Zorn's lemon?

I've set Britten's Noye's Fludde as my wake up music!

@ClarkKent Babylonian kings apparently aren't as picky about the liquids they ingest as some Belgian mathematicians here...

You apparently had too little (or too much) fire water, Clark.

Be careful, my bat mobile runs on Kryptonite.

Mine runs on anything with a high %

I'll go to bed. Good night guys!

Ding ding...

Ding.

Good idea, me too. Good night.

Good night, Michael

2u2

snore

@Chris After choice of bases: matrices = linear maps

back from work and errands :-)

hey, robjohn!

I see Jonas's bell had an echo :-)

I see the same matrix. What is the question?

@Chris all values of $a$ make that a linear map

@Chris I think I answered that :-)

@robjohn: "all values of a make that a linear map" yes you did :-) thank you!!

@Chris Can you recite the requirements for a linear map?

@Chris I think it's understood that you're working in $\mathbb{R}^4$. Now check that $A (x+y) = Ax + Ay$, etc.

@Chris in addition to tb's check, check that $A(rx)=rAx$ for all $r\in\mathbb{R}$.

@Chris true every time? for what? for all matrices?

@ClarkKent I think he is seeing that by the definition, all matrices represent linear maps.

@TheChaz I guess .blt is short for "ballot".

@MarkDominus of course! I wasn't thinking about that :-)

@MarkDominus I didn't even get a chance to read the whole comment before it disappeared :-p

That's okay, it was silly.

@Chris why isn't that a linear map? what requirement does it violate?

@Chris and $r\cdot0=0$ and $0+0=0$

@robjohn: Hmm ok, so it is a linear map even if r == 0?

@Chris Yes, for all $r\in\mathbb{R}$ you need $A(rx)=rAx$ even $r=0$

Ok then, so the x vector where it belongs? In $R^{4}$?

And if so, how do we know this?

@Chris The rank of a matrix $A$ is the same thing as the dimension of the image of the linear map $f$ it represents. No matter how big or small that image is, the map still is a linear map between the same vector spaces, hence $A$ still is a $4 \times 4$-matrix even if its rank can be smaller than $4$ for some values of $a$. You're not allowed to drop rows or columns of a matrix.

@tb: These were exactly my "doubts"! Thank you very much!

Okay, guys that's it for me today. See you soon!

I need to construct a function, given a $k \in \mathbb{N}$, a $\eta \in \mathbb{R}$ and a point $p$, which is periodic and smooth except at the point $x = P$ where it is $k$ times differentiable and the $(k+1)^{nth}$ derivative jumps by an amount $\eta$.

Okay the problem I am facing is that there is no unique way of doing this. If we construct a function $f$ and $g$ with the same property but $f$ and $g$ differ only by a smooth function, it is still not a problem. The problem is we can contsruct $f$ and $g$ where it is not possible to write $f = g+h$ where $h$ is some smooth function. I absolutely hate this problem and i need a way around.

surely if it's periodic with period T then the points $x+nT$ also have the nonsmoothness property

also, what counts as a "way around" would depend on the intended destination

Oh, elections are over? That sort of snuck up on me.

Bye @tb ! Thanks for all the help :-)

@anon : i am talking about only in one period

fundamental domain (interval), okay

I thought @robjohn was a shoo-in.

is vector space a ring?

Vectors do not necessarily come equipped with multiplication (meaning, the multiplication of two vectors, not scalar multiplication from the underlying field)

@KenLi Not necessarily, no.

Many common examples of vector spaces happen to be rings, though.

But nothing in the definition demands a ring structure on the "$V$".

I see

do they have any subset relations?

@anon : I was kind hoping that there is a unique way to get $f$ and any other function $g$ with this property can be expressed as $g = f + h$ for some smooth function $h$. But teurns out that it is not the case. There is a $g$ which cannot be written a $g = f + h$ for some smooth $h$. I will explain what i need

perhaps $h$ can in fact be any function that's smooth outside of the point P, and at the point P has a continuous k+1st derivative...

yes

but that not good enough for me

@KenLi Like, "a ring is always a vector space", or something? Afraid not.

Goodnight by me! Talk to you tomorrow!

