Mathematics - 2022-11-28

what are you doing with quaternions, akiva

put them down

ah sure the quaternions are great, imgur.com/oKDOABg

the picture was taken beside broom bridge en.wikipedia.org/wiki/Broom_Bridge

@leslietownes for Clifford the big red dog

my daughter is a huge fan of clifford the big red dog.

not so much a fan of the algebras.

well, quaternions are in her gene pool

i mean we all wake up thinking what if i square that and it becomes $-1$?

2:30 AM

@TedShifrin Hello Sir

can we say that a real sequence and an infinite dim real vector are the same object?

I dont see anything that goes against this unless i am missing something

why does it matter?

Thorgott

"an infinite dim real vector" is not really a well-defined thing

but sequences of real numbers do form an infinite-dimensional vector space, yes

Just as you can identify $\mathbb{R}^n$ as functions $\{1,...,n\} \to \mathbb{R}$ you can identify sequences with functions $\mathbb{N} \to \mathbb{R}$.

3:05 AM

I see thanks !

I am working on this problem

where sequence have finite non zero entries

and i want to show if i have a caucy seqence of such objects then they do converge

with ofcourse a distance function between two sequences as the sup of each entry

it seems like cauchy implies convergence here but it is a bit messy to write it down properly

3:42 AM

there are not enough details there to work with...

4:00 AM

I already know that $T_ISL_n(\Bbb R) = \{M\in M_n(\Bbb R):tr(M) =0\}$. If $A\in SL_n(\Bbb R)$ then I think multiplication by $A$ gives an isomorphism $A: T_ISL_n(\Bbb R)\to T_ASL_n(\Bbb R)$ so $T_ASL_n(\Bbb R) = \{AM\in M_n(\Bbb R): tr(M) = 0\}$?

Q: retraction induced homomorphism is surjective

3

I am having a hard time proving this although it looks trivial... Let $r:X\to A$ be a retraction between a topological space $X$ and $A\subset X$ such that $r(a_0)=a_0$ for $a_0\in A$ then the induced homomorphism $r_*:\pi_1(X,a_0)\to \pi_1(A,a_0)$ is surjective. I tried to prove it as follows:...

Here, suppose that r is a retraction of X to A s.t. r(a)=a for every a in A. I want to show that the induced homomorphism $r_*$ is surjective.

Since $A \subset X$ surely that is immediate?

So let [g] in the co-domain. Define $f(t):=g(t) $ for every t in [0,1]. So $r\circ f=r\circ g= g$ So $[r\circ f]=[g]= r_*([f])=[g]$

Hi @copper.hat!!

@copper.hat yeah, I think so too.

But the post is making it complicated.

@Koro Hi Koro!

@Koro not much of a consolation, but i find retractions, etc, to be confusing in isolation.

Ohh, here I was trying to proves the set of irrationals is a Baire space.

10 hours ago, by Koro

Let $B_i$'s be a countable basis of R. Then $B_i\cap Q^c$'s give a countable basis of $\mathbb Q^c$.

By Urysohn metrization theorem, $Q^c$ is metrizable.

But $Q^c$ is not complete so can't conclude that it is a Baire space.

I think one has to proceed with definition to prove that Q^c is a Baire space.

Probably the only way to prove this.

4:19 AM

does it not follow from the fact that $\mathbb{R}$ is the countable union of closed setswith empty interior?

@copper.hat You’d better retract that statement.

@TedShifrin :-) always getting some pull back there

Push-back?

:-)

@Koro urysohn metrization? Really?

4:24 AM

Hi @TedShifrin!!

Hi, Koro.

So I'll proceed with - Let U_i's a collection of dense open sets in Q^c and then show their intersection is dense.

Basically using the definition.

UMT doesn't clearly work here.

Isn’t the set of irrationals $G_\delta$?

Yes.

G_delta is Baire?

I actually didn't know this result.

I enumerate Q with r_i. And take intersection of R-{r_i} over all i's.

Ah it's an exercise in munkres. I'll try that.

Thanks @TedShifrin

4:48 AM

Basically the proof of BCT.

5:16 AM

how about math.stackexchange.com/a/637809/27978

Alessandro Codenotti

5:52 AM

@Koro you can also prove that the space is completely metrizable, but in this case this is more complicated than just checking it I think

@AlessandroCodenotti By completely metrizable, do you mean metrizable to a 'complete metric space'?

