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12:00 AM
heh, that's what I've been doing for quite a while now, @Clarinetist
I'm basically in a research position right now, but I have a bit of additional job security since I'm technically in the IT department, so if the research doesn't turn out as well, I can shift to more IT projects.
oh. Stick to research......
IT is for clowns
If I can keep up my research output in the long term, I'd like to, since I really like doing this work. However, I'm not a faculty member focusing on research, and I don't call all of the shots when it comes to how our research projects are run.

It's like I'm working as a research assistant, but I can run my own projects periodically, and I'm mainly there to provide data and stats support. I actually am about to submit my first article as first author. Fingers crossed. (It's in something completely unrelated to math/stats.)
If $\Bbb R^2$ is embedded on each face of a unit cube, and is the orthographic projection of a manifold $M$ inside the cube, then is $M=\Bbb R^3?$
what department do you do research in?
12:06 AM
I work for a medical school, not clinical trials or epidemiology (thankfully).
I'd probably be bombarded with more questions about COVID if I did. Lol.
very nice. OK, I didn't mean to be nosey..
No, I didn't get that impression at all.

My focus is primarily in medical education. The primary aim is to link educational initiatives to educational outcomes, and eventually workforce and patient outcomes.
I honestly love the work, because it uses a lot of my knowledge from my previous education jobs (in K-12 and higher ed institutional research), and it's helped me get an impression of what biostatistics work looks like.
Plus, I really appreciate how the field I'm focusing on is a natural intersection of things that really matter to people.
If $\Bbb R^2$ is embedded on each face of a unit cube, and $\Bbb R^2$ is the orthographic projection of $M,$ bounded by the cube, then is $M=\Bbb R^3?$
12:22 AM
If I knew anything about topology or geometry or whatever that question is from, I'd help you, but I don't
that's very generous of you :)
What time is it?
12:53 AM
I'm stuck a bit here:
Why $F(S)$ is that instead of just $F(a_1, \dots , a_r)$ ?
well is $\alpha_{r+1}$ algebraic over $F(\alpha_1, \ldots, \alpha_r)$?
brb, am trying to set up LaTeX.
@JoeShmo I know it should be algebraic over $F$, isn't it?
@BillyRubina you can use Overleaf instead if you don't want to manually set up LaTeX in your computer
@manooooh No, I was talking about the chat LaTeX.
it should be algebraic over $F(S)$, not $F$ (in general)
1:01 AM
@BillyRubina oh sorry, I didn't pay attention. I have an extension on Chrome called TeX All The Things, if it is worth
@JoeShmo I am bit confused because there is an inclusion in a previous part of the book:
It looks like those inclusions will keep working if we add more elements.
@manooooh I was trying to use this. My browswer is one of those troublesome browsers. :P
@BillyRubina lol, what explorer do you have?
@manooooh Opera.
Does using Opera still grants the title of "outcast"? If it doesn't, then I need to change.
What do you mean by "outcast"? Where does it appear?
@manooooh This I guess: "a person who has no place in their society or in a particular group, because the society or group refuses to accept them:"
1:16 AM
@BillyRubina sorry, I don't use Opera. If you have problems installing some extension you can contact Opera community
@manooooh No, it's not trouble with an extension. It's the MathJax thing.
I see, I thought it was easier for Opera users
MathJax should care about them
2:00 AM
@MikeMiller see the picture above. There is a symmetry $T$ (glide reflection) which preserves the pattern. Then take the quotient by $T(p)\sim p$.
2:23 AM
apperntly caffine has a ridiculous effect on my heart rate
@anakhro I'd really like a formula. If the map is badly behaved, and has strange orbits, maybe the quotient space is awful?
2:57 AM
@MikeM I guess I'm not sure what a glide reflection on the sphere is. What's a translation?
There are no fixed-point-free orientation-preserving isometries.
@anakhro I should have tagged you :)
@TedShifrin How would you explain the intuition behind choosing the length of the neighborhood $\delta$ so as to show that
$$\bigcup_{0 < |h| < 1/n}\left\{x \in \mathbb{R}:\dfrac{f(x+h)-f(x)}{h} > c + \dfrac{1}{k}\right\}$$
is an open set?
(assume $n, k \in \mathbb{N}$)
Oh, and $f$ is some continuous function $\mathbb{R} \to \mathbb{R}$
3:16 AM
Where is $\delta$? And what is $c$?