A ring R is an R-module, and a vector space over a field k is a k-module, so they are both subsets of modules

or subclass, category, whatevs

This might be something to look at: en.wikipedia.org/wiki/Algebra_over_a_field

I am looking for a theorem where in i can get the jump $\eta$ and differentiability $k$ and the point $p$ given the function $f$. But there is no unique $f$! My method should not distinguish between a $f$ and a $g$. How can i do this ? What properties of the function i should concentrate on ?

I'm just wondering because the way that ring's are defined is somewhat similar to how vector spaces are defined

well, they are usually defined by a set of axioms

I wonder if there's a question on the site about these

I would instead wonder how many questions there are

Hi @Brian

Chipotle chili is the best argument for intelligent design

??

which intelligent design @Mariano

http://en.wikipedia.org/wiki/Intelligent_design?

yes

without some such thing, nothing so perfect would have been created

I'm sold.

hopefully you got paid well

@DylanMoreland I am glad that I held hope, but definitely I realized that I lacked certain things that might be hard to overlook. My meta participation was not as great as that of Bill or Eric, nor was my reputation or seniority.

hi there

@leo Hey :-)

@DylanMoreland Things I need to work on if I am going to run again.

My computer is lagging. I think it needs a reboot. It has been months since it has been turned off. brb

I think so

@robjohn Obviously others disagree, but I don’t see meta participation (as distinct from being generally aware of what goes on there) as being particularly important. In fact, my immediate gut reaction to a high level of meta participation is negative, though that can be overridden by knowledge of the specifics.

@BrianMScott I think meta participation is essential because it is the only efficient way to understand the concerns of the community. Probably this is why SE highlighted such stats on the voting page. All of the current mods are very active on meta, and I think they represent a good cross-section of opinions in the community.

@BrianMScott It seems to be a popular opinion, and that is what wins elections :-)

@robjohn (Grrr, keyboard has a mind of its own). I'm kicking myself because I wanted to write a statement endorsing you on the election page but it slipped my mind. Perhaps it would have helped.

@BillDubuque Thanks, but it wasn't really even close. I think it would have come out the same.

:-)

@BenjaminLim: good day!

@robjohn hello!!

@robjohn may I ask are you working now or something?

@BenjaminLim I was, but I got interrupted, and I haven't restarted. What's up?

mods are active on meta because we get notified immediately of everything that happens there :)

@MarianoSuárezAlvarez Hey you know that guy that posted about any surjective map between noetherian rings must be an isomorphism

@BillDubuque Did you read what I wrote? Even if you’re right, one need only read meta to understand the concerns of the community; participation is plainly unnecessary for that.

@BenjaminLim Or were you asking if I was gainfully employed?

@BenjaminLim, hm?

@robjohn By "now" I meant as in like your daily life

@MarianoSuárezAlvarez The two questions were kinda unrelated

@BenjaminLim Yes, I am :-)

@BrianMScott But I think one really needs to participate in meta - lurking is not the same.

@robjohn What do you work as?

Ah

@BenjaminLim Currently, I am a programmer at UCLA.

@robjohn ah ok. programmer yet answering so many maths questions???

@BenjaminLim, modules, not rings, no?

@MarianoSuárezAlvarez I don't know how to do the second one

oh sorry yes on modules

but the proof is kinda the same....

@BenjaminLim I taught math at UCLA for a couple of years before Apple hired me :-)

what second one?the one I answered?

@MarianoSuárezAlvarez yes

@robjohn Ah ok. No wonder you're so pro!

@robjohn what year did you graduate?

suppose M is semisimple, so that it is a direct sum of simple modules

it is easy to see that an infinite direct sum of non-zero modules is never artinian

@MarianoSuárezAlvarez Is that Maschke's theorem?

@BenjaminLim undergrad 1981, grad 1986

@BillDubuque And I don’t. I’m exceedingly leery of non-moderators who have a high level of meta participation; as I said, in the absence of specific knowledge of what they’ve said, I view it as a negative.

no, it is more or less the definition of being semisimple

@MarianoSuárezAlvarez ok.

I don’t want activist moderators. That’s one reason I ruled you out almost immediately.

