Mathematics - 2023-08-21

Thomas Finley

02:41

I don't understand the last sentence.

How is the search for the subset is related to the question of whether or not some vector in S is a linear combination of the other vectors in S ? I can guess the validity of this statement somwhat intuitively, but not much though. Also, I don't have a rigorous proof to justify the claim made.

Ted Shifrin

Keep reading.

Faust

03:12

@TedShifrin cAN I SEND YOU A PROOF?

fken caps

Ted Shifrin

Heya Faust. Of what?

Faust

an incorrect proof of the goldbach conjecture and infintly many twin primes

Ted Shifrin

No. Not remotely my cup of tea.

Faust

i cant figure out what wrong with it

its only alittle over a page

Ted Shifrin

I know no number theory.

Faust

03:15

well i didnt really use number theory in it lol

but i get it, i will wait for my prof to have a look at it

i know one who does number theory

Koro

A vector field **X** on $\mathbb R^n$ is said to be compactly supported if there exists a compact subset K outside which $**X**$ vanishes.
(a) Prove that for such a vector field $\bold X$ we can construct a flow map $\Phi:\mathbb R^n\times J\to \mathbb R^n$, where J is an open interval containing 0.
(b) In fact, it can be proved that such an $**X**$ has a global flow.

Faust

i wanted to ask something about hyperbolic geometry too

if you have itme at some point

Koro

please ** X ** as bold X as mathjax is not rendering.

how to do (b) here?

Ted Shifrin

That is more down my line of work, although it might not be something I know. What is that question?

Faust

lol its very closely related to the proof actually

Ted Shifrin

03:18

You need to use mathbf, Koro.

Koro

A vector field $\mathbf X$ on $\mathbb R^n$ is said to be compactly supported if there exists a compact subset K outside which $\mathbf X$ vanishes.
(a) Prove that for such a vector field $\mathbf X$ we can construct a flow map $\Phi:\mathbb R^n\times J\to \mathbb R^n$, where J is an open interval containing 0.
(b) In fact, it can be proved that such an $\mathbf X$ has a global flow.

thanks.

Faust

it seems that hyperbolic geometery is is a space where you can go halfway to something, well rather you shouldnt

Ted Shifrin

Huh?

Faust

one secound let me see if i can write this out in a halfway logical way

i am trying to consider all even numbers but 2 to contain a portion of them as a zero divisior

David Raveh

When do people learn about prime numbers? I started explaining it to my brother in law, and he got insulted, but he doesn't know anything about complex numbers, never learned beyond algebra 2...I feel like prime numbers is only studied by pure mathematicians

Faust

03:22

when in hyperbolic geometry

Ted Shifrin

@Koro What makes a vector field not have a global flow?

Prime numbers appear in middle school or earlier.

Faust

well i mean what is there really to say about complex numbers?

Koro

for (a): I know that local flow exists, i.e., given any p, for any nbd. U_p of p, there exists an e_p>0 and a smooth map $\phi_p: U\times (-e_p,e_p)\to R^{n+1}$ such that (i) $\phi_p(x,0)=x$ for all x in U_p and (ii) d/dt $\phi_p(x,t)= X(\phi(x,t))$

Faust

there like a really simple version of the reals

David Raveh

There is a lot to say about complex numbers

And it is definitely not simple

Faust

03:23

there way simpler than the reals

Koro

Now I can cover K by finitely many U_p's etc and then I can define a local flow as desired in a).

@TedShifrin smoothness may be lost?

Ted Shifrin

Nah.

Koro

actually I don't know. I only know its definition and existence of local flows for smooth vector fields.

David Raveh

I can't recall learning anything about prime numbers. Most of what I know is from MSE and group theory

Faust

well prime numbers is what make all the fields

Koro

03:26

a book discusses it but in an abstract setting and currently I'm in R^{n+1} differential geometry. We'll get there but in some time.

