@EM4 Of course! it would be a shame if the Vince Lombardi trophy ended up on the Ocean floor.

LOL

Oh oh, a potential fan of the Phoenix Sunburns has entered the room! Hi @Xander!!

who you got for NBA finals?

@EM4 The Milwaukee Bucks (fear the deer!), and the Phoenix SunBathers! @Xander can better advocate for the SunBurns! ;D Lol.

All in fun, that's all.

is your star player injured?

@EM4 He had a hyperextended knee, after game 4 against the Atlanta Hawks; but two MRIs performed the evening of the injury revealed absolutely no structural damage. He sat out game 5 and game 6, with a likely return if game 7 was required, but Middleton and Holiday secured to more wins for the Bucks, who tromped the Atlanta Hawks 4-2 in best of seven.

if he plays in the final Bucks have good percentage to win.

We had no clue after his injury (The Greek Freak (term of endearment) Gianis), but there's more depth in the team than doubters realized.

@EM4 Although I'm talking big, I have a thing about not being too overconfident. Sort of a Karma thing. As always, may the best team win (but I pray the best team is the Bucks! ;D)

this Final is more new than old same team going at it or super teams.

@EM4 Indeed! I think the "final four" were, for the most part, teams that haven't enjoyed a Championship for some time. I always love underdogs. And some teams have been strong, season wise, but couldn't quite pull it off in playoffs.

it should just be the warriors, every time.

nah it should be the Celtics HAHAH.

this is Germany

Yes. This is Germany

all of you are evil.

the fireworks have not stopped since the sun went down.

@leslietownes They started here before the sun went down. I can't believe that they started so early. They are still going on.

@robjohn

sus

help me

What do I do

Help me

@FitzWatson Did you read my computation? You didn't need to write everything out, because I had all of that there. I just switched whether the student saw the seen/unseen question (so that is all you needed to say).

but yes, your computation is correct.

Had they said "old" and "new" rather than "seen" and "unseen" it would have been less confusing.

it's a simple switch... "16% that both are unseen" is switched to "16% that both are seen".

@robjohn But, this computation assumes that we ourselves ask the person setting the question paper (now referred to as paper-setter) "is atleast one of your questions unseen?". However, the question says that we "somehow find out that one of the questions is unseen". Could this not also mean that the paper-setter randomly chooses to talk about one of the questions? In that case the probability would change to 0.6

translate "we somehow find out that one of the questions is unseen" as "given that one of the questions is unseen"

translate "the probability that the other question is seen" as "the probability that at least one question is seen"

given that one of the questions is unseen, what is the probability that at least one question is seen?

The only information we have is that one of the questions is unseen.

"Could this not also mean that the paper-setter randomly chooses to talk about one of the questions?" it could, but the only thing we know is that they said that that question was unseen.

However it is we find out, we know that one of the questions is unseen.

I have seen some "Anti-Pi Rant" videos where they say that Pi being a boring number. There argument is that "Sure pi is an irrational number , but so is root 2 , root 3 , the golden ratio phi etc".My confusion is that , isn't pi also a transcendental number too ? I heard that to this day , very few numbers were proven to be transcendental.

@Prithubiswas In my opinion, I find that argument completely ridiculous; pi's irrationality is hardly it's most extraordinary feature!! I'd say that the variety of problems in which it appears on the solution to makes it extraordinary. As you say, it's also transcendental which is a very interesting property.

Leave pi

Return to pie

pi + e = pie . "pi" is a transendental number . "e" is a transendental number . So , "pie" is a edible composition of transendental number?

@Prithubiswas and yet almost all complex numbers are transcendental

@EdwardEvans I guess pi is a number which is proven to be transendental and can be visualized as the ratio of a circle's circumference to its diameter.

Not sure if other transendental numbers have nice visualizations.

