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12:10 AM
@leslietownes It isn't called that any more? I'll have to update my answer about Gibb's.
it's called that. i was just joking about runge's phenomenon being a later, uncool, major label ripoff of the indie original.
which isn't even a good comparison because runge's phenomenon is less well known.
gibbs is the beatles, runge is wings
Paul McCartney was with a band before Wings?
Who's Wings?
I felt old when I heard the first; I felt ancient when I heard the second.
they're only the band that the beatles could have been
1:01 AM
a runge band
band on the runge
1:48 AM
@robjohn Thank you Soo much
Btw Can it be seen in mathjax format only? How can I make it appear normal?
1 hour later…
3:14 AM
Am I able to take an inner product inside a inner product to the outside?
My thinking is that yes as the inner product is simply a scalar
I'm here mainly for confirmation ty in advance
3:54 AM
Suppose that f is a continuous function from R to R. Let B be a Borel set, then it is to be proven that $f^{-1}(B)$ is also a borel set.
I know that any open set will be pulled back to an open set by f. But to prove the statement in last message, I should know the complete description of elements of Borel sigma algebra.
i game: in the real case yes. in the complex case you may have to conjugate depending on which side you are working on.
That is, I think that I must consider cases like the following: 1) Let B be a countable union of open sets, 2) countable intersection of open sets, 3) countable union of (countable intersection of open sets) etc...
How do I exhaust all these cases (I have listed only 3 but I'm sure there are more.)? There seems to be no end to the number of possibilities.
mm, you don't really need to get into the borel hierarchy to do this, although that is one way of doing it. you maybe need to do a transfinite induction or at least an induction that is more than over N.
will that work, Leslie? Because Borel sigma algebra is uncountable. Is there any induction for uncountable?
if X and Y are spaces and B is a sigma algebra on Y then {f^{-1}(S): S in B} is a sigma algebra on X. as you note, if X and Y are topological spaces and f is continuous and B is the borel algebra on Y, then applying this result, we see {f^{-1}(S): S in B} is a sigma subalgebra of X. and applying your observation it is a sigma subalgebra of X that contains all open subsets of X.
now i guess it depends on how you've defined 'borel' sigma-algebra. is it clear that it is the minimal sigma algebra containing open sets? or was it defined via some inductive process?
4:03 AM
yes, Borel sigma algebra was defined as the smallest sigma algebra containing open sets.
OK. maybe i put that the wrong way around. convince yourself that {S subset Y: f^{-1}(S) is borel in X} is a sigma algebra on Y that contains the open sets.
so (by the minimality) will contain all borel subsets of Y.
nice. You used the 'good sets principle'. :)
@leslietownes I'm thinking about this. The set is non empty as it contains Y.
Check 1: Closed under countable unions- take $\{S_i\}_{i=1}^\infty \subset $ {S subset Y: f^{-1}(S) is borel in X}. $f^{-1}(\cup S_i)=\cup f^{-1}(S_i)$, this is because the inverse functions are nice.
The union $\cup f^{-1}(S_i)$ is a borel set in X because $f^{-1}(S_i)$ is a borel set for each i.
Therefore, $\cup S_i\in $ {S subset Y: f^{-1}(S) is borel in X}.
Similarly, the set is closed under complements.
Hence, a sigma algebra on Y.
The set (already proven to be a sigma algebra on Y) contains all open sets on Y (by continuity of f) and therefore must contain Borel sigma algebra on Y.
and that completes the proof.
thanks a lot @leslietownes. :)
4:24 AM
Oh good. For this term I can ignore Koro's questions :)
more to come soon, haha.
Ted is allergic to all things measure theory.
my daughter's making me take her swimming every day, and everything hurts.
4:43 AM
$(-1)^{k} * \frac{1 \cdot (1-2)\cdot (1-4) \cdot (1-6)\cdot ... \cdot (1-2(k-1)) }{2^{k}k!} = (-1)^{k} * \frac{1 \cdot (-1)\cdot (-3) \cdot (-5)\cdot ... \cdot (1-2(k-1)) }{2^{k}k!} = (-1)^{k} * \frac{ (2k)! / (k! 2^k) }{2^{k}k!} = \frac{(2k)!} {2^{2k} (k!)^{2}}$, right? However, I don't have (2k-1) in the denominator.
It's the very last part of the calculation.
2 hours later…
7:13 AM
@JaiSriKrishna Do you have ChatJax? The installation link is in this room's description.
7:41 AM
I solved it. Nevermind. HAve a good day
2 hours later…
9:18 AM
is it true that a set $A\subset\mathbb R^n$ has lebesgue measure zero implies that $A$ has $n$-measure zero, where $A$ having $n$-measure zero means that for any $\delta>0$ there exists a countable cover of $A$ consisting of rectangles $R_i$ such that the sum of their volumes is less than $\delta$?
my guess is yes, because I am aware of the result that
for a bounded function $f$ on a compact interval $I$, the Riemann integral exists iff the discontinuities of $f$ have measure zero, and also of the result that the Riemann integral exists iff the discontinuities of $f$ have $n$-measure zero
I'm interested in a direct proof of Lebesgue measure zero iff $n$-measure zero
9:34 AM
isn't what you call $n$-measure zero just saying outer measure zero?
it seems so, ya?
9:48 AM
but does outer measure zero imply $n$-measure zero?
because if that's true, then I'm done indeed
Q: Set has Measure Zero iff There is a Collection of Intervals