Faust

all finite fields are of the form p^n where p is prime and n>1 and a integer

or 1 lol

Ted Shifrin

@Koro what if there is a hole in your space where the flow wants to go? Or what if the vector field gets really big? What’s a simple exampleob $\Bbb R$ of a vector field whose flow is not defined for all time?

@Faust Not correctly stated.

Faust

there are other finite fields?

lemme think for a secound if i can come up with one

Ted Shifrin

Kids learn about prime factorization and gcd, etc., David.

Faust

no im pretty sure those are all finite fields

if it had a composite number of elements it would have zero divisiors

so it can only have p ^n elements

Ted Shifrin

03:31

Oh, number of elements. Did we say that?

Faust

wi assume it was obvious from the statement

lol

what else could i possibly meant?

Ted Shifrin

By this point, you need to make clear, correct statements. You’re not a first-year math student.

Faust

how else would you uniquely differentiate between them other than number of elements as they are all isomorphic

lol

Ted Shifrin

To me, it sounds like you’re thinking of modding out by $p^n$.

David Raveh

I guess I must have forgotten learning prime numbers

Faust

03:33

oh i have tried to do that for all p and n

it does weird things though

Ted Shifrin

Yes, not fields when $n>1$.

David Raveh

Primes didn't come up in my memory of classes until I took quantum computing

Faust

you can build the rationals that way i belive

Ted Shifrin

What country did you go to school in, David?

David Raveh

US

Ted Shifrin

03:35

I know the US curriculum.

Faust

@TedShifrin i said integers and i added one at the end, but yes it couldnt of been stated in a less clear way and been correct in any sense of the notion.

David Raveh

I'm still in university, btw

Ted Shifrin

Faust, $n=1$ is fine, too. The others are harder to show someone.

Faust

so in hyper bolic geometry, if you think about moving 3 distance its the same as moving 6 distance

but 5 distance is more than 6

so isnt it like a space where there are no even numbers but 2?

well i guess 2 isnt even in it

cause 2=1

so its like you lack uniqueness of the identitys

2+2=2 and 2a = a so half the time its also the additive identity and its always equvlent to the multiplicative one

Ted Shifrin

I have no idea what you’re saying. Where did rings come from in hyperbolic geometry? And moving distance 3 and 6 are most definitely different.

Faust

03:46

well i am trying to think about creating a group using multiplication of primes

this is fine we have inverses

but odd shit happens when we try n add things

Ted Shifrin

What you’re saying is garbage unless you say something precise and valid.

Faust

lol fair enough

i am trying to define some structure that gives hyperbolic geometry in some sense

and look at what is broken

that makes it hyperbolic]

it kind of looks like we have some zero divsiors that dont cancel everything out to 0

just trying to understand things, it doesnt need to make sense ^^

David Raveh

If it doesn't need to make sense, how could you understand it?

Faust

hyperbolic geometry clearly has some structure about it

weird things happen but its not like we cant saying anything about how that structure works

even if its not quite right, comparing something to something you understand and looking at the diffrences is well one of the ways i like to understand things

like we can create strucure of fields usings axioms

and we can amke hyperbolic geometry using axioms

why cant we make it using structure...

personally it looks like hyperbolic geometry is just a weakening of associativity

Ted Shifrin

Well, I’m out. I don’t have patience for this.

Faust

03:54

a+b = b+a isnt always valid in it

lol

sorry

well i guess i will head out then, if you cant understand the nonsense i am thinking about i doubt anyone can.

