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11:01 AM
@Krijn The first thing that comes to mind is hairy ball theorem. The foliations interpretation is that there cannot be foliation on $S^2$ by 1-manifolds.
Similarly, Poincare-Hopf theorem tells how many points you need to throw away from your ambient manifold for it to admit a foliation by 1-manifolds. (you need to throw away 2 points from $S^2$; $S^2$ minus north and south poles admits foliation by lattitudes)
Morse theory makes a lot of sense with foliations; it's really doing general position of submanifolds with an appropriate foliation.
 
I see
I'd say the torus is more interesting than $S^2$ in foliations
But that may be my naivety
 
No, you're right! Torus has lots of foliations by 1-manifolds.
Foliate $\Bbb R^2$ by parallel lines of slope $\alpha$. Then under the covering map $\Bbb R^2 \to T^2$, that gives a foliation on $T^2$.
Cool thing to think about it what happens when $\alpha$ is irrational.
 
Yeah, that'll give some nice images
When $\alpha$ is irrational you get that infinite line on the torus right
 
Yeah.
it keeps winding on and on. a dense line on T^2
 
We probably get there soon in the course on Topology that I TA
Gonna blow some minds
 
11:09 AM
kewl!
 
11:32 AM
[Explosion of ideas (Brief)]
\textrm{Last night dream} $\xrightarrow{} \cdots \xrightarrow{} $ Existence of discrete uncountable set?

It is known there are no such sets in the reals, but I wonder... are there named examples of them given that they must be quite hard to work with due to not being second countable and non separable?
 
@Secret Through some weird googling I found this on a Wikipedia Talk page
the well-ordered set of ordinal type $\omega_2$ is uncountable but discrete
 
Take the diagonal $y=-x$ in the Sorgenfrey plane, this is a discrete uncountable set
 
Ah nice, thanks guys
(The Sorgenfrey plane look so harmless on a diagram, guess that's another example of not to rely on diagrams)
Cannot believe what DHMO taught me many months ago (upper limit topology) is actually a very nontrivial topological space, without realising
 
The Sorgenfrey plane is the product of 2 copies of $\Bbb R$ with the lower limit topology
 
11:47 AM
@BalarkaSen So critical points give ones that are not in general position?
 
Starting to think... if we introduce some really complicated mathematical object to students, ask them to work through it, but never mention its name until they have worked on it enough, perhaps they will get really good at it
I wonder if the very act of naming something actually has a framing effect that make people cringe...
(But then, we have countless counterexamples in fiction where so many things have name, but no context)
so...dunno...
 
Sorgenfrey is almost a homophone to "without worries" in Dutch
 
:)
Might be where it comes from
@arctictern never mind my previous ping ^^
 
The ordinals are very useful whenever we need a well ordered uncountable structure to experiment on
 
Every well ordered set is isomorphic to an ordinal (under AC)
 
11:59 AM
@Danu No, degenerate critical points are points where the submanifold is not in general position. Suppose $(M, \mathcal{F})$ is a foliated manifold. A submanifold $N$ is in general position if it's tangent bundle $TN$ is transverse to the tangent bundle of the foliation $T\mathcal{F}$ inside $TM$.
This is just reflecting the fact that Morse theory is transversality in the tangent bundle.
 
3
Q: On reducing complex ODE's to Bessel's form using Kummer's series

Naveen BalajiI am trying to reduce the following ODE to Bessel's ODE form and solve it: $$x^{2}y''(x)+x(4x^{3}-3)y'(x)+(4x^{8}-5x^{2}+3)y(x)=0\tag{1} \, .$$ I tried to solve it via the standard method, i.e., by comparing it with a generalised ODE form and finding the solution from then on. The general form (...

 
@Krijn Sorgenfrei should work in German too
 
Most Germanic languages, I presume
It appears to be Danish
 
Makes sense
 
Sorgenfrey worries me though
not at all a hakuna matata :P
 
12:08 PM
That's understandable, it's a rather worrying space :P
 
Hmm, interesting analytic lemma. $f, g$ be two $C^2$ diffeomorphisms of $[0, 1)$ fixing $0$ and such that $f \circ g = g \circ f$. If $0$ is an isolated fixed point of $f$, then either $g$ has no fixed points or it is identity.
what does that really mean P:
Purely algebraically, it gives a description of the centralizer of contractions in $\text{Diff}^2_0[0, 1)$, I guess.
 
