Mathematics - 2022-10-26

Akiva Weinberger

00:35

I am clearly very rusty with my general topology

6

Q: Do the connected sets determine the topology in the case of manifolds?

Let $f:M\to S$ be a bijection between two spaces, at least one of which (without loss of generality, $M$) is known to be a manifold. Suppose $f$ preserves and reflects connected sets, meaning $X\subseteq M$ is connected iff $f(X)$ is connected. Must $f$ be a homeomorphism? It seems to me that the...

That's a neat counterexample that I would not have spotted

one potato two potato

01:59

17 hours ago, by one potato two potato

A smooth vector (or tensor) field $W$ is invariant under the flow of $V$ (defined on a smooth manifold) if and only if its Lie derivative $\mathcal{L}_VW = 0$. That is one importance I can consider. Any other importance? When do people take Lie derivative to observe what? One motivation given in Lee's textbook is the question: What is the directional derivative of a vector field? It's not very tempting to me.

Even Wiki says that Lie derivative evaluates the change of a tensor field along the flow defined by another vector field.

So the case $\mathcal{L}_VW =0$ is meaningful (and only the meaningful case?).

Joe Shmo

02:16

hi everyone

copper.hat

03:38

hi Joe

Seeker

Hi! @JoeShmo

Ted Shifrin

JoeShmo is long gone.

Cotton Headed Ninnymuggins

04:19

Is it ever ok to ask and answer your own questions on this site for reputation? If so, under what circumstances?

leslie townes

i'm not an expert on site policy, but if the question meets the site guidelines (well posed, not a duplicate, etc.) it seems fine to me. if it's only being done 'for reputation,' i might wonder about the quality of the self-assessment of whether something meets the site guidelines, but there's a review process for that.

if nobody could tell from the question and answer alone that the focus was to add to rep, i don't see a problem with it. the intentions of the asker/answerer don't matter as much to me as the quality of the question/answer.

copper.hat

i have answered my own question. not for rep but i thought someone might be interested.

doing it for rep might affect your vision.

leslie townes

i mean, that's basically why the feature exists, i as i understood it. it's not limited to situations where you ask, wait, don't get an answer, and then find out the answer. you can also use it just to document an interesting problem.

and maybe get other interesting answers.

as a practical matter i wonder if you can really juice your rep that much solely by answering your own questions. i do think you can accept your own answers but it would surprise me if it gets you the same rep. and if other people upvote a submission once it's out there because it's a good q/a, that's just earned rep, not self promotion.

copper.hat

i used it to communicate secret messages. mostly encoded using rot$-{ 1 \over 12}$.

leslie townes

oh, i use answers to other people's questions for the purpose of communicating with my home planet. self answers would be too obvious

copper.hat

04:30

i hide them in closed questions while publicly decrying the procress.

user4539917

Internet points do not count as credit for
a university degree

copper.hat

they do count towards the mse jumpsuit

Ted Shifrin

Jumpsuits are a pigment of your ancient imagination

user4539917

04:48

Has anyone know name of this notation:

(a, b, c) = { a, { { a, b }, { a, b, c }}}

The author says it's a definition of the ordered triple in terms of pairs.

Cotton Headed Ninnymuggins

@user4539917 Really? I'm just going to school for fun while I develop my expertise and wisdom on StackExchange and Reddit. Guess I was going about it wrong, thank you for making it clear

PseudoLooped

05:09

@CottonHeadedNinnymuggins you can’t gain rep for accepting your own answer, but there is a badge that’s titled “Self-Learner” which rewards you for answering your own question. tho, i feel like base line is that you should answer your own question with the same amount of sincerity and approach you would do for any one else’s question. the more perspectives answering a question the better; people learn differently and broadly with multiple explanations

and it’s kinda neat to watch and learn someone else’s development, gives an idea on how proper self-studying and learning should be from what is asked, and what is self-answered

Seeker

Hi, I was trying to answer someones question and I am reading something regarding that. In what I am reading, they use the phrase "vanish identically" when referring to the determinant. Here is what I am talking about. What does the phrase vanish identically mean?

