This is a strange story of mine, but:

When I was obsessed with polytopes as of 11 years old, I was drawing Coxeter-Dynkin diagrams.

One naughty friend of mine saw that and yelled, "Hey! What's with these eyeballs!?"

> You might still be clinging to the belief that you're inherently good

I wonder, do people feel that way?

I have hard time relating to this video

> Under the right conditions your morals can crumble faster than a sandcastle in a hurricane

I disagree with this, I think it's just a matter of mental preparation and the fact that most people are weak on this

If someone is decisive and strong in their conviction, then I believe that your morals, even if occasionally staggered by, say, fear, won't change. You won't be subject to any rationalizations if you put your mind into it and stop justifying yourself

@SoumikMukherjee damn, then I have no chance :(

Hi! It's my first time in chat. Sorry to interject but is anyone willing to help me understand this answer to this question? The part I don't understand is how to show that the binary expansion isn non-terminating. I've spent the last few hours banging my head against a wall! Thanks! math.stackexchange.com/a/2124557/1207792

@clee the uniqueness part?

@Jakobian Hi! No, I meant the existence part when he asks "Can you see why it is in fact a non-terminating expansion?" Sadly, my own answer to this question is "no".

@clee if it was terminating then we would have $\lceil 2^{N+k}\alpha-1\rceil = 2^k \lceil 2^N \alpha-1\rceil$ for some fixed $N$ and arbitrarily large $k$, and setting up some easy inequality and dividing by $2^k$, taking limit, we see that $2^N\alpha$ is an integer. So this equality says that $2^{N+k}\alpha-1 = 2^{N+k}\alpha - 2^k$, and this is impossible.

by the way if you're new to chat I recommend turning off notification (next to "all rooms") and enabling ChatJax (see chat description)

@Jakobian Thank you. I'm sure it was trivial for you but for me it's like a haze has been lifted. Really appreciate it.

> I'm sure it was trivial for you

I don't like this interplay of better vs worse

Its a bit presumptive as well. But your thanks are well taken

Sorry, it wasn't intentional. I see... what's your perspective? I think it's human nature to compare yourself to others you know? It's a hard habit to break at least...

it's an EASY habit for ME to break

@clee it's not the comparison itself that I find issue with, but the fact that you are assuming things about me

and in particular this might also be a cause of some internal misunderstandings

you might be misjudging that anyone experienced might approach the problem you're struggling with and instantly be able to solve it

sometimes this is indeed the case, other times it's not

i see all and know all

it's important to understand that everyone needs to put this so called "work" into something to understand it

its just that some people have already established knowledge

of course it is easier, since you get the hang of things and are more confident in what you need to do and so on, but yeah, generally speaking the need to put in some work is still needed, at all levels

The other day I was at an event of sorts, and while I am average of height, I still compared myself with others, being higher than me. I agree that it's human nature

I see... that makes sense. It's kind of a reassuring thought actually. I guess we are all just in different stages of knowledge and development. If you don't mind me asking, what has your mathematical journey been like? I realize this is a broad question, but I'm curious. It can be anything, like your mathematical interests, career, or education. By the way, I am below average height so I know that experience well lol. Such is life I guess.

@clee I don't really know how to answer this. This sounds more like a @leslietownes type of question

clee a lot of people on the chat either are, or were, focused on math as a concentration at the post secondary level (university level). some have degrees specifically in math, or degrees in subjects that potentially use a lot of math, some are working on those kinds of degrees. so, the level of experience with the work jakobian describes is probably significantly higher than the average, even the average in most university classrooms.

certainly significantly more past experience with studying math than the average asker of a question on MSE, on average. so yeah it can skew perspectives a little bit.

of course there are other folks in here too, but, if it sometimes seems like people might have phds in this stuff, it's because some people do. :)

@leslietownes I see... that's a bit intimidating. I am a mere university student as you might be able to tell lol. Are you one of those people with phds? I've actually been considering going to graduate school (not for math lol) and was wondering if you might have some general advice on the type of person who should/shouldn't pursue graduate level education. Anything helps. Thanks for your response.

@Jakobian Again, thanks for your help!

@clee Do you understand the term "work/life balance"? And do you value a healthy work/life balance?

