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12:00 AM
@AttractorNotStrangeAtAll No.
 
geometrically evident is usually a much more fuzzy notion than one would like to believe anyhow
 
And I can lecture very fast.
Oh, Thor makes a good point. 2 hours a day 5 days a week, it's feasible.
 
We have classes of 6 hours a week, 3 days 2 hours each.
 
that's doable
 
@user2103480 that would be problematic.
My homology lecture for the high school students was a bust because it was too much information for one lecture.
 
12:04 AM
I believe the analysis 1 lecture I took covered the same topics in a similar timeframe
 
@Thorgott I checked, mine too
 
Oh, so 21 hours is indeed a lot of time. Can do a lot!
 
about 4 weeks for differentiation and integration, 4 hours of lectures per week. but we did 10 weeks on real & complex numbers, sequences, series and continuity before
 
Differentiation in one variable has very little to do.
 
we even covered power series on top of that in this timeframe, iirc
 
12:06 AM
@MikeMiller but random walks in sober spaces are far less interesting ;-)
 
What's the usual convention for conjugation in S_n? Is it $ghg^{-1}$ because composition is written backwards, or do they not usually change it?
 
That's how I usually conjugate.
 
you can probably find either convention in some places
 
@Thorgott we did that before starting continuity. But only pointwise. We did uniform continuity then afterwards in the first 1, 2 weeks of analysis 2
Then the lecturer went into point-set topology and alienated basically everyone lmao. I found it great though
 
Thanks, guys. I asked because my colleagues are having a discussion in our group about the conduction of the classes.
 
12:09 AM
we only did metric and normed spaces at the start of analysis 2
 
@AttractorNotStrangeAtAll any resistance to that?
 
The discussion is that the professor spent 2 months of class talking about
introductory things like set theory and induction, when he could have started with real numbers.
 
Professors are human .... They do lots of stooopid things.
Not that I would know.
 
At the same time some people had never seen those topics before, so it was good for them as it was a gentle introduction to analysis
 
I remember, in analysis 1, we introduced the real numbers axiomatically and then defined the naturals within the reals as smallest inductive set to obtain the induction principle
 
12:13 AM
@TedShifrin: did you see the falling ladders I posted?
 
silly from a foundational pov, but surprisingly effective
 
The truth is that it is impossible to please everyone, and as the professor himself said, you can advance the content of the discipline independently, with the textbook itself
 
Real analysis is tough. Charles Pugh liked to describe it as complex analysis' evil twin.
 
Yes, @robjohn. I can write animations too! :)
 
okay, no more animations.
 
12:14 AM
@Thorgott our analysis 2 and 3 were hell. The exams were both really easy, but the topics were harsh
 
LOL, they're good for other folk here.
I animated my bicycle path question years ago. Given the path of the rear wheel, plot the front wheel too.
 
How do you do the animations?
 
Mathematica
 
Ahh. I'm to cheap too spend the $ on it.
 
I finally did and now they want more to run on the latest Mac OS.
 
12:18 AM
I used to love tech, but find the constant marginal changes (& charges) tiresome.
 
Agreed.
 
Age, I suppose (in my case).
 
I'm older.
 
@Thorgott in one week, we were introduced to metric, normed, banach and hilbert spaces, and in the next one there was a short excursion into banach algebras, and then general topological spaces were introduced with metric spaces as first examples
 
I think the analysis 1 exam I took in my first semester was probably the harshest exam I've ever written. Like 7 exercises, none of which were standard and all of which required a clever trick to solve that could potentially be really hard to spot for inexperienced first semesters. It seems harmless looking back to it, but it was hella scary back then.
 
12:19 AM
Oh that's harsh
It was all standard in our first exams, with old exams and practice exams at hand
 
It taught you to be a star in here, @Thor.
 
The problem with a lot of real analysis exams is that some questions have a trick and if you have seen it before it is straightforward, otherwise it depends on how creative you can be.
 
Undergrad real analysis isn’t as crazy as graduate.
 
Difficult problems are unironically great to see (in hindsight)
 
lol I wish
there was some weird integral on the exam I had no clue how to solve and I just randomly tried the $f=1\cdot f$ trick and partially integrated wrt $1$, it happened to work out and I felt like the luckiest person alive
 
12:22 AM
I like/hate analysis. It is relatively easy to ask a simple question or minor variation that is very difficult.
 