Ted Shifrin

@Seeker: It's a function of $x$. Vanishing identically means that it is $0$ for all values of $x$.

Seeker

@TedShifrin Oh ok. Thanks a lot for explaining!

Ted Shifrin

Sure thing :)

Oliver Díaz

Hi @copper.hat, I have a question that I think you may be able to answer. In general topological linear spaces, what is th difference between directional differentiability in the sense of Hadamard and compactly directional differentiability?

user4539917

05:23

@CottonHeadedNinnymuggins sorry if I offended you

I guess I was being overly sarcastic

Seeker

I doubt I will be able to learn enough about Wronskian is a short time frame. But this question could use some help in answering what is going wrong with the calculations in the question. I did provide a proof that contradicts the answer they got using Wronskian.

But after having read up on it and in my limited understanding, I am confused about where they are going wrong.

Ted Shifrin

05:40

@Seeker: I answered it.

Seeker

@TedShifrin Thanks! I will try to understand what is going on!

user4539917

Professor @TedShifrin what advice do you give grad students to avoid burn-out?

in The h Bar, 2 hours ago, by Sir Cumference

is it normal to feel perpetually exhausted in grad school

Cotton Headed Ninnymuggins

06:02

@user4539917 You didn't really offend me, I was being overly sarcastic back to you ;)

user4539917

:-)

Rover

06:55

Did we got the second equation by observation?

one potato two potato

09:41

I recently felt that the reason for learning smooth manifold theory is to learn Riemannian manifold theory. Lots of higher mathematics related to smooth manifolds always uses the Riemannian structure.

Yai0Phah

12:25

@leslietownes IMHO, the correct context of this is the structure theorem of finitely generated modules over a PID. Proofs that only uses linear algebra are not informative to me.

@onepotatotwopotato There is no "the Riemannian structure" on smooth manifolds.

one potato two potato

a?

Thorgott

13:12

differential topology is very much an independent subject that is distinct from riemannian geometry

riemannian structures can appear as tools as in differential topology, but they're not the object of study

Sonozaki

15:52

Let $f_n(x)=nx/(3+n^4 x^4)$ for $x \in [0,2]$. It is $f_n (x) \to 0$ as $n \to \infty$ for any $x \in [0,2]$, hence $f_n$ converges pointwise to $0$. It is correct to say that $f_n$ does not converge uniformly to $0$ in $[0,2]$ because for any $n \in \mathbb{N}$ it is $1/n \in [0,2]$ and $f_n(1/n)=1/4$, hence for $0<\epsilon \le 1/4$ the definition of uniform convergence is not satisfied?

Ted Shifrin

16:02

@Sonozaki That’s right.

copper.hat

16:15

@OliverDíaz How do you define the compact directional derivative? (So, I guess the immediate answer is I don't know :-).)

Oliver Díaz

16:34

@copper.hat I thought It would not hurt to ask. This kinds of derivatives are used in certain optimization problems where the domain space is a topological vector spaces (not necessarily metrizable).

@copper.hat: Here is more or less the general notion: In any case, suppose $X$ and $Y$ are topological vector spaces $f:X\rightarrow Y$. Consider a collection of sets $\mathcal{H}\subset $Y$. Then $f$ is $\mathcal{H}$-directional differentiable at $x$ if there is a homogenous function $A_x:X\rightarrow Y$ such that $\lim_{t\rightarrow 0}\frac{f(x+th)-f(x)}{t} = A_x(h)$ uniformly for over $h\in S$ for each $S\in\mathcal{H}$.

@copper.hat: When $\mathcal{H}$ cosists of finite sets, one get the Gateaux derivative; when $\mathcal{H}$ consist of bounded sets for example. When$\mathcal{H}$ consists of sequentially compact sets then one has the sequentially compact directional derivative. Hadamard derivative seems similar to that one.

