Please do.

I'm often in the room but not checking this chat. I'll eat a bit.

k

Generally functions are approximated with other functions converging under some norm like $L^1$ or $L^2$, etc.,

This means we are giving importance to the shape of the function and we do not care about the behavior of the function (differentiability at different points)....

this is how functions are mostly used in many known applications of mathematics like physics, etc.,

But consider functions with singularities...if we approximate them under norms like $l^1$, $L^2$, etc., we loose the significance of the singularities.

The idea here is to propose a theory where singularities are also important to the theory......and thats where the class of functions defined here come into picture and i am of opinion that they are the best candidates for such a theory...(in this theory there wouldn't be any place for functions which do not belong to this class)...

@Jonas

Is it just me, or do single-dollar TeX formulas suddenly not get jaxified on the main site?

@HenningMakholm Lemme check

Looks fine to me. The 2.0 beta takes a bit longer to respond since it does things in blocks to reduce flicker.

Hm, nothing happens for them for me, either at home (FF 3.0.6 on Linux) or at work (FF 3.6.15 / Windows).

@rob : why does the equations on chat site appear blurred on my computer ? any ideas ?

Works OK in Chrome, through.

@HenningMakholm It could be a Windows thing, I have to use Chrome to come to this site because Windows freezes on me.

@Skullpatrol Not quite, since I have the same problem on Linux at home.

@HenningMakholm Just an idea.

Does anybody know why WindowsXp freezes on this site?

Man, that was a great sleep.

any dreams ??

nightmares ??

@AsafKaragila Did it help ease the pain of your headache?

Rajesh, none. However sleeping during the day is always better sleep than sleeping during the night.

@RajeshD The font substitution in the 2.0 beta seems to be worse. I get some clipping of the bottoms of characters, too.

@RajeshD I can't imagine that such at thing doesn't already exist.

???

@Jonas : where ??

There are so many norms, and of course the $L^p$ norms get rid of the singularities.

But then you need to give an example where it is needed.

Plus many functions with singularities are not "ignored" by the $L^p$ since those integrals just blow up 8-).

@Jonas : there will not be any $L^p$ in the theory....they just aren't used...they are not part of the logic there

Mr. Personality

@Jonas : Do you like the plain idea just on the face of it ?

I still don't know why you would want to do this.

There surely are a quadzillion ways to generalize certain function spaces, but to pick one you should give a why.

yes

i'll do that

the functions should have singularities, but how many and where they should be ?.....it doesn't make sense for them to be only at so and so points..(the theory

would look ugly) and it doesn't make sense if they are finite in number ( again how many) ...Hnece let them be on a countable dense subset set !......and why should be of the specific type i mentioned........for these class of functions...(there need to be more reason for this later) ..(i feel they make intuitive sense)..but one thing i could think is .if you differentiate them successively..they still be of the same type except that they are not defined at finite number of points.....

@Jonas

@Jonas : Is it interesting enough to begin with ???

I still don't get the why. What do you want to do with these classes of functions?

@Jonas : I want to see "what can be done ?"....there certainly something that can be done !

pardon me if its a ridiculous argument

@Henning: are things still not rendering for you on the main site?

@robjohn No. Have been writing my last answer blind ... which browser does it work in for you?

Hi all. Just a quick hello from me. I'm terribly busy these days, so I'm forced to cut down my procrastination hereabouts. Sorry about this spontaneous unannounced disappearance. I'll pop up occasionally, but definitely not as frequently as before. Have a good time, you all -- Theo

Good luck @tb!

Hi @t.b. : great to see you back !

Probably that has been checked, but can the mathjax-site be reached? I think that the .js was loaded from there.

@tb nice to hear from you, good luck in your affairs

Does anybody know how inherited sigma algebra is called in the right way?

What do you mean inherited sigma algebra, Mr. iPad?