To be sincere, I like to see some of you saying that intro analysis was not easy peasy. The feeling I get is that most professors/researchers would say that intro analysis was actually very easy (so in order for me to be a researcher I would have to find it easy as well)
 
@Thorgott in a probability exam that was nothing like the exercise sheets, stackexchange saved me lmao
the prof made the exam, the assistant the sheets, that was the disparity
 
oh lol
 
and the prof had like 3 exam problems for maxima of random variables, which seem impossible to compute if you don't know the trick - and I knew it by reading SE solutions
 
Probability can be difficult in the same way as real analysis. Look up Erdos and the Monte Hall problem.
 
12:25 AM
@TedShifrin what is graduate analysis though? we don't have such a clear term here
 
Lebesgue theory, measure theory, $L^p$
 
Let me share with you one of the worst homework exercises I ever had to do (analysis 1):
"Show by differentiation that the following function is constant and determine its constant value: $2\arctan\left(\frac{1}{2}\tan\frac{x}{2}\right)-\arccos\frac{3+5\cos x}{5+3\cos x}$"
 
That's calculus, not analysis .
Meh.
 
the lecture was called analysis, what can I do
(it also had a lot of good exercises, but that one was horrible)
I hate tedious computations
 
For example, a rectifiable curve has a Lipschitz parametrisation hence differentiable ae. But does it always have a differentiable parameterisation? If not, give an example. I have struggled unsuccessfully with this for years.
 
12:30 AM
Our worst exercise was about the cauchy definition of the reals
(pdf)
 
I can't speak my native language, German is outside my realm :-(.
 
yeah, that's technical, but at least it's useful
 
we had to prove existence of the multiplicative inverse, which sounds super easy and is easy if you sweep a few obvious things under the rug, but I was still new to uni and proved that the "adapted" algorithm (additive inverse-> multiplicative inverse) converged via some epsilon.delta
 
@user2103480 they once made us prove CLT for poisson variables by hand
 
because I thought we had to prove everything in super detail
took like 2 pages
@Thorgott rip
if we share bad exercises
we had to prove that submanifolds of R^n have lebesgue measure 0 in analysis III
and weren't even given the fact that uncountable covers have countable subcovers (I think the proof needed something like that)
 
12:35 AM
that's fairly easy, though, isn't it?
 
@user2103480 That is not too bad? Just as a strict subspace has measure zero.
 
cover with countably many precompact carts, each of which is the image of a measure zero set under a Lipschitz ($C^1$ on compactum) mapping
oh, if you don't have Lindelöf...
yeah, that's more awkward
but you can just take coordinate charts around all rational points
so using separability instead
 
I think that solution was also pretty close to what I wrote
 
yeah ok, if you also had to prove that differentiable maps map zero sets to zero sets, that's a good amount of effort
@user2103480 btw, same probability lecture also had us prove the continuous mapping theorem, Slutsky's theorem and parts of the portmanteau theorem as homework
which is funny, because these theorems didn't even come up in that (undergraduate) lecture, but only came up in the graduate probability lectures
 
these are all not that bad if one is used to the concepts
But the whole problem with probability lectures is the bombardement with different types of convergence so I would have had serious problems
 
12:46 AM
I am not sure I see the value in proof regurgitation in an exam.
 
@Thorgott they came up in our undergrad probability lectures but there's not much motivation so students dont get the point
 
I remember I struggled really hard to prove that $X_n\rightarrow X$ in distribution iff $\mathbb{E}[f(X_n)]\rightarrow\mathbb{E}[f(X)]$ for every bounded Lipschitz function $f$
 
@copper.hat yeh, but tbh oral exams mostly aren't more than proof regurgitation
@Thorgott tbf, that was our definition
 
I guess I would rather than someone knows the idea rather than the gory details.
 
our definition was pointwise convergence of the cdfs at every continuity point
 
12:49 AM
Far too few folks focus on motivation which is essential for understanding in my opinion.
 
just looked it up and that + Slutsky were like 1.6 pages on my hand-in
we did probability before measure theory, which is horrible anyhow
 
Uurg.
 