I guess it's obviously open, regardless, just by continuity.
@TedShifrin So, I am trying to show that $(x - \delta, x + \delta)$ is contained in there so as to show that it's an open set, but I'm having difficulty trying to choose an appropriate $\delta$. $c$ is just a fixed real number.
Don't do that. Just think preimage of an open set under a continuous function. Fix $h$..
@TedShifrin So the preimage of an open set is open under a continuous function. If I fix $h$, this would mean that.. I'll have to think about this some more.
And the union of opens is open.
Ah, hence you can just fix $h$. Thanks for pointing that out.
3:30 AM
Is it really just this?...
Fix $0 < |h| < 1/n$. Define $g(x) = \dfrac{f(x+h)-f(x)}{h}$. $g$ is continuous over $(c + 1/k, \infty)$... oh wait, not quite there yet
That inequality after the $:$ is giving me a mental block for some reason.
No, you're taking the preimage of that interval.
@TedShifrin Right, so the preimage... oh. Is it just the fact that $(c + 1/k, \infty)$ is an open set, and because $g$ is continuous, its preimage and hence one of the sets in that union is thus open?
3:38 AM
Great. Thanks.

I keep telling myself I will take a topology course some day.
So yet another thing I'm going to have to learn on the fly... $\lim\sup$ and $\lim\inf$ for functions. Why didn't my university use Rudin? I think they're still not using Rudin.
Rudin's not a good book for beginners
Yes, it's good for people with strong background who are aiming for grad school. That’s a tiny percentage of math majors.
3:48 AM
So what, do students in that boat take an additional 1-2 semesters of intro analysis before Rudin?
I'm genuinely curious.
I don't understand where books like Understanding Analysis by Abbott fall in a math curriculum that offers classes on Rudin's text.
4:02 AM
@TedShifrin Morning!
Find some work at the local college doing a tutorial and tutoring math, was surprised they were willing to hire me; considering i dont actually have a degree.
At least it gives me something to do
4:41 AM
@MikeMiller Ah
4:58 AM
morning? @Faust
hi, a @Balarka
Hi @Ted
@Clarinet: Most schools that teach Abbott aren't going to try to use Rudin. But the very top, hardest schools in the US still use Rudin (but often offer other sections that use easier texts — e.g., MIT, Berkeley ...).
hi @Alexandru
Wow, a Romanian in the chatroom :)
(or am I mistaken?)
haha why is that special?
We have all sorts of Europeans, Indians galore, but you're the first person I recognize ... of course, when people use names like userxxyyyzzz002385 it's hard to know.
5:01 AM
So what brings you by?
oh wanted to see what the chat is about
I need to prepare a question soon for MSE related to my research
LOL, at the moment, apparently not much.
What area is your research?
non linear spline interpolation i.e. using an older definition for the term spline here, a piecewise analytic function
@Balarka: Did you ever think about my $H^1(M,\mathscr O)\otimes H^0(M,\Omega^1) \to H^{1,1}(M)$ remark for the torus?
Ah, well, there are plenty of numerical folks who answer on main, not so many in this chat.
5:04 AM
I am working on building G^2 smooth spline interpolation algorithms using clothoid segments on the plane and spherical clothoid segments on the sphere and R^3
In particular, I myself am woefully ignorant thereupon.
What are clothoid segments?
great question, curves with curvature a linear function of arc length
Oh, I knew that in the geometry context ... just didn't expect it to show up in numerical. That's a bizarre numerical setting.
yes that begs the question why use clothoids?
Yes, why?
They certainly are more difficult than piecewise linear.
5:06 AM
potentially better curves for path planning purposes for robotics
Ah, I see. Interesting. Although why should a robot want to follow clothoids? Is there some natural ODE reason?
they better approximate mechanical splines i.e. bending a stick so it passes through knots
Bizarre. OK.
yes there is a physics reason that is connected to how Euler first considered the Euler spiral as a rough approximation to the elastica problem
yes quite weird
That connection (Euler) I've never seen.
5:08 AM
another name for the clothoid is the Euler spiral
@TedShifrin Thanks for the info on that.
also Cornu spiral
Oh, I thought the Euler spiral was effectively $r=e^\theta$.
I guess I haven't computed the curvature of that. Shame on me.
