Mathematics - 2012-03-11 (page 3 of 4)

Kannappan Sampath

13:00

@anon Did you check those integrals lately?

Skullpatrol

@tb Thank you ;-)

anon

@Asaf: What does Bengt Ekerot have to do with anything?

Asaf Karagila

@anon He is Death.

t.b.

@anon: concerning your question on "Banach limits:" I'm rather skeptical that this can be achieved in the generality you asked about. There are simply too many constraints to fulfill.

anon

Oh well. I have one other question up my sleeve.

Kannappan Sampath

13:02

:/

anon

@Kanna: Nah I haven't checked in awhile, sorry.

I'm wondering if the Euler factors in more general L-functions have p-adic interpretations like the Riemann zeta's do (as seen in Tate's thesis, if I understand Tao's blog post about it well enough).

t.b.

@AsafKaragila I thought that was Chuck Schuldiner

Asaf Karagila

@tb A whole other Death! :-P

t.b.

@anon Does this answer by Matt E address what you're asking?

Daniil

Hi.

anon

13:09

:O Now I wish I had a copy of that thesis even more.

Kannappan Sampath

@tb Does this comment of mine answer the question here? I am afraid, not because Heine Borel property means every closed and bounded set is compact. But OP wants to know the name of those spaces where every bounded set is contained in some compact set.

t.b.

@KannappanSampath It's equivalent: the closure of a bounded set is bounded and the closure of a subset of a compact set is compact.

Kannappan Sampath

Actually, looks like, if every bounded set is contained in a compact set and it is closed, then we have it must be compact and hence HB.

@tb Beat me to it. I was writing something of a similar sort.

So, this answers the OP question?

t.b.

I think so.

Kannappan Sampath

Are you going to write an answer?

Daniil

13:13

@tb can you help me with Beth's definability for predicate logic please? I read through the chapter you linked me and as I've understood they've moved from FOL to propositional calculus via compactness. But I don't understand what are they doing after that.

I thought about starting a question, but I am afraid it's too localized since it's an exercise from a specific textbook

t.b.

@anon You probably want to get your hands on Cassels-Fröhlich. It's reprinted in there

@KannappanSampath No.

@Daniil Let me dig up that link again.

Kannappan Sampath

@tb Oh. Then I'll write it, : )

Daniil

i.minus.com/ibuzCFMFD7F1Wj.png ; scribd.com/sunils_139/d/50731355/…

t.b.

@Daniil thanks :)

Daniil

Thank you for your willingness to help me

t.b.

13:20

So you're fine up to formula (5.1)? (I'm not sure if I can really help you, but let's try ;))

Daniil

Yes.

I don't understand the step between (5.1) and (5.2)

t.b.

Dropping the parameters for a moment you have $(A \wedge A') \implies (P \iff P')$, right?

Daniil

Yes, that is precisely what I want to prove.

Erm

sorry

t.b.

No. That's what you obtain from compactness and introducing parameters.

You want to see that this gives $(A \wedge P) \implies (A' \implies P')$.

Or do I misunderstand you?

Daniil

That is what I have and I want to prove that if that formula holds, then $A \implies (P \iff G)$ where $G$ contains only literals which are not in $P$.

@tb as intermediate step, yes.

I checked two times and $(A \wedge P) \implies (A' \implies P')$ and $(A \wedge A') \implies (P \iff P')$ have different truth tables

t.b.

13:29

I believe you. But you only want that the second statement you mention implies the first.

Daniil

Oh, looking at my truth table I can say that this is true.

t.b.

Note that the authors only say "it follows quasi-tautologically that..."

(so they're not asserting equivalence of the statements)

Daniil

So, according to the Craig's interpolation theorem, there exists a formula $D$ such as $Vars(D) \subseteq Vars(A \wedge P) \cap Vars(A' \implies P')$

And $(A \wedge P) \implies D$ and $D \implies (A' \implies P')$

t.b.

yes.

