Here's my attempt, grateful for any feedback. $\overline{\mu}$ agrees with $\mu$ on $M$, same for $\overline{\nu}$ and $\nu$ on $N$. So $\overline{\mu}\times\overline{\nu}$ agrees with $\mu\times\nu$ on $M\times N$, and so also on $\sigma(M\times N)$ (do I have to require that the measures be $\sigma$-finite here?).

But $\overline{\overline{\mu}\times\overline{\nu}}$ agrees with $\overline{\mu}\times\overline{\nu}$ on $\sigma(\overline{M}\times\overline{N})\supset\sigma(M\times N)$ and hence $\overline{\overline{\mu}\times\overline{\nu}}$ and $\mu\times\nu$ agree on $\sigma(M\times N)$. Since completion is unique, the implication above follows.

@psie notationally awful

I'd just adopt the objectively better notation of Bogachev

In here you have the illusion that the over line means something. In Bogachev's notation the dependence on a given measure is explicit

also no one would write something as atrocious as $\sigma(M\times N)$, they would simply write $M \otimes N$

Rephrasing, the equality you are trying to show is $(M\otimes N)_{\mu\times \nu} = (M_\mu \otimes N_\nu)_{\overline{\mu}\times \overline{\nu}}$

where $M_\mu$ means completion of the sigma-algebra $M$ where $\mu$ is a measure on it

@psie To show this implication, it suffices to notice that completion of a sigma-algebra $\Sigma$ with respect to a given measure $\lambda$ is the sigma-algebra generated by $\Sigma$ and sets of the form $A\subseteq B\in \Sigma$ where $\lambda(B) = 0$. So all one has to show that the latter type of sets are contained in the completion on the right

So, lets suppose that we are given $B\in M\otimes N$ with $\mu\times \nu(B) = 0$, what is to show is that $\overline{\mu}\times \overline{\nu}(B) = 0$

@psie you do need $\sigma$-finite to obtain equality on $M\otimes N$ from the equality on the generators

But then again taking products, for them to be unique, you also need that

We could argue what if we take the Caratheodory construction for definition of $\mu\times \nu$ but eh

@psie but yeah, this is the thing that you actually want because it shows that if $A\subseteq B$ and $B$ is as above, then $\overline{\mu}\times \overline{\nu}(B) = 0$, and so $A$ is a null set of $\overline{\mu}\times \overline{\nu}$

if we go by "okay what about when they aren't sigma-finite route", then from definition $\overline{\mu}\times \overline{nu}(B) = \inf\{\overline{\mu}(A_n)\overline{\nu}(B_n): A_n\in M_\mu, B_n\in N_\nu, A\subseteq \bigcup_n A_n\times B_n\}$

we have $A_n = A_n'\setminus C_n$ and $B_n = B_n'\setminus D_n$ where $C_n, D_n$ are $\mu$ and $\nu$-null sets, so that we could replace $A_n, B_n$ by $A_n', B_n'$ in above

@Jakobian of course here I mean to sum $\sum_n \overline{\mu}(A_n)\overline{\nu}(B_n)$

either way this would show that even if we go for generality, the equality still holds

here of course $A_n'\in M, B_n'\in N$

@VladimirLysikov What is the point of learning something if you will forget it anyway?

@pie no point in anything

a point only exists with goal in mind

@VladimirLysikov @VladimirLysikovI have trouble remembering things I've learned. I even made a post on MSE about this

And this happens with topics I've studied and practiced, not just with papers I've read once.

@pie I made a github repository for my analysis / metric space notes+ cool theorems i came across from various sources + assignment solutions and so on. in this way I am allowed to forget things, its one click on your cellphone away

good chunk of that repository are filled with information jakobian gave though,xD

@AlessandroCodenotti I have what I think will be my final version of a theorem about zero-dimensional compactifications

I've shown that if $X$ is strongly zero-dimensional and $\beta X\setminus X$ is locally compact, then every compactification of $X$ is zero-dimensional if and only if $\beta X\setminus X$ is scattered.

@ThomasFinley did you get your answer?

For an example of a non compact set on which all continuous functions are uniform continuous, an infinite discrete metric space works no?

yes

hmm, N with usual metric also works

im wondering if some subset of ell_p or ell_infty also works

If you plot $x^y = y^x$ on Desmos, you get the line $x=y$ and another curve. Is there a formula that represents the other curve?

@SoumikMukherjee Nope.

@SineoftheTime yeah, but that excercise doesn't make any sense to me.

@copper.hat haha... ok, wait...

@SoumikMukherjee, @copper.hat Actually, I need some help on how to solve problems like this.