Koro

why is convolution of two L2 functions on 1-Torus (T) is continuous?

one potato two potato

and the title should be the first room-temperature proof of the Goldbach conjecture.

copper.hat

04:43

it suffices to prove it for a temperature of $-{1 \over 12}$C.

lots like chatGpt made it to the chat room.

leslie townes

05:07

david's comments above had me wondering when i first learned about prime numbers. i think at some point in late grade school they were discussed. i even hazily remember covering the gcd and lcm in terms of prime factorization. but, nothing in middle or high school or indeed in college before first semester 'abstract algebra'

ted, when prime numbers were discovered and they burned whoever did it at the stake for witchcraft, was it fun to watch?

one potato two potato

10:14

after the film Oppenheimer, some people in my college want to study nuclear physics lol

the important thing is that none of the physicists in my college do research on nuclear physics

Rajdeep Sindhu

11:02

Heya folks

We were just taught about vector spaces in class and it's a little confusing to process.
What's the "reasoning" behind all the axioms for a set being a vector space? Why those axioms, and not something else, if that makes sense?

And there's nothing to say that the "elements" of vectors in a vector space have to be numbers, right? Can they be matrices?

If we define a vector space $V$ such that

$$V=\left\{\begin{bmatrix}A_1\\A_2\\\vdots\\A_n\end{bmatrix}, A_k = \begin{bmatrix}a_{11}& a_{12}\\a_{21}& a_{22}\end{bmatrix}~\forall 1\leq k\leq n, k\in\Bbb N; a_{ij}\in\Bbb R~\forall~1\leq i,j\leq 2; i,j\in\Bbb N\right\}$$

Then, would $V$ be a vector field over $\Bbb R$?

Alessandro Codenotti

@RajdeepSindhu yes they can be matrices or anything else

Rajdeep Sindhu

Wow going from high school math to uni math is kinda wild

Alessandro Codenotti

Usually the axioms come from having some natural examples of the structures you want to axiomatize (in this case for example $\Bbb R^n$) and abstracting the properties that you want similar structures to have

Rajdeep Sindhu

@AlessandroCodenotti I see, I suppose I'll understand that better as we go through more examples of vector spaces in class

And vector addition and scalar multiplication for a vector field can essentially be anything as well, right?

Alessandro Codenotti

Yes, they only have to satisfy the axioms

Rajdeep Sindhu

11:16

Thanks! Pretty interesting stuff

I have another question.
When we say something like "$\Bbb R$ is a vector space over $\Bbb R$", then the former $\Bbb R$ refers to a vector space so it'll be a set of 1-tuples with real entries or one-dimensional vectors and the latter $\Bbb R$ will be a set of real numbers, which will act as scalars, right?

So not really the same $\Bbb R$, are they?

Soumik Mukherjee

later $\Bbb{R}$ as the field

Rajdeep Sindhu

@SoumikMukherjee Yep they act as different entities, one as a vector space and other as a field but apart from that, element-wise, they are different, right?

Soumik Mukherjee

element-wise they are same

Rajdeep Sindhu

Then, can $1$ act as both a scalar and a vector?
Won't the vector be denoted by $[1]$, for example?

Soumik Mukherjee

But as the first $\Bbb{R}$ is a vector space(and not a field), you can't multiply elements of it without further assumptions

@RajdeepSindhu yes

some may write $[1]$ for distinction, but not necessary

Rajdeep Sindhu

11:26

Got it, thanks!

one potato two potato

14:01

from kleinbottle.com/cartoons_and_limericks.htm

冥王 Hades

14:14

I’m pretty sure I posted that here first

one potato two potato

O'RLY?

冥王 Hades

15:33

Anyone else find looking at different clothes too overwhelming?

I do, and that’s why instead of multiple different clothes I keep several pairs of the exact same clothes

Semiclassical

15:47

fun self-intersecting surface of the day:

if you fill in the "craters", you get a dice with rounded corners:

(first surface is parametrized as $(u,v)\in[0,2\pi)^2\to (\cos u,\cos v,\sin(u+v))\in[-1,1]^3$)

Sourav Ghosh

16:52

Let $X$ be a Banach space and $f(Y) $ is bounded for any bounded linear functional $f\in X*$ . Then the subset $Y$ is bounded.

Here I want to use the U.B.P

$T_y(f) =f(y) $

Then by U.B.P $\|T_y\|<\infty$ for all $y\in Y$

Ted Shifrin

@Semiclassical A die.

