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12:00 AM
tries to connect death with the Killing form or annihilation
 
@anon: I think those are fairly specific tags. If you want to bump one of your old questions, you could always make a small edit or offer a bounty on it.
 
Algebraists don't die, they just meet their annihilators.
 
Set Theorists never die, they just meet their successor.
 
What if they don't have a successor?
Like axiom of choice negating sort of thing?
 
never joke with a set theorist when death is on the line...
 
12:04 AM
And never start a land war with a set theorist?
 
Not all of us are Sicilians :-)
We did have a postdoc from Sicily for the past two years though.
He was into set theoretical topology, interesting stuff.
 
I haven't had pizza for a while.
 
Me neither.
 
Ciao - time to go pick the up the kiddos.
 
Enjoy!
 
12:06 AM
I like deep dish pizza. I had the Chicago style pizza when I went to visit my son at college in Chicago. It was good, too.
@Mike: enjoy the kids!
 
I have no idea what either of those are.
 
@Asaf: Chicago style is a deep dish pizza, but it has a lot of good stuff in it, not just heaps of cheese.
 
What is a deep dish pizza?
 
meat and veggies, mushrooms etc
 
Yuck.
 
12:08 AM
Everyone's making me hungry, bah.
 
I hate mushrooms.
 
There is thin crust.
 
I worked in a pizzeria for almost 4 years.
 
Oh, I love 'em. I'll have yours.
 
Meat is great on pizza.
The one thing I hate about this city is that it doesn't have many non-Kosher places, so cheeseburgers and pepperoni pizza are scarce.
 
12:09 AM
thin crust pizza, has a thin crust (surprise) and a fairly thin layer of goodies on top. Deep Dish has a thicker crust and more non-crust things on top
 
flatbread or cheesecrust > thin crust or deep dish
 
anon: How old are you?
 
Different names for the same thing?
 
twenty
no, flatbread has different texture than thin crust, at least where I was
 
never had anything called flatbread.
at least pizza
 
12:12 AM
though textures vary wildly from place to place, so specifying 'flatbread' isn't necessarily meaningful..
 
Too much work again. None of the answers have gotten an upvote.
 
You might get that Tenacious badge :P
 
I only have 1 zero score accepted answer now. I got a lot of votes for old answers recently.
 
12:30 AM
Ah.
I wish it'll be raining already.
Or you know... at all.
 
I like rain, but I live in a desert.
 
So am I.
 
I'm at a crossroads, where I have to choose between (a) changing my initial parameters and so-far-written derivation so the end comes out simpler or (b) continue on with more complicated expressions. I'm going with the latter.
 
I am at a crossroads too. I should either (a) go to sleep, or (b) stay awake.
 
I was at that a few minutes ago. I chose (b) until I finish one answer and one question edit.
 
12:42 AM
I'll go with (a)
It's 2:45 and I didn't have any afternoon nap today.
I am shocked that I even sit here in the first place :D
I bid you adieu and good night!
 
12:58 AM
@AsafKaragila You should invest in catnip.
@tb The yellow isn't your color, I must say.
@robjohn The problem is that genuine newbies (rep 1) need additional 10 rep (2 upvotes from other people) to upvote. Ergo, if you don't upvote their question, they can't upvote your fabulous answer. But they can accept.
2
Tit-for-tat and all that.
(On that note, it took me quite a while to figure out the ropes of this software. As soon as I think I have things down pat, the overlords change a few things around, and I feel like a n00b again! Yeesh.)
 
1:21 AM
Speaking of "too much work" answers... I have to wonder what are the other 25 questions that can be asked on m.SE that will admit one-letter answers? (Of course, not as answers to multiple choice!)
 
some complicated expression that just turns out to be 'x'
 
@anon: that answer of yours was neat on its own, but the color was genuine art. ;)
 
It is indeed fun adding colors. I think it provides useful/suggestive local information in a derivation and should be used more often - be the change you want to see and all that.
Except it might really screw over some synesthetes. :)
 
Well, the probable problem is that some people like Andre and Asaf have some form of color-blindness...
(I've forgotten which sort they have, though.)
 