@Thorgott yeah same
but we had one good prof which didnt continue rushing through martingales after that
he started by 0 again, with the construction of the lebesgue integral and caretheodory's extension theorem etc
 
man, I spent all this time learning differential forms and I still have no clue what physicists are doing when they play with differentials. what on earth is "$d\vec{r}=\frac{d\vec{r}(t)}{dt}dt=\vec{v}(t)dt=d\vec{r}(t)$"??
 
Which was good since I was damaged by analysis III (my trauma regarding measure theory is gone now, the wound inflicted by differential forms is still healing)
 
12:54 AM
Let $f, g$ be maps $B_m \to B_n$. Define $f \sim g$ when we can find relabelings $\alpha, \beta$ of the codomain, domain such that $\alpha \circ g \circ \beta = f$. I am wondering if we we want $\alpha \circ g \circ \beta = f$ to be true, then does one of the two maps (alpha or beta) have to be identity maps?
 
@Thorgott total differential?
 
$\vec{r}$ is the position of a particle moving as the time $t$ varies and $\vec{v}$ is the velocity, but the calculation is tautological and literally just conjures notational dependency on $t$ up out of nowhere and doesn't accomplish anything else
it's bizarre
 
what course?
 
this is from a theoretical physics 1 lecture lol
@trivial what're $B_m,B_n$?
 
$B_k = \{1,\dots,k\}$
 
12:59 AM
No, for example you can take $m=n=2$, $f=g=\operatorname{id}_{B_2}$, but $\alpha=\beta$ the map that interchanges $1$ and $2$
 
In what sense was weak convergence of measures stronger than some other functional analytic form of convergence again?
 
which functional-analytic form of convergence
 
weak* probably
 
do measures on a space naturally form a Banach space?
 
I think so?
you can take the total variation as a norm
oh wait, that only works for measures of bounded variation
 
1:13 AM
 
when you cant pay for a full banach space and only get the ba space
 
but one could also interpret this as a form of riesz-markov on the dual space of C([0,1])
hrmpf, there was a document that included exactly the statement I'm trying to find
 
yeah i dunno what the right statement here is
or rather, weak convergence of measures is just convergence in the weak* topology on ba(\Sigma), no?
 
https://math.stackexchange.com/questions/2111245/what-is-the-weak-topology-on-a-set-of-probability-measures?rq=1
it seems both viewpoints are valid
 
1:32 AM
@robjohn please don't do that, sir
 
@Thorgott "But the weak-* topology on $C_0(X)^\ast$ doesn't give the probabilist's weak topology on $\mathcal{M}(X)$; in particular, $\mathcal{P}(X)$ is not weak-* closed in $C_0(X)^\ast$." Probably that was what was meant
 
ah, I see
 
 
2 hours later…
3:42 AM
@skullpatrol I think I hurt his feelings. :(
 
Animations?
 
yup
:(
 
Question: If a ring doesn't have identity, then it might also not have any prime ideals. Would this qualify as having Krull dimension 0? Or would it not have Krull dimension at all?
 
is this a word game
 
Where are the animations?
 
3:51 AM
@LeakyNun The definition of rings that I'm familiar with doesn't necessarily require identity. It's almost always assumed, even if not stated, though.
(Unless of course you weren't replying to me)
 
it just depends on your definition of Krull dimension
this is not a mathematical question
 
Fair point on the definitions
 
@skullpatrol much appreciated!
 
np, pal
 
4:24 AM
@skullpatrol I would never stop making animations. They can be useful in getting ideas across. For example, I think this animation shows what the 3-D shape looks like more clearly than a 2-D image would.
@Rithaniel I used to see exercises that explicitly said, "Let $R$ be a ring with unity." That sticks in my head to remind me that they don't of necessity have a unit.
 
If $R$ is a $0-$dimensional ring without identity, then given any $x\in R$, does there necessarily exist a $y\in R$ such that $yx=x$?
@robjohn Yeah, it isn't mentioned all the time, but it's easy to just say "ring (not necessarily with identity)" or "ring with identity" so I'm comfortable with them going either way
 
Like the ring of even integers
 
@robjohn thank you, sir :D
 
@TedShifrin No. My absence was due to taking the dogs to the park and getting dinner.
 