12:24 PM
Fun fact, $\lvert3\lvert x\rvert+5x\rvert=5\lvert x\rvert+3x$
(Any pair of positive numbers for $3$ and $5$ works)
 
Cute. Of course one can prove this piecewise.
 
@AkivaWeinberger Having a real hard time reading this
 
$\Big\lvert3\lvert x\rvert+5x\Big\rvert =5\lvert x\rvert+3x$
 
@BalarkaSen What do you mean, precisely, by this statement?
 
@Krijn I blinked and stared at that line for way too long
 
12:40 PM
Hi guys. Let $L\colon V\to V$ be a linear transformation of the inner product space $(V,\mathbb C,\langle{.,.}\rangle)$. Let $\beta=\langle b_1,\dots,b_n\rangle$ be an orthonormal basis of $V$. My book says that if $L(b_j)=\alpha_1b_1+\dots\alpha_nb_n$, then
$$
e_i^Tco_\beta(L(b_j))=\alpha_i=\langle b_i,L(b_j)\rangle.
$$
I don’t understand why we can also say that $\langle b_i,L(b_j)\rangle=\alpha_i$. The inner product of $V$ hasn't been defined, so how do we know that it will be multiplied terms wise?
 
@Danu Critical points of $f : M \to \Bbb R$ are points where $Df$ vanishes. Nondegenerate critical points are where $Df$, as a section of $T^*M$, is transverse to the zero section.
I hope I said that right.
 
Yeah, okay. I was hoping for something deeper ^^
 
Nah it's just a reformulation of Morse theory in terms of genericity.
 
@ShaVuklia $b_i$ form an ONB, hence $\langle b_i,b_j\rangle = 0$
 
@SteamyRoot oh, of course! why did I not see that:P
thank you:)
 
1:09 PM
@BalarkaSen Your initial sentence somehow seemed to hint at a sweeping generalization to other bundles :P
 
Show that the matrix $A$ w.r.t. an orthonormal basis of a self-adjoint linear transformation $L\colon V\to V$ is self-adjoint. So I have to show that $A^*=A$.
Let $\beta=\langle b_1,\dots,b_n\rangle$ be an orthonormal basis of $V$. We know that
$$
\langle L^*(v),w\rangle=\langle L(v),w\rangle=\langle v,L(w)\rangle.
$$
I have to make the translation to the coordinate vectors and the matrix $A$. I know that:
$$
e_i^Tco_\beta(L(b_j))=\langle b_i,L(b_j)\rangle,
$$
and I want to use this. So I guess I could write:
 
1:27 PM
The current answer by amrsa to this question proves Lemma 3 of the question, but does not answer the question. Should I edit the question to include this proof then ask amsra to delete the answer?
*of
It might be worth asking that on the meta site, eh?
 
(never mind, I figured it out!)
 
2:25 PM
hi chat
 
Can I say $$\sum_{i=1}^\infty \sum_{j=1}^\infty |a_{ij}|^2 \leq ( \sum_{j=-\infty }^\infty \sup_{i\in N} |a_{i,i+j}|)^2$$
 
hi @semi
 
I don't think so. Think I went 'too easy' on the left side to prove it.
 
$a_{ij}=0$ if $j\leq 0$, that's it
 
2:33 PM
Greetings from CPC: Can someone help in the following: How to get live preview on Mathjax on wordpress site's input field
 
2:44 PM
Hey, guys, is the decreasing intersection of triangles a triangle?
(Or, more general, the decreasing intersection of simplices a simplex)
(Points and line segments count as triangles here)
This should be easy but I'm not seeing how to do it.
Clearly, the intersection is convex.
 
3:04 PM
@Danu No, but what I am saying is, Morse theory can be done on foliated manifolds.
That's a generalization.
 
@AkivaWeinberger if the radius goes to zero the intersection is a point
 
@AlessandroCodenotti I said I'm counting that as a triangle
I think I see how to do it, now, though.
 
$\pm a x + bx=\pm(a x \pm bx)$

$\pm a x + b x = \pm(a x \pm bx)$

It works, but why...?
 
(Cont'd) Well, maybe.
@Secret $a\pm(\pm b)=a+b$
 
but absolute values don't distribute over + in general
$\pm(a x \pm bx)\neq \pm a x \pm(\pm bx)$
O wait a sec...
 
3:15 PM
@Secret No, those are equal
I'm assuming all through here that we make the same choices for all the $\pm$s.
 
hey @arctictern
 
So that $a\pm b\pm c$, for example, has two possible values and not four.
 