Hm. I thought color-blindness was only between shades, so that distinct base colors are different enough to be distinguished?
though the \color{Green} is kind of hard to tell from the usual black, and coloring the numeral '1' all by itself doesn't show up very well
 
1:34 AM
I can distinguish the green bit from everything else. OTOH, I needed to stare a bit longer than usual to see that you did color up the "1".
André apparently has difficulty with red.
 
@JM I have asked no questions. I must have gotten 10 rep somehow. I'll have to check my activity.
 
@rob: Your first answer here was upvoted, right?
 
ack, answers of course.
 
One upvoted answer means you can upvote other answers. ;)
A rather logical economy, yes?
 
yeah. I figured tha out when I looked at my first activity :-)
that's when I got my first badge (Teacher)
 
1:45 AM
I used to be annoyed at some of those bronze badges, until I figured that they can give useful forensic information.
 
I pay little attention to them. what useful forensic info do they provide?
 
e.g. "He couldn't have downvoted you; he doesn't have a "Critic" badge" or "Yep, you just hit the cap today. In fact I see you have a "Mortarboard"."
 
I've gotten 200+ a few times, but I guess only once without acceptances or bounties.
actually, I've capped twice
so the Mortarboard must be a one time deal.
 
You can peer at math.stackexchange.com/reputation if you need to do accounting.
 
yeah, someone pointed that out yesterday and I have it bookmarked.
 
1:51 AM
Yes, Mortarboard is one-time. That's meant for "hey, I was supposed to have 300 rep today from my blockbuster answer, what gives?"
 
It is up to date, unlike the SQL data
 
Oh yeah. I understand why they can't update the SQL more frequently though. The databases are hella huge!
 
If there were a note with the Mortarboard that 200 is the cap, it would keep people like me from seeking why they missed some points.
I gotta go get dinner. BBL
@J.M.: thanks for the correction on my answer :-)
 
Don't say I don't care about you, man. ;P
 
I try to get the backslashes in there, but I miss some.
later...
 
1:55 AM
just noticed this: twitter.com/#!/StackMath
 
Oh yeah, the twitter feed. For people who check twitter more often than m.SE itself.
 
For about 2 weeks straight I'd get up in the morning and check my twitter feed ... and it would take me 4-5 hours. Couldn't keep up, left my account dormant.
 
If I needed to stalk somebody, Twitter is pretty damn good. That's why I never bothered making an account there.
(that is, a lot of people seem to throw caution to the wind with this Twitter business.)
 
Eh, I used it more for information networking than social networking. Revealed little to no personal info (except IPs to the domain itself). But that's the reason I quit my facebook account three years ago.
 
Man, FB is an even leakier ship.
Zuckerberg is awash in private details of people I imagine.
 
2:22 AM
@JM Yellow? Grass green, I'd say...
 
@t.b. I suppose I'm used to different kinds of grass. :D
 
@AsafKaragila hm. could you please elaborate? :)
 
I wondered about that too. SFAICT a lot of your actual work counts as analysis...
 
@J.M. Hm: that's what we call grass green, I'm probably a yellow doesn't-count-as-an-analyst, then.
identity crisis
 
Hmm, okay. The grass that shows up a lot here is Bermuda grass. Only a bit paler than Jack's gravatar.
 
2:33 AM
Oh, I probably have to de-Gauss my screen then.
On the other screen it looks pretty green
Here I'm looking like an overcooked egg :(
 
Well, I was already entertaining the possibility that I have latent color-blindness... :D
 
Don't worry
 
(and an overcooked egg would have tinges of gray, too. :) )
 
Oh, I confused your gravatar with mine, I guess
 
Man, I like that curve in my gravatar. I have to wonder if Stieltjes ever considered this...
 
2:38 AM
It's pretty nice. Definitely grass green here. fires up screen settings
 
Wait, what? I'm confused now. Mine's just thin grayish-black lines. Jack's is the green one. Yours... I'm not as sure as I was.
 
t.b.'s is what you get when you take a yellow highlighter to Jack's avatar.
 
yes, probably. Ugly. It'll change in a few hours, though, so I don't mind too much
 
I'd have thought those Gravatar dudes would be better at picking colors...
 
Hm. I'm confused. How can a theorem with a simple proof be deep? And how is irrationality of sqrt(2) deep?
Ah, Austin said himself they weren't particularly deep
 
2:47 AM
That question is bugging me. I'm positive it's been asked at MO months before...
 
