@s.harp doesn't answer my question

It doesn't explain why are you trying to um, show? To me, that discrete spaces are realcompact

What you wrote didn't look like a question towards me, more like you're trying to explain something, maybe even questioning what I said in the first place

Doesn't matter though

More specifically, set $s=\sup E, i=\inf E$. Choose $n$ such that $(s-i)/n<\delta$. Then $E\subset \cup_{j=1}^n B_j$, where $B_j=[i+(j-1)(s-i)/n, i+j(s-i)/n]$ so that $E=\cup_{j=1}^n E_j$, where $E_j=B_j\cap E$. WLOG assume $E_j$ to consist of atleast two points. For each $j$ here, pick some $e_j\in E_j$. Then for each $j$ here, $|f(x)|<|f(e_j)|+1$ for all $x\in E_j$.

$M= \max(|f(e_j)+1|: 1\le j\le n)$ bounds $f(E)$.

$M= \max(|f(e_j)|+1: 1\le j\le n)$ bounds $f(E)$. (this fixes the typo in earlier formula for M. Too late to edit the old message now.)

@Koro, I think your proof is unnecessary complicated for a simple problem. But thanks for the idea, too.

I once asked you here, but you missed. Are you pursuing Ph.D.?

This uses only the definition of uniform continuity and no sequences etc. Take your time, think about it.

@noballpointpen no

Ok, I will screenshot it so I have it in my vault as a method that uses only the definition of uniform continuity.

chop and pick the maximum bound, that's it

yeah, right.

I've read it completely. Nice.

$E \subset \Bbb R$ is bounded iff$ \forall (x_n) \subset E$ , (x_n) has a cauchy subsequence.

Alt: Let $f(E) $ is not bounded. Then $\forall n\in\Bbb N, \exists (x_n) \subset E$ such that $|f(x_n) |>n$

$(x_n) $ has a cauchy subsequence.

Contradict that $f$ is uni.cont.

how are we still on this problem lol

not enough chopping

i was hoping to find a new question for breakfast over here :(

what's a nice way to prove zariski's lemma

ill have to cook it myself today

the rabunowwhatever trick

oh sorry thats for the nullstellensatz

remind me what zariskis lemma is again

a finitely generated algebra over a field is an algebraic extension of said field

What does finitely generated mean?

are you sure you dont mean for the finitely generated k akgebra to also be a field? i thought you want to say such a k algebra is a finite over k aa a field extension

usually I interpret it to mean its the image of a ring epimorphism from k[x_1,…,x_n] for some n

anyway, shintuku, one way to prove zariskis lemma is to apply the noether normalization theorem

What about $\Bbb Q[\pi]$?

at porridge: oops, yeah the lemma is that it is a finite field extension

ah, you need k to be algebraically closed too !

thats an important point

for the nullstellensatz yes, but not for the lemma

hmm

at ted: am thinking about it

the noether normalization theorem gives that any such k algebra will be a finitely generated k module, because it furnishes in particular finitely many k algebraically independent elements which it is integral over the k algebra generated by those elements. In this case you get that that list of elements is empty, or else k[x_1,…,x_n] would be a field for some n >=1 which is absurd

so the conclusion is definitely true (the k algebra is finite over k) without any additional hypothesis on k

Q[pi] is not a counterexample because its not even a field

because pi is transcendental

just checked atiyah macdonald, their statement requires that the $k$-algebra be a field, i think ted was giving a counterexample to my statement

so thats why this fails

yeah thats what i thought we just established after what I said..

