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00:04
Tomorrow is now today :-)
00:19
Ahhh... axiom of choice answers before going to bed. Feels good. :-D
00:30
@AsafKaragila like warm milk?
No... like a bottle of wine.
Wait no, that would be the bottle of wine that I drank by myself, save one glass for the missus.
@AsafKaragila I'll be having some chai, but no alcohol.
@robjohn Speaking of alcohol, can you give it a read and tell me if it doesn't make any sense?
@AsafKaragila As far as my limited knowledge of the subject goes, it looks good.
00:52
Well... goodnight!
@AsafKaragila Good night
It looks to be a quiet night tonight.
01:20
Hi leo
and Simon :-)
leo
leo
hi
@robjohn indeed, very nice answer
@leo David Moews? The determinant answer?
leo
leo
@robjohn yes
I like it because it is similar to some I've done. If only I'd seen it earlier :-)
It has almost made it to a "nice answer" badge for David
one more vote
leo
leo
yep
@robjohn, are there in the site questions dealing with the function $\omega_f(t)=m(f\gt t)$?
where $m$ is the Lebesgue measure
01:34
@leo I haven't seen any, but I've used it in some answers.
leo
leo
like which
?
@robjohn
@robjohn, does that functions has a name?
I don't know if it has a name, but I know it is used a lot in $L^p$ estimates
leo
leo
is used to represent some Lebesgue integrals as Stieltjes integrals
@robjohn, I don't see measure-theory in the tags in your profile :)
I've only answered 6 questions dealing with measure theory. You have to show more to see them.
leo
leo
@robjohn, I think you mean: "...then since $E^0$ is open..." here
01:51
@leo fixed, thanks
leo
leo
@robjohn no problem :)
02:05
Qiaochu says he has a more direct proof of the answer that David Moews gave. I would really like to see it if he ever digs it up.
leo
leo
yes, me too. Probably is something similar
@robjohn, specifically I'm trying to prove this right now: consider $\omega$ as here. Let $(f_k)$ a sequence with $f_k\stackrel{m}{\to} f$. I want to prove that for fixed $a$ and given $\epsilon\gt0$ then $\liminf \omega_{f_k}(a)\geq\omega_f(a+\epsilon)$
$m$ is the measure used in $\omega_f$?
leo
leo
@robjohn yes, Lebesge measure in $\mathbb{R}^d$
it looks like Fatou
$\liminf \omega_{f_k}(a)\geq \int \liminf \chi_{f_k\gt a}$
leo
leo
02:37
@robjohn, can you provide me a link to some thread dealing with such a $\omega$?
Isn't this true $$\omega_{f_k}(a)-\omega_{f}(a+\epsilon)< m\{x:|f_k(x)-f(x)|>\epsilon\}$$
Sorry, I've been away for a bit. and I have to leave again, soon.
leo
leo
@robjohn i don't see that inmidiately
@leo I'll think on it while I'm out. I'll be back in a while.
leo
leo
thank for that :)
 
1 hour later…
03:47
Hello, does anyone mind if I ask a question about Noetherian modules? (About a proof in Eisenbud.)
leo
leo
for me no problem, but i don't know about that yet
Oh no worries, I'll just throw it out there, since I don't think it warrants a question on the main site yet.
Prop 1.4 on pg. 28 of Eisenbud is proving if $R$ is a Noetherian ring and $M$ is a finitely generated $R$-module, then $M$ is Noetherian.
@yunone That's a true fact. What's the problem?
Suppose $M$ is generated by $f_1,\dots,f_t$, $N$ a submodule. $Rf_1\subset M$ is generated by one element, its submodule $N\cap Rf_1$ is finitely generated, say by $h_1,\dots,h_r$.
@DylanMoreland I don't see why $N\cap Rf_1$ is finitely generated just because $Rf_1$ is generated by one element.
I don't see an assumption that $R$ is a PID or anything, other than that I get the rest of the proof.
You shouldn't need such an assumption. Has he proved the case of $t = 1$ yet, though?
