@robjohn They will still be reversed if there are too many

this place is hoppin'

500 point bounty? GD... we shoudl have a real-money bounty site for math!

Researchers can delegate their tasks to the lower mathematicians (like me). And pay us

\o/

/o\

it's a reflection across line $y = 0$

How would you characterize a point $x\in\operatorname{int}\bar U\setminus U$ (where $U$ is open) in terms of neighborhoods?

It's $\partial U$ exactly

the set of all points in $V$ (your space )such that any nbhd of $x \in \partial U$ contains points in both $U^c$ and $U$ itself.

by defintion

The bar is closure, not complement

I know

I checked the statement

Draw your space on paper

then draw a $U$ inside of it, keep it simple

$\partial U=\bar U\setminus U$

Draw the closure

so that there is a well-defined boundary

not ${\rm int}\ \bar U\setminus U$

now take smaller and smaller circles (open balls) about a point $x$ on the paper on the boundary, should always contain points in both sets, right?

Until your open balls get sufficiently small that your pen marks distort things

The set I am describing is not the boundary

it is a subset of the boundary

If that's the case then its interior is $\varnothing$

For example, $U=$ unit ball minus the origin

the set I am talking about is $\{0\}$, the boundary is $\{0\}$ plus the unit circle

Waffle's Crazy Peanut

What a wild username

So cool

$W = {\rm int} \overline{U} \ U = \{x \in X: (\exists \text{ open } V_x \ni x : V_x \subset \overline{U} \ U) \}$. ...

Got that? @Mario? can't edit ne more

try \setminus instead of \

Hey

now

if $W \neq \varnothing$, then

there exists a smaller closed set containing $U$ than $\overline{U}$, contradiction

That characterization of $W$ is false @EnjoysMath. E.g. there is no open set containing $0$ that does not intersect the given open ball.

take $W^c$ and $\cap$ it with $\overline{U}$ to produce that

since $W^c \supset U$

I'm trying to find less trivial examples of non-regular open sets than regular open sets minus a few isolated points

A regular open set is one such that $W=\emptyset$

@Karl, what are you saying exactly? Then the center is then not in $W$ simple as that

@EnjoysMath Except it is.

points are open?

They are not.

For $U=$ ball minus 0, $W=\{0\}$

sometimes tho

Rather, I should have said, it is unnecessary for points to be open.

@KarlKronenfeld I think you misread my definition of $W$, check the latex, I use '\

but it doesn't work, should be setminus

I checked.

how is $0 \in W$ then?

please prove it

It belongs to $\bar U$, where $U\subset\mathbb R^n$ is the open unit ball intersected with the complement of $\{0\}$.

for $U=$ ball minus origin, closure $U$ is the closed ball, interior of closure $U$ is the open ball, difference is the origin

Okay, but there exists no $V_x$ open such that $V_x \ni x$ and $V_x \subset \overline{U} \setminus U$ since any open set $V_x$ in the outer space $X$ intersects $U$!!!

wher $x$ = center

yes, as I was saying, your characterization is wrong.

um no

that's the def of interiour

okay, w/e I give up

note that I'm talking about $(\mbox{int }\bar U)\setminus U$, not $\mbox{int }(\bar U\setminus U)=\emptyset$

@MarioCarneiro Are you interested in examples from any spaces in particular?

$\Bbb R^2$ for starters

really just something I can get my head around so that I can understand the difference between regular open and open

For example, is it sufficient to assume that $U^c$ has no isolated points?

to show regular open

Any proper dense, open set is not regular. Like the complement of some line.

I am not familiar with the concept either, though.

dense open sets are weird

complement cantor set is dense open

@robjohn I honestly thought that it was impossible to solve,but given your solution I was astonished. I've gotta say your answer is genius!

$({\rm int} \overline{U}) = \{ x \in X : \exists V_x \ni x \text{ open, with } V_x \subset\overline{ U}\}$

If there's $x \in ({\rm int} \ \overline{U}) \setminus U$, then since $\overline{U} \supset ({\rm int} \overline{U}) \supset U$, ...

and $\overline{U} = \partial{U} \uplus U$

$x \in \partial U$, so it's a subset of $\partial U$

and vise versa, so it's equal to $\partial U$.

had you up until the last part

when $U$ is open

Okay

there is no vice versa

but you have that it's a subset so far?

yes, it is definitely a subset of the bd

@Karl the original question is math.stackexchange.com/questions/752494/…

is it necessary to assume that $A,B$ are regular?