Daniil

I don't understand why they say that $P$ and $P'$ do not occur in $D$ tho.

Kannappan Sampath

13:35

Done. I finished writing that answer. : D

Rajesh D

Every Mathematician apart from his daily routine, there must be some question on his mind that is constantly bothering him day in and day out....is it true....Do you guys have any such things in your mind

anon

$85 for a book. Nope.

Asaf Karagila

Okay. I think I got the rest of the argument why everything is Lebesgue measurable in Solovay's model.

Rajesh D

@anon Who were you responding to ?

t.b.

^^ are these examples of what Rajesh is asking about?

anon

13:43

tb's C&F recommendation.

Though it looks very nice inside.

Rajesh D

@tb which ones ?

i did not understand "^^"

t.b.

anon's and Asafs comments after your question

Rajesh D

hhmmm....may be

@Asaf's looks like a classic

Asaf Karagila

Oh. I am bothered by which city would be the safest once the Iranians get a nuke. I mean, they won't fire at Jerusalem, right? On the other hand there's a good chance they'll conquer it by land and just kill every Israeli living there...

Rajesh D

@Asaf : You are welcome in India

anon

13:46

Upload your mind to the internet.

Rajesh D

ahhahhha.....it will be useful for lot of us

Daniil

@tb can you explain to me, please, why $P$ and $P'$ do not occur in $D$?

Rajesh D

@Asaf : I am taking your sentence as a banter...i don't think your fears are real...are they ?

Kannappan Sampath

They have never been, atleast in the past. : D

t.b.

@anon Have you checked the new edition?

Asaf Karagila

13:49

Half real. I mean, who knows what these crazy politicians will do next...

anon

That looks even nicer. Might be a good bday present...

Daniil

@AsafKaragila How are they going to conquer it by land? I though that Israeli army is very strong.

Asaf Karagila

Someone buy me the next Kunen book. My birthday is coming up and it's like 25 bucks.

@Daniil Well, yes but I'm guessing an all-out attack from neighboring Syria, Egypt and Iraqi/Iranian armies might be a bit too much (included all the Palestinians organizations which will surely be glad to rule Israel and kill each other instead of us...)

anon

Theoretically, the analysis I've heard is that Iran just wants a deterrent so US+Israel will stop bullying them so much, because in principle if they were to attack they'd be wiped off the map in retaliation. I'm not really knowledgeable about these things, not least because of the propaganda all over the place.

Asaf Karagila

@anon It's hard to tell. The point is that even if they attack first or second the chances are that Israel will be a radioactive wasteland for the next millennium or so.

t.b.

13:53

@Daniil But isn't this what the interpolation theorem provides?

Matt N.

@AsafKaragila Mossad, for example.

Daniil

@AsafKaragila Syria and Egypt has problems of their own, I believe. I really hope that the Arab spring brings more secular politicians.

Asaf Karagila

@MattN That's just our spies network...

Daniil

@tb Hm, not really. Maybe I am misunderstanding something tho.

Interpolation theorem states that $Vars(D) \subset Vars(A \wedge P) \cap Vars(A' \implies P')$

Matt N.

@AsafKaragila That's a lie. Spying is not all they do. Anyway: so you agree that they know what politicians will do next.

Asaf Karagila

13:55

@Daniil Actually the Arab spring wasn't great for Israel (in short term) because to avoid more internal problems they somewhat use Israel as a scapegoat sometimes to allow some external venting of the people's rage. In turn this means that if there is a war in the near future it's likely we'll be attacked.

@MattN No, I do not agree with your second assertion. As for the first, assassinations are classified under espionage.

Matt N.

Anyway. Now that I've mentioned the word they will probably make a file on me.

Daniil

Heh

Asaf Karagila

:-P

anon

I've always wanted to write the FBI and ask if I'm on a list, but then I figure if I wasn't on one before I'll surely be on one after.

"List of people who have asked if they are on a list."

Matt N.

