Mathematics - 2020-11-08 (page 1 of 2)

Thorgott

00:35

Does every topological manifold admit a metric such that Heine-Borel holds?

anakhro

R. metric?

Thorgott

metric as in distance function

anakhro

Does every metric that induces a topology identical to that of the topological manifold come also from a Riemannian metric?

Alessandro Codenotti

Does something go wrong with just embedding in R^n and using the induced metric?

Thorgott

I'd also be curious in the smooth case. Hopf-Rinow says that a Riemannian manifold has the Heine-Borel property iff it is complete wrt to the distance metric induced by the Riemannian metric. It's also true that every manifold is completely metrizable. But is every manifold completely metrizable by a distance metric that is induced by a Riemannian metric?

@anakhro apparently the taxicab metric doesn't come from a Riemannian metric, because it has infinitely many geodesics in artbirarily small neighborhoods

anakhro

00:42

!

Thorgott

@Alessandro I believe you need an embedding with closed image

anakhro

Yeah, isn't (0,1) a counter example?

Thorgott

yeah

anakhro

Obviously embeds in R.

Relatively closed in itself, and bounded.

Thorgott

yeah, but if you embed it in R as R (via some variation of tan), that works

Alessandro Codenotti

00:45

@Thorgott you can get that in the smooth case. I'd need to think about the topological one though

Thorgott

00:56

how do I get it in the smooth case

Mike Miller

You need a single proper map and then you do the usual argument to construct a proper embedding

You just need a complete metric whose closed balls are compact so indeed a proper embedding into R^n suffices

(AKA a closed embedding)

Look at the proof in Bredon I doubt the smoothness is crucial.

Thorgott

do topological manifolds admit continuous partitions of unity

same construction as in the smooth case should go through, right

Mike Miller

Yeah

X Hausdorff => (X paracompact iff X has continuous partitions of unity)

Thorgott

ah right, I should probably look at a proof of that in full generality some day, just so I never have to touch it again

Mike Miller

01:12

The hard part is that paracompact Hausdorff spaces are normal

I don't remember the proof

CroCo

What are the eigenvectors of A=[0 1;0 0]? The book I'm reading says it has two eigenvectors. It seems it has only one eigenvector. I've double checked an online calculator, it states it has only one eigenvector.

Alessandro Codenotti

It's similar to the proof of compact Hausdorff implies normal, but you have to suffer a bit more because you get locally finite refinements instead of finite subcovers to work with

Thorgott

topology is beautiful, eh

Alessandro Codenotti

You've been blinded by category theory

Thorgott

@CroCo you should specify over which field

CroCo

01:21

@Thorgott real values.

Thorgott

there are infinitely many eigenvectors

CroCo

@Thorgott ha

Mike Miller

@CroCo What you're looking at is the distinction between eigenspace and generalized eigenspace

Here the 0-eigenspace is 1-dimensional, spanned by (0,1)

Thorgott

you're probably meaning to ask about the dimension of the Eigenspace

Mike Miller

The generalized lambda-eigenspace is the set of vectors so that iterating A-lambda I kills them off eventually

That is, it's the set of v in the kernel of (A - lambda I)^k for some k

For your specific case note that A^2 is the zero matrix

CroCo

01:24

Mike Miller

So the generalized 0-eigenspace is 2-dimensional, spanned by (1,0) and (0,1)

CroCo

This is the answer provided by the author

Mike Miller

Note that A^2(0,1) = A(1,0) = (0,0)

CroCo

How come its eigenvectors are [1 0], [0 1]

Mike Miller

When you look at the characteristic polynomial, lambda is a zero of order k when the generalized lambda-eigenspace is k-dimensional

Here your characteristic polynomial is lambda^2, so 0 has a generalized eigenspace of dimension 2

CroCo

01:26

@MikeMiller is this a special case?

Mike Miller

But the 0-eigenspace itself is 1-dimensional

Sigh

CroCo

@MikeMiller are we able to diagonalize this matrix, assuming P is a matrix whose columns are the eigenvectors of A? if I do so, I don't get A matrix.

Mike Miller

You can only diagonalize a matrix if its generalized eigenspace matches with its eigenspace

I'm certain your textbook talked about this in the context of diagonalization (or Jordan normal form, if that's covered)

CroCo

@MikeMiller the book I'm reading in linear algebra (i.e. " Elementary Linear Algebra Applications Version 11th by Anton") never mentioned generalized eigenspace. But the book I'm reading optimal state estimation by Dan mentioned this example which confused me.