@porridgemathematics $S=\{f(x)\in \Bbb Q[x] : \Bbb Z is f Invariant\}$

right

oh ok, you only said the lemma is that “it” is a finite field extension

now the counterexample makes sense to me

yeah the proper statement is: if the finitely generated k-algebra is a field, then it is a finite field extension

right, i was thinking of nullstellensatz because this PLUS k being alg closed gives you the weak nullstellensatz

$S$ is ring.

that plus the trick by rab gives you the usual nullstellensatz

Sourav you could verify the ring operations directly I think

but isnt that ring just Z[X], lol

no it isnt, ignore that

just remembered the hilbert polynomials are examples

@porridgemathematics It's easy :) I want to verify whether $S$ is finitely generated or not .

you mean as a Z-module, right

because you opened with saying "S is a ring"

$S$ is a subring of $Q[x]$

yeah but what do you mean by finitely generated

off the top of my head i would say it doesnt seem likely.. maybe you could compute the coefficients by looking at divided differences Sourav, and then see if you can find a nice basis

but even when you get a basis, its not going to be finite over Z

@porridgemathematics Yes.

@porridgemathematics Example of a infinite LI subsets of $S$?

:)math.stackexchange.com/a/4056501/977780

just use the e

newton series

yeah

exactly.

that answer is just the newton series

i don't think i've seen someone refer to an MSE question by number in the text of an answer before

Thanks:)

it's like they're citing an article of law you're supposed to know

Expand $\sin^{10}x$ as $\sum_{n=1}^{\infty} a_n \sin nx$

that inspired me to find question 1

question 2 is the first deleted question

probably it was something like "lol nerds whats the square root of my [BEEEP]"

curiously, question 1 is chronologically the second mathSE question, while question 106814 is the first

they don't want anyone reverse engineering the question-numbering algorithm, just like the avatars

gotta switch it up to stay one step ahead of north korean cybercriminals

we must protect the counting algorithm that begins at 1

north korea would probably be the richest economy in the world if they had access to 1. i bet that explains most of their problems

How to expand $\sin^m(x) $ as $\sum_{n=1}^{\infty} a_n \sin nx $ ?

@SouravGhosh integrate.

orthogonalilty strikes again

my flatmate just gave me his ps5 so that he can revise for his qualifying exams

its a preventative measure for him

im going to take very good care of it

flatmate, revise, what is this, buckingham palace

lol, i share university housing with two other graduate students

one in CS and one in math

don't you mean mathS

ah yes

he's a spy, get him

Sic Munchkin on 'im.

no ducks today and she didn't want to hang out either

Hang out where? No fun at the pond sans canards?

yeah, no swimming, no "soccer" (kicking a ball), just wanted screen time

i had work to do so it actually worked out pretty well for me

Screen time? Is she Olivia DeHaviland?

haha no she wanted to watch something

she's actually become self conscious about being in front of the camera, at dinner she was going on one of her trademark rants about the cat and i got out my phone to video it and she stopped almost immediately

Are you posting her to tik-tok without her written permission?

no, it was not for that kind of posterity

She probably wants royalties.

What is an example of a space X with a disconnected subspace A such that for any two disjoint open sets U and V such that $X=U\cup V$, either U or V contains A?

line with two origins, the two origins?

two suns in the sky?

this is just any disconnected subspace of $X$, if $A$ is a disconnected subspace, then $X = U \cup V$ implies $A = (U \cap A) \cup (V \cap A)$, so that $U \cap A = A$ or $V \cap A = A$, or is there something im missing here

oh, sorry , my brain isnt on today. A is disconnected, not connected

we can take X= R-{0}, A= {-2,-1}.

@leslietownes what's that?

@onepotatotwopotato the original space X has to be disconnected.

(for a non trivial example.)

koro: an answer to a more interesting question (where for any disjoint open U and V such that A is a subset of U union V, either U contains A or V contains A)

@Jakobian I get the feeling you feel you are being attacked. For me this a normal way chatrooms about mathematics work

"doesnt blah blah blah" when somebody mentions something and you have a thought about it

sometimes i get confused in here because i lose track of who has a question and who is answering a question and who is just thinking out loud about alternate approaches

there was all this algebra earlier were it seemed like 20 different people were talking about versions of the nullstellensatz

completely did my head in

i never figured out if anyone kicked it off with a question or if it was just an eddy in the sea of mathematics

@leslietownes for this, shouldn't A be connected?

