General pedagogical advice: Do not use $v_i$ for vectors in $W$ when we're playing with a vector space $V$ as well.

@Thorgott One quick question : The local consistency in Hatcher means the existence of open ball $B$ such that ~~~. We can't replace this with any open ball right?

let c_1, ..., c_k be a basis for ran(T). then for any w in V there are unique scalars v_1, v_2, ..., v_k for which Tw = sum v_j c_j

the statute of limitations on smacks just expired, so i think i got away with that one

we can, at least for any ball whose embedding extends to an embedding of a slightly larger ball, but this is not relevant here

we're looking at a ball in $M$ that's not even contained in $M\setminus\{p\}$

that's why there's an argument needed

@Thorgott When we talk about the orientability of the manifold, do we always assume the manifold is connected?

@leslietownes there must be some anniversary of some sort, I have seen a few recent mentions. It still seems so unreal (not for John & others, unfortunately).

26th anniversary of the publishing of his manifesto.

seems like a third-tier anniversary to me, but whatever. maybe they forgot the 25th.

@love_sodam Typically it's only interesting to talk about connected components when dealing with orientability. One could naively take as a definition that disjoint unions of manifolds are orientable if and only if each connected component is.

@love_sodam Totally irrelevant. It's just a question component by component. If it's disconnected and orientable, then every component is orientable.

@TedShifrin. I will watch your video on implicit functions to see how it goes

@leslietownes yes, I agree. Thank you so much :)

@copper.hat great !! So basis of Im T1 have been extended to basis of W. The only difference in this and my answer is: S in your case is unique as you have defined it on basis of W. In my case however S is not unique as I have defined it explicitly only on range Im T1. Thanks a lot.

@leslietownes at 6:45?

@BalarkaSen I spent a night in the hospital Tuesday night. Good place to be in case Angelyne won the election that night.

Probably too much local information there.

and way too late

either at 6:45 or shortly after. she still hasn't done it, although i don't suppose she told us which day she would be doing it at 6:45. or am vs pm even.

so maybe to soon to say that it won't happen.

quick americans are asleep hello

the west coast isn't asleep yet. even my daughter isn't asleep yet.

found this somewhere

jokes dont need accuracy (:

I heard that the Euler characteristic of some topological space can be defined once $H_*(X;\Bbb Z)$ is finitely generated.

I wonder people don't care about the convergence issue

You heard?

Actually one of the exercise problem in Hatcher

I asked to my TA if there is no convergence issue but he just said one can define Euler characteristic in that way.

hey, Ted, real quick

still here?

He defines a "general" cantor set here, where at every iteration you remove an interval of length $d_n \in (0,1)$ times the length of each previous individual closed interval

so that (I think?) the total length at step $n$ is $(1 - d_1)\cdot \ldots \cdot (1-d_n)$, but does that necessarily $\rightarrow 0$?

its not like $(1 - \epsilon)^n \rightarrow 0$

@JoeShmo I think what you're doing is as in this question: math.stackexchange.com/questions/1491104/…

in which case, if you're removing a smaller fraction than for the usual Cantor set, you don't end up with a set of measure zero

um, no mine is even more wild I think

his is still a fixed $\alpha / 3^n$

mine are arbitrary $d_n \in (0,1)$ times the length of (one of) the previous "individual" intervals

Hi @Semiclassical.

@robjohn Hmm I see.

Get well soon.

o/

@JoeShmo oof

which actually got me wondering. is there an infinite product of numbers $< 1$ that $\not \rightarrow 0$?

like does $0.9 \cdot 0.99 \cdot 0.999 \cdot \ldots \cdot (1 - 10^{-n}) \cdot \ldots \rightarrow 0$?

you can't quite bound it with $(1 -\epsilon)^n$

I saw your remark re reductions, @Semiclassical. I will have a look at the link soon, seems interesting.

@Koro For stuff like this, most answers end up being very similar :-).

@copper.hat it seems so :)

I saw a confusing question: $f: R^3\to R^2$ defined as $f(x,y,z)=(y, x+y)$ is not a linear transformation.

Clearly $f$ is additive as: For any $u=(x_1,y_1,z_1)$ and $v=(x_2,y_2,z_2)$ in $R^3$ we have $f(u+v)=f(u)+f(v)$ and homogeneity also holds as $f(cu)=cf(u)$ also for all $c\in R$.

So I think that $f$ should be a linear transformation.

But apparently the correct answer is: f is not a linear transformation.