@TedShifrin: yes, I have now understood that the proof is indeed as that of BCT. And using this, we first note that $\mathbb Q^c=\cap_{i\in \mathbb N} (\mathbb R -\{r_i\})$, a countable intersection of open sets in $\mathbb R$, i.e., a $G_\delta$ set. So by $G_\delta$ version of BCT, it follows that $\mathbb Q^c$ is a Baire space.

what a meagre proof

@AlessandroCodenotti But I wonder in such a case what would happen to the sequence $(x_n)$, where $x_n= \frac{\sqrt 2}n+1$

Alessandro Codenotti

@Koro A topological space $(X,\tau)$ is completely metrizable if there is a complete metric on $X$ whose induced topology is $\tau$

@Koro It probably won't be Cauchy

@AlessandroCodenotti Ah okay. Thanks.

6:09 AM

If $\omega$ is a nonvanishing smooth $1$-form on a smooth 3-mfd $M$ and $Y,Z$ is a basis of $\ker(\omega_p)$ for $p\in M$ then $(\omega\wedge d\omega)(X,Y,Z) = \omega_p(X)d\omega_p(Y,Z)$ where $X,Y,Z$ is a basis of $T_pM$. Correct?

I calculated by letting $d\omega = \omega^1\wedge\omega^2$ since $d\omega$ is a $2$-form so locally linear combination of $2$-form of the form $\omega^1\wedge\omega^2$.

with the fact that $\omega^1\wedge\cdots\wedge\omega^n(v_1,\ldots,v_n) = \det(w^i(v_j))$ for covectors $\omega^i$'s.

Why is math > physics?

How could I prove that math - physics > 0?

@CottonHeadedNinnymuggins doesn't really make sense.

6:29 AM

Hi, How to prove this: If $\{X_n\}$ is a sequence of i.i.d. random variables with positive mean then the series $\sum X_i$ diverges to infinity almost surely.

any ideas?

6:40 AM

@copper.hat It was a joke

@CottonHeadedNinnymuggins i got distracted, i meant to write that you have to integrate both side first.

@PNDas What does ${ 1\over n} \sum_k X_k$ converge to? (almost shirley)

sample mean converges to population mean

Hmm got it.

@copper.hat Lol true. And then we have to treat differentials like differentials and not fractions

Another question, If $X$ is a r.v. with finite expectation then $\sum_k P\{|X|>k\}\leq E|X|<\infty$

Markov inequality doesn't help.

something along the line sof math.stackexchange.com/a/172857/27978

7:04 AM

What is the best theorem in mathematics? In my opinion, it's the Pythagorean Theorem

7:35 AM

Let X be a space in which one-point sets are closed. Suppose that $\{f_α\}_{α∈J}$ is an indexed family of continuous functions $f_α : X → \mathbb R$ satisfying the requirement that for each point $x_0$ of $X$ and each neighborhood $U$ of $x_0$, there is an index α such that $f_α$ is positive at $x_0$ and vanishes outside U. Then
the function $F : X → \mathbb R^J$ defined by $F(x) = (f_α(x))_{α∈J}$ is an imbedding of X in RJ. If fα maps X into [0, 1] for each α, then F imbeds X in
$[0, 1]^J$ .

I just want to know that in this theorem don't we also need that $f_a$'s are 1-1?

Because if f_a's are not 1-1, then can not be homeomorphism.

nvm, its proof is hidden deep in Urysohn's metrization proof.

8:19 AM

Jordan's lemma

In complex analysis, Jordan's lemma is a result frequently used in conjunction with the residue theorem to evaluate contour integrals and improper integrals. The lemma is named after the French mathematician Camille Jordan. == Statement == Consider a complex-valued, continuous function f, defined on a semicircular contour C R = { R e i θ ∣ θ ∈ [ 0 , π ]...

Jordan theorem Jordan lemma

Peter

10:30 AM

@CottonHeadedNinnymuggins When is a theorem "good" ?

schn

10:47 AM

How could I tell from a graph of a function if it is uniformly continuous or not? What specific traits do I need to look after?

Say for example $f(x)=\sin{(1/x)}$ over $(0,1)$ and $\epsilon<1$.