Also, should we treat them equally?

Mike Miller

I checked your book

It uses the names algebraic and geometric multiplicity

The geometric multiplicity is the dimension of the eigenspace

The algebraic multiplicity is the order of the zero lambda in the characteristic polynomial (the number of factors x-lambda you can factor out)

Equivalently the algebraic multiplicity is the dimension of my "generalized eigenspace" above

A matrix is diagonalizable iff algebraic multiplicity = geometric multiplicity

Here it doesn't

CroCo

01:43

@MikeMiller yes I've read geometric and algebraic multiplicities and they are lucid to me. But when the author states A matrix is diagonalizable iff algebraic multiplicity = geometric multiplicity, he didn't use the notion of generalized eigenspace. He merely used the thing you called eigenspace even though the matrix does have two eigenvectors.

Unless the eigenvectors in generalized eigenspace are not the same as of eigenvectors in the eigenspace.

Should we treat them differently?

Mike Miller

What do you mean should we treat them differently? They are different things.

The eigenspace is the kernel of A-lambda.

The generalized eigenspace is the kernel of (A - lambda)^k for large k.

The 0-eigenspace of your matrix is spanned by (1,0). It's certainly not two-dimensional because your matrix is 2 x 2 and nonzero.

However A^2 = 0, so the kernel of A^2 is everything. Thus the whole space R^2 is the generalized 0-eigenspace.

As I outlined above, (0,1) is a generalized eigenvector because A^2(0,1) = A(1,0) = (0,0). But it is not an eigenvector because A(0,1) = (1,0) is not a multiple of (0,1).

CroCo

@MikeMiller when I say differently, I mean from geometric perspective. As I know that an eigenvector is a vector that Ax=lambda x holds. I'm asking these questions because it is the first time to me to come across the notion of the generalized eigenspace.

Also, it seems the eigenvectors of generalized eigenspace have indeed some applications which I've came across one of them to compute the fundamental matrix of a system of differential equations.

Mike Miller

They are different things geometrically too. I have said the definition above. If you want to investigate you should either explore it with examples of your own (or try examples from the book using this language) or find another textbook reference.

If they're literally different subspaces they can hardly be the same thing "geometrically". :)

CroCo

@MikeMiller got it. Thanks.

Mike Miller

02:03

The wikipedia page is not bad

CommutativeAlgebraStudent

0

Q: Elementary proof of the twin prime conjecture via generator algorithm argument. Python code inside.

This is the simplest twin prime generator I could come up with. from math import sqrt from sympy import isprime def generate_twin_primes(C:int, p:int=2, n:int=3): while n < C*p: if isprime(n): if n - p == 2: yield (p, p+2) p = n ...

Thorgott

Let $f_i\colon M\rightarrow X_i,i=1,2$ be continuous maps between top spaces and assume $f_2$ is proper. For a compact $C\subset X_1\times X_2$, $\pi_2(C)\subset X_2$ is compact as continuous image of a compact, hence $f_2^{-1}(\pi_2(C))$ is compact by properness of $f_2$. By continuity, $(f_1\times f_2)^{-1}(C)$ is closed (ok, I assume the codomains are Hausdorff), and we have $(f_1\times f_2)^{-1}(C)\subset f_2^{-1}(\pi_2(C))$, so as closed subset of a compact space, it is compact. Hence $f_1\times f_2$ is proper.

Mike Miller

What are you proving

When one map is proper the product with any other map is proper?

Thorgott

yeah

Joe Shmo

anybody know of applications of combinatorics to analysis?

Rithaniel

02:13

Well, I found this pretty fast: books.google.com/…

Joe Shmo

hah, I'm looking for applications to harmonic analysis in particular..

because this stuff screams combinatorial bounds at some points

but I was hoping someone who knows specific examples could talk about them

Rithaniel

Apparently that link has stuff to do with harmonic analysis. I don't know specifics, though

Joe Shmo

yeah, the title says applications to harmonic analysis

Mike Miller

@Thorgott Yes that's correct I worked it out myself and came up with your proof

At a glance at least

Thorgott

ok, thanks

then the subtlety I'm missing in Bredon's proof must be elsewhere

but in any case, the argument also works in the continuous case

you don't get the dimension bounds by general position arguments, obviously, but you do get that proper embedding in some large R^N

Mike Miller

02:23

@Thorgott The subtlety is in getting an embedding, because you can't cover with finitely many charts

You have to construct a smooth proper function which is one of the first places you require an analysis of paracompactness

Thorgott

you can cover with finitely many charts, but that's a harder result (at least I think so, not that I know the proof)

but yeah, I follow the general strategy

there's just one particular remark in the proof whose purpose I don't see, so I'm still missing a subtlety

will have to carefully reread

Mike Miller

What do you mean by chart?