@TedShifrin Hi professor Ted, I didn't ignore any of your messages yesterday. I didn't understand your hint(s) and didn't know what to say. I truly apologize if you felt bad for my not responding.

hello

So value of angle will change even if trig value is equal ?

@Ishwaran do you mean something like $\sin(x+2\pi)=\sin(x)$?

yes like that but not exactly

yesterday I was discussing a sum

in that I got 1 = tan phi / tan theta cant we say that phi = theta ?

@Ishwaran nope, since $\tan(\theta+\pi)=\tan(\theta)$

ohoh.....

in radians

can you say how to install mathjax

i added the bookmark but it is not rendering

have you run the bookmark on this page?

what OS and browser are you using?

win 10

chrome

I believe it works there. you've installed the bookmark in your bookmarks or better in your bookmark bar?

yes i lined up all the book marks in that html page

yes

it works now

btw, where can i find the markup language for this ?

@Thorgott Ah Poincare series. Thanks.

fixed-point-free

hello my dear friends

how is life treating u?

how to simplify this $$ \frac{4Cos^2 \theta - 3}{3Sin^2\theta + 1}$$

I simplified but that went too complicated, but anyways I simplified it, I want to know if we can simplify this in few steps

i.e: within 5 steps

I wonder if my understanding is correct : The map $f:B_i\to B$ preserves or reverses local orientations means $f_*:H_n(M,M-x_i)\simeq H_n(B_i,B_i-x_i)\to H_n(N,N-y)\simeq H_n(B,B-y)$ maps $\mu_{x_i}\mapsto\mu_y$ or $-\mu_y$ respectively where $f(x_i) = y$.

yes, but it's probably more convenient to note this happens if and only if $f$ maps the image of $[M]$ under $H_n(M)\rightarrow H_n(M,M_{B_i})$ to the image of $[N]$ under $H_n(N)\rightarrow H_n(N,N-B)$ or its inverse respectively

@Thorgott About the orientability of $M$ from $M-p$, I think I can define $\mu_p\in H_n(M,M-p)$ by $(i^p)_*(\mu_B)$ for some small ball $B$ that contains $p$. Here, $\mu_B\in H_n(M,M-B)$ is chosen such that $(i^x)_*(\mu_B) = \mu_x$ for $x\in B-p$. This can be done since each $x\in B-p$ has a local orientation and by connectedness of $B-p$ (dimension).

yes, that's what I said yesterday

I think I now understand what you're saying.

@Thorgott Could you check this? It's a bit messy (as it becomes rigorous) but idea is same.

Revised one

I am writing an exam right now. Question 1 is "evaluate $\lim_{t\to\infty} \sin(t)/t$", and question 6 is "evaluate $\lim_{t\to 0} \sin(t)/t$". Is that too mean?

@Ishwaran No, $\phi$ does not equal $\theta$. That expression is an identity, so it's true for any $(\phi, \theta)$. Well, almost any, because the ratio is undefined if the denominator is zero. But that's generally not really a problem for identities of tan or cot.

@PM2Ring Got it !

:)

@XanderHenderson See if they sinc or swim...

My teacher comes with a teasing question everyday

@PM2Ring That is terrible.

TERRIBLE.

Sorry, I couldn't resist.

I'm tempted to abuse my moderator diamond to express how terrible that was.

$$(sin^2A +sin^2B +sin^2C)^2=2(sin^4A+sin^4B +sin^4C )+4sin^2A sin^2B sin^2 C$$

This is one such question

brrrrrr

I dont know if there is proof to show for angles a b c of a triangle

@Semiclassical in fact, you can have an infinite product of numbers $< 1$, that don't tend to $0$.

qr.ae/pGSFsK

yes just use logs

Oh yeah the answer linked is also good

@Ishwaran There's an important trig equation in celestial mechanics, discovered by Johannes Kepler, and named after him. $M = E - \epsilon\sin E$. It would be easy to calculate orbits if we could write $E$ in terms of $M$. But there's no way to do that using elementary functions. en.wikipedia.org/wiki/Kepler%27s_equation But you might enjoy playing with some of the other orbit equations, eg en.wikipedia.org/wiki/Eccentric_anomaly

@love_sodam I don't think proving the existence of $\mu_B$ is explicit enough here

@PM2Ring Looks interesting

Is Abstract Algebra == Algebra on Steroids?

no

well maybe if you understand it very well

for me it's just random stuff that maybe I'll know how to use in 10 years or something

I regret joining electrical engineering. Now I won't be able to study math anymore. Except some kindergarten math like differential equation.