I should quickly add an exercise to my differential geometry text. :D
ah I have an idea for a differential geometry undergrad textbook exercise
derivation for the argand diagram version of the euler spiral starting from the curvature equation
and from there applying euler's most famous formula to get the cartesian form
@TedShifrin I must admit this stuff is confusing for me so I try not to think about it. I think the right pairing is $H^1(M, \mathscr{O}) \times H^0(M, \Omega^1) \to H^1(M, \Omega)$, given by $(\{f_{ij}\}, \omega) \to \{f_{ij} \omega\}$, and then composing with the "trace map" $H^1(M, \Omega) \to \Bbb C$.
This is a nondegenerate pairing, part of the statement of Serre duality, I believe. So $H^1(M, \mathscr{O}) \cong \Omega^1(M)^*$ for Riemann surfaces.
5:11 AM
btw clothoids are always used in railways and highways as transition curves
Right, @Balarka, but on the torus we can write down obvious generators for two of em.
@TedShifrin One question for you: so $$\bigcup_{0 < |h| < \frac{1}{n}}\left\{x \in \mathbb{R}: \dfrac{f(x+h)-f(x)}{h} > c + \dfrac{1}{k} \right\}$$ is open.

Let's consider the upper derivative $\lim\sup_{h \to 0}\dfrac{f(x+h)-f(x)}{h} = f_u^{\prime}(x)$. I'm trying to figure out how to turn that open set into $\{x \in \mathbb{R}: f^{\prime}_u(x) > c\}$. I suspect that what it entails is throwing $\cap_{h=1}^{\infty}$ on that set, and then taking the $\sup$. But I'm not 100% sure with this.
@TedShifrin It's always morning somewhere
Really interesting. Do they patch more smoothly than you'd expect, @Alexandru?
@TedShifrin Well yes if you believe this then $dz$ is your generator for $\Omega^1(M)$ for $M = \Bbb C/\Lambda$
(cc @Lelouch)
5:13 AM
You want to intersect over $k$, @Clarinet, no?
they are the best tool for smooth line-circle or circle-circle transitions
@TedShifrin Oh yes
Of course
Right, @Balarka, and $dz\wedge d\bar z$ is (a scalar times) the Kähler form.
I wrote all this when you guys were talking, but I think you all ignored me.
Does that sound right @Ted, with the intersection over $k$?
now my research is more about fitting a nice curve to a general sequence of points
5:14 AM
I never remember Serre duality, Ted. Total mystery to me that $H^1(M, \mathscr{O})$ should be $\Omega^1(M)^*$
Yes. Of course that's no longer open, @Clarinet.
So I only remembered after you asked me
I like my sneak proof of $H^1(M, \mathscr{O}) \cong \Bbb C$ for the torus
@Balarka: Think about $H^{1,0}$ and $H^{0,1}$.
I'd rather not!
Well, be that way, then.
One is the conjugate of the other (with compact Kähler, which we have).
So the question is what $d\bar z$ corresponds to in Lelouch's view.
5:16 AM
@TedShifrin there is another advantage over polynomial splines, the intrinsic equations are "free" and that makes velocity planning over a clothoid spline simpler (seems paradoxical)
I have no intuition for this stuff at all, @Alexandru, but it's interesting.
Welp, im all done welding my laptop back together. now the question is do i do some more math or just goto bed?
@TedShifrin i am doing something more "pure" too, I have new expressions for a new family of sphereical curves with k_g = alpha *s
Here's a curvature/plane curves question for you that I made up years ago for my course, @Alexandru. If I tell you the path of the rear wheel of a bicycle, find the (possible) path(s) of the front wheel.
You get a differential equation that almost never can be solved by hand.
it reminds me a little bit of pursuit curves?
is it somehow related?
5:18 AM
Yes, it does.
@TedShifrin Antiholomorphic guys don't pair with holomorphic cocycles, so I am not sure if that has an interpretation
I could be wrong
It's a matter of chasing the Dolbeault isomorphism $H^1(\mathscr O) \cong H^{0,1}$, a @Balarka.
I don't understand but I believe it
$d\bar z$ of course sits in $H^{0,1}$.
I actually like this question. I've assigned stuff on $\Bbb P^1$ before, but I don't remember assigning this on $g=1$.
I should look back at my complex manifolds homeworks from 1980 :P
Might be a good exercise to construct the explicit holomorphic cocycle.