By open subsets of R^n?

Thorgott

yes

Mike Miller

Sure

Thorgott

I think I read you can cover an n-manifold with n+1 charts

it was some topology nonsense

Mike Miller

02:27

I dunno about that but I see the point about finitely many

There's a countable increasing compact exhaustion, using existence of smooth proper map and Sard's theorem

Each intermediate bit M_n \ int(M_{n-1}) can be covered by finitely many charts, and they can be trimmed so that each chart only also intersects, say, M_{n+1} or M_{n-1}

I guess you need to see boundedness on each compact bit

Thorgott

14

A: Surface where number of coordinate charts in atlas has to be infinite

If $M$ is an $n$-dimensional manifold, then it is in particular a normal space of covering dimension $n$, and Ostrand's theorem tells us that that if we start with a locally finite open covering $\mathcal U$ of $M$ consisting of coordinate charts (such a thing exists), there is a refinement $\mat...

Mike Miller

But once you have that you can cover it with 3N charts, where N is the bound on the number of open sets covering each of your compact pieces

OK, nice

Thorgott

why does the bound N exist

Mike Miller

I dunno

Thorgott

fair enough

AttractorNotStrangeAtAll

02:56

Hi, @MikeMiller, I took the liberty to see in your profile that you work with low dimensional topology.

If I want to do my PhD in this area, at a good university or institute, which classes should I ideally take before applying?

mathguy

Hello fellas!

Mike Miller

I'm not really competent to say without a lot of background information

In the US my advice would be to just study what you're interested in, deeper than the classes themselves, and make relationships with professors by having things you're interested in past the scope of the class to talk to them about

Outside the US I don't know anything

AttractorNotStrangeAtAll

03:13

Yes, I'm outside of the US.

Ok, thanks.

Ted Shifrin

05:58

@Sayan, I don't disagree with a lot of what you said. Biden was close to my last choice when this started. But the fascism under the Republican-Tromp party was threatening to undo democracy anywhere. What I object to, and continue to be pissed off at, is Balarka's calling me a moron. I don't need that from him or from you. Thank you.

skillpatrol

06:15

epic_math

07:04

In my complex analysis book, the definition of closed set is given as follows:

So can a set be both open and closed?

Sayan Chattopadhyay

07:21

@TedShifrin Absolutely, and I don't mean to call you a moron, that would be absolutely ridiculous and immature thing for me to do. Of course a return to democratic ideals has to start somewhere. In this case it starts with kicking Trump out of office. And if I was in US and could vote, I would have filled it for Biden. I just hope that what doesn't happen now is stagnation in politics. I hope that the American public, the senate presses on Biden to actually do things, and question when he fails

Rithaniel

@epic_math Yep

epic_math

Oh. Can you give an example?

Rithaniel

$\mathbb{C}$

epic_math

Thanks

Rithaniel

No problemo

But, for non-trivial examples, you probably want to look in general topological spaces

epic_math

07:25

Ya gonna learn about them but not now

Rithaniel

Any set under the discrete metric is both open and closed, for example.

epic_math

after I finish analysis, I will study topology

epic_math

07:38

Hey can I say a joke

Several scientists were all posed the following question: "What is 2 * 2 ?"
The engineer whips out his slide rule (so it's old) and shuffles it back and forth, and finally announces "3.99".
The physicist consults his technical references, sets up the problem on his computer, and announces "it lies between 3.98 and 4.02".
The mathematician cogitates for a while, then announces: "I don't know what the answer is, but I can tell you, an answer exists!".
Philosopher smiles: "But what do you mean by 2 * 2 ?"

feynhat

Imagine a Hawaiian earring, $X$. I define an action of $\mathbb{Z}$ on $X$ as, 'translation to the bigger circle'. Precisely, a typical point on $X$ looks like $(\frac1n + \frac1n\cos\phi, \frac1n\sin\phi)$, the action of $1$ on this point takes it to $(\frac{1}{n-1} + \frac{1}{n-1}\cos\phi, \frac{1}{n-1}\sin\phi)$.