You can study math as a hobby]

it's pretty easy

;_;

although kinda boring as well

But math takes a lot of time. Sexy Rudin Took me like 3 months to complete.

which one was sexy rudin?

3 months is pretty good right?

you can do 4 subjects per year?

Baby which sounds sexy lol.

and then just give it 15 years and you're sittin' pretty

@o.9 OMG

@BannedUser In my opinion , it doesn't really matter how much time a math book takes , as long as you understand the concepts and enjoy the journey. [Unless you have time constraints]

the best way to learn math is probably to be rich af and hire tutors

I heard engineering students are more busier than most math student. After looking at the first year syllabus I was pretty surprised that it seems true. Most math learned by Engineer are not as rigorous thus may lead me to fight some sloppy instructors lol.

although the farther you go you might find that many of the stuff isn't very well motivitated and kinda weird

Yeah sometime I think so. Like some author use proof by trivial or proof by is left as exercise method.

I find it weird some people are really good at algebra but can't implement a dfs

@o.9 what is the specialty of baby rudin?

what's dfs?

dfs is depth first search, an algorithm for exploring data

@Prithubiswas real analysis iirc

A classic use of Depth First Search is in printing a directory tree. We want to recurse into each directory as we find it & print its contents, and the contents of its subdirectories, like in this old SO answer of mine

what are the ways to interpret integrating the solution to a D.E.?

solution of D.E. being an integral curve

@PM2Ring. Hello! I was going to ask about similar thing. Sorry if binging you is annoying. is there any advantage behind used topological ordering of vertices

Is there any advantage behind using topological ordering of vertices prior to any shortest path algorithms for example to be more specific?

@Thorgott You mean for general $\mu_B$ or $\mu_B$ for open ball $B$ that contains $p$ ?

@Avra That's a bit vague, but we usually want to traverse a graph with some kind of spanning tree. That's generally more efficient than just visiting nodes at random. Eg, in this Python program I solve a simple maze using a breadth-first search.

You might also like this DFS code that finds connected components of a graph.

Hi everyone! Does anyone here study maths or physics as a hobby not as if they have to to score marks or working professionally.

But sometimes, we don't really know where the edges of our graph are until we start exploring the data. A classic example of that is counting the number of inversions in an array. There are numerous algorithms for that. I collected and timed a whole bunch of them here.

I first choose $\mu_B\in H_n(M,M-B)$ such that its image under $H_n(M,M-B)\to H_n(M,M-y)$ equals $\mu_y$ (already chosen). And show that such chosen $\mu_B$ has a property that if $B'\subset B$ is an open ball such that the image of $\mu_{B'}$ under $H_n(M,M-B')\to H_n(M,M-y)$ is $\mu_y$ for any $y\in B'$, then the image of $\mu_B$ under $H_n(M,M-B)\to H_n(M,M-y)$ is $\mu_y$ for any $t\in B'$. So connectedness with open-closed argument shows $\mu_B$ is unique.

Is there something I should prove / fix more?

@PM2Ring. I am really confused! If we just use topological ordering, we cut time to just linear time to $O(E+V)$ for Dijkestra to find shortest path

@PM2Ring. This sounds crazy for me

I have no idea how that works

@Avra That's virtually what I'm doing in that maze solving program. The nodes are already sorted in an appropriate way, due to the structure of the maze.

@PM2Ring. Thank you. So, how topological ordering helps please?

What will happen if you select edges at random?

it will take $O(VE)$

Why topological ordering cut it down please to $O(V+E)$?

@Avra I don't know if I can explain it any better than what I said before about the maze. We want to explore a spanning tree of the graph, so it makes sense to have the data arranged in a way that makes it easy to "grow" the tree.

Another example is a sensible cook preparing a meal. They lay out the ingredients, pots & pans, and the various utensils in a logical order so that they can move through the task in a linear way.

@PM2Ring. So, you mean it's better than just keep moving all around please? Ofcourse this should only work with DAG graphs?

I guess it does not work with undriected graphs please?

@PM2Ring. I have slides here and it's written Marshall's algorithm, it seems that this is a typo from the writer as I can only find Warshall's algorithm :/

@Avra It's almost 4 AM in my timezone. I can't think clearly enough to have this conversation. ;)

:/

@PM2Ring. Thank you

I had problem with computer science where key $in\Bbb{Z}$ are randomly chosen and key mod size is taken. They said if size is greater then there will be less likely to have collision (and collision means different key will have same key mod size). Is it true?