I.e. when you consider only measurable subsets B of a given measurable set A

@JonasTeuwen MathJaX is definitely running. It renders double-dollar displays and ignores single-dollar inline formulas.

Oh, I missed that.

@Henning did it become more expensive and one dollar is not enough?

@Ilya Relative sigma algebra?

Induced $\sigma$-algebra?

@Asaf let me check

Reduced sigma algebra?

Deduced maybe;)

@Jonas thanks, I think it should be call induced

Can you give the expression of that $\sigma$-algebra?

The Borel set of your mother.

@Jonas not with an iPad, I'll do it later:-)

@Asaf you're a Borel subset of your mother

Something like $\sigma(B)$ where $B$ is a set?

Asaf is a non-measurable subset of his mother.

@Ilya You're a measure zero subset of your mother.

He was

@JonasTeuwen This is only because my inner measure is 1 and my outer measure is 0.

Now all of us are null subsets of out mothers

I'm the empty set.

@Asaf an alien inside you?

@JonasTeuwen That is such an emo thing to say. Go Dedekind-cut yourself now.

@Ilya Like in the movie "Alien"?

@Jonas yes

Holy cow, that's scary.

I'm not pregnant, so no.

@jonas you see how fast your haircut has brought you to being emo :D

@Michael has found an counterarguement for your answer? Now you've deleted it

I haven't deleted it.

I'm thinking about how to correct it.

Hm

@Ilya :D.

I saw it when it was already very late and I was very tired and mildly not sober.

@Asaf : Is it wrong to delete answer when found wrong ??

Indeed it's on its place, misseen

@Rajesh are you Scull? He usually asks that kind of questions

shit...i can go hang myself !!!

@RajeshD When a question is far from trivial and yours is the only answer it's a good courtesy to leave your mistake for others to learn from.

k got it

@Asaf thanks for the compliment to my question

She will be happy

My questions, they are girls btw

@Rajesh don't say sh*t, use crap instead

k

You've lost o? Take mine

is t.b. here?

@Ilya So you think about them when in bed and in the shower?

@Srivatsan He passed through us, like a ghost. But he left a message which I have pinned.

@Srivatsan no, I think he has already gone again. Hi btw

@asaf the ghost of Hamlet's father

@Ilya The ghost of future past.

@Asaf I have a real person to think of in bed or in shower. Not a chicken like some of us do have

Thanks, Asaf, Ilya.

@Ilya So your question is a real person now? Tsk tsk tsk.

And hi

and the real person is @Asaf !!!

Talk about psychosis in the mathematical community.

@Asaf quite, I don't want to guess what were you doing with tsk tsk tsk sound

Nobody of us wants. Do it quite

"mathematics community" rather than "mathematical community" isn't it ?

No, it isn't.

@Rajesh "devils sleep at this time" -- What time was that?

around 6:00 PM

Ah, indeed that is an odd time to nap.

This sounds as if he's going to leave chat for good.

T_T

I hope not.

We should add this to the rules "Teddybear must not be absent more than two weeks at a time."

I'm so sad.

You can see him around campus, no?

@MattN Please elaborate? Theo?

@JonasTeuwen Yes.

@AsafKaragila No.

Yes, we're all sad because of this. :(.

@JonasTeuwen Maybe it's just me, but that doesn't sound quite right.

@Srivatsan Why not? Do tell.

I've been sad since he left and now that it's definite and for good I'm too sad.

@AsafKaragila Which campus?

Just when I started to get over it :,(

Why did he have to tell us? I'd rather not have known.

@Srivatsan In what way? Grammatically?

Or do you think it sounds sarcastic? I can imagine that but that was not my intention.

terrible loss means gone for good, and never comeback ! and its quite not right to say such things

Oh... :D.

I certainly hope he didn't 8-).

@JonasTeuwen Well, that's what people say for death. I am not sure if that phrase is used in other contexts.

:,(

@AsafKaragila No but at least not forever.

