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00:03
@Frusciante I think your postulation also makes sense as long as we consider the factorization to be part of the datum, but I think this is intuition in terms of lifts moreso than in terms of being initial. That said, I believe the condition I gave is, in general, weaker than the full condition of being a relative colimit, hmm.
anyway, good luck tomorrow!
@X4J yes, good observation
00:43
chat.stackexchange.com/transcript/message/66686577#66686577 this is not possible they would give ellipses.
01:31
@SineoftheTime Less was know, so it was easier to know most of what was known.
@Thorgott I read the proof of the sum of open ANR's is an ANR theorem. I could verify its true, but I don't have intuition for the proof. Which is my least favourite type of proof to read
It looks like a bunch of sets is being defined, in fact, I think there was around 10 to 15 different sets defined in that proof. Those have properties that eventually give us the desired result. Why we defined all of those? What was the intuition for it? I guess I won't know
Perhaps the person who came up with the proof, arrived at it by the brute force approach
Corollary: $\mathbb{R}^n$ and $S^n$ are ANR's for normal spaces
Proof: If $X$ is normal and $F\subseteq X$ is closed and $f:F\to \mathbb{R}^n$ then extend $f_i:F\to \mathbb{R}$ to $X$, then $f$ extends to $X$, so $\mathbb{R}^n$ is AE for normal spaces. Since $S^n$ is union of two open copies of $\mathbb{R}^n$, its an ANE for normal spaces.
01:56
Hi can someone guide me as to how to use baye's theorem here to solve this sum successfully
A bag contains 6 Balls of unknown colour. 3 Balls are drawn at random from the bag and are found to be all black. Find the probability that no black ball is left in the bag now. Initially assume all number of black Balls in a bag to be equally likely
There are 7 case either 0 black ball or 1 black or 2 black or 3 or 4 or 5 or 6 so probability of each is 1/7 ?
@sanya what is Baye's theorem
do you mean Bayes' theorem
also, what does it mean to "solve a sum"?
@Jakobian commutative property is not always valid :D
moreover, where is this "sum"?
@Jakobian yes
@Jakobian The problem that I wrote above?
@sanya the amount of black balls isn't random
so the probabilities in this case depend on a parameter, which is a number of black balls
02:06
Ok
02:45
@SineoftheTime lol
03:04
anyone here?
pie
pie
I've been awake for over 30 hours, and strange ideas are starting to pop into my head. One of them is that Biden looks oddly similar to Gauss.
maybe I need to sleep IDK
03:43
old people's faces converge
4
 
1 hour later…
05:12
@pie that's bad for your organs to stay awake that long
Biden is a Gauss' great great great... great grandson
 
2 hours later…
06:45
its topologin' time
 
3 hours later…
09:58
Exercise 3.26. If $\lambda$ and $\mu$ are positive, mutually singular Borel measures on $\mathbb R^n$ and $\lambda+\mu$ is regular, then so are $\lambda$ and $\mu$.
Definition A regular measure $\nu$ on $\mathbb R^n$ is i) finite on compact sets and ii) $\nu(E)=\inf\{\nu(U):U\text{ open},E\subset U\}$.
Attempt: If $\lambda +\mu$ is finite on compact sets, then it has to be the case that $\lambda$ and $\mu$ are finite on compact sets, so condition (i) holds. For condition (ii), since $\mu\perp\lambda$, suppose $N$ is $\mu$-null and $N^c$ is $\lambda$-null.
Since $\lambda+\mu$ is regular, for any $E\subset N$ that is Borel measurable, there exists an open set $U$ such that $$\lambda(U)<\lambda(U)+\mu(U)<\lambda(E)+\mu(E)+\epsilon<\lambda(E)+\epsilon$$ for any $\epsilon>0$, therefore $\lambda$ is regular (the first inequality above is simply due to monotinicity). A similar argument holds for $\mu$.
Question: Is this proof OK? I'm suspicious if it is assumed $E$ to be bounded above? I can't tell myself.
10:15
Maybe @Jakobian has something useful to add here :)
 
1 hour later…
11:22
@psie $E\subseteq N$?
@Jakobian yes, we first want to show $\lambda$ is regular, but it is null on $N^c$, so we can just look at subsets of $N$, right?
I don't really like that jump. I'd apply the definition to $E\cap N$ instead
The inequalities can be just changed from strict ones to weak ones so its not a problem with infinity
After those changed I'd say it'd be okay
Ok, thanks!
12:28
what r some interesting places in mathematics where the algebra $[A,B]=i$ shows up
it is a Lie bracket
hi
 