Not necessarily need to answer. It is pretty late so I was trying to get answer asap without thinking. May be dream will help.

BTW any tips and tricks to have better sleep schedule and keeping obsession in precaution? Like if you have a math problem then telling yourself to calm down and not to think about it while sleeping. I am sure most people suffer from this how do you handle this ;_;

Please tag ping me. Good night.

@Koro I assume that Axler has proved the basic extension lemma: If you have a linearly independent subset, you can always complete it to a basis. This requires induction (and transfinite induction in the infinite-dimensional case — so don't worry about that). So here's the sketch of what I think is the cleanest argument.

(1) Take a basis $v_1,\dots,v_k$ for the nullspace of $T_1$. Complete it to a basis $v_1,\dots,v_n$ for $V$.

(2) Consider $T_1(v_{k+1}) = w_{k+1},\dots,T_1(v_n)=w_n$. These are linearly independent in $W$ (that's what I think you were proving in your argument).

(3) Define $S(w_j) = T_2(v_j)$ for $j=k+1,\dots,n$. Extend the $w_j$ to a basis for $W$ and define $S$ arbitrarily on the remaining vectors (e.g., send them all to $0$).

I have a quick question about notation: does $\bigcap_{X\in C}\mathcal P(X)$ mean $\{x:\text{$x$ in $\mathcal P(x)$ for some $X\in C$}\}$?

for all X in C

@AlessandroCodenotti: Oh right. I was mixing up union and intersection. Thanks.

Howdy, demonic Alessandro.

Look up Fermat's Little Theorem.

You just made my day, that's a lok @TedShifrin

There are lots of simple proofs.

@love_sodam latter

Hi Ted

Long time no see, how are you doing?

@TedShifrin Do you ever worry that Fermat's Little Theorem has to seek therapy for always being called "Little"?

Does it drive a big car to compensate?

"i'm fine with little. you want big, go big. i wasn't so big that you couldn't fit me into the margin of a #$@#$@#ing book"

@leslietownes touchy are we?

drives off in a lifted ford raptor

I'd ask if you want to drag, but then I might offend any dressers in cross products.

Clearly robjohn is feeling better.

I would feel much better if I could get my dog to eat. This morning it was syringe feeding. Going to try again soon.

Poor thing.

Yeah. She's just been not eating for so long that she needs help to get started again and to get the nutrition she needs.

she had 2/3 of the syringe and then started licking the food off the end of the syringe, so we gave her the rest off of our fingers and she went for that. Progress.

I was telling my wife that I had an idea for the next dinner party craze: syringe feeding and IV fluids.

A bunch of gurneys around the table.

she didn't think it would catch on

Then I thought of a new restaurant name: "You WILL Like This!"

ARG!!! I completely forgot to mention the intermediate value theorem last week! Mother F*!

Stupid Labor Day, messing with my calendar!

@robjohn You’ll forgive me if I forgo that dinner invitation.

Labor Day was ages ago.

@TedShifrin You'll miss out on the latest craze! I tell you.

@TedShifrin Yes, but I wasn't paying attention at the beginning of the semester, and missed that there was a holiday (though I got the others on my calendar), and so I ended up a day behind in one of my classes, seemingly panicked, and left out something important in both classes. I am just realizing this now, as I am writing the exam for that material.

Yup. You can still work it in with the mean value theorem.

Maybe prove the theorem step by step as a problem. That way it gets covered.

wait, Ted, why does $\mathbb{E}(X \mathbb{1}_{A_n}) \rightarrow 0$ if $\mathbb{P}(A_n) \rightarrow 0$? Assuming $X \in L^1$. I got stuck when I tried to write the proof. Dominated convergence here can be relevant, by which we have $\lim _{n \rightarrow \infty} \mathbb{E}(X \mathbb{1}_{A_n}) = \mathbb{E}(\lim_{n \rightarrow \infty} X \mathbb{1}_{A_n})$ but $\lim_{n \rightarrow \infty} X \mathbb{1}_{A_n} \ne 0$ a.s.

initially I thought $\mathbb{P}(A_n) \rightarrow 0 \implies \mathbb{1}_{A_n} \rightarrow 0$ a.s., but that's not true. I can extract a subsequence that does converge to $0$ a.s., but can I pass to a subsequence here?