The forms chase is too obscuring
5:21 AM
That's how you actually understand the curvature interpretation of $c_1$ ... similar chase, Balarka.
The issue is that there is no "canonical" open covering to use.
The "obvious" one with just two sets probably won't be acyclic.
I'm afraid we might need to use four.
There's one with three which is acyclic.
How so?
I'm thinking of the square picture.
I don't want any set retracting to a circle.
$g+1$ is the Lusternik-Schnirelmann category of the surface of genus $g$. For the torus, throw out a meridian and a longitude, and you get one open set $U$. Do it for another essentially distinct pair of meridian and longitude, you get a second open set $V$. $U$ and $V$ cover all of the torus except two points, so you take another chart joining the two.
@TedShifrin about the bicycle thing, real life trajectory related questions get very complicated fast
There might be an explicit way to do this by embedding the torus in $\Bbb{CP}^2$ and using the affine opens.
I have not thought about this.
5:25 AM
@Alexandru Yes of course. :)
@Balarka, since we're working with $\Bbb C/\Lambda$, I'd rather not.
I don't think that'll be acyclic ... just to guess.
@TedShifrin introduce kinematic constraints and it gets really nuts
@TedShifrin What I described is, but maybe the CP^2 thing is not.
Well, that problem I gave you is a nice elementary differential geometry problem to set up, but unless there's a clever approach different from mine, it's hard to solve except in obvious cases (line, circle).
Well, it is, right? The affine opens are Stein, they have no cohomology
It's a fine acyclic cover.
I was meaning the $\Bbb P^2$ when I said that.
Anyway, I want to stick with the square.
5:27 AM
@TedShifrin this might be more physics than math but I know a lot of basic questions about bicycle stabiity are not known
Yes, that's definitely more physics. And I remember that stuff from my fabulous math-y freshman physics course at MIT.
@Balarka: I'll try to think about this question more tomorrow.
Yeah, I dunno. Maybe the holomorphic cocycle will be some terrible Weierstrass $\wp$ thing.
Not going that route.
I'd like to know the answer, let me know if you get something
Of course.
5:29 AM
@TedShifrin I will need help soon coming up with some asymptotic expansions for some 2d hypergeometric function special forms, that's the question I need to write soon
Yes, I know nothing about that stuff, @Alexandru, but there are surely people who do on the main site.
@TedShifrin do you know splines were discovered by a Romanian mathematician around 1940?
Isaac Jacob Schoenberg
very strange what a modern discovery splines are
Hmm, never heard of him, and would never have guessed Romanian :)
At my former university there was a big expert on splines, and he was doing stuff on spheres and other manifolds.
5:33 AM
me neither!
which one?
University or person?
M-J. Lai at University of Georgia
interesting page he has!
the second picture on his site is with Schumaker at Vanderbilt, a collaborator's post-doc mentor
@TedShifrin Hey
5:37 AM
Yes, famous guy.
Yes, a @Balarka?
Cech out these cocycles I have
Nvm bad joke
I'm actually writing out the Dolbeault iso now.
Since I'm old and rusty.
You're old and rusty but you remember the story much better than me
I forget it every time I read it
I wish it weren't so hard to get the Čech
Better to teach things than to read them.
Haha that's true
5:43 AM
So it's not hard to track through. If I have a $\bar\partial$-closed $1$-form $\phi$, on my $U_i$ I can write it as $\bar\partial g_i$. I look at the $1$-cocycle $f_{ij} = g_i-g_j$ and it is actually holomorphic. That's the isomorphism.
So we need to do this with your cover and my $\phi=d\bar z$.
Oh I see.
Are you just saying that? :D
No I had completely forgotten that was the isomorphism. Makes a lot of sense.
So it means solving the $\bar\partial$ problem on these contractible open sets (to get the $g_i$).
But this should be do-able with these explicit sets. As I said, I'll ponder on the morrow.
It's weird, because my form $d\bar z$ is already exact on most open subsets.
But something has to happen with your third set.
yeah that'd make sense
5:48 AM
To be continued if I get somewhere. Done for tonight!
Thanks! Good night
Night :)
6:16 AM
Is there any succinct way to say, in a directed graph, "all vertices of the same level at most 2N edges apart" and all vertices of the same level at exactly 2N edges apart" succinctly? I'm trying to think of good function names for siblings/cousins/siblings+cousins when working with a directed graph (tree/dag)
6:31 AM
@TedShifrin @Lelouch Take any noncohomologous cocycle in $H^1(M, \Bbb Z)$. This is a lattice in $H^1(M, \mathscr{O})$, so generates it.