Wasn't that MO thread about obvious facts with difficult proofs?
 
It was backwards? My mind is playing tricks on me then.
 
 
Okay, maybe I did remember wrongly. I see I upvoted the Jordan curve answer there...
 
the obvious one. but hard to prove you actually did it.
@JM I liked the most popular one, and Stanley's 1+1=2
 
2:56 AM
I must confess I needed to stare at the premise of the first answer for a bit, so it wasn't instantly "obvious" for me...
 
One thing I'll never get: why are mathematicians so obsessed about numbers?
 
I wonder too. I mean, the subject I like the most is called "numerical analysis", and yet I'm more with tweaking methods than with staring at figures...
Somehow it turns out that most of the math I like doesn't involve that much numbers.
 
Same here
I mostly don't care about "how many integer solutions does insert random equation have?"
 
I see why some people love number theory, but I've mostly found it, like algebra, more of "needed tools" than "stuff I'd immerse myself into".
 
same here, again
 
3:02 AM
Are you the older brother I've always wished I had, by any chance? :D
 
I don't know. prepares barrel of olive oil
I am an older brother, though
 
Actually, after musing over your generous offer, I thought it might be cheaper if you could bring a few chocolates. :)
That might be easier to explain than the barrel.
 
Well, that I could send you easily.
Seriously: does the address on the paper with Mike work?
 
Well, that wasn't a complete address. Just gave the city and ZIP code.
 
It looked quite minimalistic.
 
3:07 AM
If I were still part of a university, I could've been more bold and included the box at my institution... :)
 
I see.
In high school we had this game (those were the days when postmen still had the time to care about bringing the letters to their recipients, here) to send a postcard with the most minimalistic address.
 
So, what'd the winner look like?
 
No joke: T.B. 7272
(I lived in a very small village)
initials + ZIP code
 
Wow.
Nobody else had your initials? Really?
 
Well, among 80 inhabitants there aren't that many playing this silly game.
 
3:12 AM
Hmm, good point.
 
I assume the postman figured the joke out after the third or fourth minimalistic address and when he didn't know where to file a postcard he could safely assume it was for me :)
Still, you couldn't imagine how to go more minimalistic
 
I'm guessing you guys know the postman by name. :)
 
I still remember it: Jöri Kindschi
woow. that was about 17 years ago
 
Pretty memorable guy I guess.
 
Yeah. The most phlegmatic person I ever met
 
3:17 AM
Okay, probably not the kind of "memorable" I was thinking...
 
The running gag was: Jöri, if you should happen to fall flat on your face, don't forget to take the hands out of your trouser pockets
(I don't know if this translates that well, but the meaning should be understandable)
 
I got you. The mental image alone is funny.
 
3:41 AM
This was a nice conversation, as always. I gotta step out, see you later!
 
 
5 hours later…
8:26 AM
@tb You know about the axiom of choice more than the usual analyst who takes it for granted :-)
 
@Asaf: Ah, okay :) thanks for the clarification!
 
Sure. :P
 
I do have choice... I'm maybe slightly less obsessed about it than you :p.
2
 
I am not that obsessed with choice, you know. I just like it a lot :P
 
Heh, that much is obvious. I won't answer any choice questions, though, until you've got the 100 straight complete
 
8:31 AM
Hah, it's fine.
Even I can't answer them all.
 
It would be boring if you could, wouldn't it?
 
Well... no.
I find a lot of excitement in learning, but also in explaining.
I always read the end of the book when I was in the middle.
 
I never did that. Can be a huge spoiler, I assume
 
I don't mind at all to be given all the mathematical knowledge possible and then just teach some of it to others/commit suicide as an irony.
 
Ah, so you're like Erdös's arch-fascist with his Book
 
8:35 AM
I'm not familiar with that.
 
You know the story of the Book, right?
 
Partially, I guess.
 
?
 
The one in heaven with uncountably many pages, containing all the neat proofs that there are.
 
ah, I've heard of that
 
8:36 AM
Oh yeah, that one I know.
 
Every slick proof was called a book proof by E.
 
I've read it cover to cover too.
 
Then he concluded that the one owning the book must be the worst of all fascists because he occasionally let people have a peek, but didn't share that knowledge with mankind.
 