@copper.hat: Sorry for the LateX mess: Here is the notion of derivative I am taking about: $X,Y$ topological vector spaces, $f:X\rightarrow Y$ and $\mathcal{H}$ a collection os sets in $X$. $f$ is $\mathcal{H}$-directional differentiable at $x$ if there is a homegeneous map $A_x:X\rightarrow Y$ such that $\lim_{t\rightarrow0}\frac{f(x+th)-f(x)}{t}=A_x(h)$ exists uniformly over $h\in S$, $S\in\mathcal{H}$.

Koro

16:50

Does there exist a $g$ such that $g$ is continuous on $(0,1)\times (0,1)\subset \mathbb R^2$ and that $g$ is integrable with respect to $\lambda_2$ but $\int g(x,y) d\lambda (y)=\infty$?

I think no it doesn't as if it did then Fubini's theorem will be violated.

Oliver Díaz

@copper.hat: when $\mathcal{H}$ consists of finite sets, then we have Gateâux's derivative. Two other types of derivative of interest are when $\mathcal{H}$ consists of sequentially compact sets and when $\mathcal{H}$ consists of bounded sets. Hadamard's derivative seems tone of the first type (the later plus additional structure is associated top Frechet derivative). My question is related to whether Hadamard and sequentially-directional derivative are the same thing.

@Koro: if it is integrable with resect $\lambda_2$ (presumably the Lebesgue measure in $\mathbb{R}^2)$ the set where $\int g(x,y) dy=\infty$ (if it is not empty) has measure $0$ (Fubini-Tonellii's theorem).

Ted Shifrin

@Koro I was about to ask. You mean for some single $x$ or for all $x$?

Koro

Hi @TedShifrin @OliverDíaz! I am trying to understand a question. I got it now. The posting above was just a little modification of the question in which it was meant for single $x$.

Ted Shifrin

One example would be $g(x,y) = (x^2+y^2)^{-1/2}$, perhaps?

Koro

17:07

I don't think that will work because the iterated integral (in y) will look like $\log (1+\sqrt{x^2+1})-\log x$

ugh, I integrated it wrong!

Ted Shifrin

Restrict to $y=0$ and you get $\int_0^1 dx/x$.

Why is it integrable on the square?

Koro

nope, my integration was correct. And the hint given was: $y^{\phi(x)-1}=g(x,y)$ for some appropriate $\phi$.

Ted Shifrin

Huh?

Koro

this was for my earlier comment. I'm thinking about your comment now.

Ted Shifrin

I totally do not follow you.

Koro

17:18

Does there exist a $g$ such that $g$ is continuous on $(0,1)\times (0,1)\subset \mathbb R^2$ and that $g$ is integrable with respect to $\lambda_2$ but $\int g(x,y) d\lambda (y)=\infty$ for some $x\in (0,1)$.
The hint to find such $g$ was to consider $y^{\phi(x)-1}$ for some appropriate $\phi$ so I was trying to construct that.
I'm also thinking about the example that you gave, which didn't seem to work as $\int_0^1 \frac 1{(x^2+y^2)}d\lambda (y)=\log (1+\sqrt{x^2+1})-\log x$, which is $\lt \infty$ for every $x\in (0,1)$.

Thorgott

pick $x_0\in(0,1)$, set $g(x,y)=0$ if $x\neq x_0$ and $g(x,y)=\infty$ if $x=x_0$?

Ted Shifrin

My example is fine.

Koro

$\int_0^1 \frac 1{y^a}d\lambda (y)$ is integrable iff $a<1$ so I took $\phi(x)= x-2$ and then $g(x,y)=\frac 1{y^{3-x}}$. Now, trying to show if it is integrable w.r.t. $\lambda_2$.

Ted Shifrin

Oh, I was doing the closed square. Big deal.

