Mathematics - 2020-11-08 (page 2 of 2)

Shaun

14:24

in Constructive Feedback, 48 secs ago, by Shaun

Would someone confirm my answer below, please? I'm having second thoughts on how I show the order of $xy^{-1}$ divides $m$.

1

A: Question on proofs for order of elements in a non-abelian group

See my comments above for 1 and 2. As for 3, I will use the one-step subgroup test. Fix $m\in\Bbb N$. Since $e\in H$, ${\rm ord}(e)=1$ and $1\mid m$, we have $e\in Q_m$. Thus $Q_m$ is nonempty. By definition, $Q_m=\{\color{red}{x\in H}\; :\; {\rm ord}(x)\mid m\},$ so we have $Q_m\subseteq H$. Let...

epic_math

14:38

Where is mathguy?

Thorgott

$o(x)|m$ and $o(y)|m$, hence $\operatorname{lcm}(o(x),o(y))|m$. $(xy^{-1})^{\operatorname{lcm}(o(x),o(y))}=x^{\operatorname{lcm}(o(x),o(y))}(y^{\operatorname{lcm}(o(x),o(y))})^{-1}$ since the group is abelian. $o(x)$ and $o(y)$ respectively divide $\operatorname{lcm}(o(x),o(y))$, so $x^{\operatorname{lcm}(o(x),o(y)}=1=y^{\operatorname{lcm}(o(x),o(y)}$. Thus, $o(xy^{-1})|\operatorname{lcm}(o(x),o(y))|m$.

Edward Evans

F

RewCie

Fu

Edward Evans

you're missing a } on the exponent of $x$ and $y$

nise

RewCie

knise

epic_math

14:41

*nice

Thorgott

phew

epic_math

Sigh

Edward Evans

omg thanks guys, I didn't know how to spell nice

RewCie

BOOOOOOOOM

Thorgott

british person owned

RewCie

14:42

Yoo r belcom

epic_math

Helkome

Edward Evans

with facts and logic

Hi @Thorgott btw

epic_math

Is ma plusure

RewCie

phackts and lugik

epic_math

Na oone greeted me

Thorgott

14:42

noooooooo

RewCie

mah Comtupar Engynar

epic_math

Kalkulus

Thorgott

I was a couple seconds to late to edit a missing bracket in

Astyx

wth is happening to this chat

epic_math

@Astyx we are talcing

Edward Evans

14:43

idk

Thorgott

and hi

RewCie

kulukus

Comtupar Engynar invaded

hyyy

user2103480

@EdwardEvans destroyed by facts and logic AGAIN

epic_math

Mathgematiks

user2103480

fucking hell you wrote that already

RewCie

14:43

metaceieans

epic_math

I like numba thoory

Someone is missing in this room today

Edward Evans

Is it your alt account?

epic_math

I have only one account btw

Edward Evans

okay

epic_math

I don't know why you people say that I have a second account

RewCie

14:46

@EdwardEvans are you comtupar engynar?

Edward Evans

why are you writing like that

RewCie

Fun, actually!

epic_math

He is mispelting

RewCie

The scerendipty of gobbledygooky words is fun.

I'm feeling sooooo cold!

Even Hoodies aren't warming well....

Probably need to take out overcoat.

Shaun

Thank you, @Thorgott :)

epic_math

14:59

@RewCie are you a full time poet?

RewCie

Yes

epic_math

That's fInE

GiVe Me SoMe MoRe PoEtRy

RewCie

Name is enough

Thorgott

np

RewCie

P vs NP

ArtOfCode

15:04

25 messages deleted

Friendly reminder, folks: If you're going to talk Politics, you must respect those who disagree.

6

epic_math

I had told geocalc a joke I want to say it again. Can I?

RewCie

Where was politics actually?

epic_math

Why did the mathematician name his dog cauchy?