@BrianMScott For example, lurkers sometimes misread between the lines, something which is less likely if one is actively participating in discussions.

@robjohn thanks for telling me.

@BrianMScott Are there in uses of noetherian spaces in general topology apart from say algebraic geometry?

@BillDubuque I don’t believe that it is substantially less likely.

@MarianoSuárezAlvarez right.

@BenjaminLim You’ll have to remind me of what a Noetherian space is.

now, if the module is a finite direct sum of simple modules, it has finite length

because it is easy to exhibit a composition series

@BrianMScott A topological space where the open sets satisfy the ascending chain condition

@MarianoSuárezAlvarez ok

and finally, a module of finite length is automatically artinian

ok

right

@BrianMScott Based on my experience on meta there is a large amount of misunderstandings even among those actively participating. I suspect it would be much higher by those lurking. But I do realize that many folks have no taste for such matters. That's why I think it is important to keep meta matters off the main site inasmuch as possible.

@BillDubuque I’m quite prepared to believe it; I merely observe that you’ve little or no evidence of the level of misunderstanding amongst lurkers, and I see no reason to suppose that it’s significantly higher.

@robjohn What do you work on at Apple? [Are you allowed to say?]

@Bill would your reopen vote have reopened the question whether 5th or not?

@BrianMScott I'd wager much on that bet. Please keep in mind that moderators are expected to participate on meta, e.g. the SE software is designed so that every meta questions shows up in the inbox of a mod.

@BillDubuque The second sentence is a non sequitur in respect of the immediate question about misunderstandings.

They really need a better meta system.

@robjohn I think it would be binding, which is why I waited to the 5th vote.

@BrianMScott Huh?

@DylanMoreland Oh, there is no secret about that: QuickDraw GX. You can even see my name under the developers.

@BillDubuque The fact that moderators are expected to participate has nothing whatever to do with whether lurkers are more prone to misunderstanding what’s said than participants.

@RobertJohnson

@BenjaminLim That's it

@DylanMoreland Indeed! I am probably the biggest critic of the SE software platform.

It also has nothing to do with whether one can acquire an adequate knowledge of what goes on in meta.

@robjohn still doing maths huh?

@BenjaminLim in an amateur capacity :-)

@robjohn hahahaha

@BrianMScott I never meant to imply that. My point was that mods have high meta activity because that is part of their job. So you should not be leery of such.

@robjohn I should probably go now

bye all!

@BenjaminLim Later!

@BillDubuque I never said that I was. I specifically said that I was not talking about moderators, but rather about candidates. Obviously moderators are another story.

@BrianMScott By mods I mean all 10k+ mods, who have access to tools that can help them be site administrators.

@robjohn I see. Now everything is "Core ____".

@BillDubuque I think that mods should be able to cast regular, non-binding, votes so that they can participate as a normal user, but have a binding vote, as well.

@BrianMScott By mods I mean all 10k+ mods, who have access to tools that can help them do site administration.

@BillDubuque Then once again I disagree completely.

@robjohn I agree, some folks don't want to be mods due to that.

@BrianMScott I'm curious to understand your views. Could you please elaborate.

@DylanMoreland yes, Adobe wanted our project shut down since it made it too easy for smaller developers to do what Illustrator does.

@DylanMoreland I sort of hold Adobe accountable for the loss of my job :-|

The dire financial state of Apple in 1997 was also to blame.

@robjohn What type of software were you developing?

@BillDubuque I see no good reason to lump real moderators together with 10k+ users. Real moderators have obligations on that account; 10k+ users need not choose to exercise their administrative privileges, and if they do choose to exercise them, they can do so responsibly without actively participating in meta.

@BillDubuque QuickDraw GX was the next gen graphics for the Mac. I did a lot of graphics software programming, but I wrote all the math routines for it.

@robjohn If only they'd given you some stock to hold on to :-/

I see, so you program for the university now?

I'll probably be there next month. Looking forward to it. I really like the campus.

@DylanMoreland At the time I was laid off, the stock was worthless. The options I had would have cost me to exercise. If only I had exercised them earlier.