So, this moves a point to a point with the same 'angle' on the next bigger circle, and -1 moves a point back to a smaller circle.

What is the quotient space?

Is it $S^1$?

Yeah. I think it should be $S^1$.

But this is not a covering space action because the origin is a fixed point.

Balarka Sen

08:38

@TedShifrin Well yeah I didn't think you were a fan of Biden either. I don't know why you're taking issue with me calling people who are celebrating Biden's victory as stupid, because I don't think you're one of them.

I of course understand that a large proportion of reasonable people voted against Trump and not for Biden

love_sodam

In a characteristic p field, if f(x) = x^{p^k} - a^{p^k} and f can be factorized as f = gh, then if a_0 and b_0 are constant term of g and h respectively, then is it possible that a_0 or b_0 = 1 or -1?

I want to show that it's not possible but I don't know how to prove

g,h are not constant

Alessandro Codenotti

09:50

@feynhat yes

Edward Evans

10:40

"Mornin'" y'all

feynhat

Germany is UTC+what?

Edward Evans

1

epic_math

Can someone help me to prove that a A region $S$ is polygonally connected?

user2103480

10:55

@feynhat 1

Edward Evans

ty for clarifying @user2103480

feynhat

My town shares border with a country that's UTC+5:45.

Edward Evans

the fuh

user2103480

@EdwardEvans no problem, thought you might not know it

Edward Evans

rofl

user2103480

10:59

Hahaha i just didnt read properly and overlooked your one tbh

Edward Evans

Schamlos

Astyx

11:27

hi chat

Edward Evans

Hey @Astyx

user2103480

@EdwardEvans the only shameless one is my fluid dynamics prof who hasn't uploaded the exercises yet

Edward Evans

what a prick

user2103480

And told me his last course's notes are obsolete just to upload literally exactly the first part of his old notes

@EdwardEvans the people nowadays........

Edward Evans

rofl

user2103480

11:33

And the outline (the 5 topics we discuss) is also exactly the same, and the literature as well, but apparently this course will be more specialized

Which doesn't bode well for me

Edward Evans

oof

sounds like a pain

user2103480

I mean, not really

There will be lecture notes, and the prereqs are complex & stochastic analysis which is fine

Edward Evans

Alright fair hahaha, our AZT2 prof is one of the best teachers at the uni imo

user2103480

And I don't even need the crazier stochastic analysis stuff like girsanov's theorem or the martingale representation theorem, and I guess that all stochastic integrals will be with respect to brownian motions

Edward Evans

idk anything about stochastic analysis

user2103480

11:36

I don't know nearly enough and still have enough preprequisites

Whats AZT2

Ah its german

Edward Evans

hahaha fair, the main prereq for algebraische Zahlentheorie 2 is group cohomology

user2103480

Algebraische Zahlentheorie

Edward Evans

right

I mean, the main prereq is AZT1

but that's fine, it's the group cohomology that's gonna get me

user2103480

@EdwardEvans that's TOPOLOGY

Edward Evans

o O F

p-adic Hodge Theory is basically AZT3

user2103480

11:38

Those are big words

Edward Evans

yeah but I think it'll be alright, it seems like what we'll be doing is studying p-adic Galois representations

user2103480

Thank god I'm now learning physics topics, and their terminology is at least as pretentious as that of algebraists

Edward Evans

lel

user2103480

I'mma be learning 'bout that Zwanzig-Mori Projection Operator Formalism soon

Edward Evans

wtf

user2103480

11:41

Hahaha yeah

But I don't think it's as tough to understand as group cohomology 'n shut

Leaky Nun

@EdwardEvans how do you remember all the adjective endings

Edward Evans

In German?

Leaky Nun

ein(_) schwer(_) Sprach

yeah

Edward Evans

It comes with experience

as long as you learn the correct article when learning vocab you should be fine

Leaky Nun

but there are like 16 endings for ein etc and 16 endings for der/die/das/die

Edward Evans

11:43

a lot of them are repeated though

for instance in the Dative all the adjective endings are the same in the definite

dem alten Mann, der alten Frau, dem alten Auto, den alten Menschen

wait alte Kinder

and in the genitive too

des alten Mannes, der alten Frau, des alten Autos, der alten Menschen

Leaky Nun

what do you mean by alte Kinder?

Edward Evans

I wrote "den alten Kindern" which is dumb

cuz like

old children?