Because that would do it.
For that, you just have some surjective map $R \to Rf_1$, and you pull back a submodule of $Rf_1$ to get an ideal, which is f.g. because $R$ is Noetherian.
Here you apply that to $N \cap Rf_1 \subset Rf_1$.
03:56
Ah, I see now, thanks Dylan.
xkcd is good, but it's no smbc :)
04:13
I love the 1st one
leo
leo
are there in the site questions dealing with the function $ω_f(t)=m(f>t)$?
with $m$ the Lebesgue measure
someone please review and comment on this answer Please
04:31
I've just gotten the homework badge... :(
MaX
MaX
04:41
Guys, How many different rays can be formed from six collinear points?
leo
leo
what's a ray for you?
MaX
MaX
I am not sure which definition to use here, but geometrically I believe a ray is a line one one point.
A line = <------------------------------------->
A ray = --------------------------------------------->
A segment = ----------------------------------------------
leo
leo
ok
@Max : 6
or 12
?
12
MaX
MaX
I don't know the answer
04:56
i guess its infinite.........................I've learn't in elementary school that "the number of lines passing through a point in a plane is infinite"
MaX
MaX
@Rajesh: Yes that was my first guess.
I've already typed it as an answer
can someone explain the possible reason for this downvote ?
@Dylan @MaX ??
The question implicitly assumes the rays start at one of the points and pass through another, or some other restriction along those lines, so the downvoter is probably complaining that you insisted in the obviously non-interesting interpretation.
The question should be more explicit
06:14
oh noes...i have attracted the attention of an h-k
07:14
Oops, the resident topologist is not around!
@Brian, Isn't the definition of a seperated set funny?
An empty set is seperated from itself?
Please clarify @Brian.
what is a separated set?
@MarianoSuárezAlvarez Huh, I understand the intent of the question.
It should have been seperated sets.
the intent was that I don't know what you mean by separated :P
We cannot talk about seperatedness of a single set??
(I may be wrong, I'd like to learn)
the socratic method has the advantage that one can ask stupid even questions and people think one means something deep :D
I really do not know what meaning of separated you have in mind
07:27
@MarianoSuárezAlvarez That no point of a set lies in the closure of the other, then the sets are seperated!
well, with that definition, the empty set is indeed separated from itself
@MarianoSuárezAlvarez Isn't this funny?
the funny set is funny in many ways
for example, all its elements are simultaneously odd and even integers
it is also disjoint from itself, and separatedness is supposed to be a more stringent version of disjointness
changing the definition of disjoint because of that would be silly :)
True. So, I am TeXing up a set of notes. I'd like to add an example from class that says, every non-empty set is seperated from the empty set. My teacher naively said it escaped facing this question. So, can i drop the non-empty condition?
07:33
@MarianoSuárezAlvarez What does this mean? puzzled
heh
your question «can I drop the non-empty condition?» means, precisely, «is the empty set separated from itself?»
@MarianoSuárezAlvarez I am puzzled by your "can you"? So, I thought I could.
why do you think you can? IOW, why is the empty set separated from itself?
@MarianoSuárezAlvarez because it satisfies my definition.
well, then you can.
07:36
And, it does not satisfy the definition I have in hand for a connected set.
(A set is said to be connected if it cannot be written as the union of two seperated non-empty sets)
@Daniil Hi
that notion of separatedness is not exactly very useful... so no one cares much about the empty set—on the other hand, connectedness is very important, and in practice it is very useful that the empty set not be connected.
@MarianoSuárezAlvarez Thank you @Mariano.
07:38
by the way, that is an enormously ugly definition of connected!
I have been selected for a Thematic Summer programme in Finite fields. @Mariano
@MarianoSuárezAlvarez Why?
