12:01 AM
life has a weird way of forcing me into doing linear algebra and adding a dimension to whatever i need to be doing calculus on

godfather 3 gif where al pacino begins "just when i thought i was out..."

1 hour later…
1:30 AM
@RandomVariable that sounds like the rough render that happens first, before the final render. It sounds as if it might be getting stuck after that.

1:42 AM
jk
does it ever make sense to rest/not study on the day of an exam to conserve mental power

removes Obliv from his perch
math is not a subject where cramming works

@Obliv I was a bit drunk when I took my last topology exam in grad school.

I know that, I was just wondering if conserving mental stamina is a thing
Even if you're fully ready & mastered a subject does studying as a "warmup" help and does it hurt when you do too much or something

Though I leaned, about four hours before that exam, that I had passed the topology qual, so the outcome of the exam was not terribly important.

i Have said frequently in here that a good night’s sleep is optimal strategy

1:51 AM
I guess it varies from person to person but I also wonder if there is an optimal amount of time to study in general

i never studied that much for tests…

Either you know it or you don't. Cramming right before an exam isn't going to help that much.

In math most learning is hours and hours on homework

This.

and review of big theorems….

1:53 AM
I know.. but proof based courses are so much different than just working through calculus problems

proof-based courses have proof-based homework

@onepotatotwopotato Started writing some notes on the Thurston norm: witheredstumps.wordpress.com/2024/02/28/…

I'm not used to looking up and pondering the vast plains of math that each theorem & proof provide me access to.
like I might understand a theorem but not its reaches

nobody understands the reaches

1:56 AM
galois did @LeslieTownes
jk

I get the beeches. They're pretty.

for a ring $R$, $R/(0_R)$ is the same thing as $R$?
like nothing changes

"the same thing" is a phrase slightly up to interpretation. they are not literally the same, but they are morally the same. insofar as the canonical projection map $R\rightarrow R/(0_R)$ is an isomorphism.

2:16 AM
kernels of homomorphisms and ideals are intimately related :| I am trying to put my finger on it

"morally"

@BalarkaSen so your goal of that writing is to write a proof of Gabai's theorem?

quotient rings $R/I$ for ideals $I$ are subsets of $R$ that are closed under subtraction & absorb products from the rest of $R$. $R/K$ for ideals that are kernels do the same except are elements that get mapped by a homomorphism to the $0$ of the image
for the 1st isomorphism theorem let $f: R \to S$ be a surj. hom. then this map can be modified to be an isomorphism for the quotient rings $R/K$ of $R$ to $S$

Quotient rings are NOT NOT NOT subsets

if $K = (0_R)$ then we morally have an isomorphism from $R\to S$
@TedShifrin sorry, subrings*

2:28 AM
NO NO NO NO

wait I think we went over this some days back, my book defines them as subrings for some reason

Elements of a quotient ring are equivalence classes of elements

also thats why i said subset I guess

This has nothing to do with quotients

oh dang i found what u mean

2 hours later…
4:42 AM
@ChristianIvicevic Hi

2 hours later…
6:35 AM
If $T$ is a linear operator, $m_T$ its minimal polynomial such that $m_T = QS$, is Q the minimal polynomial for $T|_{KerQ(T)}$?
I know that certainly the minimal polynomial for $T|_{KerQ(T)}$ divides $Q$

4 hours later…
10:42 AM
@onepotatotwopotato I might do that. Certainly I will learn it at least.

1 hour later…
12:01 PM
anyone here familiar with SymPy? (Python library for symbolic mathematics)
ah, found the answer in the sympy documentation

12:22 PM
though I still have a question... so if anyone is familiar with it, lemeno

12:39 PM
I have an elementary question about a certain deduction. I'm reading the following proposition:
> Proposition 4.15 Suppose that $f:X\to\overline{\mathbb{R}}$ is an extended real-valued measurable function. Then $\int |f| \ d\mu=0$ if and only if $f=0$ $\mu$-a.e.
The author then claims:
> In particular it follows that if $f:X\to\overline{\mathbb{R}}$ is any measurable function, then $$\int_A f \ d\mu=0\quad \text{if }\mu(A)=0.$$
Could someone clarify how the above follows from the proposition?