Q: Estimate variable using curve formula

11:15 AM

schn: one helpful guidepost is that boundedness of the derivative (if it exists everywhere) implies uniform continuity, and unboundedness of the derivative is potentially bad news.

your f has the property that for any delta > 0 whatsoever, i can find x and y in the domain of f with |x - y| < delta and |f(x) - f(y)| = 2. this property has a reasonably visual interpretation, and implies that f is not uniformly continuous.

Raffallo

12:07 PM

2

I'm not a mathematician, but a programmer who needs to understand how I can calculate the value. My curve is determined by: $$a.\cosh(\frac{x - p}{a})$$ I need to estimate p that will meet my requirements: a parameter will be given (for example 1138) distance on x will be given (for example 77.1...

geometry differential-geometry

Does $sl_n(\Bbb R)$ surject onto $SO_n(\Bbb R)$ via $A\mapsto e^A$?

1:04 PM

Is space of positive real numbers a Baire space?

yes

Q: $\lim_{n\to \infty}f(nx)=0$ implies $\lim_{x\to \infty}f(x)=0$

18

Can anyone help me with this problem? Let $f:[0,\infty)\longrightarrow \mathbb R$ be a continuous function such that for each $x>0$, we have $\lim_{n\to \infty}f(nx)=0$. Then prove that $\lim_{x\to \infty}f(x)=0$. Our teacher told first to prove Baire's theorem, and then show that this is a...

real-analysis limits baire-category

it's an open subset of a baire space (all of R, which is baire by some version of the BCT)

This question was asked in my exam. Of course, I skipped it. I didn't know it had something to do with BCT.

I had no idea actually that BCT could be used here. I tried with limit definition but got stuck.

yeah, the way 'baire' is usually defined, you need some version of BCT just to know that a space is baire.

once you have a library of examples, putting positive reals into that library is not too much work.

1:08 PM

@leslietownes Ah right, I know this theorem.

open subspaces of Baire spaces are Baire.

But it has a very unnatural and complicated solution. I wonder how anyone who's never seen this before could do it.

this one: $\lim_n f(nx)=0\implies \lim_{x\to \infty} f(x)=0$

i don't know. on some level, all of real analysis is unnatural and complicated.

the gowers blog post linked in that thing is an attempt to make it seem natural. it sorta works.

it's a good contribution to the 'fields medalist attempts to explain something simply' discourse, anyway.

This exercise seems to be hard. I remember that I failed to solve it when I was a freshman.

i think there is a spectrum of opinion on this, sometimes people criticize me for being too formalistic, symbol-pushy, whatever. but, i internalized fairly early that my own notions of 'natural' and 'uncomplicated' were not very useful and if i wanted to make any progress i should discard them in favor of ripping off what others did without regard for whether i could have come up with it.

@Yai0Phah exercise is one thing but putting that in an exam?

Some professors prefer to include "hard exercises" in exams. It is hard to judge.

1:21 PM

internally, "could i have come up with this" just isn't a very useful question. but many great mathematicians and some not-so-great mathematicians would disagree with that.

@leslietownes complete metric spaces are Baire

@Koro well, $G_\delta$ subspaces of complete metric spaces are completely metrizable

you could also use that

i agree. i'm surprised someone hasn't offered a metric that turns (0, infty) into a complete metric space.

I also remember another exercise to show that the smooth function $f\colon\mathbb R\to\mathbb R$ is a polynomial function if, for every $x\in\mathbb R$, there exists $n\in\mathbb N$ such that the $n$-th derivative $f^{(n)}(x)$ vanishes.

@leslietownes yeah, you're right. Discarding natural and uncomplicated.

@Yai0Phah O_o

there's a whole genre of baire results that are a form of black magic.

kolmogorov at least partially made his early reputation with a proof that there's a lebesgue integrable function whose fourier series diverges everywhere. as it turns out, that's a generic phenomenon, and the proof that it's generic is considerably simpler than whatever he came up with.

but 100 years ago you make a reputation on that.

1:30 PM

What do you mean by a "generic" phenomenon?

ILikeMathematics

in the baire sense, meaning, holds on a countable intersection of open dense sets.

This exercise is delicate: remove continuity from the hypotheses and the result is not true.

my own post: math.stackexchange.com/questions/4342922/…

i'm realizing maybe you don't get kolmogorov's example with the argument i'm thinking about. but you get that a dense G-delta set of continuous functions on the circle has the property that their fourier series diverge on a dense G-delta set of real numbers. which would have been news in the 1920s but is now a boring corollary of the BCT.