It is equivalent to: a set is connected iff it cannot be written as th eunion of two non-empty disjoint open sets
defining first separatedness (requiring closures...) and using that to define connectedness, when one can do it directly in terms of open sets, is silly :)
@MarianoSuárezAlvarez You just replaced seperated by disjoint open. Didn't you?
yes, but the change is important: separatedness is a mutual relation between the two sets, while openness is a property of each set by itself
I added disjointness (which is also a mutual relation between the sets), but that is a very simple condition
Yes! Thanks for the dose of enlightenment. This shows my teacher is an ugly man!!
(:-))
07:42
he may have his reasons, who knows!
gotta run now
bye!
Alright! Now, it is 1:00 pm and lunch awaits me at the mess. GTG. Bye @Mariano
Bye all!
07:55
I have gotten 6 upvotes today (after an upvote yesterday) on a 3 month old answer. That just seems odd.
Peter updated his answer 5 hrs ago which bumped the thread onto the front page for additional exposure.
Now what you can't explain is a few days ago I edited in order to bump something like 10 of my old answers. None of them were upvoted, but in the meantime 3 other random old answers of mine were upvoted (without being bumped in anyway).
@anon I saw that :-) I just am amazed at getting 7 votes in the last couple of days since the answer only had 4 votes 3 months ago.
Yes, when I saw the thread I did in fact think "how did this get 10 votes..."
@anon I have gotten a lot of upvotes on old answers recently as well.
Of course they never upvote the one you want them to upvote. $*$pouts$*$
08:03
Indeed, the ones on which you spend hours and have what seem to be your most inspired work get 1 upvote.
Typical F# code:
1 |> (+) <| 1 |> (+) 1 |> (+) <| 1
looks like ascii art for a bomb storage room or something
@Daniil what does it do?
sums 4 ones
@Daniil Useful. We just deleted a question about 2+2. Perhaps that should have been an answer :-)
08:12
:D
The question about 2+2? The one where the OP asked how to "prove" that 2+2 equals four?
@Daniil That was the one.
Was it a duplicate or whatnot?
Still better than "Batman equation"
Aw, was that 2+2 thing recent? I don't think I was around. Still have a link?
I think the Batman equation was okay.
Oh, that one. I actually helped close it. Derp.
08:15
@anon :-)
I had to work real hard to make this look like an answer :-)
It just seems so obvious.
Why don't people vote my answers for free?
Found 17 video-lectures from the seminar on 10th Hilbert problem lectured by the guy who solved it. Now I just need to find some time and dedication.
Never have I received an upvote on a post older than 7 hours :/
@KannappanSampath They will, it only seems to happen with older answers..
@robjohn Alright. I am TeXing up a set of notes, so I won't answer questions for now.
BTW, I learnt that there is a unique projective plane of order 4!
08:22
finite geometry?
@KannappanSampath You have some over 7 hours that have gotten upvotes :-)
@anon Pick a point, any point...
that point. points
The proof is cool. An unexpected way of proving it. Take any set of 6 points and define an incidence on the structures you obtain from there.
I was reading something about finite geometry recently...
08:24
@Kannappan: Perhaps you would find this interesting uwyo.edu/moorhouse/pub/planes
@robjohn How? I am curious though!
You mean what, not how.
@KannappanSampath How what?
Unless you're asking how rob was reading. :)
@robjohn Which was that holy post?
@anon Lol!
08:26
@KannappanSampath You have at least 3 posts over 7 hours old that have gotten votes within the last 15 minutes.
check your reputation page.
@anon It is new to me. I like the link
@robjohn Yes, I did! Now, who is that? You, by any chance?
cstheory.stackexchange.com/questions/9969/… - wow, turns out in order to create a DFA which would match binary numbers divisible by n you only need n states!
@Daniil interesting.
I thought this answer by David Moews was very nice.
08:47
@robjohn Was it?
@AsafKaragila What not? =)
@KannappanSampath Which knot?
oh, sorry! whatnot?
I got booted :-(
@AsafKaragila
08:52
wtfamireading.jpg
@robjohn What happened, Rob?
@KannappanSampath All of a sudden I was asked to log back in
:3263506
@AsafKaragila yes?