@DylanMoreland really? Let me know when you'll be here :-)

Maybe I don't understand how stock works!

It's Hida's 60th birthday.

@BrianMScott The site would not run very well if no 10k+ folks exercised mod powers. To do so in an informed manner requires active meta participation, e.g. participating in close/reopen threads, policy decisions etc. Most (all?) of the current mods were very active on meta before they became mods. The SE model encourages such. Such volunteering is essential to the success of the site.

@BillDubuque No, it does not require active meta participation; I’m a counterexample.

@BrianMScott Huh? That would only be a counterexample if every 10k+ person behaved the same way.

What's a good example of something difficult/important that has been decided on meta?

@BillDubuque Come again? I’m a counterexample to your assertion that exercising 10k+ privileges ‘in an informed manner requires active meta participation’. What others do or don’t do has no bearing on that.

@DylanMoreland I'd be curious to know, too. My impression is that many policy-related things generate a huge uproar and peter out quite quickly without lasting impact.

See e.g. rude comments, homework policy recently (for the nth time).

@tb You've articulated what I wanted to say very well, thanks.

@DylanMoreland As I said, I think one really needs to actively participate in meta to really have a good sense of the pulse of the community. But I don't disagree that one can make 10k+ mod contributions without doing so, since some are independent of meta knowledge.

I think meta participation is good but it's hard to see examples of it accomplishing something. I guess it's like any political system :)

@BillDubuque And once again: meta knowledge is independent of active meta participation.

@robjohn So what do you code now? Hopefully something that puts Peoplesoft out of business.

@tb The boilerplate meta comment that Arturo (and some others) use on homework, imperative mode, etc evolved from meta discussion on a proposal I made for standardized meta comments. But there's still much that could be done there.

@BrianMScott I think we'll have to agree to disagree on that point. Thanks for explaining.

Has anyone encountered this odd use of $\circ$ before?

@DylanMoreland Indeed, it's difficult to reach any consensus in such a diverse community. But we have made progress. The early days of the site were much worse, with much more tension on the main site.

@BrianMScott Seems like a slight abuse of notation and also unnecessary if you're just going to do it with $x$, no?

It seems downright meaningless to me.

@BrianMScott Well I thought it might be something like: $(x^2 + 1) \circ f(x) = f(x)^2 + 1$.

I think that you’d need to use something like $\lambda$ notation to make that legitimate.

@BillDubuque That may be true -- or not, I'm not quite sure, I don't know Arturo's reasons to introduce that, and why the "copycats" use it. I've seen Zev using some boilerplate comments on follow-up-questions-posted-as-answers and the like. I do think your proposal has something to it in terms of establishing guidelines and I'm sure that your continuing efforts have contributed to calming the mean comments, even though I must confess that I prefer to use my own words instead of boilerplates.

I'm not saying I like it!

@DylanMoreland I have written software to help teach sentential and predicate logic. It grades the homework immediately so there is no delay in getting feedback on what the student has done incorrectly. It is interfaced with a database of submitted work for instructors to use when grading.

That's a subject where that would definitely work.

@BillDubuque and I completely forgot to congratulate you on your election. Re: The "error" by Banach I mentioned earlier today. It is the paper mentioned here I was talking about.

@DylanMoreland It has Derivation, Symbolization, Invalidity, Parsing, Recognizing Rules, and Truth Tables modules.

@ClarkKent Besides demonstrably raising students' grades, it saves the university money on TAs.

@tb Thanks! Iirc, Arturo adopted the boilerplate comment from someone else. I hope to find some time to resurrect the standardized meta comment proposal, since I think that could go a long way towards eliminating friction on the main site. And that is essential if we hope to attract more teachers - which I think is our biggest challenge.

@BrianMScott I am still unable to parse that question...

@tb are they simply talking about the identity?

Congrats @BillDubuque and @Eric :-)

@ClarkKent I'm trying to make it monochrome, but I'm having no luck...

@robjohn It sure looks like it, but I fail to see the difference between the intention of the $x \circ f(x)$ expression and the $f(x)$ expression.

@tb That makes two of us. So I simply tried to explain how to find $f$, which is clearly what was wanted.