4:40 AM
@robjohn How's your rabbit doing, sir?
 
(When you're working in a ring without identity, everything goes out the window, it seems)
 
@skullpatrol It is quite happy. It gets lots of good food.
 
(I'm trying to prove that if $R$ is $0-$dimesional and reduced, then $R$ is von neumann regular, but all proofs I've found rely on the fact that $x\in xR$, which might not be true if $R$ lacks identity. I'm sure that $x$ is in that ideal, but how do you prove it?)
 
4:59 AM
Wait, $x\in(x)$ by definition
I like arriving at that "aha" moment, but sometimes they also make you feel dumb
 
anybody here good at understanding $g=\frac{dxdy}{xy}+\frac{dudv}{v-uv}?$
can you pullback a vector space?
or is this just called an invertible linear map
 
5:22 AM
user image
2
Phroop of Infinite Primez
 
5:57 AM
@Khachatur Mirijanyan I think that we should chat about it.
 
6:40 AM
test
 
 
4 hours later…
10:47 AM
@Thorgott Bounded continuous is enough
There's a short proof by Skorokhod representation theorem, see here. Coupling for the win, once again.
 
11:05 AM
you have a very good knack of finding a winning argument
 
11:40 AM
@Balarka bounded measurable and $\mu$-a.e. continuous is enough, but who cares about technicalities
 
not a very significant generalization; lipschitz to continuous is
 
i never learned skorokhod representation
was it that thing about a convergence in distribution being realizable by an a.e. convergence or sth
 
yeah
 
what a theorem
 
the proof is a coupling trick
 
11:51 AM
Say I have a continuous thingy $\varphi\colon B\times[0,1]\rightarrow\mathbb{R}^n$ such that $\varphi(-,0)$ is the identity on $B$ (which is a subset of $\mathbb{R}^n$ btw) and $\varphi(-,1)$ is a homeomorphism $B\rightarrow B^{\prime}$ and each $\varphi(-,t)$ is also a homeomorphism onto its image. Is this what they call an ambient isotopy?
 
just an isotopy through embeddings
 
ah ok, and ambient would be the same thing but when defined on $\mathbb{R}^n\times[0,1]$ and it just restricts to a homeomorphism on $B$?
 
yeah
 
@Thorgott This is the wrong notion, all knots are isotopic by this definition
All tame knots at least or knots with a tame arc
 
already the correct notion in the smooth category though
 
11:56 AM
who cares about knots
I just want the right language to say I can nicely move one ball into another inside R^n lol
 
It's wrong for that too
This would make the horned ball equivalent to the standard ball
Obviously wrong
 
lol
(I agree with Mike to be clear)
@MikeMiller Giving a talk on conformal maps and Riemann mapping theorem in my complex analysis class
 
Prove Schoenflies!
 
boundary extension of the Riemann map is pretty annoying isnt it
 
Ok, here's what I'm doing. I have an open subset $U\subseteq\mathbb{R}^n$ and two intersecting balls $B,B^{\prime}\subset U$. Take $p$ in their intersection. First, continuously shrink the radius of $B$ so that it's smaller than the distance of $p$ to the boundary of $B\cap B^{\prime}$, then continuously move the center of $B$ to $p$, then to the center of $q$, then continuously blow up the radius to that of $B^{\prime}$. This process doesn't leave $B\cup B^{\prime}$ and hence $U$ at any point and continuously deforms $B$ into $B^{\prime}$ inside $U$.
 