Sir_Math_CatI'm stuck on the following real analysis problem. Show that a set $E$ of real numbers has Lebesgue measure zero if and only if there is a countable collection $\{I_{k}\}_{k=1}^{\infty}$ of open intervals for which each point in $E$ belongs to infinitely many of the $I_{k}$'s and $\sum_{k=1}^...

here they seem to prove it, but not for arbitrarily small sums
oh I think it works
because of the condition that each point lies in the interior of inifinitely many intervals
of not... hm, they seem to already work with Lebesgue measure zero = n-measure zero
10:16 AM
A: Jordan measure zero discontinuities a necessary condition for integrability

Dave L. RenfroLet ${\mathcal D}(f)$ be the set of points at which $f$ is not continuous. The correct result is: Theorem: Let $f:[a,b] \rightarrow {\mathbb R}$ be bounded. Then the following are equivalent: (a) $\;f$ is Riemann integrable on $[a,b].$ (b) $\;{\mathcal D}(f)$ is a countable union of closed Jordan...

hm I think this fixes it
hm, actually I'm starting to think that it's not true
Q: Lebesgue measure zero vs Jordan content zero

Zbigniew FiedorowiczIs a set of Lebesgue measure zero necessarily a countable union of sets of Jordan content zero? This was a question posed by a student in my undergraduate analysis course. I asked an analyst colleague about this and he did not have an answer off the top of his head. Here are a few thoughts about...