If $\mu(A)=0$, then $f\upharpoonright A=0$ a.e.

1:07 PM
that makes sense, thank you :)

@psie notice the subtlety of there not being an absolute value though
(that follows easily from $\int_A |f| = 0 \implies \int_A f = 0$)

yeah, I was just about to write, but good observation

1:59 PM
(in the meanwhile my SymPy Q got answered)

cool
What does a symbolic python of mathematics swallow whole?
🤔

2:43 PM
"cool. What does a symbolic python of mathematics swallow whole?"

Hmm... try, the following
$$\int_{\text{A\to End}} \frac{\text{Snake Lemma}}{\text{max (ker, coker)\circ (\{A,B,C,\dots\})}}$$
Provided the direct limits that is key to define the integration in all the relevant subspaces are defined, the integral converges and you can swallow a lot of short exact sequences with simply integration with SymPy

3:08 PM
@AlessandroCodenotti I've been thinking about this some more and I'm confused about the notation I think. When you write $f\upharpoonright A$, do you mean the restriction of $f$ to $A$ or the indicator function of $A$ times $f$? If you mean the former, I don't think the claim is correct; consider $f=1$ on the real line. If we restrict it to $\mathbb N$, then this function won't be equal to $0$ almost everywhere, since it is constantly $1$.
But if you meant the latter, then yes, the function $f\chi_A$ will be zero almost everywhere since $\mu(A)=0$.
I'm gone for a bit, but let me know what you were thinking :)

3:46 PM
@psie $\lambda\restriction_\mathbb{N}(A) = \lambda(A) = 0$ since $\lambda(\mathbb{N}) = 0$
where $\lambda$ is the Lebesgue measure
No matter which subset of $\mathbb{N}$ you take, it will be of measure zero
so $f\restriction_\mathbb{N} = 0$ on a set of full measure, that is the empty set

1 hour later…
4:59 PM
A question. AoC can be stated as "for any nonempty collection of nonempty sets, the Cartesian product of that collection of sets is nonempty"
I believe the answer is yes to the following but want to be sure: the converse, that a nonempty Cartesian product of some collection of sets implies the sets in question are nonempty is true too, right?
That is, AoC just tells us about the one nontrivial direction as it were?

@EE18 If there is $f\in \prod_{i\in I} A_i$, then $f(i)\in A_i$ so $A_i\neq\emptyset$ for all $i\in I$

5:14 PM
If $X$ is metric space and $Y$ is compact, is the compact open topology on $ContinuousMaps(X,Y)$ metrised by the $\infty$-metric?

@monoidaltransform no
but uniform topology on $C(X, Y)$ is only dependent on topology of $X$ and $Y$ in this case
(I'm assuming $Y$ is also a metric space)

yah, Y is metric space

for compact open topology on $C(X, Y)$ to coincide with uniform topology you want $X$ to be compact and not $Y$

Agreed, thank you Jakobian!

What metric can one give when X not compact but Y is?

5:21 PM
I am a bit stuck on a bit of a proof, is it OK to post here and ask about it?
these choice proofs are so slippery for a newbie

@monoidaltransform what do you mean?
for $C(X, Y)$ with compact-open topology?

I'm not sure if it has to be metrizable

assuming X is hemicompact

@monoidaltransform I don't have a reference for metric on $C(X, Y)$ inducing compact-open topology when $X$ is hemicompact
it can be easily done when $X$ is exhaustible by compact sets
however note that if $X$ is metrizable then $X$ is hemicompact iff its exhaustible by compact sets

5:29 PM
(hopefully it is ok but understand if no one wants to read the above)
I can't get the "Third" and "Fourth" points at the end
the order on the natural numbers is the usual $\in$

so in your setting of $C(X, Y)$ with $Y$ being a compact metric space, and $X$ being a metric space exhaustible by compact sets, pick compact sets $K_n\subseteq X$ such that $\text{int}(K_n)\subseteq K_{n+1}$
then define $d(f,g):= \sum_{n =0}^{\infty} \frac1{2^n} \sup_{x \in K_n} \min\left\{d_Y(f(x),g(x)),1\right\}$
the metric $d$ induces compact-open topology on $C(X, Y)$

you just need that $X$ is hemicompact and that $Y$ is metrizable and neither that $Y$ is a metric space nor that $Y$ is compact

WTH is hemicompact?