I was studying from Schaum series but the concepts of branch points and branch lines was introduced through some solved problems. I think that is precisely the reason why I am so terribly confused.

I don't understand how it does not make sense to you, it's the same exercise with a different function

@SineoftheTime which one are u talkin' about? The picture I just posted?

es. 2.42 Schaum

@SineoftheTime If you're willing then we can discuss about it in this room: chat.stackexchange.com/rooms/156168/… that I created.

I'm a bit busy rn

@SineoftheTime ok, no problem. But if you are free and if you agree to have a conversation about this with me please let me know by just replying a "hi" in that room ofc at ur convenient time. Thank you!

Applying the same procedure as here, you shouldn't have problems @Thomas

@SineoftheTime I presume you have some time to spare for me? Should I reply here or in that room. If you ask me, I think it'll be better if we discuss about it there. But whatever you say, will be ok with me.

if $f:X \to Y$ is uniform continuous function on a dense subset $D$ of $X$ , it must still be possible for $f$ to be non continuous on $X$ right?

@nickbros123 sure. Take the function which is $1$ on $\Bbb Q=D$ and $0$ on $\Bbb R \setminus \Bbb Q$

Righttt

Isn't Galois theory saying that you can not solve a infinite jigsaw puzzle with the pieces of 25 piece jigsaw puzzle?

How is Galois theory related to that?

hi

@RyderRude hi

hi

I need a hint in showing the following: Let $f:X \to Y$ (where $Y$ is complete) be uniformly continuous on a dense subset $D$ of $X$. Then $f$ carries cauchy sequences into cauchy sequences.

Let $f: G \to G'$ be a group homomorphism and $H'$ a normal subgroup of $G'$, $H = f^{-1}(H')$. Then $f^{-1}(H')$ is normal in $G$. The justification is the following: If $x \in G$, then $f(xHx^{-1}) = f(x)f(H)f(x)^{-1} \subseteq H'$, so $xHx^{-1} \subseteq H$.

How do we know that $f(x)f(H)f(x)^{-1} \subseteq H'$? It's clear that $f(H) \subseteq H'$ but we have two factors

f(x) normalizes H'=f(H)

$H'$ is normal

@SineoftheTime Oh, so $f(x)f(H)f(x)^{-1} = f(x)H'f(x)^{-1} = H'$

@nickbros123 We can even say $=$, not just $\subseteq$, no?

actually just $\subseteq$

Why?

@nickbros123 No, $=$ is correct

Your doubt is why $f(x)f(H)f(x)^{-1}\subseteq H'$ insted of $=H'$?

The other inclusion always holds

@BenSteffan I meant that my comment was wrong, equality was supposed to be replaced by subseteq

Thanks! And from $xHx^{-1} \subseteq H$ also follows that $xHx^{-1} = H$ I guess

in the definition of normalization, for the case of groups equality and inclusion result in the same thing

I had this feeling that somewhere, we will have a pure $\subseteq$ because we define $H = f^{-1}(H')$ and we don't know if multiple elements of $H'$ will map to one thing

some elements of H' needn't be mapped to at all, thats why

@nickbros123 I know that if i take a cauchy sequence and look at the set $\{x_n: n \in N\}$ I know this will be totally bounded, is this of any use? I have a feeling I may need to consider 1,1/2,1/3 .. balls of x_1,x_2,... to get a corresponding sequence in $D$ but how I would proceed from there seems unclear

since $f$'s behaviour on $X \setminus (D)$ is not known

Ok no my last message doesn't make sense, it doesn't have to do with that

@nickbros123 any Tychonoff space given largest uniform structure satisfies this

The uniform structure given by the discrete metric happens to be the largest

It might be interesting question to see when this largest uniform structure, sometimes called universal structure, is metrizable, or when given a metric $d$ on $X$, the uniformity generated by the metric $d$ and the universal structure coincide

In fact maybe I should make it a question on math.se

@nickbros123 I assume you mean the following? Let $f:X\to Y$ be continuous with $f\restriction_D$ uniformly continuous where $D\subseteq X$ and $Y$ is complete. Then $f$ carries Cauchy sequences to Cauchy sequences.