ok it's not true indeed
2 hours later…
12:27 PM
anyone could point out what the mistake in this answer could be? Thanks.
12:56 PM
@Beautifullyirrational Why do you assume that there is a mistake?
I had before posted with an analogy using wifi signals, then it got two downvotes. I was trying to figure out the reason for it. I suppose it was the wifi thing that led it to getting downvoted. Talking to Calvin Khor suggested to me against that it is probably that which led to the DVS
You assume that the downvotes are because there is a mistake in the question, then?
A downvote doesn't necessarily indicate that there is a mistake.
The hover text for the down arrow reads "This answer is not useful." Thus the downvoters are indicating that they don't find your answer useful. This doesn't mean that there is a mistake. It only means that the downvoters hold the opinion that your answer is not useful.
typically that has been my experience here
that wrong answer <=> downvote
Wrong answers often get downvotes, but there are other reasons that people downvote.
Okay, maybe I should start seeing downvotes differently
12:59 PM
For what it is worth (and I have not downvoted your answer), I don't think that you actually answered the question.
OP asks how it can be that X is a member in the topology of X itself
@Beautifullyirrational Yes. And you didn't answer that question, so far as I can tell.
so I say that topology is something put on top of the set rather than something you put inside the set
doesn't that resolve it?
Maybe. In a very non-rigorous, hand-wavy way.
But, in my opinion, you have failed to answer the question.
What you have written only makes sense to me because I already know the answer to the question. I don't see how it would be useful to someone who is struggling with the nitty-gritty of the definition.
Okay, I see. I think it's hard to think how a newbie would perceive what you are saying
feels sort of bad when an idea you got so excited about sharing turns out to be of low instructional value\
1:12 PM
@XanderHenderson People do that also when you ask a question that's obvious or obviously wrong
to the one downvoting at least
@Jakobian Are you talking about downvotes on answers? or questions?
@Beautifullyirrational You can always try to improve it.
questions. I think for answers people are less harsh
I don't think it is possible to add anything beyond what has been said
the only originality was the physical example, but seems so it didn't click with people
@Beautifullyirrational Well, I meant in general. If what you are excited about sharing turns out to not be as pedagogical as you thought, then often times you can improve it. That's what teaching is about: maybe the first time you don't explain it well enough, but then you improve for next time.
1:16 PM
@Jakobian I was only talking about answers.
The hover text for questions is different.
"This question shows no research effort. It is unclear or not useful."
I'll write it down somewhere and probably refine it when I get better ideas
That said, being flat out wrong seems to fall into the category of "unclear" and "not useful".
then shouldn't we be downvoting this question?
there's plenty of questions like these here
@Jakobian if you haven't noticed, the primary motivator for answer questions on stackexchange sites is fake internet points.
The audience that generates the most fake internet points are those looking to be spoonfed with the least amount of effort.
But people like these probably won't be barging at the mathematics door for a long time. So any effort is wasted. So it's like you said, wasted effort for points which somehow validate you
1:24 PM
Any effort is wasted in what sense? As far as most users are concerned, getting fake internet points is totally worth it.
In the sense that the people you "help" are not going to become good at mathematics
But that's not the goal most users have.
I think it depends. I had asked some relatively lazy questions back in the day and got pretty high quality answers. Sometimes the answers went way over my head but I felt an allure to it and kept pursuing it
We need more "pop math". There are many physicists who claim their gateway into it was pop phy , so similarly pop math to raise up interest and such
math is too abstract to gain large audience attention
There are good channels on youtube which do that already as they can. But it's limited because, math is abstract
There is plenty of pop math available. It's just less exciting than physics and chemistry.
1:29 PM
I can give a nice channel I found
this person has animated a full vector calculsu course like this
or 3blue1brown
I don't think abstractness is the issue. I think all math is at it's root derived from human experience, so there is something through which people can latch on to it and proceed further
maybe at highest level it is very difficult to do something like what I've shown but there is still a lot of work to be done before we reach that imo