@Ted exhausted nicely by a sequence of compact sets basically
@Jakobian I don't get what definition of hemicompact you're using

5:36 PM
I've never heard the term before in my 53+ years as a math person.

Sure, if $K_n\subseteq K_{n+1}$ is an arbitrary sequence of compact sets such that $L\subseteq K_n$ for some $n$ for any compact $L$, then we should be able to get away with $d$ defining a metric for $C(X, Y)$ inducing the compact-open topology

right, that's what I thought
@TedShifrin I guess it's more a general topology thing. Though Conway mentions it in his functional analysis book
there are an absurd amount of adjectives flying around in general topology

shrug
Of course, exhaustion by compact subsets is a common notion. I guess where I've encountered it, one just says that.

we were talking about uniform topology on $C(X, Y)$ and this does require $Y$ to be a compact metric space to define unique topology on $C(X, Y)$ independent of metric chosen

But shouldn't it still work out for the compact-open topology?

5:43 PM
I'm just saying that was the setting
$Y$ doesn't have to be compact, sure

okay fair

6:03 PM
hemicompact, $\sigma$-compact and exhaustible by compact sets are all distinct notions
it's awful
my favorite adjective combination is "totally paracompact", though

Ugh... I just can't seem to shake this cold. I've been working for about six hours (four of which was recording lectures that I didn't record last week because I was really sick), and I am completely wiped out. My voice is barely functional, and I don't have any more energy left to focus on grading or other work.
I think I'm going to go home. :/
I am completely caught up with one class, am up to last Friday on the lectures for the other two, and am all caught up on the grading everywhere, modulo one exam that I need to grade. :(

@Xander I know I shouldn't ask, but have you taken a Covid test? I came down with a mild case about two weeks ago. It's going around again.
@Xander @leslie @robjohn et al. I guess we should cry.

@TedShifrin I have taken four covid tests in the last 2 weeks, all negative.
@TedShifrin What a muddled mess.

Well, good, I guess. So it’s just an uncommon common cold. I find sinus tylenol and Mucinex DM helpful ….
It’s beyond just muddled.
Look at his last query to me in comments,

6:19 PM
@TedShifrin I took a couple of doses of generic NyQuil (which is basically tylenol and some kind of something that knocks me out), but I am not really symptomatic at this point---just tired and having a little bit of trouble keeping my voice going.
Though, honestly, Tylenol kind of worries me---it is super bad for your liver (and probably wouldn't even get FDA approval if it were a new drug today, rather than something that's been around for ages).

Yeah, sinus drip messes up the larynx big time. Lots of tea and rest.

According my phone, I have also been snoring a lot this week, which is unusual. And probably explains some of the fatigue (snoring correlates to bad sleep). Anywho, Imma go home and make some cocoa.

@XanderHenderson your phone tells you that much details?

@Jakobian It claims to. It records coughs and snoring.

did you ever notice that you were snoring?
or did someone tell you that

6:26 PM
I've never noticed myself snoring, nor has anyone ever told me that I snore.
My phone has been telling me that I have been snoring for the last two weeks (which correlates with this damnable cold).
This is a new phenomenon.

probably because of clogged nose?

Indeed. But I am leaving now.

alright, hope you recover fast
@LukasHeger I believe someone once told me that rings which aren't integral domains are uninteresting. But I think I disagree. You can obtain interesting integral domains from rings that aren't integral domains

I agree. Nilpotents are pretty important for certain stuff

E.g., linear algebra and Lie groups.

6:36 PM
or formal smoothness in AG

You do love the dual numbers :P
But, yes, nonreduced schemes do exist.

if you're doing algebraic groups in char p, nonreduced group schemes appear naturally e.g. as kernels of maps between algebraic groups

Well, I would never be caught dead doing any of that.

im not sure if this is a hot take but i just need to say that the open square bracket notation for intervals is horrid
$]x,y[$ rather than $(x,y)$ is ughhhh

Well, it's really not. How do you know if $(x,y)$ is an ordered pair or an interval?
I actually think the French notation wins, even though we are not used to it.
So, e.g., in Munkres's topology text, you're stuck having to write $x\times y$ for the ordered pair $(x,y)$ lest it look like an interval.