@arctictern I want to verify with you that $\phi_2 \pi_1 = 0$ they meant to say $\pi_1 \circ \phi_2 = 0$ in the following proposition
 
3 hours ago, by Akiva Weinberger
$\Big\lvert3\lvert x\rvert+5x\Big\rvert =5\lvert x\rvert+3x$
Hmm, I am missing something obvious here, why a=3 b=5 (or multiples of them)
 
It works for all pairs of positive numbers
I think I mentioned that a comment or two before that
@Secret By the way, a fun analogue of Euler's formula:
$e^{\pm x}=\cosh x\pm\sinh x$
 
3:19 PM
All pairs such that the one in front of $|x|$ is greater than the one in front of $x$ ?
 
That one is easy, because sin cos sinh cosh are all linear combination of exponentials, thus it follows easily
Hmm, I wonder if there exists linear analogue of sin and cos. Because consider:
 
@SteamyRoot Yeah, you want the RHS to always be positive (even when $x$ is negative) since it's the absolute value of the LHS
 
sin cos are for circles and ellipse
sinh cosh are for hyperbola
Therefore there should be a ? ? for parabolas and straight lines (the remaining two class of conic sections)
51
Q: Do "Parabolic Trigonometric Functions" exist?

ArgonThe parametric equation $$\begin{align*} x(t) &= \cos t\\ y(t) &= \sin t \end{align*}$$ traces the unit circle centered at the origin ($x^2+y^2=1$). Similarly, $$\begin{align*} x(t) &= \cosh t\\ y(t) &= \sinh t \end{align*}$$ draws the right part of a regular hyperbola ($x^2-y^2=1$). The ...

Found something: There's an underlying symmetry governing them all
 
@Akiva Say $P_n, Q_n, R_n$ are the vertices of the triangles $\Delta_n$ in the decreasing sequence. By Cantor's intersection theorem, there is a point $O \in \bigcap \Delta_n$. $d(O, P_n)$ etc are monotonically decreasing because of the containment condition, so $P_n, Q_n, R_n$ all converge to $P, Q, R$.
Let $\Delta$ be the convex hull of these; if $x \in \Delta$ then it's a linear combination of $P, Q, R$ so can be written as a limit of linear combinations of $P_n, Q_n, R_n$ - ie elements of $\Delta_n$. I think that should say $x \in \bigcap \Delta_n$.
I am having trouble putting this into a rigorous argument.
 
The question of finding $(f_i)$ such that $2^{\sum_i^{\infty}f_i}=\sum_i^{\infty}f_i$ is the same problem as solving $\prod_i^{\infty}2^{f_i}=\sum_i^{\infty}f_i$. The latter presumably can be operated using generating functions
 
3:33 PM
So what exactly is the point of the infinite sum there?
Aren't you just looking for any function $f$ such that $2^{f(x)} = f(x)$, and then you split up $f(x)$ in a bunch of terms?
 
Well I am exploring the case where $f(x)$ can be split up into a countable number of terms, so that the whole infinite sum is a fixed point (function?) under the map $2^{stuff}$
That is, I am pondering about convergent infinite series as fixed points of other maps
 
You can always split up a function $f(x)$ in a countable number of terms...
 
(and $f$ is not necessary differentiable or complex differentiable, hence taylor series or laurent series may not exist)
But yeah, this problem is pretty random considering it is inspired from a dream I had last night
and I currently don't have much thoughts on what to do with it (or other leads)
The good thing about fixed points is that they will stay invariant when you act a suitable operator on them. That can help elucidate some symmetry of a give expression which I hope I can use that to compute closed forms, integral representations, product representations or series representations
(Howeverm I am pretty sure more advanced methods already exists, but I am still miles away from understanding complex varieties)
 
Uh huh... I have no idea what you're talking about or what you're trying to do. However, there are no real solutions to the equation $2^x = x$, working with real functions won't help you at all.
And for any function $f$, $f(x) = \sum_{i=1}^\infty \frac{1}{2^i}f(x)$
 
@AkivaWeinberger i know, I meant that it works in the shrinking radius case, dont know about the general one
 
3:45 PM
Uh wait a sec, how will the above $f(x)$ satisfy $2^{f(x)}=f(x)$ cause there are not enough 2 to cancel out the $\frac{1}{2^i}$?
 