Vector calculus is the beginning of the ice berg
I do think abstractness is the problem, you won't truly understand a subject until you read about it
I mean pop math is for inculcating interest
Pop <topic> isn't meant to convey understanding, @Jakobian
@Jakobian I don't know what you mean by "shouldn't we". Who is "we"? Everyone is entitled to make their own judgements about what questions are useful, or show sufficient research, or whatnot.
How do you get people to actually understand the topic then. Interest alone won't suffice because the people that "get interested" don't look to learn about those things on their own, but rather seek videos about these topics for example so that they explain it in a simpler language
Doing math is very different from getting math explained
1:34 PM
Also, when I suggested that flat out wrong questions fall under the category of "unclear", I was thinking of questions like "How do I show that 3 is an even number?"
With no other context or information about why the asker thinks that 3 should be an even number.
@Jakobian I think the argument being presented is that if there were more pop math, then more people would be interested in math. More people interested in math means a higher chance that someone would be inclined to pursue math seriously in the sense you are referring to.
In the case of the topology question being discussed here, the underlying question is about the definition of a topology. There is confusion about that definition, which is well documented in the question.
I don't claim that it is a great question, or an exemplar of questions on Math SE, but it seems to tick the minimal boxes. In my opinion, it is not bad enough to merit downvoting.
I suspect that it might be a duplicate, but after 10 minutes of searching, I can't find a good target.
I think this channel is better than what pop math can give you. It makes you try and grab a book about the topic instead of watching videos about math youtube.com/c/DanielRubin1
I mean let's take a real life example, let's consider comic books
avengers and all that existed since a long time , but it is only in like last 10 years that they gained the amount of popularity that they did
why is that? That's coz, it got made into a much more accesible pop medium. Now that it is, there are a lot of people who got an interest in comics book/ source material due to it
I mean. That does make sense. Someone would just have to show the general populi that math can be interesting. But I doubt that'll even happen because, well, there's a difference between diving into a math book and diving into a comic.
And I suspect that a lot of people would lose interest at that point
1:41 PM
Well pop math already exists, Jakobian.
Well maybe pop math is bad. Because it offers you a nice candy and in exchange you get a ripe tomato
Maybe it's better to advertise ourselves as vegetable sells-men
I don't think there is anything inherently wrong with pop math. You can get all sorts of cool ideas across without handing people a textbook.
Yeah it's good. But maybe not the best way to invite someone into mathematics
I don't think anyone made the suggestion that it is the best.
It's just one of the many ways that one could spark an interest in mathematics.
Also can we just take a moment and all agree that Edwards' Galois theory book as an intro to abstract algebra is a horrible idea.
1:58 PM
I never read it, and I'm not interested in Galois theory, but if I still were in high school then it'd probably a book I'd be craving.
I'm never buying any apple product again.
Not even granny smith?
@anak Granny Smiths are terrible apples.
Get a good apple. Like a Honeycrisp, or a Pink Lady.
there are fruits by brand names also?
@XanderHenderson Some apples are good for eating, some good for cooking/baking. I wouldn't spend all that money on pink lady apples just to bake them into a pie.
2:09 PM
There is your problem. Apple is an inferior pie fruit.
I find that it is great.
@Koro they are varieties of apples. Like dogs, you can breed apples to have flat faces, curly fur, and pointy snouts.
On the topic of math, I was reading about the h-cobordism theorem.
I saw some cubical apples online but don't know how they taste like.
@anak I don't want to hear about your pants.
And I find that the proof of the Poincare conjecture for $n>4$ to be quite impressive.
Don't worry, I am wearing shorts today.
@anak Gross.
2:17 PM
I need to digest the proof of properties of $\mathcal{D}(\Omega)$ still, the space of smooth functions on $\Omega$ with compact support.
@Jakobian do you know if it has an atlas?
$C^\infty$ functions as a whole does, but I wonder if compact support ruins it.
in what sense?
Like, is it a manifold.
it's not metrizable
Manifolds don't need to be metrizable.
2:26 PM
A: Is every topological manifold completely metrizable?