6:40 PM
Is it French? I don't recall the Hubbards using it and AFAIK he was trained there??

Hubbard got his doctorate there, but he's American and teaches in America.

surely that can be clear from the context! lots of notation is overloaded after all

And I still hate his book.

fair enough to that
Just too much stuff in it?

Too much showing off. Exercises are poor (or at least were in the first few editions). While his Kantarovich proof of the inverse function theorem is interesting, it seems a poor pedagogical choice to me. I did turn it into two exercises in my book, though.

6:43 PM
@LukasHeger I was thinking of hyperreals and other quotients of $C(X)$

i just found it slogged along soooo slowly and had to stop. maybe ill come back to it one day when i know the stuff and can appreciate it more

I would highly recommend my book instead. :P
I had to write it when I created a course at UGA to teach out of Hubbard^2 and realized after 2 months that it was not a good book. That said, he has infinitely many editions and has made way more money than I have.
Of course, a lot of faculty opt for his book because he's a well-known and respected mathematician. I don't know too many students who like it.

If and when I come back to that genre I will definitely take a look :) for now I'm gonna chug along with some analysis text(s) and hope i make it through!
Other than being a beautiful book (aesthetically) I did not love it, but this was a few years ago

@TedShifrin which one?
oh nevermind. There's only one
those people are demanding 900$for Hubbard and Hubbard i suspect that's because it's an old version in low supply 6:51 PM I think someone added too many zeros @EE18 its not an antique, lol the 5th is much less but still expensive. but then again, the cost of books is doubtless a controversial subject! for sure, but if there's only one supplier then it's not surprising. the supplier is obviously not setting a market clearing price but still their prerogative i guess lol Hi guys @TedShifrin I like Dieudonne, Eilenberg more There's something very appealing about that book ive come to the realization that the implication implicit => inverse function thm is more natural to me than the other way round oh is Eilenberg just an editor? I like Dieudonne's Treatise on analysis, I mean would it be unreasonable to read all of those volumes actually 7:03 PM @Thorgott I don't have strong feelings on this. But I love the contraction mapping proof of the inverse. How're you going to do implicit first? @Thorgott I am quite fond of the "bloat" technique. Given$F : \Bbb R^{m+n} \to \Bbb R^m$with$DF_0$surjective, I can bloat it to$(F, \pi) : \Bbb R^{m+n} \to \Bbb R^{m+n}$where$\pi$is the coordinate projection to the last n variables. Now inverse function theorem becomes applicable. I like Dieudonné, but it's way too fancy and slick to learn anything for the first time. We were discussing a first course for college first/second year students doing linear algebra and multivariable calc/analysis. @Balarka Yes, that's the standard proof. And standard form for immersion and submersion are variants of that argument, of course. Right. Doesn't Dieudonne prove implicit function theorem first, using contractions I'm a big fan of this kind of arguments. I would often need to make a homotopy$F : X \times I \to Y$into a movie$X \times I \to Y \times I$,$(x, t) \mapsto (F(x, t), t)$. 7:06 PM I think that's called the trace of the homotopy. I learned that early on, maybe from Hirsch. That's a nice terminology. I always called it "movie". Which is not ideal really, i like a perspective along the lines in Bredon. the natural theorem is that if$F\colon\mathbb{R}^n\times\mathbb{R}^m\supseteq W\rightarrow\mathbb{R}^m$is$C^1$and$(x_0,y_0)\in W$is a point s.t.$F(x_0,y_0)=y_0$and the matrix of partials$\frac{\partial F}{\partial y}(x_0,y_0)=0$, there are small balls$U,V$around$x_0,y_0$resp. such that there is a unique function$\phi\colon U\rightarrow V$s.t.