I am just getting started with Computer graphics and was asked to convert some homogemeous coordinates into cartesian coordinates. What is the common notation for a point? I was just given a 1x4 matrix. I would guess the first number is the ratio and the other 3 are x,y,z. :/

@Srivatsan for me Jonas sounded ok

@Ilya Hmmm, well, perhaps it is ok. As I said I am not sure.

Frankly, I miss tb but I'm not sad, because I thi he will appear soon

@Ilya would you consider appending your name with 'na' ??

Why?

@Ilya He wrote that he doesn't have time to hang out in chat anymore.

@Matt it may mean smth good for himself, doesn't it?

@Ilya Yes! I can be happy for him while still being sad.

@RajeshD Why?

it reminds me of this

@Ilya hopefully.

@Sri you shouldn't be not knowing this

It's like we're all standing around someone's grave, mourning while listening to the priest and then suddenly this dude comes along asking whether we can help him solve this integral.

Not my type

Surreal.

@RajeshD Should I know this? Or shouldn't I?

@MattN This is exactly what I meant in the rules, to feel the room before asking a question.

"Ileana has no calms in exposing every bit of her body and is ready to go for anything for money. She is liberal in showing her skin and so is the leading actress of tollywood in fame and value." -- Huh??

You are from South India ...I guess you watch films so it should have been trivial for you

@RajeshD Yes, but I don't watch "Tollywood" movies.

Tollywood is the pornographic scene of Bollywood?

@AsafKaragila I wish.

I've seen Zita and Gita some time ago. That's it

ok she has done a few tamils ones....i guess not much...never mind

@AsafKaragila You're thinking of Pollywood. :)

@Matt why do you wish. There enough other porn studios

Polly? Like that Nirvana song?

(I was joking, btw)

@Ilya Click arrow to see what I was referring to.

@Matt I can't do it with iPad

@Sri yeah, tell it us now xD

Tollywood is the Telugu(south indian language spoken in A.P) language film industry and the biggest in India..makes more films than Bollywood !

@Asaf

@Ilya Sita and Gita, maybe?

@RajeshD Is that so?

Whatever

I am off............feel guilty when people around me think that i am doing some important work on laptop...bye

We also had a joke about the film Sita and Hitler

@RajeshD Haha

Hi guys!

Today classes were awful: automata theory lecture was too easy although we proved some nice stuff, probability theory lecture & seminar SUCKED

@MattN I'm actually more bummed about his spontaneous appearance than his spontaneous disappearance. He came back just for a few minutes and I (we) couldn't be here at that time.

If only we had some way of knowing when to expect him next...

Like we needed to prove some statement and its' converse so after we have proved the statement the teacher just wrote our proof backwards.

WTF

@Daniil Sometimes it's right to write the proof backwards!

@Srivatsan but he does that all the time!

Also he absolutely loves writing symmetrical definitions.

what theorem is this?

@Daniil What are "symmetrical definitions", may I ask?

Uh, I think it was about the fact that X_1 and X_2 are independent iff f_{X_1, X_2} = f_{X_1} f_{X_2}

@Tim Like definition of marginal prob density function

f_{X_1}(x_1) vs f_{X_2}(x_2)

Thanks! How is that symmetrical?

Well

Maybe not symmetrical, but very similar

@Dan I can't understand what are your complaints about

@Ilya the lecture is slow-paced, boring and unnecessary wordy

Hey, where are you, folds

@ Dan I had the same for probability in Russia

Try to think about applications

If you have an interesting problem in your head, which requires this knowledge it would help

Ask him, what does any property or theorem mean, why is it important

Call him in a Midnight and ask why should we know Fatou lemme and dominated convergence theorem

@Ilya interesting advice, thanks :D

@Daniil You could teach him about $\Leftrightarrow$, the lecturer's best friend.

@DylanMoreland Actually, I don't like it as much. It is far too succinct for me.

second @Srivatsan

It's strictly a proof-writing device, but not a proof-reading one. =)

But I agree, it might be a lecturer's friend: after all, it does save some time in writing the proof.