Ah... I see. Well, if I had the contents of the book in my head I'd teach set theory and help develop it well. Then I'd crush geometry completely, and partially analysis and algebra. Then commit suicide.
 
Okay, I won't let you peek :p
 
8:41 AM
Crush?
 
Yeah, prove only the theorems showing contradictions.
Let the mortals find the new theorems on their own.
 
>>:|
 
What?
I'll give you everything you wanna know about the axiom of choice though!
Also, things you didn't care about too!
 
9:01 AM
If E is a subset of metric-space. E' is the set of all limit points of E. How to prove that E' is closed?? Can somebody help?
 
@Ramana: how do you define closed?
via sequences ?
 
A set is closed if every limit point is a point of set
 
right. So take a sequence x_n in E' that converges to some x in your metric space X. What do you need to check?
 
I am asked to prove E' is a closed set
 
I understand that. As you said, E' is closed if every limit point of a sequence in E' is itself in E'
So if you start with a sequence x_n in E' that converges to some x in X, you need to prove that x in E', too. Right?
 
9:10 AM
Yeah how to prove that?
Yes I think
 
So, the problem is that x_n does not need to be in E itself
you only know that x_n in E'.
 
Somebody told me that it is enough to show (E')' is a subset of E' I am thinking in that way but I am stuck
 
I don't think that helps that much.
Can you prove that y is in E' if and only if for every epsilon there is z in E such that |y-z| < epsilon?
Sorry I should have said z not equal to y before
@Ramana: are you following me?
 
Yeah
I am thinking I am just a beginner so confused about things
 
So let us prove this first: y is in E' if and only if for every epsilon there is z (not equal to y) in E such that |y-z| < epsilon.
Assume that y is in E'
then there is a sequence y_n \in E of points different from y such that y_n -> y, right?
 
9:22 AM
but I defined E' as set of all limit points is it necessary is what I am thinking?/
Yes
 
This means that for every epsilon > 0 there is N such that for all n > N we have d(y_n,y) < epsilon.
But by hypothesis y_n is in E, so we may take z = y_n for n large enough, right?
 
so we have proved one direction
Now assume that y has the property that for every epsilon > 0 there is z not equal to y such that d(y,z) < epsilon
 
take epsilon = 1/n
choose y_n such that d(y,y_n) < 1/n
(and y_n different from y). But this means y_n ->y.
So...?
... y in E'
 
9:31 AM
We have proved that E' contains only limits of E
 
that holds by definition :)
Now back to your question: we have x_n in E' that converges to some x in X. We want to prove that x is itself in E'
 
Yeah
@Gortaur Hi
 
Given what we've just done, we can check that for every epsilon > 0 there is some y_n in E different from x, such that d(y_n,x) < epsilon
Hi Gortaur!
First try: given epsilon > 0 we start by picking an x_n such that d(x_n,x) < epsilon
 
hi guys ) just popped up for a second
wanna go to Rotterdam zoo
 
Is it for E or E'
 
9:37 AM
What is for E or E'
 
@tb the deadline for my conference submission is 7 am in my time zone. tonight I realized that didn't check the pdf after compilation for the numbers of equation and woke up to read all the paper again. such a mess
fortunately everything was on the place
 
@Gortaur: whooops! sounds like a nightmare
Oh, lucky you!
Ouf!
 
Given what we've just done, we can check that for every epsilon > 0 there is some y_n in E different from x, such that d(y_n,x) < epsilon
 
anyway I had time to re-upload it
 
In this statement is it E or E'
 
9:39 AM
It is E
We want to prove that y in E', so we need to find y_n in E arbitrarily close to it.
@Gortaur: sounds like re-checking if you have turned off your stove :)
 
@tb that's not about me usually ) just tired the last days
Herr Koenig told you that you good enough with AC for the analyst? )) royal mercy
 
@Gortaur: Yes, that was pretty generous :) feeling good
Looks like @Ramana left...
 
@Gortaur Yesterday I asked you a question about a theorem I have came across In that You have said that X is open for someother Reason. If you can remember all this can you tell me why
I am here I was solving some question posted in math.SE
 
@RamanaVenkata: are you working with a metric space or a topology?
@tb nice ) I will leave soon, so have a nice Saturday
 
Metric Spaces
 
9:50 AM
@Gortaur: enjoy the zoo! Hope the weather is nicer there than here.
 