Koro

ohh, will that matter?

Because {0} and {1} are of measure $0$.

Thorgott

17:25

oh, I didn't see Ted's example

that's also instructive

Koro

but doesn't work as iterated integral is finite?

Thorgott

but I supposed mine is the trivial one

Koro

Thor: your example is not continuous.

Ted Shifrin

@Koro It is not for every $x$. Did you read what I said above?

Thorgott

oh, I missed that adjective

Koro

17:30

But why should one restrict to $y=0$? I think you meant restrict to $x=0$, which seems not correct as corners are not inside the domain. :(

But as far as integration is concerned, that shouldn't matter due to measure 0 but that makes me confused.

Ted Shifrin

Mine is continuous away from the origin, so continuous on open square. Shrug.

No, I meant what I said,

The point is that a.e. Iterated integrals will be fine. Sets of measure zero matter for this.

copper.hat

18:06

@OliverDíaz Those distinctions are beyond my working knowledge, mine would be finite (Gateaux), compact (Hadamard) & bounded (Frechet).

Oliver Díaz

18:23

@copper.hat: Like I said, it would not hurt to ask... I know a paper where this things are discussed but I have no access to a good university library at the moment: Sova, M., General Theory of Differentiation in Linear Topological Spaces, Czechoslovak Mathematicsl Journal, Vol 14., pp. 485-508, 1964.

Ted Shifrin

@copper.hat You mean you can't differentiate among them?

copper.hat

@OliverDíaz my main foray in such things is in the real of non differentiable analysis but the focus there is more on the $'x$ variable changing rather than the direction.

@TedShifrin :-)

Oliver Díaz

18:47

@copper.hat: All this came up when reading some papers about non-parametric estimators where the optimization variable of the objective function is in the convex set of probability distributions, as a subspace on spaces of measures of finite variation with the weak topology. These nonparametric estimators have good asymptotic properties and I am using them at work; however I wanted to understand why they are good. In any case, I believe the authors (well known Statisticians in the Bay Area)

copper.hat

stats is way outside my range

Goku

I can't be the only one who thought that stats just seemed "odd" when compared to most "normal" mathematics.

I mean, from my narrow experience, it seemed no where near as rigorous as say, calculus 2 or something and yet at the same time it isn't exactly "easy" either

copper.hat

that is a broad generalisation, i know plenty of rigorous stats folks.

Ted Shifrin

The undergrad students, not so much. Once you start requiring linear algebra and real analysis (undergrad and grad), you’re ar the definite grad level.

leslie townes

19:06

yeah, stat has one of the widest gaps in what the experts know and what non grad students are taught.

when i was in undergrad it was kind of a fallback for people who weren't doing so well in the applied math major (which itself was a fallback for people who couldn't get into computer science)

Oliver Díaz

19:19

Not a statistician myself, but it uses a little but of everything I like: probability theory, stochastic processes, Harmonic analysis (parametric estimation in related to Groups of matrices) optimization, computer science, and even PDEs (calibration of parameters in SDEs and SPDEs).

Yai0Phah

@copper.hat Do they make sense for arbitrary topological vector spaces? I only know the definition of Fréchet for Banach spaces.

copper.hat

Banach spaces.

Yai0Phah

Gâteaux derivative makes sense for arbitrary TVS's.

Oh, no, maybe local convexity seems necessary.

Oliver Díaz

@Yai0Phah: The Gâteaux derivative makes sense in general tvs. So Hadamard's derivative which is a little more useful than Gâteaux's There is a survey by Averbuks and Smolyanov, some times a little dense for me, that discusses the theory of differentiation in tvs.

Yai0Phah

19:42

I find it unnatural to consider sequences in non-first-countable spaces.

Jakobian

19:57

@Yai0Phah huh?

there are sequential spaces which aren't first countable

Ofek

i am trying to make hyperbolic game but i can't find enough information about hyperbolic geometry. the best thing i found is this mphitchman.com/geometry/section5-1.html . im using the transformation they specify there but i don't understand how i can compose 2 transformations, does anyone know?

