Mathematics - 2023-05-26

12:20 AM

@D.C.theIII Nah. Just saw this.

12:35 AM

All good. I read the section and saw the use of the proposition. I'm just trying to put off going over series and sequences again until I do Real Analysis after this, since I know that they come up there again.

noballpointpen

12:50 AM

I think I have solved that problem. Do you want to see the proof, Ted?

Your hint was a cure.

But I am not sure why they included the hypothesis of the existense of infinums. If we reason from the point which you suggested, we have contradictions by just looking at the supremum. I know that if you don't use a hypothesis, then there is probably something wrong with your proof, so I am going to investigate until you enter the chat.

Ted Shifrin

@noballpointpen My guess is that you may need to look at the inf of the second set if the sup of the first doesn’t work.

noballpointpen

Ah yes, it might be the possibility that the one set is just empty, am I right?

Ted Shifrin

No, neither can be empty.

You need certainly to start with that point.

noballpointpen

But what if $x < f(x)$ for all $x \in S$? Then one is empty, no?

Ted Shifrin

Oh, if one is empty, then the missing endpoint must automatically work.

It can’t. Look at $b$ (the max element of $S$ if it has one)

noballpointpen

1:05 AM

Yeah, I actually just remembered about $\sup S$, which exists too. $f(\sup S) > \sup S$ is impossible.

Ok, I am going to continue. Thanks.

noballpointpen

2:15 AM

No, I reread my proof and found a mistake. It was about one case where $\sup$ belongs, and this case does not lead to a contradiction. And now it seems that even with your sets the same pattern (that one which before was a brick wall for me) appears.

I will write what I have.
So, we have $L = \{x \in S \colon x < f(x) \}$ and $R = \{x \in S \colon x > f(x)\}$.
The situation is that
$\sup L \in R$, $\sup R \in R$
$\inf L \in L$, $\inf R \in L$.

I am going to leave soon, so I think I better hit you tomorrow. Bye.

user10478

2:35 AM

Is there a name for the red equation here (math.stackexchange.com/a/4706597/109355)? It looks like a recurrence relation with respect to the subscript, but $a$ has an additional input that changes across the equation, much like a functional equation (in the narrower sense of the term). What key word can I look up to learn how these might be solved or approached?

Ted Shifrin

@noballpointpen From the fact that $f$ is monotone, you can deduce that $\sup L\le f(\sup L)$, etc.

Koro

4:17 AM

@robjohn I have started getting them only lately. Why?

Is it because of some rep threshold?

shintuku

the machine has decided that you've become a trusty and honorable member of the community koro

all hail the machine, blessed be its honored

Koro

5:45 AM

It's described here meta.stackexchange.com/a/271268/1133862. It's because of crossing a rep threshold.

shintuku

this time we can understand the machine, but soon we won't

Soumik Mukherjee

7:55 AM

Is there no room for algebraic geometry?

robjohn

8:46 AM

@Koro You got access to moderator tools when you passed 10K a couple of months ago. That might have affected your chat privileges, as well.

leslie townes

9:25 AM

remind me to be nicer to koro. :)

冥王 Hades

9:39 AM

Bro is about to suspend users who disrespect One Piece 💀

sunny

I want to find out for which $z\in\mathbb{C}$ the series $\sum_{k=1}^{\infty} \frac{2k+i}{k+2i} z^k$ converges. I have determined it converges absolutely if $|z|<1$, since the ratio test yielded $$\left|\frac{a_{n+1}}{a_{n}}\right|=\left|\frac{2(k+1)+i}{(k+1)+2i} \frac{z^{k+1}}{z^k}\frac{k+2i}{2k+i}\right|=|z|.$$ For $|z|=1$, we have that $\sum_{k=1}^{\infty}\left| \frac{2k+i}{k+2i} z^k\right|$ diverges.

However, now I am stuck on a conceptual issue; we have found out for which values the series converges absolutely, though might not some of the $|z|\geq1$ make the series converge conditionally?

Soumik Mukherjee

10:07 AM

@冥王Hades Isn't that zoro? Koro is from assassination classroom.

