Mathematics - 2024-10-21

psie

04:53

I have a basic question. In Rudin's PMA, by which theorem/proposition can you deduce that $0^p=0$ for $p>0$?

I accept that $0^p$ equals $0\times 0\times 0\times\cdots\times 0$ whenever $p>0$ is an integer, but if it's any positive real, then I'm in doubt.

Jakobian

06:28

@psie how do you define it

for instance, you can define $0^p$ as $0^p := \lim_{n\to\infty} 0^{p_n}$ where $p_n$ is any sequence of positive rational numbers with $p_n\to p$

then one verifies $0^p = 0$ for rational number $p$

looks like Rudin defines powers for numbers $ > 1$ here

which seems to be the only place where he does that, which is interesting that he half-assed that

psie

06:44

@Jakobian yeah ok, you mean part (c) right?

oh wait

sorry

Jakobian

but yeah, Rudin's PMA doesn't define what $0^p$ is

psie

yeah, doesn't look like it

that's a shame

Jakobian

yeah, I suppose Rudin's PMA is not a book which starts from the ground up

or at least it doesn't cover everything

either way should be obvious that one should adopt $0^p = 0$ for $p > 0$ as a definition

psie

ok, so you'd put it as a definition rather than something to be proved?

Jakobian

probably, yes

psie

07:00

actually, I don't know where to find this result otherwise, couldn't find it in my lecture notes either

leslie townes

psie: any proof oriented textbook would have to spend some time defining what "a^b" is before a statement like "a^b = c" becomes something meaningful in the context of that book, let alone something "to be proved." but yeah it is basically up to the author

what's going wrong here, maybe, is your expectation that rudin should define it. why? does he need to use it anywhere?

if he somehow makes an appeal to 0^p for general positive real p somewhere, okay, he should have defined it. if he doesn't, it's just.. i don't know. something it is OK for the book to be silent on. i don't imagine that it's actually that common for real or complex analysis books to have a need for 0^p for general positive real p

psie

@leslietownes I think it is important, especially in regards to his other books. Consider the $p$-norm. How do you prove $\|x\|_p=0\iff x=0$ for $x\in\mathbb R^n$ say? Well, for $\implies$ direction, we have $\left(\sum_1^n |x_j|^p\right)^{1/p}=0$. Now I'd like to raise both sides to the power of $p$, a real number between $1$ and $\infty$. But we can't, because $0^p$ is not defined.

leslie townes

he implicitly defines it for positive integers in a remark about notation in general fields (1.13 (a) in the edition i'm looking at now), where he treats it more as a notational thing than something that needs formal definition. implicitly he is assuming that the reader will be OK with the fact that this remark might be all he says about the case for positive integer powers

psie: i see that as a good case for, there certainly would have been good reasons for him to include it. but he certainly doesn't need to define it in one book because of what he might have used or done in another book. that isn't how anything works.

it's also fine for an analysis book to introduce the p-norms without defining any of this stuff. i don't imagine folland does, for example. it's just part of the assumed background for a given book, or not. which is something that varies, sometimes considerably, from book to book

Jakobian

@psie well, you should have worried about this when defining $\|x\|_p$ in the first place

either way you are stuck on something of no impact

leslie townes

"this might be referenced in another book, so it has to appear in this one" is just, no. not at all. shouldn't every real analysis book also be a complex analysis book by that measure?

psie

07:10

@Jakobian what do you mean? $\|x\|_p=\left(\sum_1^n |x_j|^p\right)^{1/p}$ is the definition (for $x\in\mathbb R^n$)

Jakobian

@psie now put $x_j = 0$ for any $j$

if you didn't define $0^p$ then $\|x\|_p$ is not well-defined

psie

I see

Jakobian

what it morally should be, is $0$, and so, regardless of someone's definition, you should use that $0^p = 0$

that's what its meant to be when defining $\|x\|_p$ for example.

Its irrelevant who defined it in what way. Lets move on

psie

@Jakobian by saying what you're saying here, there's nothing to prove in $\|x\|_p=0\impliedby x=0$. Ok, whatever.

Jakobian

does there always have to be something to prove that isn't completely obvious from the definitions?