$x\in\mbox{int }\bar U\setminus U$ iff all the points in a neighborhood of $x$ are close to $U$, but $x\notin U$

by contrast, $x\in\partial U$ iff $x$ is close to $U$ and $x\notin U$

where "close to $U$" means every neighborhood intersects $U$

@r9m Here?

Can somebody look at this?

It isn't homework!

@Sush Hi.

Does anyone know where to find a simple rpoof of (3) and (4) here, en.wikipedia.org/wiki/Jensen's_inequality#Finite_form ?

what are (3) and (4)?

It follows from the finite form of the convexity definition with $\lambda_i=1/n$

@Mike The formulas marked 3 and 4 there.

The formula was not marked, but it is the second equation in the Proof 1 section

I know what you meant, but my phone doesn't display markings.

fourth equation in section "1.1 Finite form"

The second equation is proved from the first by induction.

And the inequality comes from the second equation.

Or I should say 'the second inequality is proved from the first by induction"

I shouldn't bother while I'm on my phone.

@MarioCarneiro Thanks, I will look at this.

@MarioCarneiro An idea I have is to define, for any open set $U$ (in the space $X$), the set of "irregular points" $I(U)=\{x\in\operatorname{int}\bar U\colon x\notin U\}$. Define $Y=X\setminus I(U)$. Then, in the space $Y$, $U$ should be regular. Thus, the problem amounts to moving a solution in $Y$ to one in $X$.

One counterexample to the theorem without the regular assumption (that is, for any two disjoint open sets $A,B$ $A$ is the interior and $B$ the exterior of some set $S$), if I have done this correctly, would be to use $X=\Bbb R$ where open sets are cofinite and contain $0$, and let $A=X\setminus 1$ and $B=\emptyset$.

$T=\{0\}$ is a dense and co-dense subset

$S=A=X\setminus\{1\}$ has interior equal to $A$ and exterior equal to $\operatorname{int}(\{1\})=\emptyset$.

yeah, now I've gone and confused myself

By the way, dense open sets can't produce a counterexample, since if $A$ is dense open, then $B=\emptyset$ and $S=A$ satisfies the requirements (this is a generalization of your remark)

ah yes

If $I(A\cup B)$, using my notation above, is closed with empty interior, then I think any solution in $X\setminus I(A\cup B)$ works in $X$.

Are you sure you don't want $I(A)\cup I(B)$?

(Doubt the interior of that thing could every be nonempty)

@MarioCarneiro I'd hypothesize they are the same. Do you know a counterexample?

Can somebody humor me with this? math.stackexchange.com/questions/752270/…

@Karl They are not the same. If $A=\Bbb R\setminus C_{[0,1]}$ is a cantor set on $[0,1]$ and $B=\Bbb R\setminus C_{[1,2]}$, then $I(A)=C_{[0,1]}\setminus \{0,1\}$, $I(B)=C_{[1,2]}\setminus \{1,2\}$, $I(A\cup B)=C_{[0,1]}\cup C_{[1,2]}\setminus \{0,2\}=I(A)\cup I(B)\cup\{1\}$.

Ah, nice.

I think that relies on $A$ and $B$ intersecting though.

wait, that's not right, $I(\Bbb R\setminus C_{[0,1]})=C_{[0,1]}$

lol, I didn't notice it either, of course $I(A)$ is the complement of $A$.

oh duh

$I((0,1))=I((1,2))=\emptyset$, $I((0,1)\cup(1,2))=\{1\}$,

mm, true. Yeah, I certainly want $I(A)\cup I(B)$ instead of $I(A\cup B)$.

Note that $I(A)$ is always empty interior, but I don't think it is always closed

Hey I guys, I know you're busy, but you could you just give a quick answer?