@leslietownes Yeah, but having a PhD doesn't mean that one is necessarily smarter than the average bear. Just more stubborn, and lacking any real works skills. Most people with phds are to dumb to have done something else. :D

@leslietownes oh, I'll break YOU!

clee: yes, i have a phd. re the question, even within a very narrow sphere (e.g. "phd in pure mathematics") the relevant considerations vary considerably from person to person, both in terms of the financial impact of the commitment (even for a fully funded program, there are opportunity costs, and there are enormous variations in what 'fully funded' means in terms of leaving time for research), and the effect of commitment on other obligations (e.g. family members). it is hard to generalize.

i guess i would probably generally discourage anyone from paying out of pocket for any form of graduate school, or getting any phd premised upon an expectation of getting an academic position, or at least an academic position that resembles one attainable by people in some previous generation of academia.

other than that, go nuts.

@leslietownes Translation: grad school will drive you nuts.

i should say that my experience is specific to the USA, i have no idea what anything is like anywhere else. i would caution you against taking any one person's advice too seriously, even if their experience seems generally comparable to yours and they have experiences that suggest they know what they're talking about. there are simply way too many variables to generalize all but the broadest statements.

:P

i wouldn't say that, exactly. i just wouldn't pay out of pocket for it or go in expecting an academic position. grad school can be enjoyable and fun, within the parameters of not committing financial malpractice or ignoring the rest of your life.

On the other hand, I loved grad school, but I also didn't know what else I might have done with my life. My father had two doctorates, my mother is ABD, my grandfather had a doctorate, as did his father, my grandmother had an MFA, and my other grandparents had bachelor's degrees.

Surrounded by that... What are you supposed to do with yourself? No one ever taught me how NOT to be in academia.

@leslietownes sure. I'm just feeling cynical tonight.

i think most of the people i was in grad school with were reasonably well adjusted and not going for 'the wrong reasons' (which is not to suggest that i am any kind of judge of that in individual cases). but, a number of people my wife was in grad school with were miserable during school, and also pretty obviously going for the wrong reasons, and several of them wasted a whole lot of time and money.

Anywho, I should have gone to sleep an hour ago... Maybe two.

Night.

there were tons of folks in my wife's program who had deep seated self esteem issues, where they had no conception of value outside of getting good grades and getting praise in an academic environment, and many of them wanted to be professors at schools that were better than the one they were going to grad school at. not a recipe for good vibes.

@leslietownes ouch.

it's definitely a good sign when someone is mulling over the question, "should" i go to grad school, and not just thinking in terms of "when i get my phd, i'll..."

@leslietownes @XanderHenderson Thank you both for the valuable advice. It's a lot to think about. It seems like you should go to graduate school for the experience and not just as a means to an end. But it sounds like going to graduate school ended up being worth it for you both? Can I ask what your motivations were for going to graduate school? Also, I'm curious if there's anything that maybe surprised you or was unexpected when you first entered grad school? e.g. misleading assumptions, etc.

clee going in i just wanted to study more math, i enjoyed doing it and didn't have a lot of other responsibilities. in the background i thought that math was not so limiting that it could only be useful in an academic setting, and yet specialized enough that certain kinds of experiences you could only get in academia. which felt like a kind of insurance policy against any specific job path not working out even though i had nothing in mind.

@nickbros123 The exam doesn't require any advanced tools tbh. But the questions are little tricky. Overall the written exam is easy. I failed the first time because I am good at messing things up ^_^. And about the 2nd interview, I got nervous as usual.

by way of contrast, i like movies a lot, not just watching them but learning about the process of making them, but was never interested in majoring in film or going to graduate school in it, because i can get everything i like out of film without getting a degree in it, let alone a graduate degree in it. i don't think a phd in film makes sense for anyone, except if they want an academic position that somehow requires it.

or if you just have all the time and money in the world and just feel like goofing off (you will meet lots of those people in arts programs but not so many in math programs)

math at US universities is also generally very international, so depending on the program you might have half or more of the phd students from some other country and many not even thinking of working in america at all, let alone working in american academia. which can set a very different tone from, say, most US humanities phd programs, where the opposite might be true

i wasn't surprised by very much in grad school, it was more or less the experience i imagined i would be getting

Talking about goofing off, I know a student who did his bachelors in computer science, masters in business administration, and now int PhD in physics

someone with an odd sense of fun. if you were to take out the mba (or change the timing to later in the career) it would make somewhat more sense

@leslietownes Thank you for your insights. I'll quit pestering you now lol!