(convergence in measure $\not\Rightarrow$ convergence a.s.)

DCT applies since $|X \mathbb{1}_{A_n}| \le |X|$

@copper There are no proofs like that in Calc I. No way to prove it without the least upper bound property, etc.

@JoeShmo Why have you decided that I remember/know measure theory?

:-( cause youre smart

also you said $L^1$ was a sufficient condition I didn't know you were guessing .. :-)

Less smart on this stuff.

Well, that was a reasonable guess.

certainly true

When I was the "non-expert" writing the real analysis qualifying exam(s), I worked very hard to work the exam and convince the so-called experts that many of their questions were absurd.

Some were false. But I spent some hours doing that.

And that was almost a decade ago.

I was always astounded at how few of us took our "non-expert" role seriously. Most faculty were too f***ing lazy.

the experts hazed you by asking you to prove false questions?

that's rough

That wasn't hazing me. That was their effort at the exam questions.

(I take it you were asked to take the, say, analysis quals as a non-expert to see if the exam was reasonable?)

Obviously, I made them change things.

ugh, they weren't even thinking them through.

I was also on the grading committee. Three person committee. I also served as the algebra non-expert a few times. For complex analysis and topology I was sufficiently "expert." :D

wow I can't imagine how pissed off I would be to sit in on an impossible exam only to find out half of it was flat out wrong

It was bad exams that caused this system to be created. But, unless you take the system seriously, it's still going to fail.

I'd feel so stupid walking out of the exam room

@XanderHenderson Thanks for verifying my statement to assure a user in CURED that even you make mistakes ;D

not the same thing but during my qualifying exam i remember wanting to tell my 'outside' member "you realize you can just sit there and not ask me questions, don't you? other outside members do that."

if youre experienced enough you'd be able to chew them out with counter examples on the fly, but that can take time

no way you get through the entire thing

Yes, one of my advisees (not for the Ph.D., just the first few years) years ago wasted 2 of 3 hours trying to do a false real analysis question. Things like that led to "change."

Qualifying exams changed after my day, @leslie. In fact, I was one of the ones who instigated change.

Even though I excelled at the three one-hour oral qualifying exams on the basic subjects. It changed to the written exam on undergraduate + advanced exam that you had.

But so many grad students turned into alcoholics freaking out about the pressure of these oral exams.

What's the munchkin up to today with no ducks?

Anyhow, leslie should be able to help @JoeShmo on real analysis.

yo @leslietownes

her day care let her try out a half-day. they kept her in a stroller so they wouldn't have to pick her up. she'll be getting back soon.

Oh, they did let her go. What about bathroom visits, eating, etc?

I am wondering can someone please help with this question: math.stackexchange.com/questions/4255821/…

bathroom visits, someone has to put her on a potty. toileting is all still done in the classroom so they only have to pull her out of the stroller and onto a potty and back.

Ah, that's what I get for meddling.

Your question is very confusing, @Node.

I would never write $x<y \in \{a<b, c<d, a<d\}$, let alone things with $v_i$ and $v_j$.

Let's see more words here.

yeah dude I opened that link and my head started spinning

Compatibility is not a standard word, either.

I didn't understand a word

If I have a partial or total order on a set, I can ask whether two elements are comparable — not compatible.

I don't know if your vocabulary comes from some other language. At any rate, I'm done.

I suspect as much, too ^

Basically, I have two total orders and I want to say they are compatible, that's the essence of my question

what do you mean compatible

write it down rigorously

@leslietownes are you available for a quick analysis question?

For starters, then, do not use the same symbol for the two orders. Geez.

You have two total orders on the same set. You denote one by $<$ and the other by $<<$. What you mean by compatibility is that for all $a,b$ in the set, $a<b \iff a<<b$.

That's it

Why are your two examples compatible? They're not. $b$ and $c$ are not related in the second order.

Yes, I want to say that a < b and c << d are compatible because we can use a function f(c) = a and f(d) = b

You want to say what?

haha

you want an order preserving isomorphism

I think

if only I knew what you were asking

Oh, that is totally different from what I said (to which you said "that's it").

You need to define a bijection on the entire set, as JoeShmo said. Just saying $f(c)=a$ and $f(d)=b$ isn't going to do it.

Makes sense.

I don't think your two orders are order-isomorphic.

How?

You wrote down two values of $f$. Finish the one-to-one correspondence, please.