Uhhh... wut?
What does that have to do with anything?
 
$\sum_{i=1}^{\infty}\frac{1}{2^i}=\frac{\frac{1}{2}}{1-\frac{1}{2}}=1$ ok nvm, its a geometric series and $f(x)$ has no $i$ dependence
But I am interested in e.g. $f(x) = \sum_{i=1}^\infty \frac{1}{2^i}g_i(x)$
 
You may want to put some conditions on $g_i(x)$ then.
 
@AlessandroCodenotti I see that your question made it onto . Characterizing topological spaces with a minimal basis
 
Otherwise I can just take $g_i(x) = f(x)$ for all $i$. Or just shuffle things a bit around or whatever.
 
3:51 PM
Consider $A\in\mathbb K^{k\times k}$. We know there exists $X\in GL_n(\mathbb K)$, such that $X^{-1}AX=R$ is upper triangular, and $X=QU$, where $U$ is upper triangular. Then
$$
Q^{-1}AQ=UX^{-1}AXU^{-1}=URU^{-1}.
$$
I don’t know how to interpret this. I interpret $X$ as the matrix of basis transformation from the “upper triangular basis” (“X”) to the old basis of $A$. And $Q$ is the basis transformation from the orthogonal basis to the old basis of $A$. However, what is $U$ then? I’m guessing $U$ is the basis transformation from $X$ to orthonormal. If we look at $X=QU$, then we know that $U
 
$\forall i, g_i(x)=f(x)$ will be basically a "constant sequence of functions
Otherwise yeah, will think about the conditions on $g_i$ later. Continuity is usually a good start
 
And, still, at this point, you don't do anything with the separate terms
Your only condition is on their sum, so there's no reason not to work with the sum only in the first place
 
@BalarkaSen @AlessandroCodenotti Here's a generalized version. Suppose $P_n\to P$, $Q_n\to Q$, and $R_n\to R$, and call the associated triangles $\Delta_n$ and $\Delta$. Let $x$ be an arbitrary point. Does $\lim\limits_{n\to\infty}d(\Delta_n,x)=d(\Delta,x)$?
 
I think what I am thinking is that suppose the sum $\sum_i^{\infty}g_i$ for a sequence of functions $g_i(x)$ is known (which indirectly means that we have some known condition on the sequence $(g_i)$), I am investigating how the sum will behave as it transform under various maps such as the exponential map, and see how this transformation tell us something more about the extra constraints $(g_i)$ need to satisfy in order for the sum to be a fixed point of said map
So acting the map $2^{()}$ on the known sum will give us the equation $\prod_i^{\infty}2^{g_i}=\sum_i^{\infty}g_i$ which can then further be manipulated to work out e.g. the condition in terms of the sum it will need to satisfy
 
@BalarkaSen @AlessandroCodenotti The triangles are not necessarily decreasing here.
 
4:01 PM
As it currently stands, it is a very general problem which I have not think of imposing the concrete conditions on $g_i$ yet (only saying, that $g_i$ satisfy some property P), but I might revisit this in the future
 
@NaveenBalaji re: your comment to my answer, I can't update immediately but will do so in a few hours.
 
@MartinSleziak oh, nice, I didn't notice it
 
4:25 PM
Hi there. Does anybody perhaps know anything about galois cohomology?
I have a small question.
 
4:46 PM
Hey guys - maybe I'm missing something obvious here, but how do you prove $E[Z|X] = E[E[Z|X,Y]|X]$?
if Z,X,Y are r.v.'s, discrete
 
The proposition in probability theory known as the law of total expectation, the law of iterated expectations, the tower rule, the smoothing theorem, and Adam's Law among other names, states that if X is an integrable random variable (i.e., a random variable satisfying E( |X| ) < ∞) and Y is any random variable, not necessarily integrable, on the same probability space, then E ⁡ ( X ) = E ⁡ ( E ⁡ ( X ∣ Y ) ) , {\displaystyle...
 
Yeah I know the general law, but I can't do it in a very "clean" way
i.e. the dirty way is expand: $\sum_y(\sum_z zp_{z|x,y}(z))p_y(y) = \sum_{y,z}zp_z|x(x)$
which is equal to $E[Z|X]$
 
Hey everyone!
 
Gonna pop the question here last time (as to not bother too much):
https://math.stackexchange.com/questions/2246784/convergence-of-linear-operator-in-l2
Functional analysis, if anyone is interested :) cheers!
 