12:01 PM
take that one for granted lol
 
yeah i just never thought about it until you told me to prove schoenflies
i was gonna say "obvious corollary of Riemann ma---"
 
Too much notation can't read
I am still not sure what your goal is
 
i move one thing to another thing inside a third thing
 
Ok cool boundary extension is an exercise in Stein-Shakarchi
I'll prove it
Thanks for the tip
 
Who cares about moving things
2
@BalarkaSen It's a theorem of Caratheodary lol
 
12:05 PM
oh actually
SS says "generalize from piecewise linear things. requires further ideas"
 
Maybe the hard part is showing boundary objectivity
objectivity
injectivity
 
lol
 
AKA showing Riemann maps extend so long as your region is bounded by a Jordan curve
 
whatever, ill just call it phi
 
\o @epic_math
 
12:06 PM
Call it whatever you want but tell me why you're moving disks
 
also dude Koebe's Riemann map is so genius
its also an ex in SS, solved it yesterday
 
you don't wanna know
 
OK so it's for an exercise about perfectoid rings
 
Start with the red domain $\Omega \subset \Bbb D$ on the left
Find a point $\alpha$ such that if you bloat a disk from the origin it osculates $\partial \Omega$ at $\alpha$ first time
the blue disk
Find a hyperbolic reflection sending $\alpha$ to $0$, reflect about the green geodesic
The red domain gets sent to the orange domain, still on the left picture
A branch of $\sqrt{z}$ is defined on this orange domain, cuz its away from the purple stick, the branch cut
So take square root; that sends the orange domain on left to the orange domain on the right
Reflect about the green geodesic so $\sqrt{\alpha}$ gets sent to $0$ again
You get a new red domain $\Omega_1$
Iterate it forever, $\lim_{n \to \infty} \Omega_n = \Bbb D$
This is the Riemann map
 
@Thorgott I can say something useful if you ever tell me what the real point is lol
@BalarkaSen This is wild
 
12:17 PM
yeah its beyond genius
my picture above is not accurate but im pretty sure this is what it looks like:
@MikeMiller Apparently Riemann's idea was the modern PDE proof you like of uniformization
 
Interesting
He thought of it in terms of curvature then not complex diff
 
hm oh maybe im misremembering what you like to do. he wrote down a map $\partial \Omega \to S^1$ (this homeo coming from being a Jordan domain) and then tried to extend to the interior conformally by Dirichlet boundary value problem i think
 
Ah no I like to solve the conformal change equation to produce a metric of constant curvature
 
ahh got it
 
It's more irritating when you are noncompact since you need to ensure a complete metric
 
12:25 PM
that makes sense
 
So you usually do it as a boundary value problem on a compact exhaustion and take a limit
 
12:39 PM
theres no real point
im just showing two maps are homotopic
 
stupid
 
as per usual
 
1:11 PM
Hi all
 
 
2 hours later…
2:51 PM
I started reading papers on Neural networks and I often come across the term: "degenerative minima". But what does this term mean, what does it express? Does someone know/is able to guess?
 
it's when the Hessian is not definite (i.e. has a 0 eigenvalue)
 
@LeakyNun ah thank you :) that helps a lot
 
@RewCie Hiiii :-)
 
3:16 PM
May I ask a question on triple integrals?
 
@BalarkaSen I still don't get why we did all the partition shite when, directly afterwards, the hamiltonian is suddenly continuous on R^6N again
related to the statistical mechanics stuff
guess I'm gonna have to look up some infinitesimal lattice crap to find something that I find satisfying
Hm one could make a consistency condition that the integrals are equal due to averaging over microstates
 
4:06 PM
@satan29 Just ask, don't ask to ask
 
nevermind now, its settled
 
Does anyone have any suggestions for learning the basics of functional analysis as quickly as possible starting from a background on Rudin's PoMA?
 
What do you want to know about functional analysis
 
This is what was presented in class today, and it's gibberish to me.
In the mathematical theory of functional analysis, the Krein–Milman theorem is a proposition about compact convex sets in locally convex topological vector spaces (TVSs). This theorem generalizes to infinite-dimensional spaces and to arbitrary compact convex sets the following basic observation: a convex (i.e. "filled") triangle, including its perimeter and the area "inside of it", is equal to the convex hull of its three vertices, where these vertices are exactly the extreme points of this shape. This observation also holds for any other convex polygon in the plane ℝ2. == Statement == Throughout...
 