: )

t.b.

13:59

@Daniil It is very likely that I'm missing something. I'm not really into those things (it's a long time ago that I thought about it). But assuming that statement, is the rest fine?

Kannappan Sampath

@anon I ROFL. : D

Daniil

@tb Well, to be honest, I don't understand the transition between (5.4) and (5.5)

Skullpatrol

@tb I found the proof of the Pythagorean theorem that is understandable by a fifth grader I was talking about yesterday, do you want to see it?

anon

probably seen it a million times before, but go ahead

Asaf Karagila

@tb Do you remember my frustration yesterday regarding how people sometimes expect to intuitively understand nontrivial mathematical proofs?

Kannappan Sampath

14:03

I think I do, Asaf.

Even I faced with one today. Arghh... He never knew groups but wanted to understand conjugacy classes.

Asaf Karagila

The comments here made me go :|. (Both the question and the answer by "you'")

Kannappan Sampath

I stabbed myself and went to bed.

t.b.

@Skullpatrol go ahead

Skullpatrol

@tb Proof #2: If you make a paper cut-out model and move the pieces around

Daniil

"So I would very much appreciate as little of symbols of any kind as possible - maybe NONE?" what?

Asaf Karagila

14:06

@Daniil Keep on reading... it gets worse. :\

t.b.

@Daniil to be honest, I don't see it either, right now. Sorry. The longer I stare at this the more confused I get... :/

@Skullpatrol Thanks. :)

Kannappan Sampath

I am starting to hate the site these days.

t.b.

why?

anon

"without referring to, say, the Bible" what the brownies does that have to do with anything?

Daniil

@tb Ah, no problem, maybe I'll figure it later :)

Kannappan Sampath

14:10

Someone asks a question where he uses words he absolutely has no idea about. After we finish writing an answer, leaves a comment like the one below, I neither understand what he means, nor get a feel for what he is trying to say.
Not sure I understand what your link is explaining. If I partition a list of 6, I can have (12)(345)(6) for instance, but I could just as easily do (1)(234)(56) and yet both (12) and (56) would be considered part of the (12) conjugacy... unless I am totally misunderstanding what such classes represent here.

anon

He's pointing out that two different permutations can be representatives of the same conjugacy class (they are the same cycle type), and it seemed arbitrary to him ('why use this one but not the other?').

Kannappan Sampath

Ah, I see. Now it suddenly blossoms into me. But, if only he knew what conjugacy classes meant. sigh

anon

I think the issue is more with his articulation than with his comprehension.

Skullpatrol

@tb Np, the proof only appeals to the idea of equal areas, and that is what makes it so understandable, in my opinion.

t.b.

@Skullpatrol oh, sure it's nice :)

Kannappan Sampath

14:15

Dinner time.

t.b.

(and understandable by people who know the concept of area of polygons)

I have to step out for a while. BBL

Skullpatrol

@anon Have you seen that one before anon?

anon

49

A: Pythagorean Theorem Proof Without Words (request for words)

How about just three letters? $\;\;$ (Hat tip: Clipart Etc.)

Not only have I seen it, merely posting a picture of it got me my highest voted answer.

(Rather depressing when you think about it.)

Skullpatrol

@anon I'll take that as a yes ;-)

Matt N.

14:32

@tb See you later.

robjohn

Hi all; bye all :-) I just got up and now I have to go grocery shopping.

Daniil

Oh you :p

I need to go to the pharmacy, but there is a snowstorm outside :|

Eh, I am just going to watch Archer instead, I am such a lazy fuck.

Matt N.

: D

robjohn

14:52

@Daniil I don't think anyone will blame you for staying in during a snowstorm.

It is going to be a busy day here. I probably won't be doing much here today. We have someone coming to help us clear out some stuff around the house this afternoon. Not much fun there.