The point is you just forget about the holomorphic problem and produce a locally constant cocycle
There's no analysis here, just pure topology.
I asked my roommate and this is what he said
Should have been obvious to us
@BalarkaSen What is the "trace map" $H^1(M, \Omega) \rightarrow \mathbb{C}$ ?
@BalarkaSen And what do you mean by a "lattice" in $H^1(M, O)$ ?
6:46 AM
@Lelouch You know from my computation that it's a $\Bbb Z^2$ sitting in $\Bbb C$
In fact $\text{Pic}_0(M) \cong M$, so it's the same lattice that the torus $M$ is a quotient of $\Bbb C$ by
@Lelouch $H^1(M, \Omega)$ is smooth $2$-forms on $M$ modulo $d$ of $(1, 0)$-forms on $M$. So take a smooth $2$-form representative $\omega$, and consider $\iint_M \omega$
This is the map $H^1(M, \Omega) \to \Bbb C$
@BalarkaSen ah ok, right. Yeah so $dz$ is a generator for $\Omega^1(M)$, I see. Thanks
That's not what I claimed
Who knows what $H^1(M, \Omega)$ is
I mean $\Omega^1(M)$, sorry
we should read this as time pass
@Balarka: I had arrived at this from what I told you. All my $g_i$ were locally constants + $\bar z$, so the $1$-cocycle is just locally constant.
7:00 AM
Makes sense
$H^1(M,\Omega^1) = H^{1,1}(M)$, to answer your question?
Anyhow, now it really is bedtime for this bonzo. Night!
@BalarkaSen sure
@Lelouch Did you want to say something elsewhere? You invited me to a room.
Or was it a mistake
@BalarkaSen no lol, I didn't invite you any room
If I did, it was a mistake
I clicked on the artcile from my phone so maybe something happened then
5 hours later…
11:46 AM
If $\Bbb R^2$ is immersed on each face of a unit cube, and $\Bbb R^2$ is the orthographic projection of $M,$ bounded by the cube, then is $M=\Bbb R^3?$ The orthographic projection of $M$ from a face to its opposite is $\Bbb R^2.$
1 hour later…
12:51 PM
Assume that no point of $K_1$ belongs to every $K_{\alpha}$
what does this mean?
I am quite confused by this statement
I don't understand that why
Belongs to every $K_{\alpha}$" means "belongs to the intersection of the $K_{\alpha}$
@Alessandro @EdwardEvans youtube.com/watch?v=jvfJEj5Ao7M
ok I understand it now lol
"In set theory, an inhabited set is a set that contains an element, i.e. a set X such that ∃x,x∈X is true."
@Astyx Nice
1:08 PM
**Theorem**. If $ \{K_\alpha\} $ is a collection of compact subsets of a metric space $X$ such that the intersection of every finite subcollection of $\{K_\alpha\} $ is nonempty, then $\cap K_\alpha$ is nonempty.

*Proof*: Fix a member $K_1$ of $\{K_\alpha\}$ and put $G_\alpha=K_\alpha^c$. **Assume that no point of $K_1$ belongs to every $K_\alpha$.** Then the sets $G_\alpha$ form an open cover of $K_1$; and since $K_1$ is compact, there are finitely many indices $\alpha_1,…,\alpha_n$ such that $K_1\subset \cup_{i=1...n} G_{\alpha_i}$. But this means that $K_1\cap K_{\alpha_1}\cap\dots \ca
wierd what did it contradict
It didn't contradict anything
it said no point of $K_1$ belongs to intersection of $K_\alpha$
It contradicts the hypothesis that the intersection of every finite subcollection of $\{K_{\alpha}\}$ is nonempty
but it showed it turns out to be true
@Thorgott but it said in second last statement that finite subcollection is empty
yes, that's the contradiction
because the hypothesis said it should be non-empty
1:11 PM
I don't get it
My hypothesis is that A is true. I assume B and see that B implies not A. I conclude that B is false.
assume no point of K_1 belongs to every K_alpha means their intersection is empty no?
can you tell me which line has this pattern "I assume B and see that B implies not A"?
@BalarkaSen looool
1:15 PM
where in hypothesis did it said should be non-empty?