Usually the solution looks something like: $E[E[Z|X=x,Y=y]|X=x] = E[g(Y)|X] =...$ but I can't figure it out from there
 
4:51 PM
There's a bounty and all
 
oh, I think I figured it out: $E[Z|X,Y]$ can be expressed as "given Y" first and then "given X", as in $E[E[Z|X]|Y]$ (sorry for the bad notation). Then $E[E[E[Z|X]|Y]|X] = E[E[Z|Y]|X]$ = $E[Z|X]$ by total expectation
 
Hi, Demonark. Are you still alive?
 
Hey @Ted!
Yeah the pain is quite a lot less than before
Thanks! ('-')/
 
Probably some ibuprofen would be a good idea.
 
Yeah, I agree
Hey @PVAL!
 
5:05 PM
Oh, has @PVAL snuck in?
 
Oh also today we started a bit of geometric measure theory in analysis
We proved the Vitali covering theorem and started Besicovitch
 
Demonark: A cool observation is that Sard's Theorem has a much stronger GMT statement (in terms of Hausdorff dimension of the set of critical values). I learned that from Federer's weighty tome, but actually had to use it in one of my early papers.
 
Huh, nice
We didn't actually do any proof of Sard's in difftop
 
Right. Nor do I ever do the proof. But I'm just telling you there's a powerful statement of it.
 
So it may be worth checking out
 
5:09 PM
hi chat
 
Hey @Semi!
@Ted Is it this one?
 
Federer has a stronger result. See his book or see this from Morgan (which I found by googling, since I no longer have Federer's book).
hi @Semiclassic
 
Heya @Alessandro
 
Hey @Alessandro!
@Ted this is pretty neat
 
5:20 PM
Yeah, studying the locus where smooth maps have rank $\le k$ turns out to be an important thing for some sorts of questions. Did you guys do the exercise in G&P about $m\times n$ matrices of rank $=r$?
(Proving that's a submanifold of codimension $(m-r)(n-r)$)
 
In difftop no, but Schlag assigned that exercise to us last quarter
 
Oh right, I remember hinting you for that.
 
Honestly that pset did a good amount of the work in getting me to take difftop this quarter, it was some of the most fun I had
 
odd kid
 
Oh hey @Mike!
And lol
 
5:22 PM
G'night @MikeM
 
hi chat
 
Yo @Eric!
 
my horoscope
 
Hey, Eric. Did you ever look at that conjugate points question I linked you to?
 
'While it’s true that life often imitates art, it’s odd that your life imitates J.G. Ballards’ avant-garde fiction piece “The Assassination Of JFK Considered As A Downhill Motor Race.” '
 
5:28 PM
oop I did not @Ted
lemme look
hmm neves relegated conjugate point stuff to exercises
and we havent defined cut locus and stuff
 
ah, I never usually did that stuff in class either (just not enough time), but I thought it was a very interesting question if you've thought about Jacobi fields.
 
Hey, sorry for just dropping in out of the blank and bothering ya but is it ok if I ask a [basic] question? I'm having real difficulty understanding something simple and it's frustrating me
 
We'll decide if it's something simple, @Phase.
 
I might think about it later, I have another midterm in an hour
 
LOL, sure, @Eric. Good luck with it.
 
5:30 PM
It is, it's just about partitions and equivalence classes, for some reason I feel like I understand the basics of partitions but can't wrap my head around equivalence classes
 
Go on.
 
thanks, I'm getting serious test fatigue, this is my third big test in as many days
 
gg @Eric, civ?
 
yup, though it's latin american history, so a subject im pretty well versed in @Daminark
 
@Mike I wonder which character you are in that story
 
5:32 PM
although the idea of a history midterm is still weird to me
 
Yeah, I've been hearing that civ is very different from hum and sosc, which I guess I expected in part? But not to the extent that it seems
 
For example, would the equivalence class $E$ representing the partition of a space as circles with certain radii, be $E = [x,y \in A \text{x} A \vert \text{(x,y) a certain distance from the origin}]$?
 
So the partition is into concentric circles centered at the origin?
 
And @Ted I have acquired Federer's book looks somewhat aggressive
 
yeah
 
5:33 PM
It's extremely rugged, Demonark.
 