Skullpatrol

@anon Well, I thought about it ... and I wouldn't be depressed if I were you. Think of it as a testament to clear thought i.e. if even a fifth grader can understand it, its gotta be good ;-)

Kannappan Sampath

Ha, the joy when random answers get upvoted! : D

Skullpatrol

@robjohn Doing some early Spring cleaning are we?

Rajesh D

Surprising ...@Skull is making a lot of sense in his conversations these days.....good trend

I hope he won't rip me apart for this

Skullpatrol

Thanks @RajeshD ;-)

I would never rip you apart ... you're too nice of a person for that.

Rajesh D

15:05

thanks @Skull

Kannappan Sampath

@Skull Planning to tell us on which site in the Stackexchange network do you have account, anytime today?

Skullpatrol

@KannappanSampath Skullpatrol does not exist.

Just like my name on Santa Claus' list ;-)

Kannappan Sampath

@Skullpatrol Telling us helps you know?

robjohn

15:53

router down, router up, off to the park :-)

network calisthenics, bbl

t.b.

Hi robjohn, bye, robjohn

Matt N.

Yay! You're back! : )

t.b.

I am, apparently :)

I don't want to sound obnoxious, but have you tried verifying the statement on the discrete Fourier transform of the $\operatorname{sinc}$ function?

Matt N.

16:11

Nope.

How would I verify it? There is more than one table floating around on the internet claiming the same DTFT pair is correct.

t.b.

Oh, I'm not saying it's wrong. I'm just suggesting that it might be a worthwhile exercise to actually compute it...

Rajesh D

@Matt : could you give me a link

Matt N.

Heh. I just deleted my answer because I thought you're saying it's wrong.

@RajeshD web.eecs.umich.edu/~aey/eecs216/lectures/pdffiles/dtft.pdf

@tb I thought about it but then it looked painful...

t.b.

@MattN I would usually say so directly...

Matt N.

Now I'm thinking about undeleting it but then I'd have to actually compute it...

-> pain

Rajesh D

16:24

@Matt : What are you saying or in doubt about the expression...is it $\pi$ ?

Matt N.

But I think you're quite right. Answering a question for which I had to look up the answer in a table felt stupid enough.

@RajeshD No, never mind. : )

Rajesh D

@Matt N. : give me a link to the question...i'll try it when i am free

Matt N.

@RajeshD Here.

t.b.

@MattN It was no criticism nor did I try to induce you to delete the answer nor anything else; except trying to suggest that you try to do the computation at some point (you need not include it in the answer here).

Matt N.

@tb Yes, I know how you meant it : )

If it was easy to compute though would there be tables to look it up?

(I'm not questioning your suggestion, I'll do as I'm told, I'm just slightly worried that it will be a right pain.)

Rajesh D

16:33

@Matt : the formula is given explicitly in the link you mentioned. All the OP has to do is substitute the value for $B$....I don't think the OP is expecting a derivation...If he had then he would have given a formula for DTFT in the first place.

N3buchadnezzar

Hi =)

Matt N.

Hi.

N3buchadnezzar

16:48

Bleh, tripple integrals is no fun =(

I tried to find $$ \iiint ( 3 + 2xy ) \, \mathrm{d}V$$ over the solid hemispherical dome $D$
given by $x^2 + y^2 + z^2 < 4$ and $z>0$

After some pundering I thought about using polar coordinates, because of the obvious spherical symmetry as shown below

$$ \int_0^{2\pi} \int_0^2 \left( 3 + 2 r \cos \theta r \sin \theta \right)r \,\mathrm{d}r \, \mathrm{d}\theta $$

But this gives me $12\pi$ and not $16\pi$ which I am supposed to get, does anyone know where my rookie mistake lies? =(

Matt N.

Isn't one dimension missing?

Your integral is $dx dy dz$ in Cartesian coordinates. Then in polar coordinates it should be $dr d \varphi d \theta $, no?

N3buchadnezzar

I do not know, my book covers this topic in three pages. And no mention of switiching to polar coordinates =/

Not $dx\,dy\,dz$ = $r \,dr \,d\varphi\,d\theta$ ?