@tb That would be the transparency of the identity :-)

@ClarkKent Can’t be, when you look at it closely.

I agree.

@ClarkKent doesn't look like it.

On a completely different but somewhat related note: what is the Right Way® of putting degrees into LaTeX? The \circ looks too small when used as a superscript and I haven't found a good substitute...

But $x$, so written, isn’t even a function, let alone the identity function. In other words, that may have been the intention, but the execution is abysmal.

@ClarkKent that was my impression.

@BrianMScott well, the function $f(x)=x$ is the identity and that is probably what $x$ as a function means.

@robjohn But it requires establishing a specific (and rather unfortunate) convention to justify the usage.

@BrianMScott The "standard" argument is $x$ is what they were thinking

@BrianMScott but I agree that as a strict formulation, it sucks :-)

@robjohn Maybe, but that doesn’t square with the idea that it represents composition with the identity. (I will admit to being a bit of a prig about notation.)

@ClarkKent That doesn’t really enter into this discussion; it’s a completely separate issue, so far as I can see.

@BrianMScott most people say that $x^2+3x+1$ is a parabola or quadratic function...

That is no better.

@ClarkKent maybe those who like to see functions as subsets of $P(P(P(A\cup B)))$

@robjohn Saying that it is a parabola is no better; saying that it’s a quadratic function is fine.

@tb Some alternatives listed here. Probably won't work on this site, though. Let's see: $\degree$, $\textdegree$.

@BrianMScott I don't think saying it is a function is any better than saying $x$ is a function.

@robjohn I’ve no problem with saying that $x$ is a function; the problem is using that notation for it in a composition.

the latter is designed to be used in text mode, though

@MarianoSuárezAlvarez I've always used ^\circ

I’d accept $[(\lambda x)x]\circ g$, though I’d shudder a bit.

\textdegree takes the symbol from the font if it is there, so it is usually best

I cheat: $85°$.

gensymb's might just be \circ

@BrianMScott That would require some sophistication of the student. Perhaps this course didn't have that.

@MarianoSuárezAlvarez I am sure that a dedicated \degree would be better.

Thanks Dylan and Mariano.

@robjohn I’m sure that it didn’t. But if they must complicate matters with the unnecessary identity function, they should give it a real name, not abuse the notation: $\operatorname{id}_{\Bbb R}$, $1_{\Bbb R}$, or anything reasonable.

I’m rather curious myself. But for all I know, it’s in Turkish.

@tb \usepackage{unitsdef} and \degree

if what you want is a roman font you should simply write \mathrm{id}, @BrianMScott

\operatorname does a few other things which for the identity map are not needed, nor desirable

@MarianoSuárezAlvarez What would those be?

I don't recall

but surely the spacing will be that of an operator

a \mathop

@MarianoSuárezAlvarez That would be okay if I’m not using parentheses, no?

there would be extra space if you wrote id f

than between g f

@MarianoSuárezAlvarez I meant following it with its argument without parentheses. (No, I don’t generally want to do this; I was just curious about the effect.)

when there are no parentheses after a mathop, you get a little extra space

I tend to have multiletter names of things set in sans-serif

test: $f \circ \operatorname{id} \circ f$ versus $f \circ \mathrm{id} \circ f$

but operators in good ol'' roman

@MarianoSuárezAlvarez That’s what I thought; which is why I use it for $\operatorname{cl}$ and the like.

\circ is a \mathbin, so it cancels the \mathopness

yes, for cl, sin and friends you should use it

in fact, in actual latex source it is best to use \DeclareMathOperator

well, it looks like \circ is interpreted as the \mathop's argument, hence the spacing afterwards looks crooked

(at least the way it's rendered for me)

because \DeclareMathOperator makes the command robust, which is usually the best idea

@MarianoSuárezAlvarez And so far I’ve written almost no actual latex source at all; almost my entire knowledge of the subject has been acquired here.

ah, well :)

you kids :)

@MarianoSuárezAlvarez Just young at heart: I turned 64 two days ago.

you are an honorary kid

@BrianMScott oh, congratulations! (is this why chessmath became checkmath?)

@tb Thanks! (What’s this about chessmath?)