RewCie

Because he didn't like to name it X AE D-278

epic_math

Because he left a residue in every pole

Lol

Btw he wasn't elon musk

NoW sTaR iT fAsT bOyS

Edward Evans

15:11

wow ignoring them removed a LOT of messages

Astyx

25

Edward Evans

lol

epic_math

Ya sometimes you should ignore people.

RewCie

Ya, just as everyone ignores me here.

epic_math

"Weak people take revenge, strong people forgive, and intelligent people ignore."-albert Einstein

@RewCie no worry, I would never ignore you

RewCie

15:14

Apparently someone laughed at my Twitter Joke

Thorgott

what's a topological space $X$ such that every continuous map $\mathbb{R}\rightarrow X$ is constant, but $X$ is not totally disconnected

RewCie

Assume Topological space be a function of $f: \mathbb R \rightarrow X$

Mike Miller

@Thorgott R with the finite/countable complement topology

Alessandro Codenotti

cofinite topology is path connected

Mike Miller

Yeah I meant to just so countable complement

Was being sloppy and not remembering the argument

RewCie

15:21

such that for every $x \in \mathbb R, $ $f(x) = \vec \mu$ which is same for all $X$

My room got first compliment!

I'm happyez! :-)

in Computer Science noob tries to get better (ft. RewCie), 1 min ago, by skullpatrol

Nice room @RewCie cya pal

Com and join here:

Computer Science noob tries to get be…

Computer Science Engineering Discussions and Software/Hardware...

feynhat

@AlessandroCodenotti Can you kick people?

geocalc33

good morning

RewCie

Morning :-)

user2103480

So apparently, the pointwise limit (in a metric space) of any sequence of measurable function with separable image is measurable with separable image

Ah sorry, in the preceding it should be a Banach space, not a metric space. Anyways, since simple functions have a finite (i.e. separable) image, this immediately implies that any borel-measurable function (with values in R^d) has a separable image, right?

Alessandro Codenotti

@feynhat Only temporarily, I don't remember for how long or how it works exactly

RewCie

15:33

It first kicks them for 30 second, then 1 min, 2 min, 5 min, 15 min, 30 min, 1 hour, 2 hours only

love_sodam

Professor said that if a is a root of x^6+3\in Q[x], then Q(a)/Q is Galois. Is this true?

I found that \sqrt[6]{3}e^{(2k+1)/6\pi i} where $k = 0,1,...,5$ are roots of them

But if k = 1 then It seems that x^6+3 does not split on Q(a)

a = \sqrt[6]{3}e^{3/6\pi i}

Thorgott

Ok, that's connected cause any two open sets intersect by uncountability. The only convergent sequences in that space are eventually constant ones (otherwise there are infinitely many elements not equal to the limit and their complement is an open neighborhood of the limit that the sequence isn't eventually in).
Assume $f$ is not constant in any neighborhood of a point $x$. Then there is a sequence $x_n->x$ such that $f(x_n)\neq f(x)$ for all $n$, contradiction to continuity. So $f$ is locally constant, hence constant.

ok, thanks

geocalc33

If two 2-manifolds M=N are diffeomorphic to $\Bbb R^2$ (equivalent metrics) what can be said about $M \times N?$

Thorgott

I feel like I've hit a new record for ugly topology questions asked in the span of 24 hours

geocalc33

$M \times N$ will be a 4-manifold. But I'm hesitant to say that it will related in any way to $\Bbb R^4$

Thorgott

15:45

if $M\cong M^{\prime}$ and $N\cong N^{\prime}$, then $M\times N\cong M^{\prime}\times N^{\prime}$

geocalc33

oh sorry you told me that before

user2103480

@Thorgott maybe your own record, but I'm sure @AlessandroCodenotti asks himself more of those on any day

Thorgott

poor him

user2103480

I feel you though, all I'm doing atm is asking "is XY measurable??"