2 hours later…
14:19
What is the motivation behind the definition of coherent sheaves?
15:08
@SoumikMukherjee Well, you want to be able to distinguish them from incoherent sheaves.
15:51
If you really believe in the idea of tariffs, why not put tariffs between every country, every city, every neighbourhood, every street, every house and every human being.
@MatsGranvik In the US, that would violate the constitution.
But it certainly wouldn't be unprecedented in history.
The Commerce Clause describes an enumerated power listed in the United States Constitution (Article I, Section 8, Clause 3). The clause states that the United States Congress shall have power "to regulate Commerce with foreign Nations, and among the several States, and with the Indian Tribes". Courts and commentators have tended to discuss each of these three areas of commerce as a separate power granted to Congress. It is common to see the individual components of the Commerce Clause referred to under specific terms: the Foreign Commerce Clause, the Interstate Commerce Clause, and the Indi...
I suppose that congress could enact interstate tariffs, but... why? It would be super unpopular...
@Jakobian did you check out the relevant Figure in Sakai? the details are finicky, but I think the idea of the proof is not too strange
@Jakobian same line of thinking then shows any compact manifold is an ANR for normal spaces, interesting
@SoumikMukherjee they generalize finitely presented modules (in a very precise sense if you're working over well-behaved schemes)
Joe
Joe
16:24
In the definition of "local homeomorphism" $f:X\to Y$, we require that for all $x\in X$, there is an open neighbourhood $U$ of $x$ such that $f(U)$ is open and $f:U\to f(U)$ is a homeomorphism. Is there any particular motivation behind wanting $f(U)$ to be open in this definition?
$f(U)$ isn't open automatically, if it weren't then I wouldn't really say that $f$ really preserves local properties of $X$
the idea, to me, is that we want to preserve local properties of $X$ to local properties of $Y$
and demanding that $f(U)$ is open, is a condition so that we have something happening locally in $Y$
yeah, the condition basically ensures that $X$ and $Y$ look germinally the same
16:47
accidentally answered a homework question from the course I'm tutoring on the main site because I didn't look at the new exercise sheet 💀
the best thing is, another tutor apparently did the same thing, on a different homework question, a few weeks earlier
the asker is so sneaky too
"so I was reading these K-theory notes" no you weren't stop lying 💀
it's fine, nobody really cares. these things are not grade relevant and asking about them online is not a crime or anything
if anything they are sabotaging themselves by asking online a day after the new exercise sheet is published instead of spending more time trying to work the questions out themselves, etc.
17:04
@Jakobian hi. I messaged u on the physics chat
must be extra funny for them, too, given you're not posting pseudonymously
indeed lol
I remember one funny instance a couple years ago where somebody posted an exam question and Emily Riehl herself wrote an answer saying "meet me in my office"
4
17
A: Universal properties of $\mathbb Z$

Emily RiehlI'd be happy to discuss this in person either during or after your oral exam. Come find me in my office.