Yes, you can somehow understand this phenomenon as follows: think of $g(n,x):=f(nx)$, and we know that $\lim_{n\to\infty}g(n,x)=0$. The continuity gives some sort of continuity of $g(n,x)$ with respect to the parameter $x$, and probably also some control of this continuity, and the result is some sort of control of the "uniformity" of $\lim_{n\to\infty}g(n,x)=0$ with respect to $x$.

ILikeMathematics

2:16 PM

More detailed than in the image here:
https://math.stackexchange.com/questions/4586877/on-endpoints-not-being-able-to-be-local-extremum-after-the-definition

4:12 PM

I have a small question about compactness

so the open interval (0,1) is not compact since we can find an open cover with no finite subcover

i know this is wrong by heine borel theorem but

what if we add two open sets

and cover [0,1]

wont that be an open cover of [0,1] that has no finite subcover?

the two added open sets will cover only the endpoints

Ie the singletons {1} and {0}

it is clearly wrong argument but where does it fail ?

@robjohn @TedShifrin @LukasHeger @leslietownes any ideas ? ;)

@JackOhara the singletons {0} and {1} are not open in [0,1], so those two added open sets will cover more than just the endpoints

my intuition is that if E is a subset of K , and K is compact then so must be the subset ( since it is smaller )

@LukasHeger Oh I see good point

but what about

[1, inf ) ?

that's not compact

that is closed i think so will not work

yeah it's not open in R

4:18 PM

No I was thinking of that as a potential open set to cover just the endpoint 1

any open set containing 1 will contain an interval of the form (a,1] for some a<1

it seems a bit strange that a somewhat smaller set may be not compact using this defintion of covering , while the larger set is compact

i mean if we could cover the bigger one with finite open sets, we should be able to do the same for the subset !

@JackOhara no that's not true as fthe example shows

well, compactness is a topological property and (0,1) is homeomorphic to R and that's clearly not compact

Yes thanks ! I was just saying how I intuitively saw it, I tried to find where the flaws in the argument were

but it seemed in picking some open cover that contains 1

has to have infinitly many more than just 1 point

yes, that's a good description of where your argument goes wrong

4:22 PM

Thanks so much ! this drove me crazy for a bit =)

schn

5:51 PM

I have an integral over $\mathbb R$ which equals zero. Can we then say that the integrand must be zero?

6:15 PM

@schn no

6:28 PM

@schn Would you think that if it were on $[0,1]$?

6:50 PM

if a sum is zero, do the terms in the sum have to be zero?

But of course.

7:06 PM

If a + b = 0, then b = -a.

just giving the caveman version of schn's question.

Cavemen didn't know about negative numbers; yet, they did live in negative temperatures.

the aliens that made them build the pyramids certainly did

that reminds me of something funny. augustus de morgan's father-in-law was something of a crank who denied the existence of negative numbers. coincidentally, one of de morgan's books about arithmetic has a whole lot of detail setting out the rationale for negative numbers.

we are all a product of our times.

Arithmetic was the foundation of algebra in those times.

7:19 PM

my father in law is annoying but he isn't that annoying.

The Bourbaki took annoying to a whole new level.

(speaking of history :)

huh didn't know they were still active, according to wikipedia

History repeats.

I don't understand why Bourbaki gets so much hate. For example the commutative algebra volumes are still valuable references. Obviously the taste of the group is very algebraic, but there are some gems in the Bourbaki. Obviously it's a bad idea to teach from a Bourbaki text, say on topology. But that's not the fault of a reference work...

does bourbaki get hate? a lot of the people i worked with barely knew who bourbaki were.

7:29 PM

and let's not forget the mystique of a secret french group publishing under a pseudonym is very aesthetic

yes I think Bourbaki gets hate. For some reason people think that Bourbaki is responsible for a lot of bad didactics

i guess the closer you are to algebra, or somehow otherwise to bourbaki's 'core,' the more you have to deal with opinions about it, and i guess its influence.

i don't blame bourbaki for that, i blame french people.

6

They explicitly said their books were not meant to teach from; but, the educators didn't listen.

Sort of like R.L. Moore saying his method wasn't meant for the average (colored) student.

moore worked overtime to make sure there were other reasons not to listen to him.

that might be the true 'moore method'

He certainly worked overtime producing "disciples."