Mike Miller

@Thorgott I don't like this argument, why can you talk about continuity sequentially here

I guess your point is composite of continuous maps is continuous

I would probably have argued by observing that $f$ has countable image because $\Bbb R$ is separable ($f^{-1}(f(\Bbb Q))$ is a closed set containing $\Bbb Q$, hence $\Bbb R$) and then observing that a countable subset of $\Bbb R_{cc}$ is discrete, so $f$ must be constant

Alessandro Codenotti

15:58

@user2103480 Nah I only work with Polish spaces, no ugly stuff

Thorgott

continuity always implies sequential continuity

user2103480

@AlessandroCodenotti nice

reasonable assumptions gang gang

Thorgott

but I prefer your argument

looking at $f^{-1}(f(\mathbb{Q}))$ is nice

lmao, someone I know apparently managed to convince a prof at our uni to publish the following exercise sheet

Alessandro Codenotti

@user2103480 Well actually I look at Polish groups, but an object associated to them that I care about is often nonmetrizable so rip anyway

RewCie

You people are speaking which language? I'm unable to understand a word of it.

Thorgott

16:04

it's german, but what it actually says doesn't matter

user2103480

@AlessandroCodenotti I still find it nice that you found a cool subject in the intersection of general topology, descriptive set theory and group theory

@RewCie we're inventing terms on the fly

Just gotta say smart things like hydroxychloroquine or remdesivir, like a very very smart individual - maybe the smartest ever - would

geocalc33

locostatic epigenetics

love_sodam

isn't there anyone answer my question

?

Balarka Sen

16:19

@TedShifrin I was wondering, is it somehow true that the unit tangent field to the Hopf fibration doesn't occur as a magnetic field? I don't know any physics, but in the Maxwell stuff, the magnetic field $\mathbf{B}$ is a 2-form (thought as a vector under musical isomorphism) which has a "vector potential", i.e., $\mathbf{B} = \text{curl}\, \mathbf{A}$ where $\mathbf{A}$ is a 1-form (again thought as a vector).

Then $\int_{S^3} \mathbf{A} \wedge d\mathbf{A}$ computes the linking number. Do you know if this could be zero?

it feels like magnetic lines shouldn't link but i dont know how to explain that

I guess I'm also having trouble with more basic things. If $f : S^3 \to S^2$ is the Hopf map I would like to study if the unit tangent field to the fibers of $f$ gives rise to the vector field $\mathbf{B}$. Is this the same question as asking if $\mathbf{B}$, as a 2-form under the musical isom, the same as $f^* \text{vol}_{S^2}$? I guess

Because transverse to the fibers, sort of.

Mike Miller

@Thorgott Yeah I recognized after a moment, it's just the statement that composite of continuous is continuous

I just don't like to think about sequences in stupid spaces

Balarka Sen

16:35

But then I suppose there is no obstruction because any zero divergence field occurs as a magnetic field. I'd like to know what arrangement of magnets give rise to the Hopf fibration, then.

user2103480

Statements about metric spaces that are topological in nature extend to metrizable spaces, don't they? For example, any subset of a separable metric space is separable, and I'd guess this extends to subsets of separable metrizable spaces

Mike Miller

Yes

More or less tautologically

user2103480

Since the topology is unchanged

ok thx

Balarka Sen

just be careful about completeness because thats not a topological property

that always annoyed me

(this is why people use uniform structures)

Mike Miller

I don't like that statement

user2103480

16:38

what's a uniform structure?