Sourav Ghosh

If Y is a unbounded subset of a banach space then there exists a bounded linear functional such that f(Y) is unbounded.

$Y$ is unbounded. Then $\exists (y_n) \subset Y$ such that $\|y_n\|>n$

Now I have to define a bounded linear functional $f$ on $X$ such that $f(Y) $ is unbounded.

Can we define a bounded linear functional on $\textrm{span}(M) $ or $\textrm{span}({y_n}) $?

robjohn

17:18

@TedShifrin Grammar Police

Alessandro Codenotti

@SouravGhosh The converse to your statement is a typical application of Banach-Steinhaus. Suppose that $Y$ is such that for all functionals $\varphi$, $\varphi(Y)$ is bounded. Use Banach-Steinhaus to show that $Y$ is bounded

"weakly bounded sets are norm bounded"

Xander Henderson

17:33

@robjohn Grammer police.*

(Because why miss out on the opportunity to misspell something?)

geocalc33

That's a cool die

Sourav Ghosh

@AlessandroCodenotti Done✅ Thanks :)

Xander Henderson

The Simpsons - Die Bart Die (English)

►

"No one who speaks German could be bad!"

Sourav Ghosh

17:57

Q.2) $X$ and $Y$ be two normed spaces and $X$ , $Y$ are homeomorphic. Does this implies $X$ , $Y$ are topologically isomorphic or linearly homeomorphic?

Alessandro Codenotti

What does topologically isomorphic mean?

Sourav Ghosh

there exists a LINEAR homeomorphism

Jakobian

@AlessandroCodenotti homeomorphic

Alessandro Codenotti

@SouravGhosh so what is the difference between topologically isomorphic and linearly homeomorphic?

Sourav Ghosh

Same

Jakobian

17:59

I wouldn't call those topologically isomorphic if they're linearly homeomorphic

that's just confusing and I doubt people say that

linearly isomorphic or isomorphic as normed spaces

@SouravGhosh no it doesn't

Sourav Ghosh

31

Q: Is a normed space which is homeomorphic to a Banach space complete?

I have a normed space $(E,||\cdot||)$ which is homeomorphic (as a topological space) to a Banach space $F$. Does this imply that $(E,||\cdot||)$ is also a Banach space? I think I read something like this to be true if $E$ (and therefore also $F$) is separable, but I am not totally sure. So, al...

Jakobian

$\ell^p$ and $\ell^q$ are homeomorphic for example

For $1\leq p, q < \infty$

Alessandro Codenotti

and they are also all homeomorphic to $\Bbb R^\omega$

Jakobian

yes this is Anderson-Kadec theorem

Sourav Ghosh

Let's consider the above question.

Alessandro Codenotti

18:01

I've never looked at a proof. Geometry of Banach spaces is too hard for me

Sourav Ghosh

X, Y are homeomorphic as nls and X is Banach. Does this implies Y Banach?

Jakobian

linear isomorphism is a bi-Lipschitz map which should do it

Alessandro Codenotti

@SouravGhosh Isn't that answered in the MO question you linked a few messages ago?

Sourav Ghosh

Yes. But i am unable to follow the argument.

Jakobian

I interpreted "homeomorphic as nls" as linearly isomorphic

Alessandro Codenotti

18:04

@SouravGhosh Which part is unclear?

Sourav Ghosh

@Jakobian Then there is nothing to prove.

X, Y homeomorphic (as topological spaces)

Jakobian

Yeah, I don't see what is unclear in the proof of this given by Nate Eldredge

Sourav Ghosh

(0, 1) homeomorphic to \Bbb R

Jakobian

yes

Sourav Ghosh

Ohh. We can define a complete metric on (0, 1) which is topologically equivalent to the usual topology.

Jakobian

18:11

Same with irrationals

this is the Baire space $\omega^\omega$, one of the most important complete metric spaces

Sourav Ghosh

So the result can be extended to metric spaces.