Of course not

leslie townes

07:16

if the background motivational question "why is rudin so worried about what a^b is for some values of a and b and not other simpler values of a and b" it's because, first and foremost, rudin is addressing the cases he actually intends to use in his book (and i don't think he introduces p-norms in PMA), and second, he is focusing where he thinks his reader may need to invest time and effort to make good sense of things, and apparently in his view, 0^p is not one of those cases

Jakobian

well, Rudin is still half-assing this. He only defines $a^b$ for $a > 1$ and $b > 0$, in an exercise

leslie townes

and he's not getting into that any more than he sets out in a formal definition what a^m is when a is an element of an abstract field and m is a positive integer. he just does stuff. it's his vision of an analysis book and not whatever book you want it to be

jakobian: well, in a later chapter he uses the exponential function to define a^b for a more sensible domain (which admittedly still excludes 0)

he may even refer to how he's going to do this later in the earlier chapters

Jakobian

@leslietownes oh, he does? I browsed first 6 chapters I believe and didn't see that

leslie townes

jakobian: well, i guess browse a few more chapters :)

Jakobian

of course, I've stopped at chapter 7 and it was in chapter 8

leslie townes

07:21

i wouldn't say he is half-assing it unless he actually makes use of something that he didn't define, or defined only partially in an exercise

if he's just deferring discussion of something he doesn't actually use until later, he's just, organizing a textbook differently from what other people would have done

Jakobian

okay sure, he isn't half-assing this definition, as chapter 8 shows, he's just waiting to define it in terms of logarithms and exponentiation

leslie townes

working through the partial definition he hid in that exercise is a pretty good education in why it maybe isn't that great of a definition to work with, particularly for beginners

maybe a rare moment of pedagogical insight from walter rudin

Jakobian

its certainly the most natural one

Pizza

09:10

Hi

1

Q: Calculate the integral with the Gauss Green formula

Calculate the following double integral (directly and with the Gauss Green formula) $$\iint_D xy\,\mathrm{d}x\,\mathrm{dy}$$$$D={(x,y)|y\le x \le \sqrt{y+2}, 0\le y \le 2}$$ directly: $$\int^2_0\int^\sqrt{y+2}_y xy \ \ \mathrm{dx} \ \mathrm{dy}$$$$\int xy \ \mathrm{dx} = y \ \cdot \int x \ \mathr...

@SineoftheTime I was reviewing this exercise, but now I realized, but here the triangle wasn't supposed to be divided in 2?

psie

Maybe $0^{1/p} \neq 0, p>0$ leads to a contradiction of the field axioms?

Zac U.

Is there a way to essentially have a sequence of positive integers and determine how many structures that can remain balanced can be built if the bottom row has the first number of perhaps squares or rectangles and then the next is stacked as so forth

Thinking squares but then there's a question of whether they should be flush and additionally how would having a row of 1 affect

Jakobian

09:28

@psie field axioms don't concern exponentiation

Zac U.

If you build a rectangle of squares in my example they can function as a whole so that a middle square of a 3x1 wouldn't fall through a single gap

sai prabhav

hello

psie

@Jakobian ok

Jakobian

Let $u, v\geq 0$ and $s, t > 0$.
1. $u^tu^s = u^{t+s}$
2. $(u^t)^s = u^{ts}$
3. $u < v \iff u^t < v^t$
4. $0^s = 0$, $1^s = 1$
5. $u > 1, s > t \implies u^s > u^t$
6. $0 < u < 1, s > t \implies u^s < u^t$
7. $u^n = \prod_{i=1}^n u$
8. $u^{1/n}$ is the unique non-negative $n$th root of $u$

In the past I've came up with those conditions for what a function $(u, t)\mapsto u^t$ where $u\in R$ is an element of a partially ordered ring and $t > 0$ is a real number is ought to satisfy in order to be considered exponentation

in this post to be exact

Perhaps you can obtain $0^s = 0$ from some of the other axioms present here

@psie

psie

ok 👍

Jakobian

09:40

or maybe - all of those properties hold for a totally ordered ring with $n$th roots when $s, t$ are positive rationals

this already implies the need that perhaps this should also hold whenever we talk about real exponentation

certainly 7 and 8 imply that $0^s = 0$ for rational $s$. And 1 implies that $0^p = 0^s\cdot 0^{p-s} = 0$ where $0 < s < p$ is a rational number

so yeah, $0^s = 0$ can indeed be deduced from those axioms, namely 7, 8 and 1

I don't believe one can similarly deduce $1^s = 1$ for $s > 0$

well, for real numbers you could

so I guess axiom 4 is irrelevant if our ring is $\mathbb{R}$

Thorgott

11:11

you want exponentiation to be continuous, so $0^p=0$ for any $p>0$ it is

Jakobian

11:35

right, and $0^0 = 0$

in this case I'd be careful about justifying things as per continuity

Sine of the Time

@Pizza what do you mean? When you compute the double integral?