Leaky Nun

sicherlich kann es geben ein Gruppe Kinder, und manchmal Kinder sind alt und manchmal jung

Edward Evans

Ja gut, da hast du auch recht

Leaky Nun

manch*

Edward Evans

11:49

manche Kinder

you get to know them by using German a lot

it's hard to just memorise them imo

Leaky Nun

und was sind die Endung vom Akkusativ

Edward Evans

den alten Mann, die alte Frau, das alte Auto, die alten Menschen

so almost the same as in the nominative with the exception of the masculine

user2103480

@LeakyNun if that's your level of german before you even live in germany, you'll manage just fine

Leaky Nun

ich kann kein Deutsch sprechen

Edward Evans

tbh you can be fairly sloppy with cases and articles and it won't matter that much

and you get better with time

user2103480

11:54

@LeakyNun You mean that you're having a hard time actually speaking, instead of writing?

Edward Evans

there are lots of words where the plural and the singular are the same, and the only thing that differentiates them is the article

e.g. das Mädchen, die Mädchen

'as Moadle, d'Moadle

Leaky Nun

@EdwardEvans danke

Edward Evans

npnp

user2103480

@EdwardEvans grrrr

Leaky Nun

@user2103480 wahrlich, ich kann nicht Deutsch verstehen, weil sie sprechen zu schnell

oder sie benutzen Wörte, die ich kenn nicht

user2103480

11:57

Haha fair, I can barely read french but listening to french is even harder

Leaky Nun

gibt's viele Wörte, die ich kenn nicht

user2103480

Like several orders of magnitude

@LeakyNun das wirst du schon hinbekommen

Keine Angst

Filip98

Hello everyone, do you speak german in here? ahaha is it a maths chat?

Edward Evans

No we're just talking about German language atm, it's usually full of maths

Leaky Nun

zis is nau a Cherman chatt

Edward Evans

11:59

hahaha

user2103480

Ve are annexing zis ruum

Edward Evans

vi hef taken ofa se chät

Filip98

Are you all mathematicians?

Leaky Nun

nein, ich bin Student

Edward Evans

Elsaß-Lothring belonks tu Chörmani

Leaky Nun

11:59

oh no

user2103480

@LeakyNun is germany

@Filip98 what's your definition of a mathematician?

Filip98

Someone who has a degree ( master degree or higher) in mathematics!?

user2103480

Ok, I think that question is a dead giveaway

Nice, ok, then I can honestly say I'm not a mathematician

Filip98

I'm not a mathematician , but I tend to it ( I'm a student who is fast going to graduate !)

almost ( not fast) ** sorry I was reading in german and I said "fast"

user2103480

Do you have any particular areas of interest?

Leaky Nun

12:02

du sprichst Deutsch?

user2103480

@EdwardEvans why do I read that in schwarzeneggers voice

GET TO DA CHOPPAH

Edward Evans

hahaha Arnie wanted to be his own Synchronsprecher in Terminator and they declined because he has a strong Steiermark accent

Filip98

@user2103480 Actually I'm an undergraduate(?) student. I actually still do not know what I like more. Everything fascinates me. ( probability apart , I hate it haha)

user2103480

Damn

@Filip98 damn

Edward Evans

i bruch dei Kleidung, dei Stiafl, und dei Motorrodl

ohne Schüssl

wtf

i need your clothes, your bowl, and your motorcycle

user2103480

12:05

"Find the word that doesnt fit"

Edward Evans

rofl

Leaky Nun

@user2103480 sprichst du dialekt?

user2103480

@Filip98 next thing you'll probably tell me that you find algebra utterly fascinating

Filip98

@user2103480 ahahahah how did you know it?

user2103480

@LeakyNun I guess I have an accent, but learning kölsch (the local dialect) is pretty tough, and nobody really speaks pure kölsch

Filip98

12:07

@user2103480 Bist du grade in Deutschland!?

Alessandro Codenotti

@user2103480 thank god for that

Edward Evans

oof

user2103480

Except for some old people and there's literally a speech barrier between them and me

@Filip98 yup

Alessandro Codenotti

the only thing worse than kölsch (the dialect) is kölsch (the beer)

user2103480

@Filip98 by your distaste for probability

Alessandro Codenotti

12:08

and schwaebisch

user2103480

@AlessandroCodenotti wrong and wrong

Edward Evans

wtf is wrong with Schwäbisch

Filip98

Schwäbisch ist echt schön eigentlich

Alessandro Codenotti

Literally only the people raised in Koeln like to drink Kölsch

user2103480

Alessandro just hates every accent that he's been confronted with regularly

Filip98

12:09

Is he german?