@BrianMScott 53 8 days ago Happy Birthday :-)

as for this chat thingie, I think it is rather wasteful to write \operatorname{id} at all here :)

@robjohn I thought that you were probably a bit younger if you were reading The Hobbit as a kid.

38 here

@BrianMScott oh, there used to be this moderator candidate whose name was chessmath but now it's checkmath. The 64 kinda sorta reminded me of that.

@tb Sounds like a universal outsider!

@BrianMScott but you don't need to follow down that road, fellow.

@tb I seem to recall that there’s a story called The 64-Square Madhouse.

@tb :-)

@BrianMScott Fritz Leiber, Google informs me. See his bibliography.

@tb Yes, that was my recollection. And like most Leiber, it was a neat story.

@robjohn So we are close (I'm a couple years younger)

hey guys

@BrianMScott Meta Happy Birthday

@BillDubuque Thanks! :-)

@BrianMScott I confess complete ignorance. Never heard that name before. The only Leiber I knew before was the one of Leiber/Stoller (of Hound Dog fame among many other things). Looks like some Leiber's have a knack for a humungous output in various areas.

@tb - I have my Precalc final in a week and and a few days.

Fritz wrote some classic science fiction and fantasy. His son Justin also wrote some.

@tb I think I have finally proven the equivalences of limit point, sequential and open cover compactness for metric spaces!! :D :D

@Moshe Oh, hi, I'll be crossing my fingers!

@Moshe good luck!

@tb I'll be studying. :-) And asking in here occasionally.

Thanks!

@Moshe Good luck with it!

crap where did I put my ink

We have so many questions on Hausdorff distance this year that I'm starting to think about whether a tag is called for...

@BenjaminLim That’s almost as good as some of the odd things that I’ve heard just as an elevator door was closing.

@tb I hate the Hausdorff distance with a vengeance

@BrianMScott hahahahaha

@BrianMScott I love how in Rudin he develops using exercises the fact that limit point compact implies open cover compact :D :D

Good night. Don't let the exponents bite.

@BenjaminLim I’m not too fond of it myself. But then I don’t know why people want to muck up perfectly good topological spaces with metrics. :-)

@Moshe First time I ever heard exponents likened to bedbugs.

@BrianMScott I hate it because in our analysis course there is a random section on fractals thrown in, including the hausdorff metric in there

@BrianMScott Was being clever.

@BrianMScott hey

I'm an outsider: I do like the Hausdorff distance and the Gromov Hausdorff distance.

suppose we have a limit point compact space $X$ that has a countable basis for the topology generated by the metric

@Moshe Oh, I realized that; I just don’t meet many people who are familiar with that good night phrase.

I'm about to take Hausdorff measure, btw

@BenjaminLim Which is simply to say that you have a compact metric space.

@BrianMScott suppose we did not know that that was equivalent to the usual topological compactness

Okay ...

now we know limit point compact means that $X$ has a countable basis for the topology generated by the metric

why is it then that every open cover of $X$ has a countable subcover?

what if my cover has an uncountable number of sets?

enumerate the basis and use the definition of the basis.

Oh wait I think

Because you can refine any open cover to a cover by basic open sets, and that refinement is countable.

yes

I think because

it suffices to consider a cover by basis sets

@BrianMScott I'm stupid. The topology generated by the metric is exactly the union of all the basis elements

Or if you want to do everything from scratch, let $\mathscr{U}$ be an open cover of $X$, and let $\mathscr{B}$ be a countable base. For each $x\in X$ there is some $U(x)\in\mathscr{U}$ such that $x\in U(x)$, and there is some $B(x)\in\mathscr{B}$ such that $x\in B(x)\subseteq U(x)$. Let $\mathscr{B}'=\{B(x):x\in X\}$; clearly $\mathscr{B}'$ is countable, so enumerate it as $\{B_k:k\in\Bbb N\}$. ...

... For each $k\in\Bbb N$ there is an $x_k\in X$ such that $B_k=B(x_k)$. Finally, let $\mathscr{V}=\{U(x_k):k\in\Bbb N\}$; then $\mathscr{V}$ is a countable subcover of $\mathscr{U}$.