17:10
that is absolutely hilarious
And completely off-topic. That answer has prevented the question from being deleted for years.
@RyderRude hi
sic transit...
I did saw you saying that I might try learning GR if I want something in physics that is just mathematics at its core
I don't know though, I think it was a weird whim for me to try and search for something like that, and I'll just continue on learning topology
@Jakobian oh. u can check it out whenever u r interested
17:22
@Thorgott yeah. I wonder how could one include manifolds that are finite unions of open copies of $\mathbb{R}^n$, to weaken the compactness hypothesis
but it is mostly metric and topology doesn't play a big role
well, finite unions are always included
yeah. I mean what condition could we impose on a manifold that is weaker than compactness so that this still holds
there is something called Topological Quantum Field theory tho, which I've read is mostly pure math
i think it is just physics inspired techniques to compute topological invariants
little to do with real world physics
@Jakobian a condition that ensures it's a finite union of open Rns, you mean?
17:25
yeah
In gauge theory and mathematical physics, a topological quantum field theory (or topological field theory or TQFT) is a quantum field theory which computes topological invariants. While TQFTs were invented by physicists, they are also of mathematical interest, being related to, among other things, knot theory and the theory of four-manifolds in algebraic topology, and to the theory of moduli spaces in algebraic geometry. Donaldson, Jones, Witten, and Kontsevich have all won Fields Medals for mathematical work related to topological field theory. In condensed matter physics, topological quantum...
ah, Jakobian would have to do a different kind of topology for that :)
@BenSteffan why is it different
i think it deals with manifolds while Jakobian doesn't work with manifolds mainly
Jakobian does point-set topology, but TQFTs belong much more closely to algebraic topology and homotopy theory
17:36
maybe more geometric topology, in fact, but same conclusion
it looks far too removed from physics
really? you should ask Urs Schreiber how he feels about that :^)
(don't)
but it says it has an action, which is close to physics QFT
@BenSteffan the direction I am going seems awfully close to set-theoretic topology too
@BenSteffan i think i once came across Urs saying that this stuff would be useful for quantum gravity
17:40
Urs says a lot of stuff when the day is long
well... maybe not entirely, a lot of it is about zero sets and the like
@BenSteffan yes
but zero sets for nice spaces don't matter since they just become closed sets and so on if your space is nice enough so... the most interesting parts are in set theory examples
i kind of bought it back then. but now I've learned to be skeptical of Urs
and of the nlab crowd in general :P
yeah, approach with caution
17:42
lol
but yeah I would say that the experience I have is entirely different from the experience of someone who is an algebraic topologist
yeah, it's not really the same field at all
I believe that the things I am learning will eventually take me to something useful though
and by useful I mean, for something outside of general topology
@Jakobian actually, it's always true for connected manifolds
and disjoint unions of ANRs are ANRs
@Thorgott how do you show that?
17:49
3
A: A connected manifold can be covered by finite open subsets

Moishe KohanGood question. The answer is positive. First of all, as in my answer here, there exist $n+1$ families ${\mathcal F}_1,..., {\mathcal F}_{n+1}$ of subsets in $M$ such that: Each ${\mathcal F}_i$ is a union of closed (and tame) pairwise disjoint $n$-balls $B_{ij}, j\in J_i$, where each $J_i$ is (a...

thanks
so you just need to demand finite amount of connected components then
For $S \subseteq Y \subseteq X$, this seems like a neat little result $\operatorname{int}_Y(S)=\operatorname{int}_X(S \cup Y^C) \cap Y$
looks like someone took complements in the expression $\text{cl}_Y S = Y\cap \text{cl}_X S $
hmm, I was gonna say that arbitrary disjoint unions of ANRs are ANRs, but I suppose I don't know if this is true in your setting
it's true if ANR means ANR for metric spaces
Are you sure its true for metric spaces?
Because the proof that I saw in Hanner's article seems to work for countable unions of open subspaces, and the one for separable metric spaces uses that the separable metric spaces are Lindelof
17:59
yeah, if you want to extend $f\colon A\rightarrow\coprod_{i\in I}X_i$ from a closed subset $A\subseteq Y$, you extend $f_i=f\vert_{f^{-1}(X_i)}$ to an open subset $U_i$ of $Y$ and then all you need is to find a collection of open $V_i\subseteq U_i$ s.t. the $V_i,\,i\in I$ are pairwise disjoint and still intersect $A$ in $f^{-1}(X_i)$. this is possible if $Y$ is metric, but I don't know if it is possible if $Y$ is normal
but we are talking about disjoint unions, I suppose
sounds like something that might be true for a collectionwise normal space
oh yeah, that would do the job
18:19
@Thorgott I see, thanks
18:47
I made something in Desmos, it maps complex numbers to other complex numbers with lines, based on a function. It creates nice patterns for certain functions. Here is the graph: desmos.com/calculator/sgqkew1npa
19:36
Anybody have any insights with regard to this question? math.stackexchange.com/questions/5002192/…
@Thorgott How do I format a link like this?
just send the link of the question without text
2
Q: Counting the number of unique color combinations on a grid?

farrow90I have an 8x8 grid with a pattern involving 5 colors that can look like this: I am not sure how to describe the pattern in mathematical language, but they are supposed to create continuous borders/boundaries such that each color must be connected to least one of the same color How many unique co...

@SineoftheTime got it, thanks
@TheEmptyStringPhotographer looks like the top of my head
the desmos plot you mentioned
19:40
are you balding? :(
so we can probably simulate this
using your plot
@SineoftheTime no, but i get my hair cut to a #2 (fairly short).
i used to shave/#0 but people treat you quite differently when you have little hair
the purpose being to minimise hair care time :-) i used to do a lot more outdoor stuff
yeah, hair care is annoying
I'm not going to a barber since 2021
19:56
@Thorgott That every point-finite open cover of a normal space has a shrinking is a theorem of Lefschetz, apparently
from his book about algebraic topology
at least that's what Nagami seems to imply
that's cute, I didn't know who it was due to
20:25
when i was younger i used to play with iron filings and magnets trying to get patterns like the above, but it never worked
20:42
1
Q: Where should I begin to understand $\mathcal I_{\Gamma}$ in complex space?