The most notable counterexample I have seen so far is $A=$ unit disk minus 0, and $T=\Bbb Q^2$

I'm just a little confused on what the differential under an integral means.

@Anthony it means do the integral

Does it have a well-defined meaning, or is it just to tell us which variable we're integrating with respect to?

So it's not a a differential?

if you are in year1 calc, it is the variable you are differentiating by

there are loads of SE questions on this topic

I'm in Real Analysis!

I guarantee your question will be closed as duplicate

:(

your linked question links to 10 others

I just find it hard to believe that it's just to tell us which variable we're integrating with respect to.

that's obviously whitewashing the situation, but it suffices for most applications

Well what's the non whitewashed version? I really don't know.

there are many excellent answers on the linked questions about all the different interpretations of the $dx$

like differential forms

or infinitesimals

Why are there multiple interpretations, shouldn't there a be a definition?

the definition is that it is the variable of integration

it is then redefined depending on context, because the syntax is very suggestive

I see.

But in the most general sense, it really is just telling you the variable you are integrating with respect to.

basically

Basically?

it allows for various abuses of notation that help in the memory of e.g. the chain rule for integrals

:(

I hate that….

Thank you, though.

$\int f(x) dx=\int f(x(u)) \frac{dx}{du} du$ because $dx=\frac{dx}{du} du$

the latter bit is an abuse of notation

really it's just a convenient memory tool

Yeah.

Indeed. But so the point is writing our variable of integration as such is just convenient.

Thank you for your time, @MarioCarneiro.

the fact that the $dx$ matches with the $dx$ in a derivative is convenient that way

Thank you!

@Anthony context ALWAYS matters

@r9m Here?

How do you automatically extract a grammar from a set of example strings?

It's not easy in general

genetic algorithms probably, and techmology

@EnjoysMath tech'm'ology ?

@skullpatrol Size always matters

@N3buchadnezzar But size isn't everything

@N3buchadnezzar Talking about shoes?

Or Pizza?

@MarioCarneiro No, I don't mind that you used the details in my last paragraph. I just wrote my answer for my own enjoyment; I never thought I'd have fun doing general topology. :) Btw, the result is nice.

I didn't think it was going to generalize to any open sets, I just thought up that question today and was surprised at the generality in the end.

@robjohn You should be in bed

Finally got enough reputation points to talk on here :)

@KhallilBenyattou Welcome!

Thanks! :)

On my browser, only the LaTeX code shows. Is that the same with everyone here?

@Sawarnik Hi.

@KhallilBenyattou On the chat?

You have to run ChatJax manually, see the link on the right

math.ucla.edu/~robjohn/math/mathjax.html

@MarioCarneiro Thanks :)

I would always hesitate to say that there is only one way to do it

click on the arrow next to your post

should have "edit | delete"

I think there is a time period beyond which you cannot edit

You could always write the equation down as a system of equations and solve for $\mu,\lambda,z$

@DanielFischer The alphabeth

@N3buchadnezzar A-ha

@DanielFischer I am trying to find some historical sources, man that is hard.

Trying to figure out some of the earliest proofs of the Fresnel integrals, I can only find a few scattered articles online with zero sources.

@KhallilBenyattou yes

Greetings

@robjohn are you around? I was wondering if we can make a bit more rigorous the last step in your proof to that series involving the convergence.

@N3buchadnezzar I'm not old enough to remember the original proof, sorry.

@robjohn More precisely the point $(9)$.

Yeah, I have no problems understanding the proof =) It is very nice, but it would be nice to see who came up with it "first". Or atleast sources older than 1980..

So far the ebst I have found are a few numerical calculations in AMS from 1960

I will ask on the main, but I wanted to give it a go myself first =)

@Chris'ssis what's wrong with $(9)$?

@robjohn I didn't say it's wrong, I said that it would be nice to make that more rigorous.

@Chris'ssis are you talking about $(9)$ or the end of the proof after $(9)$?

@robjohn It's about both $(9)$ and the end of the proof, but that explains things about $(9)$. I mean that part where the series increases by approximately 2 and then decresease by approximately 2.

@Chris'ssis Okay, since $(9)$ is just $\frac1{k+1}=H_{k+1}-H_k$

@robjohn It's about the way you explain the series diverges. That part is as difficult as the series itself.