i dunno, man, you tell me

some of these nitty gritty things do become the same, e.g. in metric spaces

from the tenor of the original question i wondered if you were working through some material on the hierarchy of separation axioms and all the weird stuff that fits between the usual levels people like to work at

it got kicked off by shintuku asking about how one could prove zariskis lemma, it was followed by an incorrect reply assuming zariskis lemma was the nullstellensatz, followed by a self correction and the question 'what is zariskis lemma', which led to an incorrect statement of zariskis lemma which in turn led to a correct formulation of the lemma, then a misunderstanding of a given counterexample to the incorrect formulation, and mention of a possible proof approach (via noether normalization)

that pretty much summarises the algebra confusion episode

oh lord

anyway it felt like a russian novel for a minute

yw :)

haha

maybe it wasn't 20 people, maybe it was 4 people, each with 5 names

something like that yes

@leslietownes That reminded of a theorem that I know: if A, B form a separation for X and C(connected) is subset of X, then C is either in A or in B.

how am i supposed to know that ivan is the same person as sasha is the same person as dimitrovich

Ohh

Bunyakovsky

name, nickname, patronymic, other nickname, and then transliterated in five different ways

i was ready to go hang myself in the barn by the end of all that, but thankfully the novel finished before i did that

zariskis is zariski's greek cousin

yes, zariskis was the pseudonym for a group of eminent algebraic geometers of the age of zariski, zariski refers to the single oscar zariski

zariskis was an answer to bourbaki

i like that version better

that reminds me actually there was a restaurant named tonelli's in my hometown and i always thought it would be a great gag to one day open up a similar but slightly better restaurant named fubini's next to it

but tonelli's went out of business

Haha

tonelli's and fubini's usually work together

I wasn't a big fan of measure theory but geometric measure theory is something more interesting than a general measure theory

that's a quote for the back cover of the book

maybe the front cover

"more interesting than general measure theory"

It's the product of people's constant efforts to make measure theory fun

more like to solve the plateau problem

also geometric measure theory is pretty , measure theory like in flavor

the estimates for theorems are just a bit more geometric, e.g. for the besicovitch covering theorems

its not like differential geometry, for instance

besides of course objects like varifolds, it still feels much closer to analysis than geometry at least to me

I thought orbifold is the only fold besides manifold

lots of folds

I actually didn't read a book titled GMT but I recently learned that Hausdorff measure is a part of GMT so I presumed it's something not boring.

Sets of finite perimeter is a nice introduction to GMT

By Maggi

its not boring, i didnt mean to say it was - but i would regard hausdorff measure as still within general measure theory/analysis (also haar measure)

maybe its because im confusing general measure theory with analysis, maybe the former is supposed to be something like the caratheodory construction for measures on spaces over and over again

It might look more fun and geometric, but under all that it's still measure theory with all of its downsides

if you're going to learn geometric measure theory, I didn't really see any fully formal introduction, people usually redirect to Federer's GMT book for any technical results. So if you're a person that's bothered by unproved results and details of things, you'll be really bothered here

And I think I've browsed like 15 different books on the topic

@Koro Construct $A=\{a, b\}$ disconnected and $a, b$ are not separated by two disjoint open sets.

Goal: To construct a disconnected space $X$ and $A=\{a, b\}$ disconnected subspace and two points can be separated by two disjoint open sets.

Lets start with $\Bbb {N}$ and $A={-\infy, +\infty}$

Let $X=\Bbb N\cup A$

Here is a generalization: Suppose Y is a connected component of X. Then take disconnected set A in Y. Then, if X= U cup V, U, V are disjoint open, then A is either in U or in V.

@porridgemathematics to complete what you were trying to do earlier.

@Koro oh nice , yeah

that gives you a big class of examples

Then $X$ is totally disconnected.

$A=\{+\infty, -\infty\}$ disconnected set.

You can't separate $\infty$ and $-\infty$ by two disjoint open sets.