@SoumikMukherjee happens. I've been practicing some of their papers, scoring something like 40-50/160, but I am severely lacking in the background since I was a student of physics before. It does expect analysis, basic group rings and fields, complex variables, ODEs

Did you guys know that Teary Coqwand invented first proof assistant 🤣

There are so many puns in Math. We don't discrim

@mick I voted for you then @Shaun (second choice) as moderators :| Did you guys win?

the election is ongoing, i think. voting is still taking place, and will be for some days

@nickbros123 are you applying for MSc or for PhD?

@leslietownes though anyone who does a bachelor in computer science goes for campus placement here.

@nickbros123 This channel can be useful for your preparation

@SoumikMukherjee MSc, currently I'm in my 3rd year

@SoumikMukherjee thank you 👍

Okay, all the best

I am trying to understand the Baire Category theorem (BCT); I know the statement of the theorem and various other versions. Then it is stated in my book that

> Single-point subsets of the real numbers are nowhere dense, and the uncountability of the real numbers is just the fact that countable unions of such single-point sets cannot be the whole of the real numbers.

Why not? What would contradict the BCT if it were?

It would contradict completeness I believe...

The statement I am familiar with is e.g.

> Let $(E_n)$ be a sequence of nowhere dense subsets of a complete metric space $X$. Then $\bigcup_n E_n$ has empty interior.

Thank you, @Debug. As said above, the election is still ongoing :)

@psie countable union of countable sets is countable (in ZFC)

what is the definition of the divergence of $n$-form?

Is there an intuitive explanation of why uncountable additivity in probability theory in impossible, without needing to know measure theory?

(that is, generalizing countable additivity further to include both countable and uncountable)

What is the intuitive explanation and the formal definition of the root operators on a set, such as the real numbers?

Take for example, the multiplicative group $G=(\Bbb R_+,\times)$ and quotient over every subset $[a,b]$ for $a,b \in G$ s.t. $ab=id$

Does the operator implicitly represent the element in $n-$space as a "cube"?

meaning $a \sim b$ is the equivalence relation

This is a one point compactification of each subset of $G$ whose end intervals are multplicatively related, resulting in equivalence classes of concentric circles

@zetaspace Are you talking to me?

No I was laying the groundwork for a question @Obliv

I think my question is, how do make it so the circles are actually concentric

How do you make it so the circles are concentric?

Associate the length of the intervals to be the radius of the circles and embed the circles into $\Bbb R^3$, right?

this would render the structure to be a partition of $\Bbb R^2$ by S^1 equivalence classes

Suppose $V$ is a vector space over $\mathbb{F}$, that is finite dimensional with a basis $\{ \alpha_1,\alpha_2 \cdots, \alpha_n\}$. If this is the case, the basis for $\mathfrak{L}(V,\mathbb{F})$, the set of all linear functionals on $V$, would be found out by $f_i(\alpha_j)=\delta_{ij}$. Suppose then, I am given a basis for $\mathfrak{L}(V,\mathbb{F})$ that is $\{f_1,f_2,\cdots,f_n\}$, is there a procedure to find the basis $\alpha_j$-s in $V$ so that $f_i(\alpha_j)=\delta_{ij}$??

that gluing two functional equations question turned out better suited for mathoverflow

@user10478 so are you accepting the existence of probability measures at least?

Well, regardless, if $P(A)$ for $A\subseteq [0, 1]$ is the probability that a uniformly randomly chosen point belongs to $A$, then $P([0, 1])$ which is $1$, would have to be equal to $\sum_{x\in [0, 1]} P(\{x\})$ and since we are assuming that $P(\{x\}) = P(\{y\})$ for all $x, y\in [0, 1]$, we would need to have either $P([0, 1]) = \sum_{x\in [0, 1]} P(\{x\}) = \infty$ or $0$

but either case is impossible, so allowing for arbitrary additivity makes it so that you can't chose a point uniformly from $[0, 1]$

@Obliv for real numbers, having fixed some $r\in\mathbb{R}$, you want to solve the equation $x^n = r$, but this doesn't always have a solution, and if it does, it doesn't have to be unique. If it doesn't have a solution, or it's unique, we either don't take root or just set it to be this unique solution. If the solution isn't unique, there is exactly two of them, differing by the sign. One then takes positive one

Thus, $r^{1/n}$ is defined as the solution of $x^n = r$ when it exists and is unique, and for the positive solution of $x^n = r$ when it exists but isn't unique

here you see the relation between "algebraic roots" of polynomials, and "arithmetical roots"

I have a basic doubt. If the supremum of some nonnegative function over a nonempty set $S$ is $0$, is the function identically $0$?