Jakobian

well yes, the theorem proved is that completely metrizable normed spaces are Banach spaces

The crucial part of it is that completely metrizable subspace of a metric space is $G_\delta$ (not sure here if metric spaces are necessary, but it's irrelevant here)

And completely metrizable subspaces can actually be characterized as $G_\delta$ subsets if we are in a complete metric space, that is, the two concepts coincide

there's been a lot of new faces here, and some old faces too

@XanderHenderson what does $0_k$ mean

geocalc33

18:46

Partition $\Bbb R^3$ into 10, 3-cells and take the boundary of the non-compact 3-cell to be $K$ which has $S^1$ type singularities. Is $K$ bad?

'bad' meaning trivial fundamental group. In other words, what can be said about the fundamental group of $K.$

Of course it is trivial. It's because $K$ retracts to $S^2.$

But how do you endow $K$ with a metric that obstructs contractibility?

I don't know - maybe use a normalized geometric flow so that the volume is forced to stay constant or something.

*I mean $K$ is isotopic to $S^2$.

Ted Shifrin

19:07

@Jakobian The zero element of the field?

@robjohn I don't think that qualifies as grammar; but, yes.

@geocalc33 Whether a space is contractible has absolutely nothing to do with a metric.

Jakobian

@TedShifrin I don't know, the thesis is too long for me to check

oh wait no I found it

@XanderHenderson I don't think it should be $\mathbb{k}$

you are consider $|\cdot|$ as a function from $\mathbb{k}$ to $\mathbb{o}_+$

where $\mathbb{o}$ is an ordered field

so $0_\mathbb{o}$ is the additive identity element of the field $\mathbb{o}$

oh the second $0_\mathbb{o}$

here's me paying attention to what someone writes

Xander Henderson

19:37

@Jakobian It is the zero element of a the ring $\mathbb{k}$.

@Jakobian It should be. Just not the one you are thinking of.

robjohn

@TedShifrin mixing plural with singular; I think that is erroneous grammar.

We need less grammatical errors ;-)

Jakobian

@XanderHenderson yeah, I've noticed that. Sorry

Ted Shifrin

@robjohn The quantity of errors is inacceptable.

I think we were trying to explain number versus amount to Jakobian last week.

Semiclassical

dice being plural and die being singular never seems to sit in my head right

Ted Shifrin

19:55

It is a dicey subject.

Remember that dice are like mice.

Semiclassical

Mie?!?

also, that frustrating moment when you spot a hole in your proof

is frustrating

Ted Shifrin

I spent much of my career discovering that there were holes in my research proofs ...

Semiclassical

oof

Ted Shifrin

One bugaboo in joint work with two other (smart) authors kept creeping in in various disguises, despite that the fact that we were on the lookout for it.

冥王 Hades

@TedShifrin no they’re like hice

Ted Shifrin

19:59

At one point we thought we had it, and I was on the airplane on my way to talk about the stuff at a conference in New Jersey ... only to realize on the plane that it had fooled us yet again.

@冥王Hades That'll just confuzle Semiclassic.

Semiclassical

reminds me of how much of physics is spent trying to catch sign errors

Ted Shifrin

Complex geometry is plagued not only by $\pm$ but more seriously by $\pm i$.

冥王 Hades

Me when the velocity of a projectile comes out to $-300,000 ms^-1$

Semiclassical

(including the rather famous case of asymptotic freedom, where a Nobel Prize hinged on whether it was plus or minus at the end of the calculation)

Ted Shifrin

Whoa. That's cataclysmic.

Semiclassical

20:02

"In high excitement, Politzer phoned Coleman, who was spending the spring of 1973
on sabbatical at Princeton, to report his “stupendous” discovery. But Coleman put a
temporary damper on Politzer’s enthusiasm. Gross, still under the influence of a
momentary sign blunder, had just assured him that Yang–Mills theories cannot be
asymptotically free. But Wilczek soon found the offending sign error and Politzer,
having rechecked his own calculation, was confident in the negative leading term of
the β function.""