Whenever $X=U \cup V$ where $U, V$ be two disjoint open sets, $A\subset U$ or $A\subset V$.

Personally if I ever would like to try learning GMT, I'd just read Federer's book, it's a big terse book so I'd need lots of time for it though

unless I'd be neglectful and take what Federer says for granted

probably best way to learn the subject would be to just take a course on it

Eat \\ Sleep\\Math\\Repeat

@SouravGhosh yes that works too. Thanks :).

Applications of Weierstrass approximation theorem ?

1)$( C[a, b] ,\|•\|_{\infty})$ separable.

If K is compact Hausdorff, then K is metrizable iff C(K) is separable

your application is a special case of this

Is this known as Stone-Weierstrass theorem?

It follows from Stone-Weierstrass theorem about dense subalgebras, I'm not sure if it's called the same

which Weierstrass approximation theorem is a special case of

@Jakobian Can we prove this theorem using the Weierstrass approximation theorem?

@Jakobian Algebra of real continuous functions on a compact set. The algebra is non-vanishing and separates points.

@SouravGhosh Weierstrass approximation theorem works here because it makes sense to talk about polynomials in $C[a, b]$

Weierstrass approximation theorem: $\forall f\in C[a, b]$ there exists $(p_n) \subset \mathcal{P}[a, b]$ such that $\|p_n-f\|_{\infty}\to 0$ as $n\to \infty$

in general case this is replaced by subalgebras, so the proof doesn't really follow from Weierstrass approximation theorem

(polynomials in $C[a, b]$ is a $\mathbb{R}$-subalgebra of $C[a, b]$ generated by the identity function $x\mapsto x$)

Any other applications?

Of Stone-Weierstrass or Weierstrass approximation theorem?

Weierstrass approximation theorem

Analytic/smooth functions are dense in $C[a, b]$?

Polynomials are real analytic.

maybe by switching to Fourier transform, we can say something interesting and non-trivial

Trigonometric polynomials

Q. 1) Prove that a concave function with $f(0) \ge 0$ is subadditive over $[0, \infty) $.

Q. 2) Prove that a concave function on [a, b] is lower semicontinuous but not upper semicontinuous.

Q. 1) is done!

$f[(1-t) x+ty]\ge (1-t) f(x) +f(y) $ for all $t\in [0, 1]$

For $x=0$ , $f(ty) \ge t f(y) $

@SouravGhosh what is the source of these questions?

where do they come from?

@Koro I am trying to verify one of my old question. A metric on a linear space where no ball is convex.

To prove $\ell_p$ for $0<p<1$ is a linear space, I need to prove an inequality: $x, y\ge 0$ then (x+y) ^p\le x^p+y^p$ for $0<p<1$

Here $f(t) =t^p$ concave and $f(0) =0$

But to prove the inequality, we only need the concavity of $f$ and $f(0) \ge 0$

Generalized result: $f$ concave and $f(0) \ge 0$ the $f$ is subadditive over $[0, \infty) $

@porridgemathematics Help me to prove Q. 2

I need to show that the sub-level set is closed.

For any $\alpha\in [a, b]$ , S_{\alpha}=\{x\in [a, b]: f(x) \le \alpha\}$ is closed.

$f[(1-t) x+ty]\ge (1-t) f(x) +f(y) $

@Koro 🌌

@SouravGhosh finite valued convex/concave functions are a lot more regular than only lower semicontinuous

*semicontinuous (upper/lower resp)

although if you allow convex functions to take values in [-infinity,+infinity) then you only get upper semicontinuity

much like subharmonic/superharmonic functions in the plane

Fun fact: $\sqrt[n]{i}=\cos\left(\frac\pi n\right)+i\sin\left(\frac\pi n\right)=e^{i\frac\pi n}$

@PlaceReporter99 Euler formula

@SineoftheTime i thought euler formula was $e^{in}=\cos(n)+i\sin(n)$

Let $G$ be a lie group. Let's say you have the data of the generators of $T_{\text{id}}G$, i.e. the directional derivatives of $G$ at $\text{id}$. Is there a reference which talks about how to use this data to compute the directional derivatives at an arbitrary point $g \in G$?