Mike Miller

A topological property should be something stated in the language of topological spaces, and completeness is not a property with a truth-value you assign to a topospace

But rather to a metric space

The natural topological property is completely metrizable

user2103480

@MikeMiller that would've been my next question

Balarka Sen

@MikeMiller yeah

Mike Miller

I actually think it's very hard to come up with properties of topological spaces that are not topological properties

Balarka Sen

thats fair but i mean i think you run into real issues because homeomorphic metric spaces with one of them being complete doesnt imply the other one is

im trying to come up with a good example lets see

Mike Miller

16:40

A good example of a real issue you mean

Balarka Sen

yeah

Thorgott

just goes to show that continuous maps aren't the right choice of morphism in the category of metric spaces

user2103480

@Thorgott shush

Mike Miller

Algebra brain

2

Go tell Banach space theorists that they should only study isometric Banach spaces because short maps are the right morphism in that category

user2103480

@MikeMiller in my functional analysis class last semester I was actually quite confused what the right notion of morphism should be, since both linear homeomorphisms and linear isometries came up

Mike Miller

16:46

Algebra brain

2

You use whatever is interesting at the moment

Thorgott

short maps are good, because they make the unit ball functor reflect isos

user2103480

Yeah, no use putting or losing structure where it isn't needed

Mike Miller

Nobody cares about the unit ball functor, what is wrong with your brain

user2103480

@MikeMiller Algebra brain

2

Mike Miller

It's a mild amusement that Banach spaces with short maps have free Banach spaces (corresp to forgetful map to unit ball)

Thorgott

16:48

it's representable by R

clearly you care about R

feynhat

Hello people

Balarka Sen

I cannot formulate my example but here is what I had in mind. Suppose $X$ is a noncompact metric space and I want lots of proper function on it; I embed $X$ inside some complete metric space $Y$ as an open subspace and take a point $\infty$ on the frotiner of $X \subset Y$, and look at $1/d(x, \infty) : X \to \Bbb R$

user2103480

@MikeMiller how exactly is that meant

feynhat

@AlessandroCodenotti Know your powers man.

user2103480

Taking a set X and forming first the linear combinations, then taking the completion?

Balarka Sen

16:49

This prescription gives lots of proper functions but I don't know how to do this stuff without metric

Mike Miller

You have to tell me the metric to define the completion

Balarka Sen

Sorry if bad example

Alessandro Codenotti

@feynhat I never needed to use them luckily

Also boundedness and total boundedness are metric properties, not topological ones

Mike Miller

But the free Banach space on a set $X$ is $\ell^\infty(X)$ (in the category of short maps with forgetful map the unit ball as a set)

Alessandro Codenotti

Fun fact: for a metrizable space being separable is equivalent to admitting a totally bounded metric

Mike Miller

16:50

Or something like this

Maybe I mean $\ell^1$

I don't really care

Balarka Sen

algebra brain

2

"free banach space" jesus dude

Mike Miller

Exactly

[Reading Thorgott talk about the geometry of Banach spaces]

user2103480

@MikeMiller ah shoot true

I'll just go ahead and invent a metric using X as an orthogonal basis

@MikeMiller lmao

And since we're all about the ugly stuff, I'm talking hamel basis

feynhat

Is there a simply connected space which is not CW?

user2103480

are CW always deloopable?

Mike Miller

17:00

@feynhat Bruh

Why would every simply connected space be CW

Take like the 2-dim version of the Hawaiian earrings eg

Or the cone on the Cantor set

feynhat

oh yeah. Right. bunch of S^2's with decreasing radii

Mike Miller

Every CW cpx is locally contractible

Fuck it take Hatcher Ex 0.6's picture

user2103480

@MikeMiller do you know that book by heart wtf

feynhat

Yes he does.

Mike Miller

I just assigned that exercise lol

Thorgott

17:03

every mathematician should memorize all of Hatcher once in a while

Balarka Sen

I can give examples of contractible spaces which does not have homotopy type of a CW complex but why will I, is the point.

Obviously such things exist

Because there's no reason for it not to

Thorgott

brilliant

Let $U$ be an open subset of some Euclidean space. I want to argue there exist an unbounded real-valued $C^1$-function on $U$. $x\mapsto1/d(x,\partial U)$ works continuously, but I believe it's generally not $C^1$. How do I get something smooth? Visually, it's obviously possible.