$\phi(x_0)=y_0$and$F(x,\phi(x))=\phi(x)$for all$x\in U$, and this$\phi$is$C^k$if$F$is$C^k$. if i attempt to prove the inverse function theorem directly, i always mess up cause i never see how to "naturally" make the right estimates to apply the contraction principle @BalarkaSen Hirsch calls it "track" somewhere in Hirsch is the wrong claim that "the track of an isotopy is an embedding" Oh, I remembered trace. Nobody asked, but here's an application of this. Suppose$f, g : S^2 \to S^5$are two embeddings, I claim they are isotopic through embeddings (so there are no nontrivial$2$-knots in the$5$-sphere). Well, they certainly are homotopic as maps, let$F : S^2 \times I \to S^5$be the homotopy. Take the trace$H : S^2 \times I \to S^5 \times I$. Approximate$H$by an immersion rel boundary. As$2 + 1 = 3$is half of$5 + 1 = 6$, the resulting immersion only has transverse clean double points as self-intersections. For each double point$p$, choose an embedded loop$\gamma$starting at$p$, going along one branch of the sheet coming at$p$, and coming back along the other branch and ending at$p$.$S^5$is simply connected, so$\gamma$bounds a topological (not necessarily embedded) disk. But we can approximate this by an immersed hence embedded disk, as$2 + 2 = 4 < 5$and the ambient is$5$-dimensional. Treat this as a Whitney disk, cancel the double points. This shows there's an embedded$S^2 \times I$sitting in$S^5 \times I$connecting$f, g$. This may not itself be a movie of a homotopy; instead it's what is known as a "concordance". Apply the relative h-cobordism theorem on$(S^5, S^2)$to conclude that it can be made into the trivial concordance i.e. an isotopy. Oh, I never tried to write a proof of implicit directly with contraction mapping. It's in the function space, whereas inverse is just in the original Banach space. Pedagogically, for beginning students, I'll stick to$\Bbb R^n$and doing inverse first. The function space stuff is great for ODE and integral equations, too, but I don't want to be doing Banach spaces with first-year students. And there's no time. 7:21 PM @Thorgott Your quip is noncompactness, I assume? The track of an isotopy is certainly an injective immersion yeah That's a little silly, because who on earth would do nonproper isotopies @Thorgott sounds like the argument in Dieudonne the claim is correctly stated elsewhere in the book, he just forgets it at that point a couple years ago, i figured out a counter-example and discussed it in here with Mike I remember it from a course, not his book. Yeah, even the pros screw up sometimes. 7:22 PM @Jakobian im not familiar with that book its a great book, does most things with Banach spaces. This proof is in chapter X, section 2 Yeah, I'd forgotten that Dieudonné doesn't do the inverse function theorem. He does the rank theorem after implicit, and then ODE stuff. Again, NOT the approach for beginners. @TedShifrin that's 10.2.5. Its slightly after, but its there Well, it's a special case of the rank theorem. Oh, no, you're right. He does state it, with no fanfare, before. For you Banach manifold inverse function theorem enthusiasts, here's an exercise. Let$M$be an infinite-dimensional Banach manifold and$G$be an infinite-dimensional Banach Lie group acting freely on$M$. When is$M/G$a Banach manifold? 7:29 PM For geometers and differential topologists, the inverse function theorem is a big deal. @BalarkaSen are they over the same Banach space? no thanks i dont like Banach manifolds But you love nilpotents :P @Jakobian Let's say all my Banach spaces are separable so yeah @Thorgott This is the 0-level question while trying to define stuff like Gromov-Witten invariants and whatnot Or gauge theory @BalarkaSen I don't think all separable Banach spaces are diffeomorphic though 7:34 PM The finite-dimensional proof that$M/G$is a manifold when$G$acts .... is something I messed up the only time I tried to do it in the graduate geometry course. honestly, i never proved the finite-dimensional