Akiva Weinberger

20:16

@Ofek Plug the output of one into the input of the next

Shaun

Hi all. I just want to draw some attention to the following question:

1

Q: A morphism of affine varieties $\phi: X\to Y$ is an isomorphism iff the algebra homomorphism $\phi^*$ is an isomorphism.

This is Exercise 1.4.8(3) of Springer's book, "Linear Algebraic Groups (Second Edition)". The Question: A morphism of affine varieties $\phi: X\to Y$ is an isomorphism if and only if the algebra homomorphism $\phi^*$ is an isomorphism. The Details: Since definitions vary: A topological space $...

abstract-algebra general-topology algebraic-geometry affine-varieties zariski-topology

I'm keen to find an answer . . .

Ofek

@AkivaWeinberger i tried that, but it doesn't give me the same structure of the transformation

the best i could do is to extract the $e^{i\theta}$ because geometrically $e^{i\theta}T(z)=T(e^{i\theta}z)$

Thorgott

@Shaun try showing that a map $\phi\colon X\rightarrow Y$ is a morphism if and only if its coordinates are regular functions on $X$

Overtherainbow

20:42

Let $X=C([0,1])$ the space with the norm $||f||_\infty=\max|f|$. Then for a fixed $\phi$ define $L:X\rightarrow \Bbb{R}$ such that $f\mapsto \int_0^1 \phi(t)f(t)dt$. I need to show that $||L||=\int_0^1 |\phi(t)|$.

I did the following: We know that for $f\in X$ with $||f||_\infty \leq 1$ we have $||L(f)||=|\int_0^1 \phi(t)f(t)dt|\leq \int_0^1 |\phi(t)| |f(t)|dt \leq \int_0^1 |\phi(t)| \max|f|dt\leq \int_0^1 |\phi(t)|dt$ so by definition we know that $||L||\leq \int_0^1 |\phi(t)|dt$. Now I wanted to find $f$ such that $||L(f)||\geq \int_0^1 |\phi(t)|dt$ is this the correct idea?

leslie townes

yes. i might use | | instead of || || around L(f) which is after all only a real number.

i might also write $\|f\|_{\infty}$ where you have written $\max |f|$, to make the relation between that and the hypothesis $\|f\|_{\infty} \leq 1$ more explicit.

but those are notational nits.

Overtherainbow

right, sorry. But now could you give me a hint how to find $f$ so I first tried to take $f=1$ but then I always get to the problem that when I take the absolut value inside the integral it gets bigger. So I tried another way. My next idea was somehow to use that we have a continuous function on a compact interval, i.e. it attains it's maximum but also there I get some troubles with the absolute value.

leslie townes

i guess i should say, given $\epsilon > 0$ you want to find $f$ so that $\|f\|_{\infty} \leq 1$ and $|L(f)| > \int_0^1 \phi(t) \, dt - \epsilon$. this is a slightly easier problem than what you have set out.

the f you wrote up above would be an f that attains the norm of $L$, and while such f may exist for specific phi, you may not generally know that such an f exists for all phi, and even if it does exist for some phi, finding f for a given phi might be more difficult than showing only that the norm can't be smaller than $\int_0^1 |\phi(t)| \, dt$.

f = 1 certainly works for some phi (which ones)? how would you handle the case $\phi(x) = \sin(2 \pi x)$?

Overtherainbow

My first question is why does it helps me to show $|L(f)| > \int_0^1 \phi(t) \, dt - \epsilon$ or in particular $|L(f)| > \int_0^1 \phi(t) \, dt$? I don't see why this should be useful

Oliver Díaz

@Overtherainbow: the idea is to approximate the function $g(x)=\operatorname{sign}(x)$ by continuous functions in the metric $\|f-g\|_1=\int_{[0,1]}|f-g|$. A little Lebesgue integration would do it, but If that is not under your belt, it might still be possible, but a little harder.