The reason I'm asking is because I am considering the metric space $(X,d)$ and the metric on $X^S$ given by $$\rho \left(f{,}g\right)=\sup_{s\in S}\min \left\{1{,}d\left(f\left(s\right){,}g\left(s\right)\right)\right\}.$$I am trying to verify the property $\rho(f,g)=0\implies f=g$.

If the function $\min\{\ldots\}$ is identically $0$, then that would mean $f=g$ I think.

@psie yes

thanks!

If $k$ is a non-negative function, then $0\leq k(x)\leq \sup k = 0$ for any $x$ in its domain i.e. $k(x) = 0$

ah nice 👍

@nickbros123 by procedure, what exactly do you mean. Surely we can find $\alpha_j$'s if we, given $g\in V^{**}$ are able to find corresponding $\alpha\in V$ representing it

so is your question about some method of finding such $\alpha\in V$, given $g\in V^{**}$?

@Jakobian Interesting. The only situations where $x^n = r$ doesn't have a solution is when the solution is in $\mathbb{C}$ i.e., when $x \in \mathbb{R}_{<0}$ and $n>0$ right? Also, is $n \in \mathbb{N}$?

I know if $n \in \mathbb{Q}$, the denominator denotes the 'root' and the numerator indicates the exponent which we raise the base to before taking the root usually

well maybe this is the same thing, anyway. We would probably like to first establish some basis $\beta_j$ of $V$ as a starting point. Then what is to solve for is $\alpha_j = \sum a_{ji}\beta_i$ where $\delta_{ij} = f_i(\alpha_j) = \sum a_{jk}f_i(\beta_k)$ which then becomes a system of linear equations

once it's solved, you should be able to obtain those $\alpha_j$

@Obliv when $n$ is even and $x$ is negative

@zetaspace what do you mean

i have been out of mathematics for a while, can you remind me of something, when some object $ T: V^k \times W^m $ does it mean that the arugments are $T((v_1,\cdots v_k),(w_1, \cdots w_m)) $ or $T(v_1,\cdots v_k,w_1, \cdots w_m)$ (The difference would be two arguments or m+k. I am trying to learn about tensors, where i encountered this question.

If $X$ is second countable and Hausdorff and $q:X\rightarrow M$ is a covering map, where $M$ is a manifold, does this imply that $X$ is a manifold?

this is true right?

@Mad People are sloppy about such things.

@Jakobian I mean, let $G$ be the multiplicative group of positive real numbers and let $a,b \in G$. For one example I was talking about take $a=1/2$ and $b=2$. We have that $ab=1$ since $(1/2) \times (2)=1.$ Take the interval $[1/2,2]$ and define an equivalence relation on the endpoints of the interval i.e. $1/2 \sim 2$. Topologically this is $S^1$.

@copper.hat i see, do you reccon you know what is exactly the meaning in a tensor context?

But I want to take not just the ex. of $a=1/2$ and $b=2$ but EVERY possible $a,b$ s.t. $ab=1$.

Part of getting back into math is relearning what you thought you knew before.

That, in itself is a valuable skill.

Say I am working with a real analytic foliation of an open manifold, but the leaves are also technically unnormalized probability distributions and also Schwartz functions. Now these involve 3 different realms of mathematics. But each has something I need. I want to avoid mixing these terms up because it makes for more difficult reading

Read very slowly and carefully.

perhaps i can find out if there is a analogue of the Fisher metric (info geometry) in foliation theory

@Mad your question is too broad to give an answer.

I guess Ted is not coming back.

😞

☹️

I don't understand why he left chat

He was always hinting at it.

wdym?

He would say stuff to indicate he wouldn't be around for much longer.

oh

didn't notice that

I was not around when he left

👋

hi, who wants a drawing of their profile picture?

hi

@BinkyMcSquigglebottom me

@Pizza okay ,it will be sent tomorrow

@SineoftheTime Unfortunately I won't be able to take the analysis 2 exam in September, but there is a physics 2 date, so I'm doing this subject

physics 2? Electromagnetism?

it would be electromagnetism

@SineoftheTime yes

you engineers are fast :)

have you already done it as a subject?

yes, but at the second year

Note that if you have mathematical methods, you have to know very well analysis 2

Yes, however, you have to give analysis 2 first to then be able to give methods, I think

can you do also geometry?