冥王 Hades

10:49 AM

@SoumikMukherjee Not Koro-sensei, this Koro loves One Piece

leslie townes

11:00 AM

sunny: this does not happen. if |z| >= 1 then the absolute value of the kth summand of that series does not go to 0 as k goes to infinity (it goes to 1 if |z| = 1 and to infinity if |z| > 1). this implies that the series diverges for all such z.

sunny: you can run into subtler examples, where you do need to worry about that kind of thing, in circumstances where the radius of convergence is 1, but the coefficient of z^k in the power series go to 0 as k goes to infinity. one example might be sum_k (1/k) z^k, where you have divergence at z = 1 convergence at z = -1 and maybe you need to think for a minute about about what might be happening elsewhere on the unit circle and analyze that separately.

you can definitely run into examples where the radius of convergence is 1, and the series converges for some collection of z with |z| = 1 and diverges for other z with |z| = 1, and where the convergence is not absolute when it exists.

Prithu Biswas

11:30 AM

Let k(n) be the sequence: sin(1),sin(sin(1/2),sin(sin(sin(1/3))),sin(sin(sin(sin(1/4)))),...
Does Σk(n) converge or diverge?

@Jakobian Oh ok. Thanks for the link =)

Soumik Mukherjee

@冥王Hades ohh

sunny

@leslietownes thank you, I am probably misunderstanding something fundamentally about absolute convergence. To summarize my confusion; if $\sum_{k=1}^{\infty}\left| \frac{2k+i}{k+2i} z^k\right|$ diverges for $|z|\geq 1$, does then $\sum_{k=1}^{\infty} \frac{2k+i}{k+2i} z^k$ diverge for the same $z$'s?

leslie townes

11:56 AM

if lim |c_k| fails to exist or is nonzero, then sum c_k (note the lack of absolute values) diverges. or equivalently, if sum c_k converges (without absolute values), then lim |c_k| (with absolute values) has to be zero. that is what is being used here.

sunny

Ok.

leslie townes

things can get much subtler if lim |c_k| is 0, but thankfully in your example, it isn't.

sunny

12:16 PM

@leslietownes I think this is known as the divergence test, although I have not seen it with absolute values, maybe that's a stronger statement and allows the summand to be complex; here (if you scroll down a bit) they write if lim c_k is non-zero, then sum c_k diverges

leslie townes

12:58 PM

it's not really a stronger statement but it is easier to work with in your example, for example because when |z| = 1 the absolute value |(2k+i)/(k+2i) z^k| does not depend on z, making it maybe slightly easier to understand the limiting behavior.

@sunny the way they've written it is maybe a tiny bit suboptimal. they have actually shown "if it is not the case that lim c_k = 0, then sum c_k diverges," but have written "if lim c_k is nonzero, then sum c_k diverges," which is a weaker statement because it arguably suggests that lim c_k has to exist for the test to yield information. (and for example in your case when z = -1 the limit (2k+i)/(k+2i) z^k fails to exist)

this is unrelated to your point about absolute values, i'm just tossing it out there

Shaun

Hi :) I just thought I'd advertise a bounty here:

0

Q: If a hom. $\phi:G\to H$ of diagonalisable linear algebraic groups is injective, then the induced hom. $\phi^*:X^*(H)\to X^*(G)$ is surjective

This is Exercise 3.2.10(2) of Springer's book, "Linear Algebraic Groups (Second Edition)". According to Approach0, it is new to MSE. The Question: Let $\phi:G\to H$ be a homomorphism of diagonalisable (linear algebraic) groups (over an algebraically closed field $k$). Denote by $\phi^*$ the indu...

It got a downvote :/

The downvote just now might be due to context. I'm not sure what else to add though.

Silly Goose

1:14 PM

Other than "group theory is the study of groups" How would you answer the question what is group theory about? Excluding any answer which only mentions applications of group theory or whose main point is to present applications of group theory

I don't really understand this blurb from wikipedia. A representation is a homomorphism, so the structure is only preserved along the direction from the Group into the Target. What is an example in which one gleans properties of the abstract group itself by representing it? An example also in which we are only interested in the group qua group. Not, say, SU(2) in the context of quantum mechanics.