ModularMindsetMotivation: What is the largest surface of revolution with constant positive Gaussian curvature that can be embedded in $X^3=[0,1]^3$ with a pair of antipodal corners as cone points? It can be shown that there is a unique surface solution to this question. Therefore, replicating this surface thre...

Any advice for me on how to pursue in complex space?
Also I had the thought that maybe I shouldn't deal with an embedding at all and do things intrinsically
21:06
It's neat that in real space the Gamma set gives rise to an augmented multigraph structure comprised of the six quartics - orthogonal pairs lying on coordinate planes. And the restriction of the normal bundles to these quartics gives a nice line bundle which vanishes at six points corresponding to the vertices of the graph 🤔
21:23
I dislike how Grothendieck writes
why name anything when you can just refer to everything as a "canonical map" in some capacity
maybe his parents fed him a bowl of lambda calculus for breakfast every day when he was young
@BenSteffan I do that too tbh
boo :(
my average commutative diagram has at least half the arrows unlabeled cause they are canonical
Ever since attending a Schwede seminar I am scared of the word canonical
and of the word "technical" I guess
just blast the word canonical
21:36
on an unrelated note, Waldhausen has passed away
Suppose that $\delta$ is a real-valued, non-negative function such that $\delta(x, z)\leq A\max(\delta(x, y), \delta(y, z))$. Can we prove that $$\delta(x, y) \leq A\delta(x, x_1)+A\delta(x_n, y) + 2A\sum_{k=1}^{n-1} \delta(x_k, x_{k+1})?$$
21:52
@copper.hat You need a canonical cannon for that...
the canon being the need for a canonical cannon
Canonically.
shoot, you're right
@Jakobian i would imagine that you would need some powers of $A$ in there?
Look at me! I'm Grothendieck! Canon canon canon!
he was homotypical
21:57
Grothendieck? not so much, but he was always scheming
albeit i have no idea what homothetical means
@copper.hat no, and the inequality seems to hold
what symbol do you guys use for the Lebesgue measure? $\mu$ or $m_n$?
$\lambda$
@SineoftheTime $\lambda$.
22:00
$m, \lambda$
Or $\lambda^n$ for $n$-dimensional Lebesgue measure.
Sometimes $m$ if I am feeling metalic.
I was staring at this but it makes sense, $D(x, y)\leq 2D(x_1, y)$ and the final inequality makes sense because of induction
A number of posts/answer I've made, some from many months ago, some closed, just got an enormous amount of downvotes out of nowhere.

Does SE have the problem with people downvoting a ton of other things because they disapprove of one thing you've said, or is this just a strange coincidence?
they seem to do periodic sweeps for pattern voting.
22:14
@copper.hat No, not really.
@NewViewsMath I don't see that reflected in your profile
Generally, we need a flag.
@copper.hat doesn't seem to be in this case
@NewViewsMath I don't see anything actionable, but if it continues, please raise a flag.
"Enormous" also seems to be a bit of an exaggeration.
the posts that got downvoted seem to also have a lot of votes for closure
22:15
https://math.stackexchange.com/questions/4851750/determining-which-characters-split-or-fuse/4904474#4904474

This has gotten three downvotes in the last day, which is strange. It also has a lot of votes for closure, but is closed
If I'm not contributing well, I understand the downvoting, but this has been closed for months
@NewViewsMath Like I said, I don't see anything actionable. It looks, to me, like someone saw the post and voted to close it. When this happens, the post is bumped into the close votes review queue, which brings it extra attention, which can lead to downvotes and votes to close.
@NewViewsMath This post in particular contains no work of your own, it is a PSQ, a problem-statement question
@NewViewsMath I don't know what you mean by "this has been closed for months". It hasn't. That's why it is capable of getting close votes now.
I marked it as having an accepted answer
And you have called extra attention to the question by asking about it here. Which may attract more downvotes.
22:18
The one I provided in the hopes it would help future people struggling with the same issue
@NewViewsMath That is very different from being closed...
I agree many of these posts as PSQ, usually a "where does one learn this" Is there a better way/place to ask those sort of questions?