Can you please explain more

Do a graph. Your first order is a tree. Your second one definitely is not.

Makes sense, I need to prove one-one first

You need to DEFINE a correspondence, and you can't do it.

If one is a tree (a<b<c<d), the other has to be as well, and it isn't.

Hello again,

I have one question please, the weights of a strongly connected graph is finite?

Answer it seems to be yes.

The question is, why the weights of connected undirected graph ? Is it finite or not please?

@Thorgott chat.stackexchange.com/transcript/message/59190604#59190604

@amWhy :P

I thought I proved if I choose $\mu_B$ that matches $\mu_y$ for some $y\in B-p$ after the restriction, then it matches all $y\in B-p$.

i like the extern system in ireland/uk. not sure it could scale here, but it has some visibility which leads to accountability.

you're not being explicit about the connectedness argument

also, the point is proving $\mu_B$ with the property exists, uniqueness is trivial

update: I think a stronger DCT applies here (convergence in measure suffices. You don't need convergence a.s.)

Anyhow, @Xander, I recommend folding the IVT into your MVT lesson :)

@Thorgott Once I choose $\mu_B$ as I stated in the previous chat, isn't that shows $\mu_B$ exists? I showed such $\mu_B$ satisfied local consistency for any $y\in B-p$.

And for the connectedness argument actually I didn't write explicitly. I thought it's a usual open-closed set argument.

@TedShifrin Too late. I spent 20 minutes on it today. I typically assign a mini-project on the method of bisection early on, then give them a more substantial project on Newton's method. If they haven't seen bisection, then Newton isn't nearly so impressive.

Ah, indeed.

Yeah, ok. The point of the argument is to prove that it is consistent with any $y\in B-p$.

@Xander You probably didn't intend to delve into Darboux's Theorem about how the derivative has the IVP :P

well, your argument shows that the set of $y\in B-p$ which are consistent with $\mu_B$ is open. you have not yet argued that it's closed

@TedShifrin I really really want to, as I think that is one of the niftier results in real analysis, but I think that it is probably a bit out of scope for this class. If I ever get the time to rejigger the curriculum, it is something I want to mention.

I would also really like to introduce the Darboux integral (rather than the Riemann integral), but there are some difficulties there, too.

i think darboux is nicer and more natural.

It's assuredly beyond the scope. I assigned the exercise when I taught the Spivak Calc Theory course, but never regular calc. Indeed, I find the content of the standard Calc course on the mean value theorem to be appalling. My first year at UGA I made up exercises to make the MVT a bit more meaningful by giving them velocity functions (that no one could possibly antidifferentiate) and asking them to give upper and lower bounds on the distance traveled.

Yes, I always did the Darboux integral in the theory courses (including multivariable), but otherwise I just followed the textbook and gave the theory short shrift.

I actually was proud of those bounding exercises because (a) it gives the correct spirit of the MVT (not the idiotic "find the c" problems), (b) it gave them practice differentiating to find max/min on a closed interval.

I think the standard curriculum most places doesn't emphasize max/min word problems enough, anyhow. To me that's the main point of the course.

Plus justifying (whether closed interval or open interval) to your boss that you actually did find the max.

@TedShifrin That sounds like a really nice exercise. I might have to incorporate something similar.

That would please me (but maybe not your students).

@TedShifrin Yeah, those "find $c$" problems are silly. I typically use MVT as a way of introducing antiderivatives, and spend almost no time on "MVT exercises". But upper and lower bounds on distance traveled via the MVT is a nice idea, and provides a nice segue into antidifferentiation.

I did that because we had uniform final exams back in the 80s when I arrived at UGA. If I was going to teach the MVT fir the uniform final, I was determined that they understand the point.

Nice.

So much horridly bad stuff in textbooks because of tradition and idiotic reviewers.

@Thorgott Maybe similar argument shows that the set of $y\in B-p$ which is not consistent with $B-p$ is open (I think everything is same except putting $-$ on $\mu_y$'s).

Yeah, it makes me crazy. And I have to at least gesture in the direction of a lot of that crap, because we have articulation agreements with the state universities which require us to put it on the syllabus.

Yup. I know.

Smart move, @love_sodam

the point of the MVT isn't to find c?

ducks

In any event, I've been at work for just under 11 hours now. I think that it is time to go home. Later, all.

Night!

@love_sodam yup, that's what I wanted to hear

good job

I'll revise the proof and upload here later. Hoping there is no more error.