Balarka Sen

d(x, del U) is smooth away from del U right

and inverse of nonzero smooth function is smooth

feynhat

2d Hawaiian earring because is a nice counter example because it is semi locally simply connected but not CW.

Thorgott

take $U$ to be the open set above the graph of the absolute value function, then $d((y,0),\partial U)=|y|/\sqrt{2}$

unless you mean like outside a neighborhood of by away from

user2103480

17:16

@Thorgott maybe you can integrate in a smart way, if it's a star domain

Balarka Sen

im not gonna think about your ctrex but there is a smooth function on R^n vanishing on any closed set

just take that if d doesnt work

this is an exercise in Guillemin Pollack so you should do it

user2103480

e^-x^2 something something

Mike Miller

If you just want unbounded and not proper just take y in the boundary and do 1/d(x, y) lol

Thorgott

I... hadn't thought of that

Balarka Sen

algebra brain

2

Thorgott

17:22

you still might not get smoothness in a neighborhood of y

but take y to be in the interior of the complement

Mike Miller

That's not unbounded on U

Balarka Sen

what

what crap

Thorgott

wait no

Mike Miller

You absolutely do get smoothness in a nbhd of y in R^n, sqrt(x_1^2 + ... + x_n^2) is smooth away from 0

Just not at y

But who cares about y

Balarka Sen

smoothness on an nbhd of y makes no sense

smoothness of 1/x at 0 lmao

Thorgott

17:24

youre right

Balarka Sen

anyway but d(x, A) probably isnt smooth away from A for an arbitrary closed set of R^n but not because of what you told me. its only loc. lipschitz but idk example

Thorgott

@BalarkaSen corollary of whitney ez

feynhat

@MikeMiller What is a 'Graduate Course'?

(serious question)

Mike Miller

??

Thorgott

a course for graduate students?

Mike Miller

17:31

A course whose primary audience is graduate students?

Thorgott

sniped

feynhat

I see.

Balarka Sen

(Take $A = \partial [0, 1]^2$, consider distance to $A$. Not smooth)

Is there a simple argument that the open Mobius strip cannot be properly embedded in $\Bbb R^3$? No smooth stuff allowed no Alexander duality allowed

Guess not because one point compactifying makes it a problem about embedding RP^2 in S^3

Which is the same as RP^2 in R^3

Which is hard

Rainb

what is smooth stuff?

Balarka Sen

differential topology i guess

Thorgott

17:42

that's fundamentally the same as what I gave, no

one corner of the square looks like the corner of the graph of the absolute value function

Balarka Sen

oh maybe i wasn't following yeah

you get problem at corners

but then the problem doesnt go away outside an nbhd of the square either it persists along the diagonals

so i was confused

Thorgott

ok, I'm gonna ask a very silly question

how do I argue the boundary of an open subset of Euclidean space can't be a singleton

Balarka Sen

it can be take R^2 - 0

Thorgott

wait I'm fucking stupid

jfc

Balarka Sen

nah top boundary is confusing as shit

Q in R has bd R

wild

Thorgott

17:55

ok, but say I have two disjoint non-empty open sets

I can find two distinct points, each belonging to the boundary of one of those sets respectively, right

Mike Miller

do you need this

Balarka Sen

youre saying theres no example of open sets U, V of R^n, with U and V disjoint such that bd U = bd V = {x}?

Thorgott

ye

to both

Balarka Sen

take (-infty, 0) and (0, infty)

doesnt happen for n > 1

because R^n \ x connected

@MikeMiller he's giving you content man why do you want him to stop

topology 101 homework problems

list em down quick

Mike Miller

we're in the fundamental group part of my course now i dont ever need to go back to this junk

Balarka Sen

18:00

final exam

and what if one of your student decides to do a project on lakes of wada

Fs in chats

Mike Miller

someone would have to tell them about it first

Balarka Sen

i have a subliminal message written on the RP2 thing

youre screwed

Mike Miller

Oof

Bet nobody read that

feynhat

@Thorgott How is this not smooth though?