Shaun

I have added context to the question:

-1

Q: If a hom. $\phi:G\to H$ of diagonalisable linear algebraic groups is injective, then the induced hom. $\phi^*:X^*(H)\to X^*(G)$ is surjective

This is Exercise 3.2.10(2) of Springer's book, "Linear Algebraic Groups (Second Edition)". According to Approach0, it is new to MSE. The Question: Let $\phi:G\to H$ be a homomorphism of diagonalisable (linear algebraic) groups (over an algebraically closed field $k$). Denote by $\phi^*$ the indu...

linear-algebra group-theory abelian-groups algebraic-groups characters

Group theory could, in some sense, be described as the study of symmetry and its abstractions.

Silly Goose

Hm...well perhaps an analogous situation like using cat theory for Brouwer fixed point theorem in which one translates toplogy into algebra to yield a contradition exists in which one translates group theory into linear algebra to yield a contradiction and prove a theorem or something.

does anyone know of an explicit example as above ^ :0

noballpointpen

1:39 PM

I am not doing math right now, but as looking at what I also had yesterday, we have
$\inf L \leq \inf R < \sup L \leq \sup R$,
$\inf R < f(\inf R) < f(\sup L) < \sup L$.
When I was away from my desk I thought that I could deduce a contradiction from the second inequalities.

As I said, I already witnessed this and other patterns when tried to solve the exercise back in time. Maybe the definitions of your sets will help here, I will work on it today.

Silly Goose

2:06 PM

Jakobian

@SillyGoose usually people say it's the study of symmetries

Silly Goose

Hm what do you all think of this. It is meant to be like a prelude paragraph to give the reader a unifying theme to guide them in learning some basic group theory concepts. the audience is for an udnergrad physics person but who is pretty mathematically inclined.

I suppose I should also mention that the notes are meant to really teach the bare minimum finite group theory concepts and go quickly into rep theory/lie theory stuff. but i think the basic concepts of finite group theory are most important to get across for this particular situation

noballpointpen

4:12 PM

Hm, from $f(\sup L) < \sup L$, we could have $e \in L$ with $f(\sup L) < e < \sup L$, by definition of $\sup$ and the fact that $\sup L \in R$. But then $e<f(e)$ and $f(e) < f(\sup L)$, which is a contradiction. But then I do not see why infinum makes sense here, besides that we can infer the same contradiction with it.

Do you think this contradiction furnishes the proof? (To Ted.)

This problem is nuts.

noballpointpen

4:34 PM

Uh, I googled how to post images in this chat. Seems like you need 100 rep for an "upload" button to appear. Ok. Then I will leave a plain link to the image, if you want to see a full proof.

leslie townes

heh, it's like jeopardy. a proof is there, and we guess the theorem? :)

noballpointpen

Hello, leslie! It was meant to Ted, he knows which theorem :) But I see you are curious: Let $S$ be a nonempty ordered set such that any its subset has a supremum and an infinum. Suppose $f \colon S \to S$ is monotically increasing. Prove that there is $x \in S$ with $f(x)=x$.

"any nonempty subset"*

Do you think my proof is correct?

leslie townes

4:49 PM

why is L nonempty?

noballpointpen

$\inf S \in S$, so $\inf S < f(\inf S)$,
$\sup S \in S$, so $f(\sup S) < \sup S$.
It follows from $S \subseteq S$.

leslie townes

why is inf S < f(inf S)?

noballpointpen

If $\inf S = f(\inf S)$, we are done. The other direction is just not possible.

leslie townes

i'll stop being ted for a minute. the hypotheses do not rule out S having only one element, where f is the only function it can be, and in this situation L and R are both empty.

noballpointpen

But then $f(x) = x$ for this lone element, no?

leslie townes

4:53 PM

yes. i'm not saying that the result fails, only that the first line of your proof, "Both sets are not empty," is using stuff that hasn't been stated about S.

noballpointpen

Heh, thanks for pointing out. I should be more accurate ruling out other cases.