Shaun

20:55

@Thorgott Okay, thank you. I'll think about that for a while . . .

leslie townes

overtherainbow: think about the example phi(x) = sin(2 pi x)

Overtherainbow

@leslietownes I would take $f(x)=2$ for $x\in [0,1/2]$ and zero else then I would get $|L(f)|=\frac{2}{\pi}=\int_0^1 |\phi(t)|dt$ right?

Avv

Hello Guys,

I am trying just to figure out limit below:

$
\underset{1\rightarrow 0}{\lim}f\left( x+\text{1,}y \right) -f\left( x,y \right)
$

Instead of $h$ in the limit we have 1, is that okay?

Also is the answer should be $
f\left( x+\text{1,}y \right) -f\left( x,y \right)
$

I am I right please?

leslie townes

over: recall that to get lower bounds on the norm of $L$ you need to look at $|L(f)|$ for $\|f\|_{\infty} \leq 1$ (or $\|f\|_{\infty} = 1$, same thing), or otherwise consider $|L(f)|/\|f\|_{\infty}$ (if $\|f\|_{\infty} > 1$). i mention this because your example $f$ has $\|f\|_{\infty} = 2$.

Overtherainbow

oh right I forgot about this condition sorry. So yes my $f$ does not work

leslie townes

21:07

per oliver's suggestion we might want to look for f for which the product $f(x) \sin(2\pi x)$ is pointwise equal to $|\sin(2 \pi x)|$, at least a whole lot of the time. we can't do this with $\pm 1$-valued $f$ because such $f$ will not be continuous. but we can sneak up on that.

Overtherainbow

does then $f(x)=1$ for $x\in [0,1/2]$ and $f(x)=-1$ else work?

leslie townes

it's definitely the right idea. it's what we'd want to use, only it isn't in $X$.

Avv

@leslietownes. What do you think please about my question?

Overtherainbow

oh sure, hmmm so we need to sneak up on that

robjohn

@Avv $\lim\limits_{1\to0}$ does not make sense. $1$ is not a variable; its value is fixed and cannot tend to $0$.

Avv

21:15

@robjohn. Thank you.

what if we have it like this:

$
\underset{h\rightarrow 0}{\lim} \frac{f\left( x+\text{1,}y \right) -f\left( x,y \right)}{h}
$

I got the answer is $f(x+1, y) - f(x,y)$

Overtherainbow

@leslietownes sorry could you maybe give me a hint in which direction I need to think?

leslie townes

how about f(x) = 1 for x in [0, 1/2 - small number] and f(x) = -1 for x in [1/2 + small number, 1], and a straight line connecting the dots on [1/2 - small number, 1/2 + small number] to make it continuous?

if that small number is small enough, you can make $\int f \phi$ as close as you like to $\int |\phi|$

showing that $\|L\| = \sup \{|L(f)|: f \in X, \|f\| \leq 1\}$ is at least $\int |\phi|$ even though this approach does not exhibit any specific $f$ in $X$ for which $L(f) = \int |\phi|$ holds

Overtherainbow

ah okey yes I see. But I mean in your case it is easy to see that $1/2$ is a crucial point, but for a general $\phi$ it could be that I have to define $f$ on multiple intervals with the same method?

leslie townes

yes, this train of thought heads back to oliver's comment

Overtherainbow

The problem is that I don't understand his comment. Why do I have to deal with the signum?

leslie townes

21:25

although it is not in general continuous, sign(phi) is a function of sup norm 1 that makes the pointwise product sign(phi) phi equal to |phi|

Overtherainbow

So does he want to tell me that this small intervalls on which we construct the "straight" line has Lebesgue measure zero?