@Srivatsan I used to not be a fan as well.

How did you change? :)

And do Bill's answers have anything to do with this change? =)

But I think it's good in the sense that, and I really only noticed this when I started seeing Matt E do it all the time, that when you have some statements that you want to connect it's useful to just get in the habit of writing down some equivalent statements, even if they seem trivially different. And the notation removes some barriers to that.

And when you don't have a compact way of saying it then I think there's a tendency to leave out quick intermediate steps and I'm not sure that this helps your audience. Because that's not how people prove statements in the wild.

@DylanMoreland True that.

In the wild. Proving in the wild. Kolmogorov and Perelman were proving stuff in the wild, in the deep Russian forest, among bears

They were running out of vodka and getting cold, but Mendeleev was passing by

It sounds to me as an interesting plot for the novel. See you later

@Srivatsan Sometimes I have a hard trouble following Bill's $\Leftrightarrow$s. The effort is always rewarded in the end, but I guess that isn't what I'm talking about.

If $X$ is a compact (not Hausdorff though) space and $f:X\to Y$ is a continuous map into another non-Hausdorff space; is the image compact?

Yes as I remember. Compactness is preserved by continuity and hence it is a topological property

Then I can finally finish this homework assignment due two weeks ago!

Why do you think of non Hausdorff spaces?

Spectras.

If $\{V_i\}$ is an open cover of $f(X)$, then $\{f^{-1}(V_i)\}$ is an open cover of $X$.. then the result follows from $f(\cup V_i) = \cup f(V_i)$.

Right.

If I show that this continuous image is also injective does that imply that it is a homeomorphism?

I would guess no, since $f^{-1}$ is not necessarily continuous..

So in this case you do need Hausdorff-ness.

Either way I solved the problem without needing that.

Excuse me1

Could someone say me if I can use taylor expansion when the limit is to $\pm\infty$?

Or I can use taylor only for the limits that points to a finite number?

@Dylan: Are you around?

@Asaf even if you show that it is bike give it is not enough

Huh?!

Consider an image exp(I *x):[0,2pi)->S1

It is continuous, bijective but the inverse is not continuous. To prove the latter fact you can consider e.g. a compactness argument

Which book do you use to study topology?

is [0,2pi) compact?

Guess?

No, since it's not closed.

In Asaf's case we had $f: X \rightarrow Y$ with $X$ compact, but I suppose there should be an example for that too

Aha. Ah, so you cannot use that the continuous image of a compact is compact

I thought he already proved it

planetmath.org/?op=getobj&from=objects&id=4689

@Asaf the proof should be there but I cannot open the link

I meant that there should be a bijective, continuous map $f: X \rightarrow Y$ with $X$ compact, but $f^{-1}$ not continuous

On iPad. Hope it works for you

I have given up on planetmath. It just refuses to work

dyinglovegrape.wordpress.com/2010/10/03/…

Try this

What could be more simple to type "continuous image of a compact set" in google for a modern starting topologist?

I read that as "dying loving rape"

Sounds like the title of the song Asaf would listen too

Not really.

Abyssic Hate - Depression I

@AsafKaragila Is that a course you are going to be teaching?

@robjohn No, it's a song.

@AsafKaragila Ah, the I following it made me think a course number :-)

Well, there's just a second part to this.

I would hope so, since it was numbered

However Terrapin Station Part 1 has no part 2.

@Ilya how are things?

@AsafKaragila What's up?

I'm trying to come up with an example of a module whose support is not closed in Spec(A). I already know that the module should either not be finitely generated.

I was hoping for some hint[s].

@Rob fine. Mobile view is better on iPad but I warned about my vision in ten years

@AsafKaragila I can't think of anything off the top of my head. Hm.