Rajesh D

@N3bu : are you in Engineering ?

Matt N.

Well, something like that yes. But I was criticising that you had dropped one dimension completely.

@N3buchadnezzar Show me your transformation.

Like $x = r \cos \varphi \sin \theta$ etc.

N3buchadnezzar

@RajeshD Master mathematics, barely started.

I was simply thinking
$x = r \cos \theta $
$y = r \sin \varphi $
$r = \sqrt{x^2 + y^2} $

Matt N.

17:02

No. That's for two dimensions!

But you're in $R^3$ here.

You want to look at this: en.wikipedia.org/wiki/List_of_common_coordinate_transformations

There you will find the transformation of $x,y,z$ into $r, \varphi, \theta$.

Then all you need to do is compute the determinant of the Jacobian of your transformation matrix and this will give you the thing you have to multiply with.

N3buchadnezzar

Seems difficult to do it this way to calculate the integral

Matt N.

Well, this is how it's done using polar coordinates...

N3buchadnezzar

Sigh =(

I will give it another spinn, thanks =)

Matt N.

Np : ) Maybe there is an easier way to do it.

N3buchadnezzar

well

Is there any reason why we an not use cylindrical coordinates?

Matt N.

17:10

I don't know. : )

Well, we can always swap and you compute this for me:
$$ 3 \sum_{n = - \infty }^\infty \frac{\sin(\frac{3}{4} \pi n)}{\pi n} (\cos(-\omega n) + j \sin (- \omega n))$$
: )

N3buchadnezzar

Maple chrashed

Jeff

@anon (old topic) Mind you, my post said it's not really a formal proof. Also, I've had some profs who would accept that (as would I, in some circumstances).

@N3buchadnezzar Maple does that! Sometimes, Maple won't let me save sheets and I have to copy and paste an entire worksheet into a new one and save it with a temporary name, then restart Maple. :(

@Matt Ever get an answer to that (3 starred) integral on the right?

N3buchadnezzar

bah

I still can not figure out that integral

Matt N.

@N3buchadnezzar Is the integral really that difficult in cartesian coordinates? You have $0 < z < \infty$ and $- \sqrt{4 - z^2 - y^2} < x < \sqrt{4 - z^2 - y^2}$. Then you see that $0 < y^2 < 4 -z^2$ and therefore $-\sqrt{4 - z^2 } < y < \sqrt{4 - z^2 }$.

Or something like that. I have a feeling that I'm counting a part twice.

But you get the idea.

@Jeff That starred integral has an antiderivative of Bessel functions. I posted it mainly because I wanted to change topic at that point : ) Thanks for asking.

N3buchadnezzar

Why do we have $0 < z < \infty$ ? Is not z restricted by the half sphere? Sorry for asking so many silly questions.

I guess I will read a bit more before trying to chew on these.

Matt N.

17:20

@N3buchadnezzar That's given in the question, I thought.

N3buchadnezzar

It is saying z>0, but i thought because of the sphere that $0<z<2$

Matt N.

That seems plausible. Why don't you compute it with $0< z < 2$ and see what you get?

N3buchadnezzar

That is what maple gives me for your sum btw

^^

Matt N.

: D

Yikes

N3buchadnezzar

I do not want asaf on my butt again, removed =)

Matt N.

17:23

: )

N3buchadnezzar

Warning: Nostalgia incomming

►

OH, and maple was not able to solve that integral

Jonas Teuwen

17:37

What integral?

Matt N.

This one.

Hello Jonas : )

I'm starting to feel lonely.

N3buchadnezzar

I am tarzan jungle man

Matt N.

Oh hi, I thought I was the only non-ghost avatar.

Hi Kannappan.

Kannappan Sampath

Hi @Matt

@Matt Can you point me to a good MaCBook that I could ask my father to buy for me?

Matt N.

Sorry I was afk.