09:01
Just wondering why you were referring back to here
Why not?
Who cares?
Hookers. They care.
(Asaf knows this intimately all too well.)
@anon I also use "Your mother" often enough, do I know most people's mothers all too well? Does that imply an intersection between these two things I know intimately all too well?
09:05
@robjohn John L. Hooker?
@AsafKaragila one of the most famous Hookers :-)
And boy did he care...
are you speaking in code?
09:15
Are you eavesdropping?
no, that would be gutteral
this reminds me of when i was very young. my grandfather lived in a big farmhouse. he had a large family, and Sunday dinner was always quite the production. even though the dining room table was substantial, it was not large enough to seat everyone.
How does this remind you that?
so the youngest children has to sit at portable card-tables set up in the den. when i was no longer among the four youngest, i got to sit at "the big table". of course, there were all these people talking about what-not, and i hadn't a clue as to what they were on about, for the most part.
That makes more sense. We too talk about whatnot all the time.
here's the thing: i know just enough math to be dangerous, and not quite enough to be useful.
2
09:22
Dangerous? You are building geometrical traps?
what a marvelous idea!
you cannot leave! the dorway, you see, is not normal!
*doorway
Ok, I am going to protests now, bbl
i still maintain you're speaking in code
cryptography is not my strong suit....too much computation, but....well i supose even an evil motivation is better than no motivation at all
09:37
I still maintain that you're new here and unfamiliar with the chatroom and its strange social rules.
well, duh
Steganography is the art and science of writing hidden messages in such a way that no one, apart from the sender and intended recipient, suspects the existence of the message, a form of security through obscurity. The word steganography is of Greek origin and means "concealed writing" from the Greek words steganos (στεγανός) meaning "covered or protected", and graphei (γραφή) meaning "writing". The first recorded use of the term was in 1499 by Johannes Trithemius in his Steganographia, a treatise on cryptography and steganography disguised as a book on magic. Generally, messages will appea...
ooh, like the digitized numbers for bank accounts hidden in a digital photo as "background" noise on that CSI episode, right?
@AsafKaragila actually, i take the more-easily defensible position that you are unfamilar with me and my strange social interaction patterns
I don't watch CSI, but sure. I remember I used to hear some fuss about terrorists uploading images to usenet with hidden logistical communications.
(this was never true, naturally.)
@DavidWheeler I maintain that I am the Koenig of this chatroom, and you're wrong.
09:48
Shouldn’t that be melek (or something similar)?
Melech, but I was given the title of Koenig by tb, if my bad memory serves me right for once.
I have to study for the exam in algebraic topology now. $\stackrel{\large\circ\ \circ}{\huge\sim}$
all that comes to mind is Pvel Ivanisovich
*Pavel
I never cared for algebraic topology: right objects of study, wrong methods. Meaning, of course, methods that didn’t appeal to me.
I never cared for it either, however because we're such a small department and almost all research students are in algebra-related things I was forced to take more algebra than I needed, and less set theory than I needed. Jerks.
why should i care about algebraic topology?
09:52
You shouldn't. You should care about set theoretical topology.
@AsafKaragila That wasn’t a problem at Madison: when I was there we had Mary Ellen Rudin, Ken Kunen, and Jerry Keisler, and for my last year István Juhász was a visitor.
We actually have four tenured logicians. Just no students to open courses.
why should i care about set-theoretic topology?
Because it’s fun.
@DavidWheeler Because your Koenig hath commanded so!
09:54
give an example of how it is fun
a bit subjective
That’s a matter of individual taste.
@anon Not at all. I am the objective man, therefore if I think it's fun then it really is fun.
Also, set theoretic topology is fun.
that sounds a bit impredicative
@DavidWheeler No it's not.
09:57
that could be true, but i'd insist on proof
@DavidWheeler I am the objective man, so my words are the objective truth, therefore the claim that I am the objective man is (extensionally) true.
that certainly looks fun...but the connection with topology ...i don't see it
@David: Are you referring to me?