You have a vision for ten years? I have vision of infinite obscurity. The darkness is coming, and as an ominous sound interleaved with calls of ravens is booming from afar and slowly gets infinitely louder... this is the song that ends the earth.

@DylanMoreland Some googleous labor got me one of two options: $\mathbb Q/\mathbb Z$ as a $\mathbb Z$-modules; or $\bigoplus\mathbb Z/(p^k)$ over $\mathbb Z$.

I was trying to think of something like the second one, yeah.

@AsafKaragila Have you had your coffee yet today? :-)

I'm not sure why the support isn't closed, though.

@robjohn I didn't go to the university today so I didn't drink any coffee...

No, that's not clear to me either.

@AsafKaragila That explains your uplifting vision.

Um.

Oh, right.

The first one seems to me a bit strange, since $\mathbb Q_p/\mathbb Z_p = \mathbb Q/\mathbb Z_p\neq 0$ for all nonzero ideals.

So the support is $\mathrm{Spec}(\mathbb Z)\setminus\{0\}$.

Oh. That would do. That isn't closed.

Why?

The generator of an ideal contained in every $p\mathbf Z$ is divisible by every prime number. So it has to be zero.

No, that part is clear. Why isn't it closed?

...or the product of all prime numbers

I think in the second example the support is $\{(0), p\mathbf Z\}$, which isn't closed either. Cool stuff.

@AsafKaragila For that set to be closed, you would need such an ideal $\mathfrak a$. But we've shown that necessarily $(0) \in V(\mathfrak a)$

Ohhhh right. :-)

Maybe a less confusing way is to just say what the closed sets in $\operatorname{Spec}\mathbf Z$ are, using the fact that you know what all the ideals are. You have $\varnothing$, $\operatorname{Spec}\mathbf Z$, and finite collections of non-zero primes, the divisors of $V(n)$ for $n \neq 0$.

what does this mean

Simple addiction?

@anon I left a comment. Anyway it will be closed soon. Maybe sculpatrol has a new account

Maybe even too soon after an answer by Ben

Drama Eternal

For the first time in a long time, I have capped. :-)

@anon If I have 2 apples and you have 2 oranges, what is the mass of the sun?

Are you going to eat the apples? I haven't had breakfast yet.

@Ilya: did you change locations, or did you lose your connection?

I lose it

or both

@robjohn If i know the wieight of the fruits, and the distance from the sun to the fruits. I can then calculate the force, and from that the mass of the sun is an easy matter to find.

Does anybody know linear operator theory on cones?

I wonder if there any uniqueness conditions similar for those on the linear spaces

@Ilya Would that be $\rm{SNO}_3$ Cones?

What that?

With the cone I mean the space where you can add and multiply by nonnegative scalarz

@Ilya It was a bad attempt at a reference to $\rm{SO}_3$

Like the space of all measurable functions with values in nonnegative extended real halfline

@Ilya Ah, I was wondering what linear operators on cones were :-)

$\rm{SNO}_3$ looks like something chemical

True that xD

@leo It does, but I meant to refer to snocones and $\rm{SO}(3)$

@robjohn yes. just an observation.

A simple example is a fix point equation Pf = f where P has nonnegative elements, bounded by 1, and f is a vector of n components each lying in [0,infinity]

@leo I agree, when I typed it, I thought of some chemical.

P is nxn matrix

@Ilya I would think that the difficulty would be in existence, not uniqueness.

I can solve it, honestly. The solution is zero

0*infinity = 0, take it as an axiom. I think the reversed product is not needed here

@Ilya Then how would there be a second solution in the restricted case of the cone when there is none in $\mathbb{R}^n$?

UHF. You also cannot collapse it as one operator, because you have infinities

And you cannot subtract

UHF is "uhu"

@robjohn, $\mathrm{SNO_3}$. Not related at all

Here were some thoughts

Maybe it's time to put some reputation there

@leo $\rm{SnO}_3^{2-}$ appears in this table. I don't see $\rm{SNO}_3$. I think Wolfram is just blindly combining atoms :-)

Hey guys

Does the term "infinity" and "does and exist" are interchangeable in case of limits?