Edward Evans

@Filip98 danke, endlich jemand mit der richtigen Meinung

Alessandro Codenotti

That's not true, the Swiss accent is very cute

user2103480

Nah, kölsch's purpose is that you can down it quickly, and then again and again

@AlessandroCodenotti yeah I count that as regional allegiance

Swiss german is ridonculous

Edward Evans

ridonklacray

user2103480

Exactly

Edward Evans

12:10

luckily I lived on the Swiss-Austrian border for a few years

user2103480

And you can drink litres of kölsch faster than almost any other beer. Which is the cologne karneval in nutshell

@Filip98 yep

Also, the sächsische akzent is the worst by popular opinion

Bavarian's stupid as well

Edward Evans

looool Klaus from American Dad

Pfälzisch is pretty funny

My gf is from Rheinland-Pfalz

user2103480

Northern germany's accents are delightful, and the ones in rheinland pfalz. BUT NOT PFÄLZISCH

Alessandro Codenotti

@Filip98 No but I live in Germany

user2103480

I'm talking bout Trierer Platt

Edward Evans

12:13

lel

Actually she's from the Alzey area

user2103480

Have you been there? Rhineland Pfalz is pretty nice

Edward Evans

Yeah it's really nice, except for Ludwigshafen and Worms

user2103480

Especially in comparison to godforsaken nordrhein-westfalen (don't get me wrong, I wouldn't want to live outside the urban nightmare)

Edward Evans

lol I've only been to NRW for like.. a school trip years and years ago

user2103480

But plastic trees probably count as nature around here

Alessandro Codenotti

12:17

I'm glad I've lived at the two extremes of NRW skipping the long stretch of enverending cities in the middle

Edward Evans

My experience of the DACH region is Vorarlberg, Rheinland-Pfalz, and Bavaria

user2103480

@AlessandroCodenotti but thats the good part

Filip98

Alessandro mi sembri italiano

user2103480

Everything else you find in other parts of germany, only that it"s better there. The only counterpart to our post-industrial wasteland here in NRW is Berlin

@Filip98 what semester are you in?

Filip98

In Italy it is divided by years, so I am in my third year.Actually I had to end it in September but I couldn't, so maybe next semester will be the last !

user2103480

12:23

Do you plan to specialize in an area?

Alessandro Codenotti

Where are you studying if you don't ming sharing it here?

Unrelated @Edward but do you know brass against?

Edward Evans

no but listening now

user2103480

@AlessandroCodenotti I thought they'd do rise against instead of rage against the machine

Edward Evans

very cool rofl

oof they have the pot by tool

Alessandro Codenotti

They do covers of heavier groups (they did a lot of rage against the machines and tool) with brass instruments and they are great

@EdwardEvans That one is also really good

Edward Evans

12:26

I haven't played guitar in weeks

:(

Filip98

@user2103480 I do not know in which one. But I am completely fond of Criptography ( I had just an exam but it was so cool) and Numerical Analisys. If I could study these somehow it would be great !

user2103480

Did you already have finite difference & galerkin methods?

Filip98

@AlessandroCodenotti Right now in Rome, but I plan to study somewhere else next year! Do not know where exactly :(

user2103480

I've heard that numerical analysis gets somehow interesting when that starts. I did some numerics for pde course but that was literally just taylor expansion

Alessandro Codenotti

@Filip98 If you like cryptography there's a masters program in Trento focussing on it

(of course my opinion is biased since I did my bachelor in Trento)

user2103480

12:30

@AlessandroCodenotti that doesnt necessarily mean it's a positive bias

Filip98

Thanks a lot , I will give it look !!

user2103480

I would advise against doing one's master's in cologne, unless you are really into numerical analysis and can handle rather unpleasant professors (in that specific area)

for context, I did my bachelor's there

@Filip98 I also didn't like probability at first, it's pretty dry and unmotivated at the start

4 types of convergence coming out of the blue, apparently useless lemmas like borel-cantelli, and all the measure theory fuckery on top of that

user2103480

13:22

@AlessandroCodenotti if f and g are measurable functions with a separable image, does their concatenation also have a separable image?

RewCie

13:43

aello

anyone alive online?

Leaky Nun

I'm unfortunately dead

RewCie

@LeakyNun Sad to know, but you could still vote in US Elections.

Transcript for

$Mathematics$ Mathematics