leslie townes

in the toy example up above, the f we were trying to sneak up on was sign(sin(2 pi x))

the sign of the sine, ha ha

Overtherainbow

But I mean wouldn't sign(sin(2 pi x)) be continuous i.e. wouldn't we have a problem?

leslie townes

oliver's idea in general is to come up with continuous approximants to sign(phi) in general, perhaps not exactly in the way that we just 'wrote down' continuous approximants to the sine example

how measure theoretic this gets might depend on what tools you have access to

Overtherainbow

So I had a starter course in measure theory.

leslie townes

21:30

it doesn't strike me as something that necessarily requires too much measure theory, but we're in the realm of making choices in how you approximate a function, and different people might make different choices

robjohn

@Avv The answer to that would be undefined if $f(x+1,y)\ne f(x,y)$ and $0$ if $f(x+1,y)=f(x,y)$

leslie townes

as a start maybe think about solving the problem for constant sign functions max(phi,0) and min(phi,0) in a way that would allow you to combine them together

but this is already getting into the realm of, here's my sequence of arbitrary choices for attacking this problem

Overtherainbow

Okey it seems not the easiest problem.
I will think about it

leslie townes

the shortest answer i can think of would make use of at least the monotone convergence theorem

Overtherainbow

right so I know this one

so you mean that when we have a monotone increasing sequence of positive measurable functions then we can swap the limit and the integral (roughly said)

And in our case measurable can be exchanged with continuous

Ted Shifrin

22:23

Anyone else using the SE app on iOS? I am unable to post comments on there today, but can post fine from my desktop.

I just posted on meta asking about this.

Well, the apps are intended to be worthless and not functional. I got a downvote and an answer.

leslie townes

22:47

haha

seems like they could have pushed a notification of this to users of the app, instead of just a generic 'ask meta' error message

Ted Shifrin

You think?

Oh, they removed the downvote.

Trying to figure out how to see my favorite tags from web browser on ipad. Yuck.

Ah, they show up if I turn it … now to try the phone.

No go on the phone. Maybe @robjohn has sage advice.

Oliver Díaz

23:04

@Overtherainbow Consider the set $\{|\phi|>0\}$. Being open, it is the disjoint union of countable open sets, say $(a_n,b_n)$ (to avoid problems, extend $\phi$ outside $[0,1]$ in the obvious way. Now, on each $(a_n,b_n)$ you can construct a function $\phi_n$ that I one in middle subinterval of length $(1-2^{-n})|b_n-a_n|$, zero outside $(a_n,b_n)$ an linear otherwise.

@Overtherainbow The function $\psi=\sum_n\operatorname{sign}(x)\phi_n(x)$ is continuous and should approximate the sign function nicely in the metric $\|f-g\|_1$.

robjohn

@TedShifrin The suggestions on meta that I've seen say, "don't use the mobile app."

sorry, that is about the state of things.

not very sagacious

Ted Shifrin

Yeah, I’ve switched to browser, but on the phone can’t see the important column on the right.

robjohn

Yeah, it doth suck

Ted Shifrin

I figured you were sagaciouser than I.

Thorgott

@TedShifrin do you know if the infinite genus surface you proposed is a counter-example to that recent topology question or was it an open question?

Ted Shifrin

23:17

I have no idea. It just leapt to mind.

Thorgott

fair, I have no idea either. it does seem natural to consider

I tried throwing all the algebraic topology tools I know at the situation, but all that tells me is that if I can find such an embedding, the complement of the image has to be immensely ugly

user unkown

23:53

Let $X$ be an LCH space (locally compact Hausdorff), $f:X \to \mathbb{R}^n$ a proper function and $c \in \mathbb{R}^n$. Suppose there is an open set $U$ in $X$ with $f^{−1}(c) \subseteq U$. Prove that there is $\epsilon>0$ so that $f^{-1}(B_{\epsilon}(c)) \subseteq U$.

@TedShifrin @Thorgott How do I do this?

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$Mathematics$ Mathematics