If you fix $y$, it's fine how you solve it. If you fix $x$ you have to divide the triangle

Thorgott

@Jakobian there's nothing else to go by anyway

monotony is just continuity in disguise

the fact that non-positive exponents cause well-understood problems notwithstanding

Jakobian

it does follow from my axioms 1, 7 and 8 above (note no monotonicity used)

7 and 8 are basically defining the rational exponents, while 1 is the property used

Pizza

@SineoftheTime Oh okay

Sine of the Time

the domain is $\{a\le y\le b, \, f(y)\le x \le g(y) \}$ so you don't have problems

Pizza

11:50

Oh yes I wrote it like that

Sine of the Time

clear?

Jakobian

maybe its a bit... sketchy, but if you are content that it satisfies some algebraic properties, you can derive it too

Gian's Pizzeria

I took the test today

Pizza

@SineoftheTime Yes

Sine of the Time

@Gian'sPizzeria how was it?

@Pizza If you have other questions you can ping me but now I have to go

Gian's Pizzeria

11:55

@SineoftheTime We don't talk about it

Thorgott

@Jakobian what's the trick?

Jakobian

If $t > 0$ then one can write $0^t = 0^s \cdot 0^{t-s}$ where $0 < s < t$ is a rational number

and since 7, 8 imply that $0^s = 0$ for rationals, it must hold for any $t > 0$

Jakobian

12:17

I really worked hard on those axioms, I think I've extracted the essence of real exponentiation with them though

Silly Goose

12:46

my text (Mathematical Physics by Arfken) defines a singular point of a complex variable function $f(z)$ as a point $z_0$ such that $\frac{df}{dz}\lvert_{z=z_0}$ does not exist.

The text later goes on to describe "singularities" which are points at which a complex variable function $f(z)$ is not holomorphic.

Are singular points then just special cases of singularities? By the definitions in this book it seems so, but the definitions do not seem very well organized.

ModularMindset

13:15

Honest question: Do things have to be "proven" in a physics paper?

In math(s) clearly the proof is one's duty

unimaginable consequences presumably emerge if one calls oneself a mathematician yet has no concern for the duty of proof.

I figured it out: depends on the audience

Has anyone invented path dependent holonomy yet?

sorry meant to say basepoint dependent holonomy. What that means to me is the holonomy depends on the starting point $p$.

and like different points give rise to different holonomy (from the standpoint of how much the vector is rotated after traversing a closed loop)

Xander Henderson

13:37

@ModularMindset Do you know what "epistemology" means?

Sine of the Time

Xander Henderson

And do you understand that math and physics have very different epistemological requirements?

think_meaning_builds

Doesn't that just mean existence?

Xander Henderson

@think_meaning_builds No.

think_meaning_builds

Ok, what then.

ModularMindset

13:40

It means @XanderHenderson that physics and math have different requirements for proof

Xander Henderson

Epistemology is the study of knowledge---how do know things, how is knowledge justified, what qualifies as sufficient evidence, etc.

@ModularMindset No.

ModularMindset

Actually that seems pretty useful

epistemology

think_meaning_builds

Ok, what exists as knowledge.

Xander Henderson

Physics is empirical. The epistemological foundation of physics (and all sciences) is that nothing can ever be known for certain, and that all conclusions are tentative and subject to overturning or refinement based on observations in the world. There is no such thing as proof in physics.

ModularMindset

what about mathematical physics?

Xander Henderson

13:42

@think_meaning_builds That is reductive to the point of misunderstanding, I think.

think_meaning_builds

What is understanding, then?

Xander Henderson

@ModularMindset It depends on whether the work is being done more for the mathematics, or for the physics.