Xander Henderson

@Semiclassical It follows the same pattern as "mice": mice is plural, mie is singular.

Ditto "lie" and "lice". As in "The boy told many lice after he broke the window with a baseball."

Ted Shifrin

en français on a du pain de mie.

Semiclassical

oic

Ted Shifrin

Xander is adding to the illustration of the phenomenon that English is random and arbitrary.

Semiclassical

a line I like: “English is not a language, it's three languages wearing a trench coat pretending to be one.” – Gugulethu Mhlungu

Ted Shifrin

20:05

The trench coat pretends to be one?

Dangling participial phrase ... :D

Semiclassical

roll

Ted Shifrin

@robjohn Now that is grammar. :D

冥王 Hades

I’m the 21 year old here

Ted Shifrin

@冥王Hades And your point ... ?

冥王 Hades

I forgor

monoidaltransform

20:15

Is every compact subset of euclidean space have a simplicial complex structure?

Ted Shifrin

I sure doubt it.

Try the Alexander horned sphere?

sunny

21:56

> Corollary 10.23. If two power series $$\sum_{n=0}^{\infty} a_{n}(x-c)^{n}, \quad \sum_{n=0}^{\infty} b_{n}(x-c)^{n}$$ have nonzero-radius of convergence and are equal in some neighborhood of $0$, then $a_{n}=b_{n}$ for every $n=0,1,2, \ldots$.

Why neighborhood around $0$?

I would have thought neighborhood around $c$.

leslie townes

given the amount of commentary around it (namely none), that "0" this is probably a typo for c. look at the beginning of the proof of theorem 10.22 for why the author may have made this mistake

sunny

@leslietownes alright, thanks for the advice :)

leslie townes

e.g. they were stating results for power series about an arbitrary point but already proving them only for c = 0 because it was enough to handle that case. then they forgot to even state the result as intended for an arbitrary point :D

Xander Henderson

@leslietownes I've seen several quotes from that text over the last week or so to feel like maybe I would recommend a different book.

Have you heard the Good Word™ of Baby Rudin?

leslie townes

did you know that walter rudin sacrificed all exposition, motivating examples, etc.... for you?

sunny

22:08

If there was a newer version of Baby Rudin, with Tex from the 21st century, it would be more appealing to me.

leslie townes

it wouldn't surprise me if someone somewhere has attempted a modern latex version of rudin, but you might have to go underground for that. there was that group a while ago going around typesetting older books, whether they had the rights to redistribute them or not. they seemed to stay away from 'hot' properties for obvious reasons but someone must have done it by now.

also, outside of extreme examples, choosing books based on what typesetting they use is goofy.

Xander Henderson

And lo, Walter Rudin came upon a class of analysis students who were lost in the Mean Value Theorem. They struggled with a proof. It is then that Walter Rudin produced a divine function from the void. The mighty theorem fell to differentiation of the divine function, and was proved.

Jakobian

@leslietownes Does that mean he has a big crush on me?

leslie townes

his treatment of measure theory is nightmarish enough to be a stand-in for the book of revelation

@Jakobian yes

Xander Henderson

@leslietownes I'm pretty sure that 90% of math texts out there fall apart in the last chapter or two. For Baby Rudin, I feel like the wheels start to loosen when he gets to the multivariable stuff, and completely fall off in the next chapter.

Jakobian

22:12

oh, we are referencing the bible
my bad

leslie townes

jakobian, on that topic, you know who else has a big crush on you? (hint: he was a carpenter born a long time ago a little south of jerusalem)

Jakobian

Does he also live in the clouds to spy on me

Xander Henderson

I was taught out of Dangell and Seyfried. The book is essentially unremarkable, but they include the Riemann Rearrangement Theorem, which instantly puts it near the top of my list of great undergraduate real analysis texts.

leslie townes

we try to downplay that, but yes

Ted Shifrin

@leslietownes I hate to agree with leslie, but …

Jakobian

22:17

@leslietownes I disagree. I want a book that looks pretty for maximal enjoyment

Ted Shifrin

What book is this that is full of errors?