case either @Jakobian Yeah good point. Alright, let's assume they are modelled on the same Banach space except when$G$is finite I was secretly thinking Hilbert space the generalities of when quotients are manifolds are very subtle 7:37 PM average Duistermaat-Kolk reader @Thorgott Das stimmt. In the below, I can't follow how to bound the second part. i.e. I know that$\frac{1}{1 + \beta/\Tilde{y}| \geq 1 + |\beta/\tilde{y}| = 1+ \delta(\tilde{y})$and$\frac{1}{1+ \delta(\tilde{y})} \leq \frac{1}{1- \delta(\tilde{y})}$, but these don't help me get the desired result since both inequalities are same direction$\ell^1$and$\ell^2$are not diffeomorphic, right? 18 There is a theorem stating that there is essentially one smooth structure on$R^n$for every n other than 4. Does anybody know where i could find the proof of this? Not so much of what happens in dimension four, where there are infinitely many, but of the uniqueness in other dimensions? Thanks! contrary to finite-dimensional case damn, my latex skills are sucking right now I have no idea, @Jakobian. I've never contemplated it. 7:44 PM If they were diffeomorphic, then taking the derivative at a point would say that they were also continuously isomorphic Trying again: I know that$\frac{1}{|1 + \beta/\tilde{y}|} \geq \frac{1}{1+|\beta/\tilde{y}|} = \frac{1}{1+ \delta(\tilde{y})}$and$\frac{1}{1+ \delta(\tilde{y})} \leq \frac{1}{1- \delta(\tilde{y})}$, but these don't help me get the desired result since both inequalities are same direction hallelujah are they? @BalarkaSen you mean homeomorphic? EE18 The very first inequality is obviously incorrect. @Jakobian I mean the derivative at a point is a bounded linear map 7:45 PM Oh you mean linearly yeah, thanks it was an error, should be correct now no Ted? It's just triangle inequality and then taking reciprocals well, the finite-dimensional case is also not uniform darn$n=4$petition to make$4$an infinite number? the issue with that is that$5$still ought to be greater than$4$Oh, right, never mind, EE. .... I need to leave to I will leave it to you. 7:47 PM @EE18 your LaTeX is broken see my second post above i redid it should be clean now, i couldnt edit the first in time which one is the second part @Jakobian My most recent message above this one sorry, doesnt let me reply to my own no, in the screenshot oh screenshot is from the text what's broken there? 7:52 PM not that "bound the second part" I've moved on from the broken LaTeX ah ok i understand what I mean is that I would like to be able to argue as follows:$\frac{1}{|1 + \beta/\tilde{y}|} \leq \frac{1}{1+|\beta/\tilde{y}|} = \frac{1}{1+ \delta(\tilde{y})} \leq \frac{1}{1- \delta(\tilde{y})}$so that i can come to Zorich's conclusion but the first is not true which one is the second part i'm not really sure what i was referring to at the time now that i think about it lol, but hopefully my reformulation above is somewhat clearer Basically I want to be able to show$\frac{1}{|1 + \beta/\tilde{y}|} \leq \frac{1}{1- \delta(\tilde{y})}$but my current strategy doesn't work$\Delta \tilde y = \beta$?$\Delta(\tilde{y}) := |y-\tilde{y}| = |\beta|$8:01 PM so you're trying to show$1\leq |1+z|+|z|$for$z = \beta/\tilde y$is what this boils down to which is just triangle inequality sorry just thinking about that for a moment Or$|1-|z||\leq |1+z|$if you prefer sometimes people write$||a|-|b||\leq |a-b|$Yes that's it exactly the absolute value on the inside ok will think about that yeah so you were just trying to apply wrong inequality not$|a+b|\leq |a|+|b|$but$||a|-|b||\leq |a+b|$is needed here from below ah this is "reverse triangle inequality" ? 8:05 PM it might be called that but its just a consequence of triangle inequality doesn't matter how its called... dang, bad that i forgot about that. thanks jakobian. sometimes names help (me anyway) remember things. clearly not one of those instances here :) another netflix L Hello chat. I've encountered the following two definitions of essential supremum and was wondering if someone could clarify how/if they are equivalent, i.e. discuss a proof sketch. The first one is$\operatorname*{ess\,sup}_X f=\inf\{a\in\mathbb R: \mu\{x\in X:f(x)>a\}=0\}$and the second is$\operatorname*{ess\,sup}_X f=\inf \left\{\sup_X g:g=f\ \text{pointwise a.e.