Balarka Sen

lmao yeah

Thorgott

18:03

whatever, smoothness is a spook

I'm thinking about BOUNDARIES now

Balarka Sen

lol

cant stop laughing

Thorgott

ah, I get it, if $\partial U=\{x\}$, then $U$ is clopen in $\mathbb{R}^n\setminus\{x\}$, hence equal thereto

so that's literally the only example

Balarka Sen

ye

geocalc33

I have a calculus question

nvm

If $\Bbb R^1$ has a change of coordinates $x \mapsto e^x$ then define the manifold $\Bbb G^1$ to be the manifold with new coordinates $u=e^x.$ What is the product manifold $\Bbb G^1 \times \Bbb G^1$?

Thorgott

18:23

(0,\infty)^2

geocalc33

yes that's correct!

but where do you define the origin? (0,0) or (1,1)

Thorgott

whats origin

geocalc33

the origin lol

it's the point of reference

the origin of R^2 is (0,0) don't you agree?

Thorgott

the Wolverine movie?

Balarka Sen

lol

geocalc33

18:27

I can't

I've never seen the Wolverine movie actually.

Thorgott

rid yourself of the shackles of origin normativity

henceforth, all spaces shall be affine

geocalc33

but suppose one wishes to furnish the space with a metric

then it matters where the origin is

Tobias Kildetoft

@geocalc33 No, it doesn't

geocalc33

So I've calculated the metric in $(u,v)$ coordinates

Tobias Kildetoft

metrics don't care or know about origins

geocalc33

18:32

with origin (0,0)

@TobiasKildetoft so I can define the origin to be (1,1) if I so choose?

anyway the metric I got in $u-v$ was of course $ds^2=\frac{1}{u^2}du^2 + \frac{1}{v^2}dv^2$

which actually reminded me of the hyperbolic metric actually

but I digress.

okay figured it all out. (1,1) has to be the origin because, (0,0) is not even included in the manifold

derp.

Thorgott

I think the origin should be (\infty,\infty)

Tobias Kildetoft

@geocalc33 What does "origin" even mean for a manifold?

geocalc33

it's the reference point

or so I've been told

Tobias Kildetoft

reference for what?

geocalc33

so you can purposely pick the origin to simplify the mathematics

any origin will work, but oftentimes it's easier to choose a convienient one

Thorgott

18:47

it's probably more convenient to work with a probabilistic origin

geocalc33

you bring up a good point. maybe for some problems yes

Alessandro Codenotti

@Thorgott Yes but if you want them in the symmetric difference of the boundaries then no

Ah but that's obvious, $\Bbb R^2$ minus a line

I was thinking about lakes of Wada, but there's no need for such weird examples here

Thorgott

yeah

Sayan Chattopadhyay

19:45

Suppose $V \subseteq k^n$ is an algebraic irreducible set. We have that the coordinate ring of $V$, $\Gamma(V)$ is a finite dimensional vector space over $k$. I want to show that $dim(V) = 0$. I was thinking that if I can show that $\operatorname{trdeg}(k(X)/k) = 0$, where $k(X)$ is the fraction field $\Gamma(X)$, then I am done. For this I have to show that for any $a/b \in k(X)$, it satisfies some polynomial in $k[X]$. Maybe I have to use that it is a finite dimensional vector space somehow.

It's not clear to me as to how

Tobias Kildetoft

yes, you surely need to use finite dimensionality, since that it the only assumption you have that matters

Sayan Chattopadhyay

Yeah but the way that I am proceeding, I don't seem to find a way to use finite dimensionality

And the hint in the question suggests that I need to specifically use the fact that $V$ is irreducible. The only way I could think of was transcendence degree, because I don't see anything apparent in the Krull dimension way of defining dimension which would use irreducibility

Thorgott

20:05

are star-shaped sets simply connected?