@Jakobian this was an assigmnent question, see the above image part b

No I don't mean for you to give me a screenshot

u r saying the problem has less hypothesis then needed?

its just that some of the hypotheses aren't written, they are assumed

right.. Ill think about this with continuity on X

sure, go ahead

yeah got it. Thanks. Basically we can take 1-ball around $x_1$ and get $z_1 \in D$, $1/2$ ball of $x_2$ and get $z_2 \in D$ and so on. $z_n$ would also be a cauchy sequence from triangle ineuqliaty, and from uniform continuity on D, $f(z_n)$ would also be cauchy. If we fix an $\varepsilon$ and look at $d(f(x_n),f(x_m))$, we can apply triangle ineuqlity to get $d(f(x_n),f(x_m)) \leq d(f(x_m),f(z_n))+d(f(z_n),f(x_n)) \leq 2/n$ for sufficiently large m and n

@nickbros123 math.stackexchange.com/a/1292656/476484

I found this post by the way, which gives precise condition for when every continuous function $X\to Y$ where $Y$ is a metric space, is uniformly continuous

I think the extension lemma in itself given in the above screen shot may not require f to be continuous on X, but part b of that question does. is this correct?

The condition is that every closed discrete $A\subseteq X$ is such that there exists $\delta > 0$ with $d(x, y)\geq \delta$ for each $x\neq y, x, y\in A$

I might go follow the proof to see how far it generalizes to uniform spaces

@nickbros123 well honestly now I see there are serious problems with the screenshot, namely $f$ is defined on $D$

@Jakobian I see. That makes sense.

The terminology I know is that those kind of subsets are called $d$-discrete

So that the condition is that $A\subseteq X$ is $d$-discrete iff closed and discrete

it does make sense, and I think this is in fact equivalent to $d$ generating the universal uniformity on $X$

(got to check those things later)

My theorem from earlier is probably final, but I'm still curious if one can find such space with $\beta X\setminus X$ not locally compact: mathoverflow.net/questions/482174/…

I feel like I'm stretching it thin with this one

that's not what stretching it thin means

I mean that I'm at a point where saying much more seems borderline impossible

Hi

@Pizza hi

Are you free ? @SineoftheTime

@Pizza not now, but you can type the problem

I'm talking a stroll

Ok, I had some doubts about this physics exercise, more than anything the part about the integral, if you can help

I modified the image

So I basically wrote $r = \sqrt{l^2+l^2} = \sqrt{2}l$

$dEy = k \frac{dq}{r^2} \cos(\theta)$ , where $k = \frac{1}{4\pi\epsilon_0}$

$\lambda = q/2l, dq =\lambda dl$

$\cos(\theta) = l/r$

So to find $E$ , I wrote $\int^{l}_{-l} dEy$

$2\int^{l}_{0} dEy$

$2 \int^{l}_{0} k \frac{\lambda dl}{2l^2} \frac{l}{\sqrt{2}l}$

$2k \int^{l}_{0} \frac{q}{4l^3} \frac{l}{\sqrt{2}l} dl$

$\frac{kq}{2\sqrt{2}}\int^{l}_{0} \frac{1}{l^3} dl$

So it comes out like 1/0...

Mmm

What is dl equal to?

@Pizza Its wrong

r varies, so this cannot be the case

What you are writing makes no sense

@Pizza Why then does a l disappear afterwards?

@Binky It rappresents an infinitesimal element of length along the charged side

dl=

?

To integrate on the whole side

@Binky ?

"an infinitesimal element of length" Never understood what physicists mean with that

@Pizza What is the analytical expression of dl

@Binky dl = dx?

dl=dx=?

(also riemannian metrics given in the form $ds^2=\ldots$ or something, is always so confusing)

@Binky i dont know :(

Okay

Do I have to do the whole exercise?

No, I would like to understand where I went wrong :(

Okay!

@Pizza This is wrong here

r is not constant

Ah ok

But in a similar exercise I did it like this and I found correct , in the sense I wrote r like this

Can you show the exercise?

I have to go, bye

I'm curious about which composite numbers pass a probable prime test I've developped.

@Binky

can I see the solution?

@Pizza the integral here is divergent

@SineoftheTime yes, but it must have a finite value...

@Binky I can send the photo of my work and send it here

@Pizza what's the formula for $E_0$?

@SineoftheTime Do you mean the basic formula of the electric field?

$E = k q/r^2$

@Pizza matematicamente.it/forum/problema-di-elettrostatica-t80414.html

Try to see this

It's been I while since I've done those things :(

Oh ok thanks, don't worry anyway in the end this is a room about mathematics not physics

@Binky what is $\theta_1$?

Plot $z<z$ on Desmos 3D graphing calculator.

why?

you see the face of God when you do it

at least i hope that's God

@TheEmptyStringPhotographer Everything got red

Why are there so many different notations and terminology in mathematics? Why can't mathematicians organize a conference or meeting to standardize these notations and terms? It’s so frustrating, especially when you're reading two books (not necessarily on the same subject) and each one uses different notation and terminology.