7:50 PM

i don't blame moore for that, i blame the uni of texas for allowing him to continue for so long

Q: Estimate variable using curve formula

Does anyone know Bogachev's Real and functional analysis?

I'm tempted to give it a look

Raffallo

8:06 PM

2

I'm not a mathematician, but a programmer who needs to understand how I can calculate the value. My curve is determined by: $$a.\cosh(\frac{x - p}{a})$$ I need to estimate p that will meet my requirements: a parameter will be given (for example 1138) distance on x will be given (for example 77.1...

geometry differential-geometry

$C_0(\mathbb R)$ is the set of bounded continuous function vanishing at infinity. Then $C_c(\mathbb R)$ is dense in $C_0$. I want to take the sequence to be $f\chi_[-n,n]$ and then join the both ends by linear functions to x axis then it is constantly zero.

Is there any flaw you can see?

that will work, if the idea is that if f in C_0 is given, your recipe produces a sequence of functions in C_c that converge to f in sup norm.

Hmm that's what I wanted to do.

In my phone keyboard, both the forward and back slashes are represented by forward slash. The symbol for black slash is a little bit longer though.

And I kept looking for the back slash for a few minutes like a stupid.

it might help to be a little clearer about how you finish those functions off. e.g. you could do that straight line thing in [-(n+1), -n) and (n, n+1]. from the verbal description it isn't clear exactly how your approximants have compact support.

if i were super nit picky i would also desire clarity as to what ||f - f_n|| is bounded by in concrete terms.

but i am not.

Support is the closure of the set on which function is non zero

So as long as it is bounded it is compact

8:16 PM

So, I've seen the proof of Nagata-Smirnov metrization theorem few years ago already. But I realized that there was another metrization theorem, Bing's metrization theorem, the proof I haven't seen. So I wonder, how do I show that a metric space has a sigma-discrete basis.

@leslietownes hmm I should give a clearer argument.

you have the right idea, and that is enough for me. :D

but yes, if i were in nitpick mode, it would be helpful for the argument to exhibit, given n, an interval I_n on which the approximant f_n is supported, and a value of ||f - f_n|| that somehow makes clear it goes to zero.

Now that I am writing it down it seems difficult to do.

8:34 PM

So suppose that we have some locally finite basis

and I suppose the trick would be to for each point from a countable dense set to find a nbd which intersects a finite amount of those buggers

hmm... I guess that wouldn't exactly work

8:48 PM

This is consequence of Stone theorem 4.4.1 in Engelking, which says not only that metrizable spaces are paracompact, but also that the refinement can be taken to be sigma-discrete

Engelking, what a surname

primary reference for general topology

PM 2Ring

10:31 PM

@Raffallo Sorry, that question isn't very clear. How can y=7.05 when a=1138? $\cosh(x) \ge 1$ for all real $x$.

@user2236 Kind of. In those days, geometry was the foundation of all mathematics, including arithmetic. Algebra was merely a bunch of clever tricks that could be used to assist in solving geometry problems.

Even by the time of Euler (~ a century before Morgan), negative numbers were still treated as a bit of a novelty. Although mathematicians had been using negative numbers for centuries, they weren't considered to be legitimate numbers. They were just a notational trick. If you have a directed line segment of length -3 heading east, that's just a funny name for a line segment of length 3 heading west.

Is there a difference if for the partition of unity $f_\lambda$ we demand that $\text{supp}(f_\lambda)$ is locally finite vs $\{f_\lambda \neq 0\}$ is locally finite?

PM 2Ring

Similarly, expressions like $x^2 + x^3$ don't make sense geometrically, unless you include an extra factor of 1 with the $x^2$ to make it 3 dimensional.

I see it now that I wrote it

wikipedia cites something that I'm not sure is true
> In fact, a T1 space is Hausdorff and paracompact if and only if it admits partitions of unity subordinate to any open cover (see below)

$\implies$ is clear but I doubt the other direction is true

PM 2Ring

10:52 PM

@Raffallo OTOH, if you just want to find $p$ such that $\Delta y = a\cdot\cosh\left(\frac{x-p}{a}\right) - a\cdot\cosh\left(\frac{x}{a}\right)$, given $a, x, \Delta y$, then that's straightforward, and doesn't require Newton's method.

okay I got confused. But there seems to be a proof in Dugundji using nerves of covers

Thorgott

@Jakobian no, a collection is locally finite if and only if the collection of the respective closures is locally finite