Nate EldredgeYes. The proof I know is a little roundabout. Let $M$ be your manifold. It is locally compact and Hausdorff, so it has a one-point compactification $M^*$ which is compact Hausdorff. Now $M^*$ is again second countable (see One point compactification is second contable), and (locally) compac...

then what do you mean?
Unless I am mistaken, you need paracompactness.
Though I wonder if that is sufficient in the infinite dimensional case.
Well, the definition of a manifold I know is that of Lee, of a second countable, locally Euclidean, Hausdorff space
Well if we were going by Lee's definition, then I think just based on the fact that it has to be locally homeomorphic to R^n for some finite $n$ is enough. :P
If they're not locally Euclidean, then I'm not sure what kind of manifolds are we considering here. I heard of infinite dimensional manifolds in the book by van Mill, but I didn't have the chance to study them yet.
2:47 PM
Which book is that? Is it any good?
Wacky question
Q: Does this knot collinearity condition have a nontrivial solution?

Akiva WeinbergerRecall that a link is an embedding of some number of circles in space up to isotopy. (Think "a knot but maybe multiple pieces.") Three oriented links form a skein triple if they have identical diagrams except for at one place, where they differ as in this image: (This image should be familiar to...

@BalarkaSen Thoughts?
@AkivaWeinberger do you have any nice way of imagining such functions $F$?
I have a bunch of points in space and each point has a label that's a link
Like is there a typical way of sending a link to a vector in R^n that inspires this problem?
Nah it was just a random thought
2:57 PM
@XanderHenderson Granny Smith apples make the best pies.
Is there any reason to believe this has anything to do with the actual knots? I just am wondering if the presence of "knot" in it is a red herring or not.
I mean, knots are links
Links are just more general than knots so that's why I used them
Yes, I know, but I am wondering what thinking about it in terms of knots/links gets you.
(also if $L_-$ and $L_+$ are knots then $L_0$ is a link 'cause it disconnects it)
Like if it were just {knots} instead of {links}, then you could phrase it in terms of sequences of {-1,0,1}.
Probably with some equivalence relation, necessarily. :P
3:03 PM
Not sure I understand
"sequences of -1, 0, 1"?
Like binary sequences but can take on -1
By the way, I'm working with links rather than link diagrams
and the same link can have lots of different diagrams
On second thought you need something quite a bit more complicated, I guess.
I am just thinking something like the planar diagram presentation used by knot theorists.
Maybe it is the simplest way of putting it. It just doesn't seem to be a "knot theory" question so much as it is a combinatorics one.
@anak Here's an example
So I was wondering if there was a specific reason knots come into play.
3:08 PM
(unknot, unlink, unknot) form a skein triple
because if we start with that figure-eight diagram of the unknot as $L_-$, then $L_+$ is also the unknot and $L_0$ is two unknots
(Of course this doesn't help us because this just means $F(O)$, $F(OO)$, and $F(O)$ lie on a line, which is immediate since two of them are the same point)
A better example would be to start with a trefoil
This shows us that F(trefoil), F(hopf link), and F(unknot) are collinear
So if you defined F according to knot diagrams, you'd need it to be a link invariant, and this SCC property, right?
Maybe there is an existing link invariant that has a useful skein relation that can be exploited to gain the SCC? Probably something you have already thought of.
I mean, HOMFLY is the closest thing
I am unaware of any nice skein relations for Khovanov.
3:17 PM
Up to a change of variables, Homfly is basically defined by $P(O)=1$ and $xP(L_-)+yP(L_0)+zP(L_+)=0$
where $P$ maps oriented links to polynomials in $x$, $y$, and $z$
It's nonobvious that $P$ exists and is unique
but not too hard to show
3:30 PM
@robjohn Again, apples are an inferior pie fruit.
@anak The Infinite-Dimensional Topology of Function Spaces
Personally it has so much different fields of topology in it that it's never a waste to read through it
and it's fun
I kind of stopped reading but, but I've done the whole appendix together with exercises in it
I remember I've stopped reading at simplicial stuff, I guess it must be some kind of fear of mine about simplexes.
Maybe it's because the statements are so combinatorial that it's hard for me to sometimes formalize the ideas and I struggle to understand it without such things.
4:19 PM
@Jakobian I find drawing pictures to be of special importance to dealing with simplicial stuff.
4:51 PM
@anak Don't let Leslie hear this.
1 hour later…
6:14 PM
Leslie is a duck-sympathizer.
his favourite movie is The Mighty Ducks (1992)
@anak Not Howard the Duck?
Or DuckTales: The Movie: Treasure of the Lost Lamp?
No, he works in law and the closest thing you can get to golf and ducks is The Mighty Ducks
@TedShifrin I am having troubles understanding the "geodesic spray". Are you familiar with it?
Isn’t that just the geodesic flow on the (unit?) tangent bundle?
I think the analogy is that the geodesic spray is the Hamiltonian vector field whose Hamiltonian flow coincides with the geodesic flow.
I think
With that analogy, it sounds like geodesics can be determined by the geodesic spray.
That it's just the ODE version of geodesics.
The answer is that you’ve thought about it more than I have. But geodesics are always ODEs, so I don’t know what that remark means.
6:31 PM
I am wondering if they do this in particular for infinite dimensional manifolds. Geodesics are ODEs in a local chart, but in infinite dimensional manifolds the chart stuff starts to go funny.
In particular this paper claims that computing Christoffel symbols is hopeless in this particular infinite dimensional case they are considering.
So I think that it's at least not easy to express geodesics as solutions to ODEs in a chart.
OK, that sounds reasonable. Totally not my pay grade.
Hi @Ted
Hi Lukas
@anak Maybe Sternberg’s diff geo book will help.
I've been reading some stuff you would like. I think you recommended Forster's Riemann surface book to me ages ago
Thanks Ted, I will have a look!
6:45 PM
Yeah, I have taught out of Forster several times.
Hi everyone!
Let $Q=\{9,11,13\}, S=\{11,13\}$. The statement $S \subset Q$ is true. When thinking about $S \nsubsteq Q$, is it True or False. I mean it is false that it is not a subset, but here when the equal condition in subsets $\subseteq$ is also considered is there any difference?
It allows for equality. It does not mandate it.
@Hasini do you mean $\subsetneq$ or $\not \subseteq$?
I personally write $\subset$ for $\subseteq$.
Oh, missed that typo n …
@TedShifrin Okay... Then $S \not \subseteq Q$ is false right
@TedShifrin :) it wasn't a typo
6:48 PM
Yes, false.
I wanted to mean $\not \subseteq$
Still a MathJax typo.
@LukasHeger Thanks @LukasHeger yes I wanted to mean this
@TedShifrin Aaa yes :)
You are negating an “or” statement. Understand it that way.
@TedShifrin Thanks @TedShifrin
6:49 PM
@Ted I read somewhere that you can use Hermitian metrics to prove the finiteness of $H^1(X,\mathcal O)$ on a compact Riemann surface. That's different from how Forster does it. Do you know the idea of the proof using Hermitian metrics?
@TedShifrin I didn't exactly get this...
Hodge theory?
How it is related to or statement?
@Hasini $\subseteq$ means $=$ or $\subsetneq$
@Hasini NOT (proper subset OR equal)
6:52 PM
Aaa okay. Then it can't be written as $\nsubseteq$ because then the not will not get applied to both parts?
Anyway, I'm sorry. I'll think about it more and understand it. Thank you very much @TedShifrin and @LukasHeger
Have a nice day!
You too
@TedShifrin yes this is seems to be the proof that Wells uses (for more general compact complex manifolds of any dimension)
seems like you can't avoid functional analysis for this result
7:08 PM
Q: Prove or disprove : If every functions with closed graph are continuous then the target space is compact.