Xander Henderson

@TedShifrin Even a blind squirrel finds an acorn every once in a while? (and yeah, agreeing with @leslietownes sucks)

leslie townes

ted: the question above was from hunter's "introduction to real analysis." i don't know if it is full of errors, the thing above was along the lines of a typo. it appears to be online notes (not a printed book) whose advantage of free availability is not nothing.

Jakobian

there's always those illegal sites where you can get almost any book

but a lot of copies are also offered by universities

completely legally

leslie townes

such as multivariablemathematics dot ebooks dot free ted shifrin books dot biz dot ru

Ted Shifrin

22:31

Have no knowledge thereof. I agree it’s not nothing, but sometimes — my diff geo text being an obvious exception — you get what you pay for.

Jakobian

I've got introduced to those kind of sites by a doctor

I like electronic versions, and in English too

Ted Shifrin

23:11

@XanderHenderson Never heard of the university where the first author taught. His RMP page is shockingly horrid, too.

Xander Henderson

23:23

@TedShifrin RMP?

Ted Shifrin

Rate My Professors

Xander Henderson

Ah.

Yeah, like I said, the book is totally unremarkable. Except that it includes the Riemann Rearrangement Theorem, which is one of my favorite results in undergraduate real analysis. And, to be honest, I probably haven't opened it since 2004, so it might actually be terrible. I wouldn't know.

But it earns a couple of bonus points for including my favorite theorem. ;)

Ted Shifrin

23:38

Isn’t that in every analysis book? It’s in Spivak and I covered it every time (not pedantically).

Xander Henderson

@TedShifrin I do not believe that it is in either Rudin nor Folland.

sunny

It is in Rudin's.

Xander Henderson

@sunny Are you sure? I remember trying to find it Rudin a few years ago and not being able to.

sunny

@XanderHenderson It's Theorem 3.54 I believe :)

冥王 Hades

Why are there onions in my egg?

Ted Shifrin

23:43

It’s in Rudin. Folland’s which? It doesn’t belong in grad real analysis.

Xander Henderson

@TedShifrin No, the undergraduate Folland.

Ted Shifrin

@冥王Hades why not?

leslie townes

yeah, rudin proves a slight generalization of the rearrangement theorem - that you can rearrange so the partial sums have an arbitrary liminf and limsup subject only to the obvious restriction. it's peak rudin, generalize so that it's impenetrable, but somehow also the proof is shorter.

Ted Shifrin

Advanced calculus, then. I don’t know that at all.

冥王 Hades

@TedShifrin I mean. Did the chicken eat onions? Is that why?

Xander Henderson

23:44

@leslietownes Ah, okay. I don't like that quite so much. It isn't nearly as "obvious" (to me) what the point is. I like the clean "You can rearrange the series to whatever you like."

Ted Shifrin

I think you should revoke the extra credit, Xander.

@冥王Hades Do shallots instead, then.

Xander Henderson

Which is corollary of the result involving limsup and liminf, but not as nicely stated (in my opinion).

冥王 Hades

Oh so that’s what it is. Shallot is some kind of food?

Ted Shifrin

It’s a sophisticated, subtler onion.

Xander Henderson

I'm not seeing it in Apostol, either.

Ted Shifrin

23:48

I have hardly any books, so I can’t play this game.

sunny

@XanderHenderson I believe it's in there too, Rudin stylish

Xander Henderson

@sunny If it is, it is not in the chapter on sequences and series, and nowhere near the definition of "conditionally convergent".

In any event, I need to go home.

G'night.

Shaun

Hi :) Quick question: The axiom of choice (AC) is equivalent to the statement that every set is equivalent to an ordinal number; but in what sense are those sets equivalent?

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$Mathematics$ Mathematics