@Chris'ssis okay, that is because we are simply partitioning $[n\pi,(n+1)\pi]$ into small segments with divisions at $H_k$.

@Chris'ssis no, this is much simpler.

@robjohn I know your point, but that is not obvious. I was wondering if we can find a way to clearly explain that.

@robjohn or maybe I miss something there that is obvious.

@Chris'ssis I am just approximating $\int_{(n+1/2)\pi}^{(n+3/2)\pi}\cos(x)\,\mathrm{d}x$ using non-uniform, but very small, intervals.

:14948690 Since $H_k$ takes longer and longer to cover each period of $\pi$, that counteracts the fact that $k+1$ increases

hi @Chris'ssis

@Chris'ssis The sum cannot be bounded either... If the sum were bounded, Dirichlet would say that $\sum\frac{\cos(H_k)}{kH_k}$ converges

@Charlie Hi great cat! I didn't see you for a long period of time!!! :-)

@robjohn I see.

@Chris'ssis :D yes, very long, how are you?

@ParthKohli Hi.

@Charlie Hi.

@Charlie I'm in trouble with a series I created. How about you? :-)

@Chris'ssis I'm in trouble with a problem I created. studying math.

@Sawarnik hi

@Charlie haha! same with me.

Wow.

@robjohn Yeah.

@DanielFischer Some people seem to be walking the border line of sanity. "please do not omit the space from my name; my name is not "CarlMummert".".

Oh. My. Gawd.

@Sawarnik we're not alone.math people, unite!

I don't know whether to be amused or whether to think this is creepy.

I think I'll go with creepy.

Le /me shrugs.

I'll see you later, earthlings.

later

@MattN. lol. Considers doing that from now on.

@MattN. I am an alien, don't you know?

@KarlKronenfeld Hello.

@MattN. I guess it wouldn't have happened if the other part of the conversation was an at least quarter-reasonable person, and not who it was. MK has a way of annoying people beyond moderation.

@PedroTamaroff noooooooooooooooooooo.

@KarlKronenfeld sadly walks away

@KarlK ronenfeld, what's the problem?

It hurts me deeply, @Denial Fischer

@KarlKronenfeld Karl.

@DanielFischer Yes I know about that but I can't understand it.

@Karl Much as I like that spelling, it has the unfortunate effect of not pinging me.

@MattN. Boo.

Boo too.

@MattN. You know the expression "drives me nuts", I suppose. Sometimes, it really happens.

@DanielFischer Will you hunt me down if I ping you in comments using DanielFischer instead of Daniel Fischer?

@MattN. If I can prove that you do that by typing out the name manually, and not by autocompleting, I may.

@DanielFischer Oops. Now... SCARED!

: D

Actually, I click on it. In chat, at least.

@MattN. I have not yet snuck a keylogger into your computer. For the time being, you're safe.

@Daniel Fischer On a more serious note: do you actually care about the space in your name?

@MattN. That depends where. In a hand-written note/letter, yes.

@DanielFischer No, I'm talking about an online forum like this one here.

Wonders if there is a "non-whitespace" unicode character that looks like a space.

Uh-oh:

@KarlKronenfeld Can I discuss some stuff with you?

About modules.

@PedroTamaroff ok

k o

"on the other hand, I am disappointed to see you (a moderator) intentionally misspell my name as well."

@KarlKronenfeld That would still count as mis-spelling... : )

@DanielFischer

lol

@KarlKronenfeld Well, let's start with the beginning.

@PedroTamaroff If we ever play a Tennis match, that is going to bite you ;)

@MattN. True, I was thinking there may be a way to enforce the system to insert that space automatically with such a unicode character. I tried, but it does not allow you to put that crap in your display name. :(

@DanielFischer Takes note.

@PedroTamaroff I've never played Tennis, though, so the probability is - not high.