@anon I'm sorry, the correct formulation which we were looking for is "Are you talkin' to me??"
10:01
i suppose...i can't say Asaf's set-theoretical sketch of a proof that something he says is true, is true, has anything of the smell of fun about it. srs bidness that.
how goes the studying, asaf? ;)
and no, the correct formulation is: "Are you talking to ME?" although trying to simulate a Robert DeNiro-esque Brooklyn accent in type-face is beyond my meager capabilities.
would $\color{Blue}{colors}$ help?
@anon Guess.
for some reason (probably some setting i cannot fathom) tags don't render correctly for me in this chat. i see dollar signs, but i ain't gettin' rich
10:05
You should check the Chat Rules, from which you can continue to locating the ChatJax! script.
Or that... either way works. You should read the rules too, by the way.
confession: I never bothered to read the rules :P
$$\stackrel{\Huge\circ.\circ}{\Huge\sim}$$
i don't understand how to implement the bookmark
10:10
what browser do you use
a friend of mine from college claimed to have invented bookmarklets...i wonder if this is true
just drag "render MathJax" from here onto the bookmarks bar, assuming you use chrome or ff
um, that microsoft one...oh, what's it's name? Express? Excursion? Expedition?
Exploiter.
then you probably have more serious issues than latex rendering in chat...
10:12
something like that...i'm not very good with names. will it offend thee, if i mistakenly refer to you as yusef sometimes?
other than latent schizophrenia...don't think so...
yusef? why does that name sound familiar to me.. like I've heard it in a childhood anime of some sort...
i'll get to you anon. oh wait...
Nah, I'm thinking of yousuke aramaki, nevermind then.
Yusuf is an Arabic name derived from Yoseph, which is the origin of Joseph, and its derivatives.
right...which might mis-interpret as a slur upon thee, rather than the middle-onset alzheimer's it actually is
ok now the tags work. thanks for the help.
10:23
tags? you're probably thinking of html. this is $\LaTeX$.
Goodmorning!
i eat my food in tiny pieces. it's what happens when the teeth go south.
apparently "Asaf" means "gatherer"
It's morning where "I" am....
Asaf means "collected" or "gathered"; it's a past simple third person singular and masculine form of the verb. However in Hebrew there is nikkud to help distinguish between similar words, and the name is actually the name of a biblical poet Asaph.
@DavidWheeler You're in this chatroom, and indeed it is morning in the chatroom.
10:31
the speaker of objective truth speaks. so it's f as in alef null?
@DavidWheeler where are "you"?
No, it's Aleph.
@RajeshD 'allo
What's wrong with my answer here ?
10:33
that's not what i meant and you don't know that
@RajeshD You’ve misunderstood the question; see Isaac’s answer for the correct interpretation.
I haven't misunderstood @Brian : Its Isaac who misunderstood !
No, he’s right. Can be made from in this context means are determined by.
@RajeshD: Well, the OP accepted his answer..
Why do I have a Bob Seger song stuck in my head?!
10:47
Well ..How many rays can be made from one point ? How would you answer that...just think about it
@m. k.
which one? and i agree with Isaac. sorry, raj.
I have visited math.SE for 550 days, 550 consecutive days.
next year, you'll have to be it a present.
*buy...bleh! why can't Johnny type?
Because David and Johnny are not the same one?
you know how, if you have 2 morphisms in a category: $f:A \to B$ and $g:B \to C$ to get a 3rd one $h:A \to C$?
well, i'm like that...i drop the "B" a lot, and just go straight to h.
10:58
you mean straight to C
objects are irrelevant, it's the maps that matter
In $C$ there are no objects to begin with...
sigh the thing is, all this nit-picking about syntax means my intended message gets lost in the permutation. my verbal processing center works like a free-associator that only prints statements of a given parity. odd, right?
 
1 hour later…
12:11
@Asaf: If you're around, a quick question: The problem with $\beth_\omega$ yesterday was that we could take a countable union to leave $\beth_\omega$. But what about $\beth_{\omega_1}$? I think the class of all sets of cardinality $< \beth_{\omega_1}$ should be closed under countable products and countable unions.