@RamanaVenkata set A is open iff any point x belongs to A together with some open ball B(x,r)
any open ball by definition belongs to X since it is a universal space
 
@Ramana: you need to tell me what you've understood so far. I'm not giving you the answer just like that.
 
so, any x belongs to X together with any open ball B(x,r)
that's the reason why X is open if (X,d) is a metric space
if you work with topology, there X is open by definition ;)
@tb thanks, it's +7 here
sunny. Crap
 
@tb Can you ask a question I'll answer it
@Gortaur Okay Thanks
 
I was saying that y is in E' if and only if there is a point x (different from y) in E arbitrarily close to it. Do you see why?
@Gortaur: nice. It's very cloudy and freezing here. First time the temperatures went below zero... :s
 
10:00 AM
No
 
That's just a re-statement of what we proved above: "y is in E' if and only if for every epsilon there is z (not equal to y) in E such that |y-z| < epsilon.", right?
 
Yeah I can see that
 
Once again: we want to prove E' is closed. So we take a sequence x_n in E' such that x_n -> x and we want to prove that x is in E'.
So we already know that x_n comes arbitrarily close to x. The problem is that x_n is not in E but only in E'.
So take epsilon > 0 and choose n so large that d(x,x_n) < epsilon/2
if we knew that x_n is in E, we would be done already, right?
 
but we only know that x_n is in E'
this means that we can find a y_n in E such that d(x_n,y_n) < epsilon/2
Hi, @JM
 
10:17 AM
@AsafKaragila That's what anybody who's obsessed would say. ;P
Hi t.b.!
 
Wow. I stop to answer a question and 127 messages fly by.
 
@robjohn: It was a longish answer :)
 
It was a delta-epsilon proof.
 
"ish"? That isn't "ish"... :)
 
sometimes they take a while to get all the details straight.
 
10:20 AM
Only the notation makes it look short.
 
@t.b.: you've already read the answer?
 
@robjohn: not yet, I'll have a closer look at it in a moment
 
Ah, it sounded as if you had, or perhaps you were just commenting on the duration of my absence. I also had dinner and some family time, too.
 
@anon: it seems coloring got you a "Good Answer". :D
 
indeed. :)
 
10:37 AM
@anon: there, I put you over 70% of the way to 10K from 9K :-)
 
heh, thanks.
 
You were just a few rep short.
 
@anon: it seems my estimate back then was too pessimistic. :) I'm wagering you'll be part of the 10k club by Monday or so...
 
It's only saturday. I'd give it Tues or Wed, because blockbuster answers don't come every weekend.
 
@robjohn: I was commenting on the duration of your absence.
 
10:41 AM
@t.b.: I figured, since you hadn't read the answer.
 
I've read your answer now. It looks good, but I don't see the main point of it just yet.
 
10:52 AM
@t.b.: the derivative exists for x in an open set and its value is close to y when x is close to x_0. For x in the complement of the open set, we have that the difference quotients are also close to y, but we need to show that the derivative at x_0 is y. Most of the cases can be handled with the MVT or the difference quotients, but there is one case that requires a bit of extra care.
It's one of those kinds of problems that is a bit painful to think about, but it is good exercise.
 
@robjohn: thanks for the summary! I just needed to convince myself that the extra care is needed. Drawing a picture helped to achieve that :)
 
Yeah, it is almost obvious, but for that one case.
 
I think it's one of those things that seem so clear that you need to get your hands a bit dirty to see that it's not as clear as it seems at first.
 
Indeed.
That being the case, it probably won't get too many upvotes.
 
I would be surprised if it gets a lot of upvotes. You have mine already.On the other hand there are 5 upvotes on the question, so maybe a few people tried and failed. Your solution is very clean, as usual.
 
11:06 AM
The fact that it sat for most of yesterday, I tend to agree that it has been tried, but the details are hard to get right. One of the real simplifiers is using |x-a|+|a-x_0|=|x-x_0| as the condition for betweeness.
That allows one to use the converse of the triangle inequality
 
Yes, that's quite a slick move.
 
11:24 AM
@JM And that's what anybody who's accusing the obsessed of being obsessed would say! :-P
 
I was about to reply, but I don't want to risk a stack overflow with this particular recursion...
 