ModularMindset

Interesting :)

Xander Henderson

A mathematician can prove that a statement is true within the framework of mathematics. But that doesn't prove anything to be true outside of that framework, and there is no way of proving that the "real world" matches the axiomatic framework of the mathematics.

@think_meaning_builds That is one of the questions of epistemology. Where does knowledge come from? How do we generate new knowledge? How do we verify that the things we believe are true are, in fact, true.

It is not about whether or not knowledge "exists" (I'm not even sure what that statement means).

Beyond that, I would suggest that a quick Google search for the word "epistemology" would likely be much more helpful to you. I'll bet that Wikipedia has a good article on it, and the Stanford encyclopedia of philosophy says useful things, too.

ModularMindset

Oh that's revolutionary. I can write a math paper and then relate it to physics at the end and then publish it as a physics paper because no proofs!

Xander Henderson

13:48

@ModularMindset Not really...

ModularMindset

@XanderHenderson wait why not though?

Xander Henderson

While physicists can't prove anything, they can verify whether or not the predictions of your paper align with observation.

And if you aren't making predictions, then you aren't doing science.

think_meaning_builds

If you haven't memorized it, then it does not exist to be tested on during exams @XanderHenderson :-)

ModularMindset

true Xander^^

that puts string theory in an awkward spot

Xander Henderson

@think_meaning_builds If that is your attitude, then you would likely not ever earn more than a C on the exams I write. A and B students need to generate new knowledge, beyond what can be memorized.

think_meaning_builds

13:51

Do you give open book exams?

Xander Henderson

@ModularMindset The "correct" use of mathematics in physics is generally "Assume some set of axioms which (hopefully) capture some important facet of the real world; prove some theorems in mathematics; run an experiment to see if those theorems correspond to real world observations".

@think_meaning_builds Not generally, no, because memorization (or routinization) is an important part of knowledge, and you can get a C by simply memorizing a bunch of stuff.

In any event, I need to head into the office now.

think_meaning_builds

Ok, thanks for the chat.

Ryder Rude

14:32

Re Animator has some very disturbing content, but the non disturbing parts are a great movie

Jakobian

@XanderHenderson I'd argue that a mathematician can't truly prove the validity of their own proof, so they suffer from the same issue. Its just more nuanced. But I get that's not your point

Xander Henderson

@Jakobian You are going to have to expand on that.

Mathematicians make a certain set of assumed true statements (axioms) and define a process by which they demonstrate that new true statements hold (some system of logic). Within that artificial framework, it is certainly possible to prove true statements.

Jakobian

In mathematics, reading and verifying a proof is equivalent to doing an experiment in physics which would prove the validity of something. Both of these processes are subject to leading to false results

Xander Henderson

@Jakobian Um... what?

You seem to be invoking some notion of human error, but that isn't quite the epistemological argument that I am making.

Ryder Rude

i think mathematicians can compute that their proof is valid. but they can't show that their axiomatic system is consistent

but I would say this is different from physics experiments

Jakobian

14:39

@RyderRude not with 100% certainty, and neither can a computer

Xander Henderson

In the epistemology of mathematics, it is possible to make a statement and say, with absolute certainty, that the statement is either true or false, within the framework of whatever axioms have been assumed and whatever logical system is being used.

Ryder Rude

@Jakobian but the proof is finite and u can just check if each step follows from the previous steps

Xander Henderson

Contrast this to the sciences, where nothing is ever assumed to be definitively true.

Jakobian

@RyderRude but then you'd have to check that you checked correctly, and check that you checked that you checked correctly, and so on

so its also probabilistic

Xander Henderson

We can quibble about human error, but that is a distinct issue from the underlying epistemology. Everything done by humans is subject to error. But that is a distinct claim from the nature of mathematical knowledge as compared to scientific knowledge.

Ryder Rude

14:42

@Jakobian good point

Xander Henderson

The mathematician claims to have proved a statement with certainty; the physicist claims that their experiment is consistent with their theoretical framework, and is therefore evidence that their theoretical framework does a good job of modeling reality.

think_meaning_builds

5

Q: How Exactly Do You Define Truth?