It's rather difficult to include my own work trying to determine what a definition means
Perhaps at university
lol, my advisor has told me multiple times "I don
Online, I don't know of any place which would do that, at which you can get sufficient expertise on the topic
22:22
't know, it's your job to figure it out"
you just need to make more effort towards your questions
here (as in, in this chatroom) you can ask questions, but you can't expect answers
our interests rarely intersect
oh certainly, that's why I've answered a number of my own questions, once I determine it, hopefully it helps somebody else. Other times, I've been able to get solutions. (One recently was very straightforward, and I hadn't realized)

I struggle to see a better way to say "these three papers say XX is a commonly known thing, but no text of the subject seems to describe XX, where do I go learn?" Should I just not direct such questions here?
You can certainly post them here if you put the work into your questions to include such things as background etc.
for example this is my own question, we can analyze what I did here
I've defined something and mentioned this is common in the literature, then I proceeded to define something that I actually wanted
then I defined another thing common in literature
I gave some examples
and lastly I asked question and partial solutions to it
so I was clear about the concepts, I've provided how it works, and when it works, that I know of
this is bullet-proof question to me
no one would close that, I think
alright
could I send one and we see what I might do to improve similar things in the future?
You just have to dig into what you can add, some of it is creativity, some of it is a learned skill
but you need to indicate, that this is not just another low-effort question somehow
you need to work on the contents of it
I don't know how else to help you, good luck
22:38
ok
thank you
23:14
some of the clarifying comments that you added to your own question in response to people asking you about it in february, are the kind of thing that it would normally make sense to edit into the question and not leave in comments. the comments from january from people are offering you examples of things that were missing from the OP and are still missing because you didn't edit them in
it also helps to broaden the audience for the question as much as possible, e.g. if this isn't specifically a question about ATLAS but is actually just a question about what you can and cannot deduce from a character table, it would be helpful to phrase the question like that in the first instance, as doing otherwise (1) makes it seem like people have to do their own research to figure out what the question is, and (2) makes it seem like the question is only about some specific reference
so if possible in this example i would edit the question to explain what "case G.2" is (what in the world would make anybody think that this was standard terminology? why would you want to narrow your audience to only those people who are going to click through and solve the mystery about that it is) and offer citations to ATLAS only as an example. quotations of ATLAS would also be preferable to citations for this purpose
it isn't obvious at first glance that "ATLAS" isn't some software package, which might also have pissed people off. it isn't, but how would a prospective voter know that. questions about software packages are very routinely downvoted
23:29
if I need to study uniform conv of a sequence of functions on $I = [0,c]$ and the sequence is defined separately on $I_1 = [0,a), I_2 = [a,b), I_3 = [b,c], \bigcup_{i=1}^{3}I_i = I$
I can then study uniform convergence separately on each $I_i$
@Claudio is it a question or a statement?
23:46
a question hahah
can I then study...?
you can study anything you want :)
anyways I got the right answer so I guess the answer is yes
You can study the UC over I by studying the UC over I_j
@BenSteffan hahah, thanks I needed to hear that
@SineoftheTime perfect
thanks
23:49
@SineoftheTime What is it that you are suggesting I do at UC Irvine?
@SineoftheTime we started the brief measure theory part in the last lecture on friday
I understood maybe a good 10%
I've never studied measure theory
@XanderHenderson study UC?
Seems tautological...
@SineoftheTime Oh, I thought you did my bad :p anyways I might to have to lower to 8% now that I'm rereading my notes
I've only "studied" Lebesgue measure
23:52
@Claudio at least the subset of things you understood is not a null set
I've finally understood why fully normal is equivalent to paracompact + normal, by showing that a normal open cover (i.e. it admits a sequence of star-refinements) is numerable (i.e. it admits a subordinated partition of unity). My proof is that a normal open cover induces a certain continuous pseudometric, and since metric spaces are paracompact Hausdorff, this then induces a subordinated partition by taking Kolmogorov quotient.
@SineoftheTime yeah that's what we're gonna focus on as well
did you know that lebesgue is french for "the besgue"
the best guy
:)
@BenSteffan lol
I might have to use that next time
23:56
Bro invented a whole theory on integration in his thesis
Without lebesgue measures we'd all be running around like chickens
@SineoftheTime Nah, I'mma do my own thing, Spiderman
Turing did some stuff too, man, like crack the Nazi's encryption
So did Emmy Noether, the list goes on...

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