@SineoftheTime the maximum angle between the hypotenuse and the major leg of the upper bar

Why?

it is a language, and languages evolve to meet local needs. except for $\mathbb{N}$ which should never include zero, that most unnatural of numbers.

Also, Einstein notation should be banned.

@copper.hat Just choose any one of them and just decide not to use the rest, simple.

@copper.hat Can't agree more

simpe for you

Yes, everyone does exactly that

then there's the whole $\pi$ vs $\tau$ silliness.

@copper.hat I am reading a book on multivariate analysis and linear algebra and the notations difference drives me crazy

@copper.hat I can't understand why this even exist, use $\pi$ or $\tau$ or whatever just let some mathematicians choose one and just ignore the rest.

@copper.hat The first time I saw it, it was on wolfram world and it was something related to matrices but I don't remember what exactly, I spend +2 hours trying to understand why this work and redoing my calculations over and over until I finally realised this should be a sum.

@pie I don't remember that much notation difference in multivariate analysis, maybe $\frac{\partial F}{\partial x}$ vs $F'_x$ and $\nabla F$ vs $\operatorname{grad} F$

And linear algebra is very standardized

Could you give an example?

The dot product vs inner product, ker() and N(), A^T vs A^t etc

dot product vs. inner product...?

what about the kernel

@VladimirLysikov Hoffman and kunze never used the name kernel and I din't know its meaning until I searched for some solution on mse and see people use it

@VladimirLysikov I completely disagree

@VladimirLysikov wait what? <x,y> don't equal x.y?

@pie $x \cdot y$ is only defined in $\mathbb{R}^n$

$\langle x, y \rangle$ is used more generally to some inner product on some vector space

the standard inner product on $\mathbb{R}^n$ is the dot product

@BenSteffan I am talking about analysis... so yeah it is defined but some books use <x,y> and some x.y

there are other inner products in analysis

both uses are defensible

also why do everyone write latex? do latex appear for you?

read the chat description :)

WOW

That is a lot better

@Binky and what's its value?

@VladimirLysikov There's not much difference, but if you have no experience like me and don't understand something in one book, then trying to find explanations in other books would likely require familiarity with the notation. That happened to me when I tried to read a section from Stephen H Friedberg's book and coming from Hoffman and kunze

@SineoftheTime I considered it generic

I didn't consider its value

So how did you get the result not as a function of $\theta_1$?

I think all variables in a paper should be $\alpha_1,\alpha_2,...$

@copper.hat That is a terrible idea

it's a joke.

I think all variables in paper should be $\mathfrak{a}_1, \mathfrak{a}_2, \ldots$

mathfrak for the people

what the frak?

I only know frac

personally i am terribly dependent on certain notation

one of the worst things about $\infty$-categories is that sometimes you have to write out a $\mathfrak{C}$ by hand

i struggle with writing $\cal F$, etc

@SineoftheTime because $\sin(\theta_1)=\frac{l}{r}=\frac{l}{\sqrt{2l^2}}=\frac{1}{\sqrt{2}}$

so i avoid classes at all costs

the ambiguity of English is wonderful

most of us do

if you want to work with classes you have to think about foundations ec.

@SineoftheTime I made a change

Ok, then $\theta_1$ is not generic

it's $\pi/4$

I had written wrong

@XanderHenderson Matt, my hero...

a long time ago, Hilbert spaces were assumed to be separable.

@SineoftheTime ah okay, I didn't really think about how much theta was worth

you're integrating between $-\theta_1$ and $\theta_1$, so...

@copper.hat Because they obviously are.

Except for some BS examples that are BS.

i just deal with $\mathbb{R}^n$.

@SineoftheTime but when I integrated I didn't write π/4

that's my point. I don't understand how you found the final result without knowing what is $\theta_1$

$\sin(\theta_1)=\frac{l}{r}=\frac{l}{\sqrt{2l^2}}=\frac{1}{\sqrt{2}}$ I wrote this after having done the integral

I found sin(theta) not theta

@Jakobian I think I understand everything you say, except a small detail. Why can we write $$\overline{\mu}\times \overline{\nu}(E) = \inf\left\{\sum_1^\infty\overline{\mu}(A_n)\overline{\nu}(B_n): A_n\in\overline{M},B_n\in\overline{N}, E\subset \bigcup_1^\infty A_n\times B_n\right\}?$$ The formula I'm used to is $$\mu^\ast(E)=\inf\left\{\sum_1^\infty \mu_0(A_j):A_j\in\mathcal A,E\subset\bigcup_1^\infty A_j\right\}.$$

Here $\mu_0$ is the premeasure on the algebra $\mathcal A$ that induces the outer measure $\mu^\ast$.