I've been trying to learn about the multiple theories of truth and I've taken a look at the popular Stanford article. The first section has this to say about the neo-classical theories of truth: These theories all attempt to directly answer the nature question: what is the nature of truth? They ...

epistemology truth

Ryder Rude

@Jakobian i guess physicists not only have to check that they computed their prediction correctly, but also that it agrees with nature. mathematicians only have to do the former

Jakobian

@XanderHenderson ah. Sorry, I think I get it now

Ryder Rude

like, a mathematician checking that their proof is correct is synonymous with a physicist checking that they computed their prediction correctly

but physicists do an extra step too, checking the prediction with nature

also, i do think that math is like physics in other aspects, in the sense that mathematics also makes predictions

predictions that u have to check

e.g. mathematician shows that there is no largest prime number. this makes a prediction that a computer will always keep finding prime numbers

and one CAN doubt the prediction. e.g. some mathematicians doubt if Peano arithmetic is consistent, which means they dont trust its results

im not sure if my example about prime numbers works, but in general, it is possible to doubt if a mathematics result holds for a real world computer

think_meaning_builds

14:52

Gottlob Frege - 19th and 20th Century Philosophy

►

Jakobian

eh. I'm tired of those logic and philosophy topics

Xander Henderson

@RyderRude So are the disturbing parts. They are essential to the movie.

Ryder Rude

@XanderHenderson yeah.. the bad disturbing part I meant is the rape bit

the other disturbing parts are fine

they could've made the movie without that part

@XanderHenderson what are some other movies like this? i have Rise of the Living dead on my list

i give Re Animator 8 or 9/10. it's fast paced and twist-y

Gian's Pizzeria

15:11

@SineoftheTime ...the test went badly

Ryder Rude

@Gian'sPizzeria :(

The Empty String Photographer

15:36

BREAKING NEWS:

in The Nineteenth Byte, 39 mins ago, by caird coinheringaahin g

Babe, wake up, they found a new largest known prime number

Wait this was 9 days ago!

Sine of the Time

@Gian'sPizzeria sorry to hear that

Jakobian

16:26

@psie math.stackexchange.com/questions/4987744/…

we were talking about the exact same thing the other day

ModularMindset

@RyderRude is it true that it takes a full 720 to get a fermion back to it's original orientation. might have said that loosely?

Xander Henderson

@RyderRude What do you mean by "like this"? Are you looking for more splatter horror? If so, I'm not sure that I have a lot of recommendations---it is not a genre I like very much. Early Peter Jackson films (Braindead (or Dead Alive in the US) in particular) might be to your taste.

Evil Dead and Evil Dead 2, perhaps?

A little less gloopy, but still in the realm of body horror are Carpenter's The Thing, Scanners, and Cronenberg's The Fly.

ModularMindset

16:42

I think I'm becoming better at math

I wrote a 21 page paper in 2 months lol

Xander Henderson

(1) Number of pages is not a great metric, particularly in mathematics, where brevity is fetishized.

(2) The actual writing (as in typing it out) is kind of irrelevant. When you say that it took you 2 months, are you talking about the writing, or about the entire process from "Is this theorem true?" to typesetting?

(I can bang out 20 pages of math in an afternoon, but that isn't really where the work is.)

ModularMindset

@XanderHenderson It's a draft right now - have to consolidate sections - have someone review it - other edits

It's a very good project for me actually

because I've never written an actual paper before

I'm learning a lot

ModularMindset

17:03

learning organizational skills mainly

how to not use the same letter for different things

@XanderHenderson I'm curious - where is the work is? Is it in proving the main theorems decisively?

Xander Henderson

@ModularMindset Asking the right question, phrasing the correct theorem, proving that theorem, etc.

psie

@Jakobian yup

@XanderHenderson to bang...is that appropriate? :) Sorry, but that use of the word just caught my attention.

Xander Henderson

@psie Why wouldn't it be?

psie

Sure, whatever floats your boat.

Xander Henderson

"to bang out" is a pretty common English phrase. While I suspect that I know what you are thinking, that simply isn't what the phrase means.

psie

17:18

ah, ok, learning something new then :) I will try to use it henceforth.

ModularMindset

17:28

I am officially naming my future daughter Zeta-space. If twins then, Zeta-space and Zeta-time

Holonomic-detractor1 actually

Xander Henderson

When working on my masters degree, a colleague had twins. I bought them a pair of onsies, each with a single word on the front. One read "epsilon", and the other "delta".