Guys, I hope it's not rude that I'm asking a question for the second time, but I've rephrased it a little bit, hoping it's clearer this way;
Let $A\in\mathbb K^{k\times k}$. There exists $X\in GL_n(\mathbb K)$, such that $X^{-1}AX=R$ is upper triangular, and $X=QU$, where $U$ is upper triangular. Then
$$
Q^{-1}AQ=UX^{-1}AXU^{-1}=URU^{-1}.
$$
Is there a way to make this more intuitive in terms of basis transformations? It seems that $Q$ is a basis transformation from an orthonormal basis to the standard basis. What kind of transformation is $U$ tho? Is it from the basis that belongs to the t
 
So this is a partition of the plane, @Phase. So two points $x$ and $y$ in the plane are equivalent if they lie in the same element of the partition, i.e., $x\sim y$ iff $\|x\|=\|y\|$.
 
@TedShifrin Do you know if there's a name for the set of things below a given point in a poset?
(Or above, doesn't really matter)
 
@Daminark His writing style feels incoherent and extremely complicated to me
 
Maybe sub- or super-level, DogAteMy? I don't work with posets.
 
5:35 PM
it doesn't help that the subject is already super technical, the bad writing does it no favors imo
 
@Eric: It's highly technical and not great exposition, but there is some meaty, important mathematics in there.
 
yeah this is the impression ive gotten
 
@TedShifrin Oh ok thanks, I think I was misunderstanding it. The thing I'm not really sure about is the whole $x$ ~ $y$ is like an inductive process, would you see if two elements fit the relation, then use $y$ to test $z$ etc?
 
No, no, @Phase, too hard. You're just stating what criterion divides you into the partitioning...
Two points are on the same circle (centered at the origin) if they have the same length.
 
I was speaking with neves about these big existence and regularity black boxes used in geometry and his response was like "yeah one day you'll read federer and be very confused before eventually sorting it out"
 
5:37 PM
Oh ok, thanks! I'm gonna try proving the equality / disjoint thing now
I've just started reading Munkre's topology, and it's strange to me but cool. I haven't really covered sets properly before, I'm a first year Physics student so most of the stuff in the course is uh.. less than rigorous
 
@Sha If you're in an arbitrary field, you can't talk about orthonormality
 
@Sha: So $Q^{-1}AQ$ is the representation of the original linear map ($A$) with respect to the orthonormal basis (columns of $Q$). We're saying that if we look with respect to the basis given by the columns of $U$, then we get the matrix $R$.
 
@Daminark Should I state that the field is algebraically closed?
 
@TedShifrin yep thanks, proving it was as trivial as it was meant to be in the first place now, cheers for helping me understand it
 
No, you need it to be $\Bbb R$ or $\Bbb C$ or something where there's an inner product.
You're welcome, @Phase.
 
5:40 PM
Call the set of things below a given point in a poset the point's shadow. (For lack of a better word.) A decreasing sequence of points can be identified with a decreasing sequence of shadows.
Suppose our poset has a topology, such that each shadow is compact. Must the intersection of a decreasing sequence of shadows be the shadow of some point?
 
As Ted said
 
Speaking of shadows, there's a wonderful thing called the Rising Sun lemma (problem in Spivak). Have you seen it, DogAteMy?
 
It's clearly nonempty and downwards-closed.
@TedShifrin I have not.
 
@Ted hm ok, I think it just needs to sink it. thanks
 
@AkivaWeinberger isn't this called an ideal or a lower set?
 
5:43 PM
@AlessandroCodenotti I have no idea.
@TedShifrin (There is a house in New Orleans…?)
 
I think it can be called an initial segment even if the order is not total
 
DogAteMy: I quote (you can draw a picture). "Let $f$ be a continuous function on $\Bbb R$. $x$ is called a shadow point of $f$ if there is $y>x$ with $f(y)>f(x)$. Suppose that all points of $(a,b)$ are shadow points but that $a$ and $b$ are not. Clearly $f(a)\ge f(b)$.
(i) Suppose $f(a)>f(b)$. Show that the point where $f$ takes on its maximum on $[a,b]$ must be $a$. (ii) Show that this leads to a contradiction, so that $f(a)=f(b)$." Draw pictures! (The sun is rising at the horizon to the east.)
 
@AlessandroCodenotti Initial segment sounds more like "downwards-closed set" than what I want. Like, in $\Bbb R$, $(-\infty,0)$ is downwards closed, but it's not the "shadow" of any point.
 
I'll let Alessandro think about the posets. I'm trying to do a weird diff geo question on main.
 