@KannappanSampath What are you going to use it for?

Kannappan Sampath

17:53

For my personal use. Not much specific purpose, I'd say.

Matt N.

But you might be carrying it around a lot?

Kannappan Sampath

No, I don't think I'll have to.

N3buchadnezzar

@KannappanSampath why a mac ?

Kannappan Sampath

@N3buchadnezzar I have always longed to have one. : D

Matt N.

@KannappanSampath Then all you need to find out is whether you're happy with a small screen (13 inch). I have 15 inch but I used to have an iBook and that was handy to fit into my backpack but the screen made it painful to work with.

I haven't tried Mac Book Air but that might also be an option (they're slightly cheaper than Mac Books)

Kannappan Sampath

17:59

@MattN I'd personally like a bigger screen, my present laptop has a bigger screen as well. So, I'd think I have to go with a 15' screen.

Matt N.

Ok, then there is not much choice : ) (there are no 15 inch Mac Book Air)

But I don't think I am qualified to help you choose a laptop.

Kannappan Sampath

No, you're. Else, why would I ask you. : D

Matt N.

The problem is that it really depends on what you're going to do with it. Also, have you given thought to getting a ThinkPad with Linux on it?

I mean, I'd never do that. But the Think Pad might last much longer and the Mac might last about 6 years (with intermediate investments such as replacing fans and battery etc.)

Kannappan Sampath

@MattN No. Let me google for the prices. I did not do that

Matt N.

@KannappanSampath Here.

As you can see, they not only last longer but they're also much cheaper.

Kannappan Sampath

18:06

@MattN Thank You. I am just taking a look.

Matt N.

@KannappanSampath There is also a thing called Hackintosh which is a version of OS X that will run on some non-Apple devices. If you wanted a mac for the OS then you might want to try that. I don't know how geeky you are though the installation of it might be painful (depending on how much it's improved since the last time I tried)

I tried to install it on an old Dell computer I had and I couldn't make it work.

The best is probably to go with Ubuntu.

Kannappan Sampath

I am not all that geeky when it comes to computers. : D

Matt N.

Come to think of it: I think I'll start contributing to Hackintosh after handing in my BSc's : ) That sounds like a hobby with a good purpose.

@KannappanSampath I'm not geeky when it comes to anything, unfortunately. : (

Kannappan Sampath

@MattN Oh. I see. Then, you're an expert programmer in a previous Avatar?

@MattN Ah, I am not ready to believe this. : P

Matt N.

@KannappanSampath No, not at all. And contributing to open source projects is not for expert programmers only. : )

@KannappanSampath But maybe I can work on it : )

Kannappan Sampath

18:14

Finally, I am now having tough time having to decide.

Matt N.

Do you have the possibility to try both and then decide?

I know you can go to the mac store to try the mac. What about the thinkpad?

Kannappan Sampath

@MattN There are shopping Malls out here but I am very lazy to go out shopping.

Matt N.

@KannappanSampath I can relate to that. OTOH, you don't want to make a wrong decision.

Skullpatrol

@MattN Have you ever gone to the chat room at the Physics Forum Matt?

Matt N.

:3780259 But that's missing the point! You will have to use the bloody thing, not him : )

@Skullpatrol No.

Kannappan Sampath

18:19

@MattN I get it, I'll walk out sometime in the next two days and do this thing. : D

Skullpatrol

@MattN You should check it out it is very slick & busy

Matt N.

@KannappanSampath Good : )

@Skullpatrol I like it here and I don't know any physics.

Kannappan Sampath

I am back to bed. Bye, folks.

Matt N.

Good night.

N3buchadnezzar

hya

I saw someone solving $\int_0^\pi \sin^2x \, \mathrm{d}x$ using the beta function, that made me giggle

Sivaram Ambikasaran

18:54

@AsafKaragila Thanks for the clarification. I need to revisit ZF axioms and understand them better. I will buzz you and kanna regarding some more questions on set theory.