Perhaps I acquired this bad habit from Rudin's short proofs.

leslie townes

there's a good lesson from rudin in here somewhere, which is, maybe there's some way of writing the argument that doesn't involve handling the case where S has one element separately from the other cases.

if rudin could find such an argument, he'd probably print that one, and forget that he ever thought about the one-element case as its own thing.

noballpointpen

Yes.

leslie townes

maybe even tell others that he'd never think of such a thing, and why would anybody else think of that.

noballpointpen

:D

I found the other side of the coin when I was dealing with logic. The proofs of semantic satisfaction are technical and precise, and my leaps resulted in profound gaps in my proofs, which do not teach you to understand why something holds there.

You didn't say anything about the proof in overall. Does it work?

leslie townes

5:02 PM

if you assume for purposes of a contradiction that f has no fixed points, then L union R = S certainly follows from that, and then the fact that S is nonempty tells you that at least one of L, R is nonempty. but i don't think it immediately tells you that both L and R are nonempty. there still seems to be an argument missing there.

the weird thing of course being that if the result is true and we're assuming it's not for purposes of finding a contradiction, absolutely any goofy thing can happen.

noballpointpen

Doesn't it immediately follow from this?

leslie townes

well, yes, but you see how this is coming out in fits and starts? i've got "assume for purposes of contradiction that f has no fixed points" down here, and up there you're implicitly using that hypothesis to deduce that inf(S) is in L and sup(S) is in R. but you don't say it up there.

noballpointpen

So you're again pointing out that I should fill in the details?

leslie townes

inf(S) in S and f being a function into S implies that inf(S) <= f(inf(S)). to get < is where you're assuming "well, if we have = we're done and there is nothing to prove, so focus on the interesting case," or equivalently that f has no fixed points for purposes of looking for a contradiction.

noballpointpen

Ah, I see. You mean that the sentence stating the nonemptiness is before the other that must be before?

leslie townes

5:11 PM

so yeah, it helps to add more detail. a proof is a series of statements, all of which are true. but a series of true statements doesn't feel like a proof unless those statements go in an order that makes sense, and where there's some explicit hint as to where the hypotheses are being used that make the true things true.

if i just hop into a proof and someone says "inf(S) < f(inf(S))" without filling me in on what they've done to get there, i get confused, or i begin thinking about one-point sets where that couldn't be true, and where and how we left the world of one-point sets.

so yeah, i agree that if you assume f has no fixed points, then your sets L and R are both nonempty, for example, because inf(S) is in L and sup(S) is in R (and note that this also proves in passing that inf(S) and sup(S) are different and that S has more than one element, whether or not we decide to notice that).

with those provisos, i agree with your first two sentences :). your third sentence, assume L union R = S, is that assumption (for purposes of argument) that f has no fixed points. that should be the second sentence, because it's what makes your current second sentence, that L and R are both nonempty, true. (i'd also maybe add some explanation of why L and R are both nonempty, e.g. because you've exhibited elements of both of them).

noballpointpen

Yeah, I noticed this incorrect order of sentences. Now corrected :)

leslie townes

i agree that if x is in L then so is f(x), and that if x is in R then so is f(x), and that both of those are close enough to how L and R are defined that maybe they don't need explanation. if i had to explain i would just say it's by the definition of those sets.

you say "it follows that sup L in R, inf R in L." i'm guessing at this point that it does, but i'm also guessing that you're using some of the hypotheses here, or some reasoning, and it would help to mention which hypotheses or what reasoning. "it follows that" by itself forces me to do literally all of the work of figuring that out, and i'm lazy and don't wanna.

noballpointpen

Now added details in this point.

shintuku

5:27 PM

let $rR$ be an ideal in $R$, $S$ multiplicatively closed subset of $R$. clearly $rRS^{-1}R = rS^{-1}R$, is there a proof of this in terms of kernels of homomorphisms?

leslie townes

but yeah i agree as to that. and i agree that then we must have inf R < f(inf R) and f(sup L) > sup L. in fact, that's so equivalent to "inf R in L and sup L in R" that i wonder if it's necessary to write out or prove both of them. couldn't we just write out or prove the one we end up using.