It seems the mapping one wants to use is left translation by $g$; for $h \in T_{\text{id}}G$ then $g \cdot h \in T_gG$. But i would like to make sure that infinitesimal generators get mapped to infinitesimal generators and see exactly which ones get mapped to which ones, if that makes sense

there's no such thing as "the" generators, and in general I'm not sure what exactly you're asking

the derivative of left translation is an isomorphism (because left translation is a homeomorphism), so it maps a basis to a basis, if that's what you're asking

oh okay the latter is what i am asking. i am asking in general: given a choice of generators for a lie group (equivalently, directional derivatives at the identity), how do you find the directional derivatives of the lie group at an arbitrary point of the group

e.g. for SU(2) given the pauli matrices, how do you find the directional derivatives of $SU(2)$ at any arbitrary point $g \in SU(2)$

and then in particular let us call each pauli matrix a "direction" associated with a parameter $\theta_i$. So, $\theta_x \sigma_x$ is associated with "moving in direction x". Then, will $g \cdot (\sigma_x)$ be mapped to the directional derivative at $g$ in the $x$ direction?

I want to integrate $\sin x^2$ on $\Bbb {R}$ using contour integration.

$\sin x^2= im( e^{ix^2}) $

Let $f(z) =e^{z^2}$

Then $f$ is an entire function, so integral over a contour will be $0$.

How to choose the contour?

Semicircle is not working:)

Is there any special contour for this integral?

@robjohn sir

what kind of a thing do you expect the integral of sin(x^2) to be? this function is not integrable

Wow :-)

:math.stackexchange.com/questions/4711397/…

@s.harp not Lebesgue integrable, but it's integrable

$\int_{-\infty}^{\infty}\sin(x^2)\mathrm{d}x = \lim_{n, m\to\infty} \int_{-n}^m \sin(x^2)\mathrm{d}x$ exists

so it exists as an improper Riemann integral @SouravGhosh

Try integrating it along the line $\arg(z) = \pi/4$

I think that's how I did it in the past

a slice of cheese

why this is nice is because square of something lying on the ray $\arg(z) = \pi/4$ lies on the ray $\arg(z) = \pi/2$ i.e. is a multiple of i

@SouravGhosh You can view this deleted answer. I will see if I can find another.

@Jakobian I am aware about the fact that f is improperly R-integrable.

@Jakobian Fresnel Contour ()

it didn't look like you were aware of it

@Jakobian I am trying to prove that $f(x) =\sin x^2 $ is not uniformly continuous function but $\int_{\Bbb R} f $ exists.

$(f_n) \in C(\Bbb R, \Bbb R) $ such that $\|f_n-f\|_{\infty}\to 0$ but $\int_{\Bbb R} f_n$ doesn't converge to $\int_{\Bbb R} f$

@SouravGhosh You don’t need to compute it to prove that. Make the obvious substitution.

@TedShifrin Yes sir. But I want to be more specific.

@Jakobian Please one more step

@SouravGhosh here is a better example.

at least it uses contour integration

but I believe I have a better example somewhere

@robjohn with his treasure trove of thousands of posts !

I want to find one with a nice diagram, but I may not have one for a Fresnel integral

@robjohn I hope you’re not having too frenesltic a time.

@TedShifrin nah, the frenesites are not biting

I botched my punny spelling. Oh well. Comes from typing on my phone.

your jab at a pun... what one might call a pun-jab?

Oy ... I want a new deal-hi.

Just my Lucknow, I can't think of a come back

Well, you did mumbai lot.

What's in bag dad?

<_<

>_>

In dia here?

I deserve a Nobel Peace prize for stomaching those puns

You and Donald Tromp, yes.

His re-election wouldn't be a surprise, to me.

The end of democracy as we know it. Among other things. Putin will rule …

Yup, peace prize.