Sourav Ghosh$(X, \tau_X) $ and $(Y, \tau_Y) $ be two topological spaces. $\forall f\in Y^X$ with $\text{Gr}(f) $ is closed implies $f\in C(X, Y) $. Question : Does this implies $(Y, \tau_Y) $ is compact? Notation: $Y^X$: Set of all functions from $X$ to $Y$. $C(X, Y) =\{f\in Y^X: f \text{ is continuous }\}$...

Any non trivial example?
@LukasHeger right, unless you appeal to algebra and GAGA, maybe
@TedShifrin I don't see how I don't see how you can get that compact Riemann surfaces are algebraic without some form of functional analysis. Once you have that you can appeal to GAGA of course
7:32 PM
True, but even then finite-dimensionality of coherent sheaf cohomology isn’t easy? Honestly, I have no recollection.
functional analysis is a necessary evil
@TedShifrin I think you use that Cech cohomology vanishes if the degree is high enough and use some Noetherian induction argument
but I'm not sure if I remember this correctly
7:56 PM
2 messages moved to ­Trash
2 hours later…
10:17 PM
better than nonfunctional analysis
10:34 PM
Speaking of popularizing mathematics, National Geographic released a series of some kind of books (?) in Poland called "Świat jest matematyczny".
They even had an ad in the tv
The title means "The world is mathematical"
10:50 PM
I canceled my National Geographic subscription as soon as I found out that they published a digitally altered cover without indicating that it was so.
11:05 PM
what is digitally altered cover
@Jakobian I think there are many levels of popularization
too much popularization oriented makes subject material lose all meaning
too little is a textbook in itself

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