"by definitions of sup and inf and above relations" feels like it's covering a lot of ground. this can all probably eventually be shortened but at the moment i think we would want to write more of this out.

Ted Shifrin

@leslie That problem comes from your favorite, George Bergman.

leslie townes

haha. did he put in his problem sets when he taught 104?

noballpointpen

I found it in his supplements for Rudin.

leslie townes

i graded homework for him at least once. it wasn't fun for me at all and i think many of his students hated him. but i like his exercises.

shintuku

5:48 PM

@shintuku forget isomorphisms, the proper proof involves proving an ideal equals its contraction followed by extension

noballpointpen

5:59 PM

@leslietownes ok, here is one with the details filled: click. Ted, does the contradiction I mentioned earlier furnish the proof? Here, if you missed. Or just click the image.

I am going to leave right now. Will be back in a hour or so. Thanks for the feedback, leslie!

Ted Shifrin

6:25 PM

@noballpointpen I don't see why $\sup L\in L$ is impossible. I claim that the fact that $f$ is monotonic implies that $\sup L\le f(\sup L)$ in the first place. So I don't need most of your argument. But you dismissed as obvious something I don't yet see.

So I think we need to imagine $\alpha=\sup L < \beta=\inf R$ with $\{\alpha<x<\beta\} = \emptyset$. Why is this not possible?

This is where I went off the rails the other day and made up an impossible counterexample.

Ah, but $f\colon S\to S$ and so then $\alpha<f(\alpha)<f(\beta)<\beta$ is a contradiction to the fact that $(\alpha,\beta)$ is empty. So we're done.

Jakobian

6:48 PM

@AlessandroCodenotti hey, would you help me with understanding an argument for realcompact spaces?

I am trying to understand the proof that the Hewitt realcompactification $\upsilon X$ is the smallest realcompact space between $X$ and $\beta X$, where $\upsilon X$ is defined as the set of those points of $\beta X$ for which the extension of $f\in C(X)$ to $X\cup \{p\}$, $p\in \beta X$, treated as a function $f:X\to \mathbb{R}\cup \{\infty\}$ is a function into $\mathbb{R}$.

So if $T$ is some realcompact space, $X\subseteq T\subseteq\beta X$, then apparently $T$ is $C$-embedded in $T\cup \upsilon X$, but how is it clear?

oh wait I think I see

Given $f\in C(T)$, there is some $X_0\subseteq \beta X$ which is the largest subspace to which $f\restriction_X$ admits a continuous extension. Since both $f$ and the extension of $f\restriction_X$ to $\upsilon X$ are such extensions, we see $T\cup \upsilon X\subseteq X_0$.

using that $X$ is dense in $\beta X$

Not sure how much of an elementary argument it is though

I understand why it works is what matters I guess

冥王 Hades

7:47 PM

Don't mind me. I'm on my phone at 4:48 AM and I have classes in 3 hours.

geocalc33

8:06 PM

Anyone here?

noballpointpen

@TedShifrin if $x\in L$ then $x < f(x)$, but then $f(x) \in L$, since $f(x) < f(f(x))$. We cannot have $\sup L \in L$, since it would imply that $f(\sup L) \in L$, but $\sup L$ is an upper bound. This is why we must have $\sup L \in R$.

We also are not interested in the non-strict inequality $\sup L \leq f(\sup L)$, since $\sup L = f(\sup L)$ would do the job.

@TedShifrin when we arrive at the facts that $\inf R \in L$ and $\sup L \in R$, your inequality would contradict the upper/lower bound properties of each.

Ted Shifrin

9:00 PM

An upper bound can be in the set!

noballpointpen

I mean, $\sup L < f(\sup L)$. This clearly is not possible. No?