Asaf Karagila

No problem.

Sivaram Ambikasaran

@AsafKaragila So the axiom of choice comes into play only when we need to make choose, without any property, an element from an infinite collection of sets.

Asaf Karagila

Well, this is almost correct. It is without any uniform property.

Matt N.

What about an uncountable set that doesn't have an order on it?

Sivaram Ambikasaran

Hi @KannappanSampath and @MattN

Asaf Karagila

18:59

It is perfectly reasonable that every set has an associated structure, or even that the structures are somehow similar. If there is no property which defines a selection from every set, then you need some form of choice to assert such selection.

Matt N.

Hi Sivaram : )

Asaf Karagila

@MattN You can always give it "trivial" structure somehow.

Matt N.

@AsafKaragila Well if it's countable I see how (take the first element of the enumeration) but if it's uncountable what sort of structure can you put on it? You can't necessarily well-order it since that would be exactly what AC is saying.

Asaf Karagila

(Also a set which cannot be ordered has to be uncountable, since countable sets can be mapped into $\omega$ and thus inherit its order by a transfer of structure.)

Matt N.

@AsafKaragila That's exactly why I wrote uncountable up there : )

Asaf Karagila

19:01

First it is possible that you can endow the set with a group structure (even without the axiom of choice, very strange sets can be made into groups); or you can perhaps consider it as a discrete space...

Sivaram Ambikasaran

So if we have a finite collection, say $\{A_i\}_{i=1}^{n}$ what axiom are we using when we say choose $a_i \in A_i$, without mentioning how we choose $a_i$.

Asaf Karagila

No axiom.

You use the fact that in the meta-language you have induction, and you can write formulae as long as you want (but finite!).

Sivaram Ambikasaran

oh ok. I am beginning to understand this.

Asaf Karagila

Have you seen my answer about finite choices without AC?

Sivaram Ambikasaran

this one math.stackexchange.com/questions/85153/…?

Asaf Karagila

19:04

The other one I linked.

This one.

Sivaram Ambikasaran

@AsafKaragila Ok. I think I get it. But I think I need to read in detail about set theory.

Would Halmos be a good starting point?

Asaf Karagila

I heard that Enderton is good, Jech's Set Theory is a nice start, although a bit concise at times. I think that the first part should be reasonable (at least for someone experienced with mathematics).

Sivaram Ambikasaran

Actually I started with Goldrei. But stopped at the end of the second chapter.

Matt N.

Did you see chat.stackexchange.com/rooms/2318/matts-universe ? That's a room about a set theory book.

Asaf Karagila

Never heard of it.

Sivaram Ambikasaran

19:13

books.google.com/books/about/…

Asaf Karagila

The description doesn't sound too hot, I have to admit.

Sivaram Ambikasaran

ok :).

Off to lunch now bye.

Matt N.

Byee.

Asaf Karagila

I am going home. It's about friggin time!

Matt N.

19:35

Teddy?

t.b.

@MattN Yes?

Matt N.

Ello : ) Sorry, just seeing if you're there.

leo

hi there!

Matt N.

Hi leo.

t.b.

Hi, leo

leo

19:44

hard exercises in Stein & Shakarchi book

Matt N.

Which one? Jonas thinks this book is for people who like pain.

Asaf Karagila

Okay. I'm back home.

Matt N.

Hello Asaf.

leo

for example: give and example of an open set with the following property the boundary of the closure of the set has positive Lebesgue measure

t.b.

@leo take the complement of a fat Cantor set

leo

19:50

@tb yes. I was trying so. That's the hint. I'm trying to figure out why the closure is positive

t.b.

Well, a fat Cantor set is nowhere dense. See also here

leo

@tb thanks

Matt N.

I'm so lucky. I don't have lectures next week, it's all cancelled (by coincidence).

It's too quiet in here : (

t.b.

@MattN sounds nice :)

Transcript for

$Mathematics$ Mathematics