SoG

18:36

How to upload image here( chat)?

Sine of the Time

18:48

@SoG near send there's upload and then you can paste the link of the image or browse it on the computer

Claudio

19:32

how would you guys try to show that $$ f(x,y) = \frac{x^2y}{|x|^3+y^2}, (x,y) \ne (0,0) \text{ and } f(x,y) = 0, (x,y) = (0,0) $$ is continuous at the origin?

Ok, I've tried with polar coords but I ended up with $\rho g(\theta,\rho), g = \frac{\cos^2 \theta |\sin\theta|}{\left\vert\rho|\cos\theta|^3+\sin^2\theta\right\vert}$ but $0<|g|<1/\sin^2\theta$ which diverges for $\theta = 0 + k\pi$

Sine of the Time

19:52

maybe something like $x^2y\le \frac 12 (x^4+y^2)$ can be useful (?)

Claudio

but that inequality isn't obvious is it?

Sine of the Time

you get it from $(a-b)^2\ge 0$

Xander Henderson

@Claudio $(x^2 \pm y)^2 = ...$?

and squares of real numbers are nonnegative?

Claudio

oh yeah

sku

Hi All, I have a question. The Bolzano-Weierstrass theorem states that "every bounded sequence contains a convergent subsequence". My question is that the theorem does not say that the given bounded sequence is convergent. Is this correct? Suppose we have a sequence (1,-1,1,-1,...). This is bounded but it has a convergent subsequence? May be I am misunderstanding this theorem.

Xander Henderson

19:57

(Basically, the same thing @SineoftheTime said, but I'm trying to be more cryptic.) :D

Sine of the Time

it's the AM-GM inequality, isn't it?

for positive numbers

leslie townes

sku: you are correct. and that bounded example that you give does indeed have many convergent subsequences

Xander Henderson

@sku That is correct. A bounded sequence must have a convergent subsequence. The sequence itself needn't converge.

Ben Steffan

Any subsequence of a convergent sequence is also convergent, so that would be a rather boring statement :)

Claudio

@XanderHenderson hahah yeah thanks anyways

Xander Henderson

19:59

Note that $(1,1,1,1,1,1,1,1,1,1,1,1,1,1,\dotsc)$ (for example) is a subsequence of $(1,-1,1,-1,1,-1,1,-1,1,-1,\dotsc)$.

Soumik Mukherjee

You really have patience to type that many 1s

Sine of the Time

Note that $(-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,\dots)$
(for example) is a subsequence of $(1,−1,1,−1,1,−1,1,−1,1,−1,\dots)$.

Soumik Mukherjee

You really have patience to type that many -1s

Xander Henderson

Note, also, that this is a statement about the real numbers, and that it doesn't necessarily hold in a more general setting. For example, $\mathbb{Q}$ does not have this property (consider the sequence $$ a_n = \left( \left(1+\frac{1}{n}\right)^n\right), $$ which is bounded in $\mathbb{Q}$ and does not converge in $\mathbb{Q}$. It also has no convergent subsequences in $\mathbb{Q}$ (though it does converge in $\mathbb{R}$).

sku

@XanderHenderson for the sequence of 1s, this means that in the proof of the B-W theorem by choosing successive halfs, somehow we will end up with a sequence of 1s. This part does not seem to come out from the standard proof of B-W.

I did not know that the B-W was for reals.

Xander Henderson

20:05

@sku It isn't just about the reals, but the version you have to hand is, almost certainly, about the reals.

@sku I don't know what you mean.

The proof should produce a convergent subsequence. This is not necessarily the most obvious subsequence.

But I don't understand "choosing successive halfs".

Sine of the Time

probably sku is referring to the proof with nested intervals

Claudio

ok yeah I give up, Ill check the solutions :P

sku

The proof I have from Stephen Abbott shows, consider [-M,M]. Make two halves [-M,0] and [0,M]. One of them will have infinite points. Choose that half and choose a number. Now divide this into two halves. Again choose a half with infinite points and choose a number that comes after the first number in the original sequence. And keep doing this.