Lovely theorem
 
4:16 PM
and its applications to measure theory
 
The two texts I prefer are Rudin's functional book (less terse than his previous books are) and Conway's functional book
Conway is a rather dry writer --- one of the driest I know --- but he is at least clear
 
Thanks for those suggestions
 
This is John B Conway, not the more famous H Conway, who it is hard to imagine being a dry writer
 
@MikeMiller omg I used Conway's book plenty of times and never realized this until now
 
It's way too dull to be H lol
Still though he is very clear
 
4:21 PM
I think Conway's book is beautiful, but my favourite for a first introduction remains Brezis (Conway covers also much more advanced stuff)
 
Does Brezis get to KM?
 
I don't remember tbh
 
I get the impression from Amazon reviews that Rudin's would not be recommended if you're coming straight from a PoMA background; would you agree?
 
Is KM important for PDEs? I would guess so and I would also guess that it is covered by Brezis in that case
 
KM shows up in construction of Haar measure
@Clarinetist That was not my experience, but YMMV
I would not blame anybody for reading something other than Rudin
 
4:23 PM
I might just get Conway
 
I dislike his writing
I think FA is still clearer than the rest but I do not like his writing lol
Conway's book is nice because in his chapter on weak topologies he (unlike every other author) gives like 6 applications
 
@MikeMiller It is stated but not proved in the comments to chapter 1, just checked, the proof is left as a guided Problem (Problem 1 at the end of the book)
 
RIP
I have someone doing a project on Banach-Alaoglu and he wants applications so I pointed him to KM and Stone-Weierstrass
 
Some exercises in Brezis book are quite hardcore, this one has a lot of hints though, it seems doable
 
how do you prove Stone-Weierstrass from Banach-Alaoglu?
 
4:34 PM
@MikeMiller Does (local) compactness of the space of characters of a commutative C*-algebra count as an application or is Gelfand duality going too far from the intended topic?
 
@AlessandroCodenotti Already gave him that ref, up to him if he decides to pursue that or not
It's more difficult since he'd have to do some basics on multiplicative functions and calculating that there is one C* field etc
 
I see, but I think that Gelfand duality is pretty cool. What kind of project is this?
 
@LukasHeger You need Krein-Milman as well
You have a closed C*-subalgebra of C(X) that separates points. You want to show that it is all of C(X) --- equivalently that the set of functionals on C(X) which vanish on A is trivial
Call that A'. The unit ball in A' is weak* compact, Krein-Milman guarantees that the unit ball has an extreme point, which gives a measure on X for which everything in A integrates to zero
I don't really remember how it goes from here lol
But you use that it's an extreme point to derive a contradiction
 
I see, thanks
 
I think you maybe show that the whole space has mu-measure zero
@Lukas I checked. You set K = support of mu (the extreme point in this unit ball). Because this is not the zero measure by assumption (towards a contradiction), the support is nonempty. The goal is to show that it is a point-mass, which will contradict that 1 in A and int f dmu = 0 for f in A
 
4:49 PM
@Clarinetist A book I liked for my first round was "Introductory Real Analysis" by Kolmogorov & Fomin. Not as terse as Rudin (which I find hard going) and some older conventions such as all Hilbert spaces being separable but readable. I would suggest getting a few books rather than relying on one if you can. Another one I liked for its treatment of fixed points was "Functional Analysis" by Kantorovich & Akilov.
 
Pick a point x in the support and a point y other than x. You can use that A separates points to construct a function f vanishing at y and nonzero at x, and show that f mu and (1-f) mu are both in A'. You do some irritating crap --- finally using that mu is an extreme point --- to then show that y is not in the support
So the support is one point as desired
There is surely content to this stuff I called irritating but I don't see it and I need to write lecture noters
 
thanks
 
I never learned a "good" proof of SW. Do you know one
 
There's the proof that uses the Taylor series of $\sqrt{x}$, that's pretty elementary
 
Oh, to be clear this is the statement that an algebra of functions on a compact metric space which separates points, is conjugation-closed, and has 1, is in fact C(X)
 
5:00 PM
don't you need that it is closed as well? or you just get that it is dense in C(X)
I learned a proof of the real version in my intro analysis course
I think if you have the real version, you can also deduce the complex version
 
is putting a norm on a family of functions trivial from a functional analysis point of view
all the examples I've seen deal with more complex function spaces
like the space of C^0 functions on [-1,1]
with supremum norm
 
user486313
5:16 PM
is operator theory a dead field?
 
user486313
what about functional analysis?
 