@Asaf: Did you see the announcement of the Exploring the Frontiers of Incompleteness series? Looks like a few interesting talks will be appearing there.
 
@robjohn Can you tell me how Z is closed in R?
 
@Ramana: Z is discrete and so has no limit points in R. therefore, it contains all of its limit points (none).
 
Okay I thought of that just for conformation I asked it Thanks
 
11:35 AM
Do you see that {0} is open in Z?
 
which functions from Z to R are continuous?
 
i don't know
 
Do you know of the characterization of continuous functions: a function is continuous iff the inverse image of each open set is open?
equivalently, a function is continuous iff the inverse image of each closed set is closed
 
@robjohn No Never heard of it
 
11:46 AM
ah, that is the topological definition of continuity. It extends the limit version on the Reals. However, it works in the Reals as well.
 
I am 1st year Undergrad and I have taken my first course on Real analysis I don't know much
 
using this characterization, it becomes immediate that ALL functions from Z to R are continuous.
All functions from Z to anywhere are continuous since all subsets of Z are open.
think about lim_{x\to a}f(x). how can x tend to a in Z?
only by eventually being a forever on.
 
and in that case f(x) becomes constant also, so lim_{x\to a}f(x) = f(a) for any function
almost vacuous.
afk for a bit.
 
@robjohn Is {0,1} is open in Z
 
12:00 PM
indeed :-)
Since {0} was open, and for the same reasons, {1} is open. and any union of open sets is open
just as any intersection of closed sets is closed.
Thus the set of prime numbers is open in Z
 
@anon, you there?
 
more or less
 
I didn't want to clutter that particular thread, so: "my", not "I's". ;)
 
...take it to english.se :)
 
Yeah. It'd be too much to do it on the main site. ;)
 
12:17 PM
@robjohn If E is a subset of a metric space and E' represents set of all Limit points then How to prove that E' is closed?
 
Suppose x is a limit point of E', it is also a limit point of E, and therefore in E', thus E' contains its limit points.
Can you show that a limit point of limit points of E is a limit point of E?
 
@robjohn: I answered that one already...
 
@t.b.: the question I just asked?
 
@t.b. But I am bit confused with that answer
 
Can anybody tell me if the [hyperbolic-geometry] tag in here is supposed to be there? Maybe I'm missing something...
 
12:21 PM
@Ramana: That's because you decided to multitask instead of thinking about what I said
 
@Ramana: so what are you still confused about?
in regards to the latest problem :-)
 
I think I can prove it
now
 
@JM: It's a bit of a stretch, I think, but not completely inappropriate. The \lambda-invariant is a universal covering map D -> \mathbb C \ {0,1}, (C \ {0,1} has negative Euler characteristic) see Matt E's answer here. This gives C \ {0,1} a hyperbolic metric
 
12:36 PM
Okay, thanks. Good thing I hesitated.
 
@robjohn If x is a limit point of E' for every neighbourhood of x N(x) then there exists a p belongs to E' , not equal to x such that p belongs to N(x). Since N(x) is a open set there is N(p) which is contained in N(x) since p is a limit point of E there is a q belongs to E, not equal to p contained in N(p) That implies for every N(x) there is q belong to E, not equal to x such that q belongs to N(x) So X is Limit point of E Is this a correct proof??
 
@Ramana: yes, that's correct. I expressed that same argument with the help of the triangle inequality.
 
@tb The only thing which I am feeling not obvious in the proof is x not equal to p and p not equal to q then x not equal to q is what i am writing . Is it necessary that it should happen??
 
@Ramana: prove that every neighborhood of a limit point x of E has infinitely many points from E in it.
 
@tb I know this theorem I can prove it and it answers my question thank you
 
12:51 PM
@Ramana: great! You're welcome.
 
@Ramana: do you have a test next week?
 
No I am reading topology on my own just because of my math interest
 
Very nice.
 
But only thing I am lacking confidence sometimes which de-motivates me
 
1:10 PM
confidence often comes with experience. give it some time, and keep on learning.
3
 
1:48 PM
Copying links from Google searches is getting more annoying by the day.
 
@JM I noticed that, too. Fortunately they still display the direct links in green, so I copy those and add http(s)://
But that of course only works for short ones like those to Wikipedia.
 

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