}\right\}$. I have a hard time seeing their equivalence. 8:23 PM If$\mu\{f > a\} = 0$, then$h := \min(f, a) = f$a.e. And$\sup_X h \leq a$And if$h = f$a.e., then$\mu\{f > \sup_X h\} = 0$@Jakobian ok, what does this show, i.e$\sup_X h \leq a$? write it out pen and paper 8:50 PM @Jakobian it shows that$\sup_X h$is a lower bound to the set$A=\{a\in\mathbb R: \mu\{x\in X:f(x)>a\}=0\}$, or? So$\sup_X h\leq\inf A$? I'm confused about what it is one needs to show to show equivalence. Call the other set$B=\left\{\sup_X g:g=f\ \text{pointwise a.e.}\right\}$. Are you then trying to show that$\inf A\leq \inf B$and$\inf B\leq \inf A$? 9:17 PM @psie$A = B$9:41 PM guys, what happens to the length of a closed simple curve if I consider$\oint_{\gamma}|d\zeta|$instead of$\oint_{\gamma} d\zeta$maybe I should be more precise: in my book it is written something along these lines: $$\left|\oint_{\gamma} \frac{f(\zeta)-f(z)}{\zeta -z} d\zeta\right|\le\oint_{\gamma}\left| \frac{f(\zeta)-f(z)}{\zeta -z}\right||d\zeta|$$ But I thought the inequality only worked after I parametrize the curve (namely, only when I deal with standard real-variable integrals) I guess I have two questions now :P @Jakobian how did you show that$a\in A \implies a\in B$, i.e.$A\subset B$? You showed that if$a\in A$, then there exists a function$h$such that$a\geq\sup_X h\in B $. 10:03 PM @ClaudioMenchinelli The first integral is the length and the second is$0$. @psie you are right the sets don't have to be equal in general, probably I went too far with claiming that we can say that they're equal the argument still shows that$\inf(A) = \inf(B)$on one hand$B\subseteq A$on the other hand, for all$a\in A$we can find$b\in B$with$b\leq a$ok, makes sense So a lower bound for$A$is a lower bound for$B$, and vice versa @ted: so, uh, can i solve it for when dS/dxdy is 1 degree and just multiply the answer by 52? 10:24 PM @TedShifrin Because of the closed assumption. I see, then I did my computations correctly. Thanks 10:40 PM @Jakobian just to be certain, by$\min(f,a)$you mean the minimum of$a$and$f(x)$for all$x\in X$right? the reason I'm asking is because I'm used to see the minimum taken of sets here to me it is a bit unclear what set is meant 10:53 PM oh wait, I think you just mean$h(x)$equals the minimum of$f(x)$and$a$, because then it makes sense to take the sup of an actual function consider$\mathbb{R}\rightarrow S^1,\,t\mapsto e^{2\pi i\alpha t}$, where$\alpha$is irrational. is this a homeomorphism restricted to$\mathbb{Q}$? @Thorgott No because there is a sequence$t_n\to \infty$with$e^{2\pi \alpha t_n}\to 1$ah, that was the argument, thanks do you know an example of a nice function on$\mathbb{R}$that's a homeomorphism when restricted to$\mathbb{Q}$and not itself a homeomorphism? @psie When I write minimum here, I don't mean minimum of$f$and$a$, but indeed the function$h(x) = \min(f(x), a)$11:13 PM @Thorgott nvm this doesn't happen in the examples im actually interested in @BalarkaSen oh I thought you already read it! I'm also planning to read that paper after I finish teichmuller polynomial paper @Thorgott glad that worked out for you, I couldn't think of any examples the question was poorly constructed what im really looking for is an example of a smooth map$f\colon M\rightarrow N$between smooth manifolds and a subset$X\subseteq M$s.t.$f\vert_X$is an embedding,$df_p$is invertible for all$p\in X$, but there is no open neighborhood of$X$mapped diffeomorphically by$f$to an open neighborhood of$f(X)$there has to be a counter-example, but it also has to be quite ugly (i can show$X$is necessarily not$\sigma\$-compact)
(the second "has" is a factual claim, the first "has" is an opinion)