@AkivaWeinberger I don't understand your definition of shadow then
 
5:51 PM
@AlessandroCodenotti Oh! I meant the set of things less than or equal to a given point.
 
That's an initial segment if your order is not strict then
(As orders usually are in the definition of a poset)
 
@AlessandroCodenotti So which of $(-\infty,0]$ and $(-\infty,0)$ is an initial segment?
By "shadow" I only mean the first one (the closed interval).
 
I'd call that a lower set because initial segment sounds like a total order if not a well order. I also think that both conventions can be found depending on the author, but usually $y$ is in the lower set it determines
 
Ah, OK.
 
See wiki for example, they include $x$ in the principal lower set it determines
 
5:59 PM
In any case, if lower sets are compact, is the decreasing intersection of lower sets a lower set?
(Autocorrect wants to write "power set")
 
Hm, intersections and unions of lower sets are always lower sets, but you're only interested in principal ones right?
Principal means of the form $\{y:y\le x\}$ for some $x$
 
@AlessandroCodenotti Right, yes
 
Hmmm, it still looks like compactness isn't needed?
 
@MikeMiller rofl where's that from
I want to read J G Ballard's High-rise at some point
 
G'day @Balarka
 
6:10 PM
@AlessandroCodenotti Without compactness it might be empty
 
Hey @Balarka!
 
Sounds like DogAteMy is looking like a poset version of the finite intersection property characterizing compactness.
 
Alternatively, consider $[0,\infty)\cup\{0'\}$ with $0$ and $0'$ incomparable.
 
hi @Ted, @Daminark
 
@AkivaWeinberger oh, right
 
6:11 PM
Hm, that actually provides a counterexample since the lower sets are compact… I guess I need Hausdorffness?
 
Hmm I'm not sure, but it's dinner time now, I'll think about it later
 
Buono appetito, @Alessandro.
 
@AkivaWeinberger $\Bbb R$ looks quite Hausdorff to me
 
LOL
It's always amusing when someone pulls a formula unknown to me out of a paper (which gives no explanation and an obtuse reference) and then the formula cannot be right. Ugh.
 
@AlessandroCodenotti Oh, I was thinking that the topology on $[0,\infty)\cup\{0'\}$ would have to be the line-with-two-origins thing.
 
6:14 PM
why do you say amusing and ugh simultaneously
 
But it doesn't, does it…
 
@Balarka: Surely you know me well enough to recognize sarcasm.
 
Ah ok
 
I love the line with two origins. Too bad I've not been following DogAteMy.
 
Oh, maybe, but you can just take $\Bbb R$ with the standard topology as a poset with the order topology and noncompact lower sets
Yet it is Hausdorff
 
6:16 PM
@TedShifrin Here's a question. Does every 1-manifold (not necessarily Hausdorff) appear as leaf space of a codimension 1 foliation?
 
@AlessandroCodenotti But I wanted compact lower sets…?
 
No, wait, I think I misread what you said. How are the lower sets compact in the line with $2$ origins?
(Is its topology even induced by an order?)
 
@Balarka: If it's not Hausdorff, then I don't call it a manifold.
 
Hmm, can that even be done with the line with two origins?
 
It's induced by the partial order, no? @AlessandroCodenotti
 
6:17 PM
I was answering to your "I guess I need Hausdorffness", I don't know if it's needed, but it's not enough
@AkivaWeinberger yes, I think so
 
$[0,a]\cup\{0'\}$ is compact in the line with two origins
 
@Ted What do you call it? I just mentioned the not necessarily Hausdorff condition!
 
As are $\{0\}$ and $\{0'\}$
 
In any case leaf spaces are most of the time not Hausdorff
 
They're not a lower sets though?
 
6:19 PM
@AlessandroCodenotti Yes they are…
 
@Balarka: I would guess the answer to your original question is no. When I wrote my final exam question for my two topology students, I was thinking of a certain non-Hausdorff leaf space and I had the topology wrong. Can you get a leaf space with a dense point?
Wait, let me make sure I'm not screwing it up again.
 
@TedShifrin Boundary leaves of the Reeb foliation on the annulus should be dense points on the leaf space.
 
I think I said it right. I want a point that's in every open set around every other point. (As opposed to every neighborhood of the point containing all the other points.)
Ugh, this confuzles me.
My exercise was a two-sided version of Reeb (graphs of tangent).
 
@AkivaWeinberger ah, your line doesn't include the negative numbers
 
@Alessandro: I thought you were dining.
 