It is not that important but I was somehow wondering if somehow this halving method will somehow produce a sequence ending with infinite 1s for the (1,-1,1,-1,...) sequence. But it is fine. I dont need to understand that.

leslie townes

sure. that argument will provide a convergent subsequence in this case. the language of the summary of the argument maybe makes it sound like no choice is involved - i.e., that there is a unique choice of convergent subsequence, corresponding to whatever choices you "have to" make at each step. but that isn't actually what happens.

Sine of the Time

@Claudio you might like this post

sku

20:12

Thank you all for the clarifications!

leslie townes

sku: at each step of that argument, it might be that both halves of whatever interval you're in contain infinitely many elements of the sequence. in that case you can choose either half. if that helps. i'm not sure if this relates to your difficulty

Claudio

@SineoftheTime that's of great help, thanks a lot

leslie townes

e.g. if you started with M = 10, you could initially choose whether to begin narrowing to [-10,0] or [0,10]. once you've done that, you are forced to choose whichever subinterval -1 is in (if you initially chose the left half) or 1 is in (if you initially chose the right half).

Sine of the Time

:)

leslie townes

so this argument doesn't really come close to suggesting the full variety of convergent subsequences of your example. it chooses constant sequences only

sku

20:15

please say more on what constant sequences mean

leslie townes

your example is a fun (if pretty rare) example of a situation where it is possible to identify both the set of all subsequences, and the set of all convergent subsequences, in "fairly concrete terms"

i.e. any sequence of -1s and 1s whatsoever is a subsequence of your sequence, and the convergent ones are the ones that are eventually constant

sku

Ha.. I see. Got it.

leslie townes

sku there's something a little weird about the way that argument is phrased. "one of them will have infinite points" - here "infinite" refers to the set of positive integers n for which x_n is in the given half of the interval, not to the set of real numbers x_n that lie in that interval (which might always be only finite, as it is in your example)

so, of the infinite bag of positive integers for which x_n is in [0,M], you choose one, n_1. then you look at the sets of positive integers n larger than n_1 for which x_n is in [0, M/2] or for which x_n is in [M/2, M], respectively. one of those two bags will also be infinite. your selections from those infinite bags are indices n_1, n_2, .. that you use to form the subsequence, not the values x_{n_1}, x_{n_2}, etc. of the sequence itself

ModularMindset

@XanderHenderson May I propose a theorem to you now?

it is related to what we were discussing cordially earlier today

Just want to get feedback :)

Maybe it's a first revision of a potential theorem I want to pursue

Xander Henderson

20:31

@sku The sequence $(1,-1,1,-1,1,-1,\dotsc) = ((-1)^n)$ is bounded between $-1$ and $1$, so $M=1$. In the first step, note that there are infinitely many terms in at least one of $[M,0]$ and $[0,M]$ (both, in fact). In particular, $a_2 = 1 \in [0,M]=[0,1]$. Take $b_1 = a_2 = 1$.

Then there are infinitely many terms of the sequence in either $[0,1/2]$ or $[1/2, 1]$ (this time, only the latter contains infinitely many terms). So choose some $a_k \in [1/2,1]$. I don't know... maybe $a_{10} = 1 \in [1/2, 1]$. Take $b_2 = a_{10} = 1$.

Now, there are infinitely many terms of $a$ in either $[1/2,3/4]$ or $[3/4,1]$ (again, only the latter this time). So choose some $k$ so that $a_k \in [3/4,1]$. How about $a_{48}=1$? Then $b_3 = a_{48} = 1$.

Keep going. Eventually, you will get the sequence $b$ defined by $b_k = a_{n_k} = 1$.

ModularMindset

I wonder what the implications are for co-vectors honestly

they seem almost useless

just my opinion though

leslie townes

21:59

0

Q: Operator strong $(p,p)$ implies adjoint strong $(p',p')$

Let $T$ be a linear and continuous operator $T: L^{2}(\mathbb{R}^{d}) \to L^{2}(\mathbb{R}^{d})$. I wonder to what extent can we conclude if $T$ is strong $(p,p)$, we will obtain that $T'$, the adjoint of $T$, is strong $(p', p')$. For simplicity, we can take strong $(p,p)$ to be restricted to $S...

real-analysis functional-analysis measure-theory fourier-analysis harmonic-analysis

my daughter would die laughing at the title of this post

Ben Steffan

chuckling

Xander Henderson

22:14

@leslietownes I larfed.

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$Mathematics$ Mathematics