@LukasHeger Yeah, sorry
 
function analysis is not dead
operator theory is still operating
 
@MikeMiller the proof of the real version went like this: if we let $K$ be a compact metric space and $B$ is a lattice of functions on $K$ that separtes points (where a lattice means it's a $\Bbb R$-vector space and satisfies $f \in B \Rightarrow |f| \in B$), then $B$ is dense.
Note that a lattice is closed under taking minima and maxima due to formulas: $\max(f,0)=\frac{1}{2}(f + |f|)$, $\max(f,g)=g+\max(f-g,0)$ and $\min(f,0)=\frac{1}{2}(f-|f|)$, $\min(f,g)=g+\min(f-g,0)$
Now the proof that a point-separating sublattice of $C(K)$ is dense:
 
@Clarinetist pro tip:
use lecture notes instead of books
 
5:25 PM
@MikeMiller law of large numbers
 
anyone listen to non mainstream hip hop music??
 
oh by SW do you mean the general thing about algebra separating points being dense?
then i dont know
i know for polynomials only
 
just europop & electronica here
 
@geocalc33 a little
 
@copper.hat what a great niche
(seriously)
 
5:27 PM
@user2103480 drives my 17 & 19 yo kids mad
 
lol
 
they have no taste

i'M 24 aNd I liStEN tO ThIs
 
was better when i could dance to it, need some replacement parts in the hip area unfortunately :-)
 
Go a step further and play krautrock on maximum volume
 
does hooverphonic classify as europop
 
5:30 PM
I was taught functional analysis by an entertaining fellow by the name of Shoshichi Kobayashi, he had a wonderful way of putting things, for example, "sets are not like doros, they can be open and closed at the same time".
 
@BalarkaSen google says it qualifies as electronica, so it's still within the acceptable range
 
mmkay
 
Now that we have the lattice-version of Stone-Weierstrass we can prove the version for algebras: if $X$ is compact and $A \subset C(X)$ is a $\Bbb R$-subalgebra that separtes points, then $A$ is dense in $C(X)$.
Proof: it suffices, by the lattice-version of Stone-Weierstrass to prove that $\overline{A}$ is a lattice. wlog $A$ is closed. For $f \in A$, we have $f^2 \in A$. Now because $\sqrt{f^2}=|f|$, it suffices to show that we can take square roots of positive elements in $A$. Let $h \in A$ with $h \geq 0$. Set $g=1-h$ so that $h=1-g$. By rescaling we can assume $\|g\|_{\infty} < 1$, but
 
@copper.hat is this actually the same kobayashi as kobayashi-nomizu
 
5:31 PM
Yup. Nice fellow.
 
damn
 
@MikeMiller I won't say that this proof is pretty, but it's pretty elementary and down-to-earth, you don't need any functional analysis
 
i just listen to what i enjoy, i like jazz in the satchmo, ella fitzgerald, etc mode.
i suppose bach is no surprise, early electronica in my opinion :-)
 
also the appearance of the power series expansion for $\sqrt{1-x}$ is an oddity which I find amusing
 
haha
nice description
 
5:34 PM
Is electronica like Martin Garrix?
 
@geocalc33 nope
 
Kygo?
 
more like M83
 
but electronica is vast and cannot be defined by one band, so take that with a grain of salt
 
5:36 PM
not really, but whatever rocks your boat. i just use genres to find other stuff i like.
i find the repetiveness nice when i am trying to focus on something.
 
@copper.hat yeah thats really what cataloguing is useful for
good takes
 
@BalarkaSen wait, it's not just a reason to be elitist?
WILD
 
hah
 
@BalarkaSen i suppose there is less information content (in a shannon sense) in the music i listen to when focusing :-)
 
lmao
 
user486313
5:42 PM
anyone here like Dave Matthews Band ...
 
user486313
@geocalc33 wild insight, thanks bro
 
you're welcome brother
 
some friends politely describe my music taste as eclectic, more direct friends use the term crap :-).
 
user486313
@copper.hat what bands / music?
 
5:55 PM
i like stuff like ncs, capella, the blue man group for example, edward maya, stuff like that. i also like bailando, panama, etc :-)
 

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