6:23 PM
So every nonempty intersection of compact lower sets is a lower set in this case
 
No, wait. If $L, L'$ are the boundary leaves, then even if all the interior leaves accumulate to $L$ and $L'$ individually, $L'$ does not accumulate to $L$. So closure of $L$ is the whole leaf space minus the point corresponding to $L'$. Not dense.
In any case, take the 3d Reeb foliation. The boundary leaf is a torus (connected) so you're fine.
 
@AlessandroCodenotti But not principal lower sets, since $\bigcap_n[0,1/n]\cup\{0'\}=\{0,0'\}$ which isn't a principal lower set.
Which is why I said this is a counterexample.
 
Well, I wanted one where $L$ is in every neighborhood of every other point.
(So I had the answer to my own question wrong in my head.)
 
@TedShifrin The boundary torus leaf is in every neighborhood of every other leaf, right?
In 3d Reeb
 
6:27 PM
No, I don't think so.
 
I think you get counterexamples of this kind whenever you have an order which is dense "right above" a pair of incomparable elements
Ok, I'm really dining now, bye everyone
 
Take a neighborhood coming from the region between your given leaf and a preceding and succeeding leaf. :)
Hi @Mats.
 
@TedShifrin Hi. Thanks for pinging me.
 
Haven't seen you in ages!
 
@TedShifrin Are you still teaching math courses?
 
6:31 PM
Nope, Mats, retired. Although I have applied to teach a little at a school for really smart kids through high school next year.
 
@Ted I don't understand. The 2-plane leaves get arbitrarily close to the torus leaf.
 
Completely unrelated: Is every measurable set equivalent to an open set modulo measure zero stuff?
Like, for every measurable set, does there exist an open set such that the symmetric difference of the two has measure zero.
Hm, wait, fat Cantor?
 
@BalarkaSen Everything accumulates to the boundary leaf. The opposite is not true.
 
What if I change "open" to "closed"?
 
@MikeMiller Yes, I agree. Ted's saying the boundary leaf is not in every neighborhood of every other leaf.
 
6:34 PM
…Complement of fat Cantor in $[0,1]$?
 
I don't get how that and "everything accumulates to the boundary leaf" can be simultaneously true.
 
What if I change it to "open or closed"?
…Disjoint union of the two, as a subset of $[0,1]\cup[2,3]$.
Well, damn.
 
@Balarka: Take the region between two hyperbolas, each asymptotic to a line. That region is disjoint from the asymptote [not from a neighborhood of the asymptote, but so what?].
This is exactly what I messed up in my head with my final exam question.
My two students explained to me I was full of ****.
 
@Ted Oh. Well, that should be easy to fix by looking at the Reeb foliation on both solid torus in S^3 in the complement of T^2.
 
I don't see how.
 
6:38 PM
No, nevermind that.
Ok, I see. But I don't see how this has anything to do with the boundary torus in the Reeb foliation being or not being dense in the leaf space.
 
I asked a different topology. I asked to have a single point that was in every neighborhood of each of all the other points.
You're giving me a point every neighborhood of which ...
Sorry, DogAteMy ... we're preoccupied with a different conundrum. Do think about the Rising Sun Lemma. It gets used, surprisingly, in some measure theory stuff.
 
Hya
 
Salut, @Astyx.
 
Finally home
 
Before your expedition to Poitiers? :D
 
6:41 PM
yup :p
 
@Ted I see now.
 
This confuzled me quite a while when I was grading my topology final and I kept thinking the same thing you were, @Balarka.
So, anyhow, I'm giving you a topological space to try to construct a foliation for :P
 
It's actually just the point Mike raised. I apologize for the confusion.
Yeah, let's think.
 
No apology needed.
 
@TedShifrin (Re: rising sun) Hm, if I flip the Cantor staircase horizontally and replace each flat bit with a little parabolic "dip", it's almost a counterexample.
(It fails only on the Cantor set.)
@TedShifrin Maybe I look at the preimage of $\frac12(f(a)+f(b))$ and take the maximum
(It has a maximum since the preimage of a closed set is closed)
(And IVT says its preimage is nonempty)
It needs to be in the shadow of something, call that something "c"
But, between $c$ and $b$, IVT says it needs to hit something in the preimage of $\frac12(f(a)+f(b))$ again, which is a contradiction since we already have that $c$ is greater than the maximum value in that preimage.
 

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