B="the intersection of $\{K_{\alpha}\}$ is empty". This is assumed in the first line of the proof ("Assume...").
Then, from this, the proof derives not A, where A="Every finite subcollection of $\{K_{\alpha}\}$ has non-empty intersection". This is the last line ("But this means..").
@EdwardEvans seven dogs are howling at the river
atmospheric bark metal
"such that the intersection of every finite subcollection of $\{K_{\alpha}\}$ is nonempty"
Huskies in the throne room
1:16 PM
@Thorgott XD now I understand
why rudin try so hard to be tricky
thanks bruh
Rudin usually aims for the most elegant proof, not necessarily the most accessible one
I think in order to understand rudin you must be very good at proof.
Ted was right
1:42 PM
I'm trying to show that $\Bbb{Z}_m$ is an injective $\Bbb{Z}_m$-module using Baer's criterion. Every ideal in $\Bbb{Z}_m$ is of the form $\overline{d} \Bbb{Z}_m$, where $d$ divides $m$. Let $f : \overline{d} \Bbb{Z}_m \to \Bbb{Z}_m$ be some morphism. I believe I want to find $\varphi : \Bbb{Z}_m \to \Bbb{Z}_m$ such that $\varphi \iota = f$, where $\iota : \overline{d} \Bbb{Z}_m \to \Bbb{Z}_m$ is the inclusion map.
I was able to show that $Im ~f \subseteq \overline{d} \Bbb{Z}_m$.
Initially, I thought $\varphi = id$ might work, but $f$ could move elements in the ideal around.
if $C \subset \mathbb{C}^2$ is an affine plane curve defined as the zero set of a non-constant polynomial in $\mathbb{C}[x,y]$ and $\overline{C}$ is its projectivisation, defined by the homogenization of the polynomial defining $C$ (but not dependent on that polynomial, by the nullstellensatz), is there an accessible way of showing that the projectivisation $\overline{C}$ is in fact the closure of $C$ ?
by closure of $C$ of course we mean as a curve embedded in $\mathbb{P}^2_{\mathbb{C}}$ via the open embedding $\phi_2 : \mathbb{C}^2 \rightarrow \mathbb{P}^2__{\mathbb{C}}$ say, where $\phi_2(x,y) = [x,y,1]$? Without too heavy knowledge of multiple complex variables? The standard proof is apparently via the Weierstrass preparation theorem which is a bit beyond me as i'm pretty rusty with complex analysis beyond the fundamentals
@Thorgott wait a seecccc... You are hypothesizeing A is true but it might be false. You assume B and see B implies not A then conclude B is false? We don't know about A so you can't conclude B is false
* $\phi_2 : \mathbb{C}^2 \rightarrow \mathbb{P}^2_{\mathbb{C}} $, sorry
@Stupidquestioninc we assumed $A$ in the statement of the theorem
Anyone have any hints on how to define $\varphi$?
1:46 PM
umm ok I will try to think more clearly
@user193319 prove that f(d) is a multiple of d
Yeah, that's what I did, and then used that to conclude $Im ~ f \subseteq d \Bbb{Z}_m$.
@Stupidquestioninc it contradicted finite intersection property, since a finite intersection of that subcollection has nonempty intersection, but you just showed that a finite intersection was empty
then just let phi(1) = f(d)/d
Should I define $\varphi (1)$ to be equal to that multiple?
Oh, I see...
1:49 PM
@Stupidquestioninc its actually a characterization of compactness of any topological space, $X$, to say that any collection of closed subsets with finite intersection property (such that any finite subcollection has nonempty intersection) has nonempty intersection
@Stupidquestioninc this characterization was famously used to show tychonoffs theorem, for instance, which shows that the arbitrary product of compact spaces is compact
makes sense will try to digest it
(Topology is a fun topic)
@porridgemathematics I'm surprised that this is hard. It's clear that the closure of $C$ is contained in the projectivization of $C$. The projectivization is obtained from adding points at infinity $\Bbb{CP}^1_\infty \subset \Bbb{CP}^2$ of $C$, so removing any of those from the projectivization of $C$ will make the resulting object non-closed, no?
It seems almost tautological. If you write down the formula for projectivization it should be direct that the added points are limit points of $C$ in $\Bbb{CP}^2$
i think the heart of the proof will be proving the points of infinity are limit points of $\phi_2(C)$
I am not sure how Weierstrass preparation theorem is relevant. We're not working with analytic varieties, just algebraic varieties.