It's after the supremum, which is the least upper bound.

shintuku

so it is just an upper bound, not the least one

noballpointpen

If we had $\sup L \in L$, the image would be in L, too. Then look that this image is after the supremum of the set that they both belong to.

shintuku

the least upper bound is the least among the infinity of upper bounds, not the only upper bound

noballpointpen

Did you read the problem and the proof, shin?

shintuku

9:07 PM

no, only the statement that $\sup L < f(\sup L)$ is impossible because the RHS comes after the supremum

Ted Shifrin

I’m saying $x\le\alpha\implies x<f[x)\le f(\alpha)$ for all $x\in L$, so $\alpha\le f(\alpha)$.

I discussed above why there can’t be a gap in $S$ from $\sup L$ to $\inf R$. That is crucial.

noballpointpen

I think $\alpha$ can very well be after $f(\alpha)$. Why not? I don't see an immediate contradiction.

@shintuku if you read the problem and the proof you will understand why I think so.

shintuku

9:24 PM

what's the problem number in rudin

noballpointpen

It is not in Rudin, it is in the supplementes by George Bergman for Rudin. Number 1.2:4.

My proof attempt is here.

Wait, is $\alpha = \sup L$? I thought it is an arbitrary object. Just noticed in your discussion.

Hm..

Now it makes more sense.

Now I see why your argument works better.

But I am curious, would my argument work, too? Like we discussed yesterday, there are different approaches to the same problem; I must know if my result would work too or not and I messed it to the point of being flawed and incorrect.

noballpointpen

9:51 PM

Now since this is done, I am free to see the solution on MSE.

robjohn

10:11 PM

@Ted: Have you ever computed the mean of the reciprocals of the distances between uniformly distributed points in a sphere?

Ted Shifrin

10:32 PM

@robjohn No. is there a reason this isxa natural thing to look at? By symmetry, can we not do it with one point fixed, say the north pole?

Distance measured in the sphere?

robjohn

Two points inside the sphere.

Ted Shifrin

Oh, “in” …

Still, why is this interesting?

robjohn

I did the stereo sphere rotating animation to test my random point generator

D.C. the III

and with a contraction mapping we can establish that such a mean exists as a "fixed point" ...... Q.E.D 😊

Ted Shifrin

Seems yucky to integrate!

robjohn

10:35 PM

@TedShifrin It was a question, I don't know if it is otherwise interesting

Ted Shifrin

is there a natural physics application with gravitational/electric potential?

robjohn

@TedShifrin I guess it would be the mean potential of the points inside the sphere

Ted Shifrin

Assuming constant density …

robjohn

yes

Ted Shifrin

So does harmonicity make this easy?

robjohn

10:40 PM

I think the fact that we are not always integrating over whole spheres might kill that.

shintuku

does picturing the spectrum of an arbitrary ring allow us to see the ordering w.r.t. to inclusion of the different forms of prime and maximal ideals?

Ted Shifrin

Yeah.

shintuku

was that to me or robjohn

Ted Shifrin

Not you

shintuku

:(

noballpointpen

10:59 PM

So, Ted, do you think my proof would work too? You missed my message in the discussion with rob. I think I should know if my solution is correct to learn a lesson.

robjohn

11:20 PM

Off to the park.

user 85795

Cya.

noballpointpen

As leslie said "all goofy things can happen when we assume a contradictory thing".
Just wanted to point out: you said "An upper bound can be in the set!" Of course, I know! But notice that this is not the thing I want to show you.
Indeed, if $\sup L \in L$, then $\sup L < f(\sup L)$, by our definition. Then we have $f(\sup L) < f(f(\sup L))$, which means that $f(\sup L) \in L$! But $\sup L < f(\sup L)$, as we witnessed! This is impossible. Which must mean that $\sup L \in R$.
@Ted, I think we can start to discuss correctness of my proof by me showing you that I was correct here.

Going to beat some other problem in Rudin. Don't worry to hit me when you come.

shintuku

11:37 PM

@shintuku yes! dimension of spectrum as topological space = krull dimension!

noballpointpen

Hey, leslie.

D.C. the III

@shintuku I noticed this is the second time you've talked to yourself in chat........everything ok bud? 🤣

shintuku

it's mostly so someone else doesn't start making an effort to answer

in case i happen to have figured it out myself

D.C. the III