Alessandro Codenotti

They are contractible

Rithaniel

Trying to prove that a $R$ is a Prüfer domain if and only if any overring of $R$ is integrally closed. Not sure how to start.

Thorgott

ah, of course

Rainb

if $\mathcal A$ is a $\sigma$-algebra over $\Omega$ is $\sigma(\{\Omega\})=\{\emptyset,\Omega\} $ or $ \mathcal P (\Omega)$ ?

Alessandro Codenotti

What do you think?

Rainb

20:19

$\{\emptyset,\Omega\}$

:) ?

Alessandro Codenotti

Why?

Rainb

because it's the smallest one containing it?

Alessandro Codenotti

yeah that's right

Rainb

yeah I was just confused for a lil bit

because it's the same one as if you feed the empty set

so just needed to make sure

geocalc33

Piggy backing off of my previous question...Consider $\Bbb R^n$ with coordinates $\eta=(x^1,...,x^n).$ Another manifold $\zeta^n$ has coordinates $\Psi=(e^{x^1},...,e^{x^n}).$ And it's clear that $\Bbb R^n \cong \zeta^n$

now $\Bbb R^n$ can be identified with $\Bbb C^{n-2}.$

what can $\zeta^n$ be identified with? Still $\Bbb C^{n-2}$?

Tobias Kildetoft

20:31

@SayanChattopadhyay Ahh, right. You need to use that since it is irreducible, the coordinate algebra is an integral domain. So it is a domain which is finite dimensional over a field.

geocalc33

sorry I meant $\Bbb C$ can be identified with $\Bbb R^2$

by $(a+bi)\mapsto (a,b)$

Sayan Chattopadhyay

Yeah @TobiasKildetoft, any hint as to what I might do next?

Tobias Kildetoft

@SayanChattopadhyay Then you apply the well-known result that a finite dimensional domain is itself a field

Rithaniel

20:52

Anyone know how to show that any overring of a Prufer domain is an intersection of localizations of that ring?

robjohn

21:06

@Thorgott hard to determine the direction to $(0,0)$ from there...

or $(1,1)$

hmm... perhaps not, using that metric.

Sayan Chattopadhyay

@TobiasKildetoft Right yes. So the $\Gamma(X)$ is a field, and its a finite dimensional vector space of $k$. So it is a finite field extension, hence algebraic. But all throughout we have had the assumption that $k$ is algebraically closed. So then does this imply $\Gamma(X) = k$?

I have a feeling I am being amazingly stupid

Thorgott

21:22

direction is an archaic concept, we shall abolish it

geocalc33

22:10

What is this? $w=(e^a+ie^b)$

for $a,b \in \Bbb R$

guess it's a complex number

user2103480

22:30

what's the weakest condition for singletons to be measurable in the borel sigma-algebra? hausdorffness?

Ah yeah should be at the least, since closed sets are measurable

okay, that condition is just the T_1 separation axiom

Thorgott

22:50

is it not possible that singletons are measurable, but not necessarily closed

Alessandro Codenotti

projecteuclid.org/download/pdf_1/euclid.pjm/1102785960 @user2103480

Thorgott

interesting

user2103480

23:28

thanks!

@Thorgott fair point, thanks as well!

amWhy

23:44

Next challenge, @robjohn (you always seem to rise to the occasion), for Thanksgiving, transform yourself into a Turkey, or wild Turkey, or Angry Bird, or Mean Tom (Turkey). The morale of this site is depending on you! That is your mission. This tape will expire upon reading this!

^^^^ @robjohn I realize I am asking a lot, but accept the challenge to rise to the occasion in the way you see fitting!

geocalc33

23:56

example of a diffeomorphism $f:\Bbb C\to \Bbb C^+$?

Computer Science noob tries to get be…

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$Mathematics$ Mathematics