Oh, but maybe you want to prepare anyway to write it as a polynomial in one variable with coefficients in polynomials on the other variable.
Yeah, that makes sense.
1:54 PM
*points at infinity, sorry
@LeakyNun @porridgemathematics @Thorgott Thanks I was blind to not look at hypothesis which theorem stated
btw now I understand the proof i want to know how mathematician observed this theorem
is there any intutive way to see it? I mean like drawing it or like some special examples?
2:10 PM
It's essentially just the definition of compactness dualized by taking complements
"every open cover has a finite subcover" =take complements=> "every collection of closed subsets with empty intersection has a finite subcollection with empty intersection"
and taking complements again reverses the implication
granted, the theorem as stated in Rudin is the contraposition of that latter statement
You can also observe that this must hold by phrasing everything in terms of filter, since families with the fip are exactly those that can be extended to a filter
oh well, and Rudin does everything in an ambient space
but I really don't like that Rudin does things in an ambient space
it's one of the biggest pedagogical errors in the book imo
2:25 PM
The biggest two being not using filters for everything, and using metric spaces instead of uniform ones, clearly
mutants man
Uniform spaces are actually very reasonable
I keep having to deal with them lately and they are not too bad
and that he never mentions infinite-dimensional manifolds in the text
also thought it was kinda weird he didn't include a proof of Fermat's last theorem
@Thorgott lol those also came up recently in some stuff I was reading
yet I'm the crazy one when I ask a basic category question smh
2:34 PM
infinite dimensional manifolds are fine as long as they are Banach
Good afternoon people, sorry to interrupt. Can someone help me with this question?

I always thought **cylindrical waves** were **3D** waves, but recently I came to know they share the same properties with circular waves (which are **2D**) because they are basically the same thing, only the cylindrical wave is “trivially extended” into the **third dimension**.

Is it correct to say that? What topics of math cover this concept?
granted, I've been spending the last like hour dealing with set-theoretical issues in category theory and am starting to feel crazy
hi! anyone here good with special functions?
@Thorgott What kind of issues
@Thorgott "does any of the stuff I do actually hold??"
2:45 PM
just trying to figure out how to make the "2-category of all (in a certain sense smaller) categories" precise with the help of two nested Grothendieck universes
That does that doesnt sound too bad
thorgott wants to validate the kind of math he does but in the process of doing so he does the kind of math that is invalid
it shouldn't be too bad, but I'm kinda dumb
I've thought about stuff in terms of classes till now, but apparently I have to make the concession that Grothendieck universes are better
youre not good enough for category theory khorgokk
switch to topology
2:47 PM
Can anyone help me with my question? Even a simple answer is gonna help.
@Thorgott call elements on the universe sets and subsets of the universe classes and it's the same thing
@Thorgott as long as you do the usual zfc operations you shouldn't reach size limits in a grothendieck universe
god why is everyone a mutant
I did some "topology" recently, if classifying the families of topological manifolds that admit a product in the category of topological manifolds counts
Okay, don't consider a space of functions whose images have unbounded ranks
But that's, like, hard to do
2:51 PM
@Alessandro I thought so, but it's apparently not. Say you have two universes $ U\in V$. If $\mathcal{C},\mathcal{D}$ are $U$-categories, the functor category $\mathcal{D}^{\mathcal{C}}$ is a $V$-category. But if you do it in terms of classes, then the functor category cannot be constructed unless $\mathcal{C}$ is small. Or so I heard.
I think you misinterpreted what alessandro meant
He doesn't mean actual classes w.r.t. the "actual" universe
Why is hyperbolic distribution named so?
is there a conventional definition of {0,1}^0 ?
But that when you have a universes, and want to work in it, then instead of thinking about all the fuss with classes defined by this and that formula, not being actual elements of the whole universe etc
You can just consider sets to be the elements of the grothendieck universe, and for all practical purposes classes are subsets of that grothendieck uni🅱erse
uni🅱erse hahaha
2:56 PM
ah yeah, this is what is meant by saying that a universe gives a "model" of NBG, yes? (tho I don't know what a model is, but I don't wanna know either)
I'm not sure tbh
But it is indeed the case that with a universe, you can treat things that normally would be classes as normal objects
Quantify over 'em and such
But only w.r.t. the grothendieck universe of course
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