They say $2! = 2$ is true, but 2!=2 is false.

@psie we talked about it already. This is the definition

I'm familiar with the fact that $\overline{\mu}\times \overline{\nu}$ is the restriction of the outer measure induced by the premeasure $\pi(E)=\sum_1^n\overline{\mu}(A_n)\overline{\nu}(B_n)$ on the algebra of finite disjoint union of rectangles, which includes the set $\{A\times B: A\in\overline{M},B\in\overline{N}\}$.

And I guess the second formula somehow reduces to the formula you gave under those observations.

what you're probably trying to object to is that $A_n\times B_n$ should be disjoint unions of measurable rectangles, something that's irrelevant

and something that we discussed already

@Jakobian repetition is important :)

@Jakobian that's exactly it :)

@psie then as an exercise in repetition convince yourself that this is true

anyone know if there is a name for a "spherical" tetrahedron in which the base is a spherical triangle and there is only one other face conecting the vertex at the apex?

ok, well, I'd simply argue that that a sequence of disjoint unions of rectangles is a sequence of rectangles and vice versa

sounds about right

@Jakobian I noticed something else

something more subtle

in the case of non $\sigma$-finite measures

You are saying $$\overline{\mu}\times \overline{\nu}(E) = \inf\left\{\sum_1^\infty\overline{\mu}(A_n)\overline{\nu}(B_n): A_n\in\overline{M},B_n\in\overline{N}, E\subset \bigcup_1^\infty A_n\times B_n\right\}$$ and then that we can replace $A_n,B_n$ with corresponding sets in $M,N$, however, will these sets still cover $E$, @Jakobian?

@psie yes

we took differences so the sets in $M, N$ are supersets of the ones in the completions

oh, ok. I see. So if we went with $A_n=E_n\cup F_n$, I'd guess we'd simply take the null set in $M$ that contains $F_n$, and so on. Great.

so $\mathcal A\subset\mathcal B\implies\overline{\mathcal A}\subset\mathcal{\overline{B}}$ holds regardless of $\sigma$-finiteness or not. Spectacular.

@psie I suppose, that is possible

@psie again, this over line is confusing and what you wrote is meaningless without any context

yes, I will bother you about that

yeah, I agree the notation is not optimal, but...you used it yourself in $(M\otimes N)_{\mu\times \nu} = (M_\mu \otimes N_\nu)_{\overline{\mu}\times \overline{\nu}}$ :P

where

see the measures :)

the over line is on the measures, not on the sigma-algebras

that's different

ah ok haha, I see

When you write $\overline{\mu}$ that's okay, the notation is not ambiguous

however when you write $\overline{M}$, that's meaningless without context

that's why its better to explicitly state the dependence on the measure: $M_\mu$

the completion of the sigma-algebra $M$ with respect to the measure $\mu$

otherwise it can and will get confusing

yeah, fair. I will try to adopt it henceforth.

the over line on sigma-algebras is susceptible to working with multiple measures

as it was slightly confusing in this problem - if you state the dependence on measure explicitly then you can easily see what is that you want to prove

Guys, if I know that on a function attains its max value on the boundary of a closed set $D$ which lies in $y\ge 0 $ and $f|_{\partial D}=0$. The function, which is everywhere $C^\infty$ presents only one stationary point $\mathbf{x}_0$ in the interior of such set and $f(\mathbf{x}_0)<0$

Would it be easy to prove that $\mathbf{x_0}$ is a global minimum of $f$ in $D$?

I'm not allowed to prove it by computing the hessian of $f$ :p

Oh and $\lim_{\lVert (x,y)\rVert \to \infty} f = 0$

My idea was: If I restrict to a closed ball $B_r(\mathbf{x}_0), r>0$, which is compact since it's a closed and bounded set, then my function attains absolute minimum inside of it (due to what we called Weierstrass theorem)

but my function is $C^{1}$ inside the ball so if $x_0$ is a minimum point then it must be a stationary point, namely $\nabla f(x_0) =0$

so $x_0$ is indeed a global minimum in $\overline{B_r(x_0)}$

the problem is that I think I cannot extend this to the whole set $D$ easily since I only know the behavior of $f$ very far away from the origin and $x_0$ is near the origin

Can't think of how to make the above reasoning work arbitrarily far from x_0 maybe it only works locally... I'll go to sleep, my brain is fried :)

looks like a groin protector

@Claudio Surely there could be a global $\min$ on the boundary?