Lang defines a projective module to be one such that whenever we have an arrow $f:P\to M''$ and a surjection $g:M'\to M''$, we can factor $f$ through $g$.

mhm

I proved this implies that every exact sequence $0\to M'\to M''\to P\to 0$ splits, which as a short nice proof using the splitting lemma. This in turn easily shows $P$ is a direct summand of a free module.

yeah

Now, to prove that ${\rm Hom}_A(F,-)$ is exact when $F$ is free things got a little non-slick. But then I used that a family of sequences $0\longrightarrow M_i\longrightarrow_{f_i} M_i'\longrightarrow_{g_i} M_i''\longrightarrow 0$ is exact iff the sequence $0\longrightarrow \bigoplus M_i\longrightarrow_{\oplus f_i} \bigoplus M_i'\longrightarrow_{\oplus g_i} \bigoplus M_i''\longrightarrow 0$ is.

To show ${\rm Hom}_A(P,-)$ is exact when $P$ is proyective.

@KarlKronenfeld How would you show the $\rm Hom$ is exact for $F$ free?

I can show you my proof if you want to, but it's a bit ugly.

ok, feel free to leave out details

@KarlKronenfeld Say $P=F$ is free, with base $B$. Given $\psi:F\to M''$, finding $\phi:F\to M'$ so that $f\phi=\psi$, amounts to finding a map $\phi_0:B\to M'$ with $f\phi_0(b)=\psi(b)$. Since $f$ is onto, given $\psi(b)\in M'$ there is $x_b$ for which $f(x_b)=\psi(b)$. Define $\phi_0:B\to M'$ as $\phi_0(b)=x_b$.

hm..

@KarlKronenfeld ?

Just thinking, I think I may have seen a cooler way to do it before.

Yes, there must be a nicer way.

@PedroTamaroff I looked it up, it's actually a way to go from the statement that $P$ splits exact sequences of that one form to the statement that Hom(P,-) is exact.

@PedroTamaroff Do you know the fibered product from category theory?

@KarlKronenfeld That is my proof that if $P$ is a direct summand of a free module, then ${\rm Hom}(P,-)$ is exact.

@KarlKronenfeld Not particularly.

If you take a peer at Lang's section of Projective modules, I am proving P1->P2->P3->P4.

@PedroTamaroff If you want to say the same thing in categorical language, you can do it as follows. It is easy to prove that $\operatorname{Hom}_A(A,-)$ is exact. Then let $h:M\to N$ be a surjection. Suppose we have a map $f:\oplus_{i\in\mathcal I}R\to N$. It is uniquely determined by an $\mathcal I$-indexed family of maps $f_i:R\to N$ which determines a bunch of $g_i:R\to M$ such that $hg_i=f_i$. To finish, use the universal property of coproducts again.

@KarlKronenfeld =D

Cool beans.

Hi all! May you take a look at this? It's on the convexity of a certain family of integrals. Thanks a lot in advance!

wow pretty quiet in here this morn'

@robjohn thanks, it should delete itself then :)

poor little baby shrimp

@meer2kat Yeah.

Everyone is just waking up.

@pedrotamaroff :( what time is it there?

@meer2kat Almost 11 a.m.

@PedroTamaroff people should be awake :(

I woke at 7:30 a.m. or so.

morning folks

@meer2kat You bored?

@PedroTamaroff i woke up at 6:30

@meer2kat What's your GMT?

@PedroTamaroff a little. my projects are mostly wrapped up

@PedroTamaroff my what?

I woke up at 6:45, which was 5 minutes ago

Hour zone?

@PedroTamaroff US Eastern (EST)

@Mike Wow, well you must like us a lot to be online so early

@Mike Drink your coffee first.

@meer2kaf time zone (GMT is Greenwich mean time; EST is GMT -5; PST is GMT -8)

@Mike I am at -3:00.

@Mike thanks i can do math :P

@meer2kat Didn't know if you knew what GMT meant.

@Pedro ae you busy?

@Charlie Nope.

@Mike yeah he just said what is "your" GMT. i figured he meant time zone

@PedroTamaroff can you help me with some simple eigenvector problem?

gotcha

@Charlie OK.

@Pedro I get to give a lecture this quarter.

@Mike ORLY

Linalg prof is out for a day this quarter, he's letting me lecture that day.

@Mike What topic?

@Charlie ??

@PedroTamaroff just a second, I'll type properly