for code golf too in many cases

Xander Henderson

@PM2Ring Heh... I kind of like outersection.

That's cute.

PM 2Ring

21:55

Oh yes. Although in non-golf contexts I tend to prefer using the method syntax rather than the operator syntax, especially for the less common set operations. And the set methods take any (valid) iterable as an arg, whereas if you use the operator syntax, both operands must be sets.

@XanderHenderson Thanks. :) I thought of it a few years ago, but I don't think I've ever mentioned it in public before.

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$Mathematics$ Mathematics