Yes, of course. But what about $\aleph_1$ many sets in your union?
If $\omega_1<\beth_{\omega_1}$ (and it is) then a product with index of $\omega_1$ and unions over families of this cardinalities must also exist in your model.
If you want to start changing axioms, you might as well just use Zermelo's original system+Foundation and take $V_{\omega+\omega}$ as a model.
True. But what I was thinking about was a model in which "countable mathematics" could be done without giving up replacement.
Either way, I have to go to the laundromat now. I'll be back in a while...
Countable mathematics is not enough.
You want power set? Sure.
You cannot guarantee that the union of $2^{\aleph_0}$ many sets is in the universe any more.
And the real numbers are very much "countable mathematics".
I didn't mean a universe where every set is countable, obviously. :p
I gotta go, see you later :-)
12:34
@Jonas: aha
@ZhenLin I understand that. You want to ensure, however, that no matter which sets you're going to take the unions are to be in the universe. In particular this means that if you only limit to countable unions then things are going to break.
If you look at Bourbaki's work they live in a Zermelo-ean framework. They do not heed to the calls of well-founded epsilon relations; to the cries of incompleteness theorems; or to the desolations of replacements.
Indeed for most mathematicians working in Z+minor fixups would probably be satisfactory.
And as I said before $V_{\omega+\omega}$ is a model for Z.
Time for The Balvenie!
Yes, for much of mathematics full separation is not needed. But I'm not sure if it's easy to do without replacement...
12:49
Separation you mean subset, right?
Yes. By full separation I mean the formation of $\{ x : \varphi (x) \}$ where $\varphi$ may have unbounded quantifiers.
Replacement is more complicated and you can throw it away for most parts.
Again, just read Bourbaki.
Hmmm... if you say so.
@Ilya Are you interested in a seminar about tensor products of Banach spaces?
At any rate, my problem was to find a set-sized universe which was closed under countable operations. I don't like assuming much more than Con(ZFC) to do "ordinary" mathematics, but apparently ordinary mathematics nowadays likes having decent approximations of the universe to play with...
13:03
@ZhenLin Well, take the natural numbers and just run them through everything $omega_1$ many times and take the limit.
Borel set kind of process.
I don't understand. :-/
You can take $H(\lambda)$ for some large enough $\lambda$. This is a model for ZFC-Power set, but if you take $\lambda$ high enough then power sets become irrelevant.
Hello everyone!
I have one small issue:
in the definition of the totient elements in one book, it is stated that one element x is a totient element if its annihilator contains some nonzerodivisor of that ring on which M is a module containing x.
How can this be true?
@Asaf: $H$ being... Hartogs number?
Does that nonzerodivisor mean one such in the ring?
13:09
Oh, $H(\lambda)$ is the set of all sets which are hereditarily smaller than $\lambda$.
Ah. I've never thought about that before.
Powerset seems like a very essential thing to me... can we still form products of sets?
If you take $\lambda$ to be very very very high then you can ensure that countably many power sets operations would still remain "small".
For example $\lambda=\beth_{\omega_4}$.
(there is no special reason for 4, by the way :-))
Thanks!
$HC = H(\aleph_1)$ in your notation, I guess?
I never saw $HC$ before...
It's right there in François's post...
13:17
Oh :P
Yes, exactly.
But I'm slightly confused. Why isn't $H(\beth_\omega)$ a model for powerset? Obviously it can't be a model of ZFC, but I thought we agreed the problem was either replacement or union.
@ZhenLin It's not a regular cardinal.
$\beth_\omega$ is not regular.
$H(\kappa)$ is a model of the nice axioms when $\kappa$ is regular.
Ah, I see. Right.
And then regular + strong limit => strong inaccessible, and we're back to Grothendieck universes again, fun.
13:29
Yup.
Sorry for wasting your time...
13:43
Well, I am going back home. See you later.

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