@Pizza sure, but some professors are more flexible and they let you do the exam and then register it after doing analysis 2

@SineoftheTime yes but as you say it wouldn't make much sense to do it before analysis 2

sure

@SineoftheTime in studying yes, only that these dates are in a period where I can't reach the university, so I can't take the exam and I will have to wait until October

instead physics is on a date where I can go to university, so I better do it well

in a couple of months I won't have this problem anymore fortunately

october is good

is your house far from the uni?

Yes it takes about 1 hour and 10 minutes by train, but that's not the problem.

ah ok

If you don't mind sharing, in which uni do you study?

@SineoftheTime yes, unina

I don't know it too much

You can search on Google

sure

but I mean

Your maybe the first person I "know" from that uni

I try to keep in touch with different people from different unis to have more infos

Ah okay!

where do you attend have students from other universities moved?

not in general

except for a couple of students

but there are a lot of students from other cities

expecially from southern Italy

I mean for example if someone from another city came there

ah yes then

I thought you meant the change uni

no, that's what I meant sorry

usally people choose unis in northern Italy

so, being from there, I see a lot of people coming outside my region

@SineoftheTime yes true

I think if someone is near to the uni and the uni it's not that bad it's ok not to move

maybe I'm too lazy

@SineoftheTime Yes that's right

What is uni ?

university

do you mean the education models?

what do you mean?

I mean university/college

Ok, what university do you attend?

you mean what faculty or the name of the university?

Faculty

maths

Have you done quantum mechanics?

no

But will you do it or is it not planned?

not planned as a mandatory exam

but there are math exams like mathematical method for QM

still, we can choose electives from other faculties

👍

you study on your own right?

@BinkyMcSquigglebottom I took a quarter of QM in grad school. It is basically the only physics class I've ever taken (aside from one semester of a two semester sequence of high school physics).

@SineoftheTime Yes

$\ddot{\smile}$

@XanderHenderson 👍

I received a warning that the computer is overheating, I urgently need to turn it off

Bye

bye

@zetaspace then you phrased it wrong. You want to take $(0, \infty)/\sim$ where $x\sim y$ iff $x = y$ or $x = y^{-1}$

A metric $d:X\times X\to \mathbb R$ is Lipschitz, right? I'm reading a proof of the fact that it is continuous, but I believe this proof is just showing $d$ is Lipschitz without explicitly saying so.

@psie yes, in both arguments

right, and uniformly continuous in a single argument :)

$|d(x, y)-d(t, z)|\leq d(x, t)+d(y, z)$

@psie Lipschitz implies uniformly continuous

ah, ok

of course the metric I am taking here on $X\times X$ is $d_0((x, y), (t, z)) = d(x, t)+d(y, z)$

if you take supremum metric or something, this should still be Lipschitz but with different constant

@Jakobian right, in my book, on products of metric space, they give a couple of example of metrics that one can equip the product with. At the end of the section, they just write that for the remaining of the book, the metric on the product space will not be specified further, as long as it satisfies two properties:

1) a sequence $\{x^{(j)}=(x_k^{(j)})\}_{j=1}^\infty$ converges to $x=(x_1,\ldots,x_n)$ in $X$ iff for each $k$ the sequence of component entries $\{x_k^{(j)}\}_{j=1}^\infty$ converges to $x_k$ in $X_k$ and

2) $d_k(x_k,y_k)\leq d(x,y)$ for $x,y\in X$ and $1\leq k\leq n$.

Here $X=X_1\times\ldots\times X_n$, and $d$ is the metric on $X$, and $d_k$ the metric on $X_k$.

Is there a name for the metrics that satisfy 1) and 2)?

@psie 1) is a condition that it's the same as the topological product

and 2) that the projection $\pi_k:X\to X_k$ is Lipschitz with constant $1$

ah ok, didn't realize this

I don't know any name for such metrics and never needed to consider them

condition 2) can be equivalently phrased that $\max_k d_k(x_k, y_k) \leq d(x, y)$

then the left side is a metric on $X$ satisfying 1)

however its not clear if say, $d$ is uniformly equivalent to it

I suspect it isn't

ok, well, I think I need to call it a day. A bug was just in my room and frightened the sh*t out of me :) but all's well, I survived. It's the time of the year when bug's fly in to my room. Last year, a bug flew in to my frying pan...while it was frying! I couldn't help but witness a brutal death.

@psie get a mosquito net for your window

yeah, probably I should