@Ilya I have not seen anything like that.

@MaX no

for example...

@robjohn:For example this limit, I think this limit exists and it is infinity,but we can't say it's DNE, Isn't ?

@MaX $\lim\limits_{x\to0}\frac{1}{x^2}=\infty$ but $\lim\limits_{x\to0}\sin\left(\frac{1}{x^2}\right)$ does not exist

@MaX I think an infinite limit might be considered not to exist in some contexts, yet to exist in others.

@robjohn: What about the one I posted?

@Ilya, in the case when $n=2$ that leads to something like $$\begin{matrix} -b_1&= (a_{1,1}-1)x_1+a_{1,2}x_2\\ -b_2&= a_{2,1}x_1+(a_{2,2}-1)x_2\end{matrix}$$

@MaX It would depend on the context of the question. That is, on how the limit is to be used.

@Ilya, and we can say something about the uniqueness of the solutions of that system

@MaX just looking at that function, I would say it has a limit of infinity, but for the purposes of some questions, only finite limits might be applicable, and so an infinite limit might be considered not to exist.

now I see. That's the bounded case

Good time of day, everyone.

What is the way to write an arrow in LaTeX and something above it?

Like q_1 -- x --> q_2

$\stackrel{x}{\rightarrow}$

@robjohn: Thanks

thanks @leo!

Can I make an arrow longer tho?

@Daniil $\stackrel{x}{\longrightarrow}$

Thanks!

@Daniil no problem

I should make a macro for that

@Daniil this is the right place for this sort of things

@leo You are right, sorry.

no problem :)

$$\stackrel{-\!\!-\!\!-\!\!-\!\!-\!\!-\!\!-\!\!-\!\!-\!\!-\!\!-\!\!-\!\!\longrightarrow}{x}$$

Heh, it worked.

$$\large \stackrel{-\!\!-\!\!-\!\!-\!\!-\!\!-\!\!-\!\!-\!\!-\!\!-\!\!-\!\!-\!\!\longrightarrow}{\stackrel{x}{-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!\longrightarrow}}$$

Okay, enough fooling around

@Daniil you can use $\xrightarrow{\rm{something}}$ or $q_1\xrightarrow{x}q_2$

@Daniil that is nice because it matches the size of the arrow to the text above.

Indeed, thanks.

Dammit, I broke my tex configuration, now emacs refuses to enable tex-mode at all

time to lunch

cya!!!

@leo l8r

@robjohn Have you ever heard the phrase "To subtract b, add the opposite of b." used as a definition of subtraction sir?

@Skullpatrol I have heard, "To subtract $b$, add the negative of $b$."

@robjohn So does that mean you haven't heard of "To subtract b, add the opposite of b."?

@Skullpatrol I can't say that I've never heard it. I don't remember ever hearing it.

Nor would I say it is wrong, since the negative is the opposite in usual language.

@robjohn Thank you for trying to remember.

@robjohn In the so called "Definition of Subtraction" do you think the symbols a-b should be read as "a subtract b" so that you use the thing you're trying to define in symbolic form.

@Skullpatrol I usually say "a minus b". "a less b" is used in some contexts. However, "a subtract b" sounds odd to me.

@robjohn I agree it sounds odd to me to, but if you're going to say "To subtract b, add the opposite (negative) of b." shouldn't you say also "a subtract b" to be consistent with your language?

subtract is a transitive verb, something subtracts something else, however, neither $a$ nor $b$ are doing anything, the operator or the person doing the subtraction is doing the operation. This is why "a subtract b" sounds wrong; it is grammatically wrong.

When you say "To subtract $b$," you are giving instructions to the person doing the subtraction. I see not need to use that same word to describe the expression "$a$ minus $b$"