You can visualise them as numbers that are not algebraic

@EdwardEvans I meant geometric visualization.

algebraicity is geometric in some sense

Can somebody tell me what is the formula for mean deviation for normal distribution $\lambda$?

hi, im sort of lost on how to show the highlighted part of this imgur.com/a/DkhLq4l , or really what exactly the highlighted part is saying (the stuff above the highlighted part is fairly standard)

it seems like its saying that if $f : [a,b] \rightarrow [0,1 - \epsilon]$ is a $C^1$ function, such that $f(a) = 0$ and $f(b) = 1 - \epsilon$, then there are intervals $I_1, I_2 , ... I_k$ that are disjoint, with left endpoints in increasing order according to indexing, and $f \restriction I_j$ is monotone increasing, and $sup_{t \in I_j} f(t) = inf_{t \in I_{j+1}} f(t)$, and $inf_{t \in I_1} f(t) = 0$, $sup_{t \in I_k} f(t) = 1- \epsilon$

or something like this..

sounds very wrong

yeah.. i was thinking something like $x^3 sin(\frac{1}{x})$ should disprove what its trying to say

ah wait no, it says something void

they're not saying it should map onto $[0,1-\epsilon]$, but that has to be the intent

I guess they're allowing for $A$ to be a disjoint union of intervals? hard to tell without the rest

oh right sorry, its not much more, ill include it just a sec

the rest imgur.com/a/nECJw0n

looks like a typo in the second-to-last line, but ok, I get it now

except I don't think it works

seems like they implicitly assume $\eta_1$ to be locally injective

yeah that should be $\eta_1$

as in $\eta_1'(t) \neq 0$?

i mean nonzero

yeah

your counter-example doesn't quite work yet, because you can still construct $A$ away from zero

but you should be able to construct a function that has a singularity like that, say, everywhere along a cantor set

hmm

ok, but if we assume $\eta_1'$ is nonzero, then what they're saying works? im guessing via a change of variables?

yes

there should be some way to fix this, though

actually I think there is an easier way than what they are proposing...

$\rho(t) = \frac{1}{1 - t^2}$, so $\rho(\eta_1(t))\eta_1'(t) =\frac{1}{2} [log(\frac{1 + \eta_1(t)}{1 - \eta_1(t)})]' $ and $\int_{a}^b \rho(\eta_1(t)) \eta_1'(t) = \frac{1}{2} log(\frac{2}{\epsilon} - 1)$ directly

unless ive fudged somethnig up

but if what i said is fine then it goes pretty against the grain of their argument

its not really patching what they are trying to say

i just used FTC and $\eta_1(a) = 0, \eta_1(b) = 1 - \epsilon$, and $\eta_1$ is $C^1$

no, what you're saying is 100% the correct approach, whatever they're saying is wack

except I think you're missing a factor of 2 in $\rho$? or maybe I'm mixing the conventions up

for some reason the poincare metric as these notes state it doesn't have the factor, but evidently it should because on wikipedia the area associated to $\rho$ is supposed to be $4 / (1 - |z|^2)^2 dx dy$

yeah, that's the expression I know

but I'm not very familiar with the Poincaré model, I much prefer working in the half-plane model

ah well in any case, i've confirmed whatever they've written doesn't seem up to par

thanks @Thorgott

np

probably a dumb question, if $\rho_2 : \Omega_2 \rightarrow [0,+\infty)$ is $C^1$ and $\Omega_2 \subset \mathbb{C}$, is connected and open, and $f : \Omega_1 \rightarrow \Omega_2$ is holomorphic ($\Omega_1$ is also a connected open), why is $(\rho_2 \circ f ) |f'|$ necessarily $C^1$?

I can see that its $C^1$ on $\Omega_1 \setminus {f'}^{-1}(0)$

it isn't? take $\rho_2=|\cdot|$, $f=\mathrm{id}$, then $(\rho_2\circ f)|f^{\prime}$ is just the norm function

ugh, my notes are so bad

@porridgemathematics here is a good way to remember this. the constant curvature +1 metric on C is 4/(1+|z|^2)^2 dzdzbar; this is what you get when you pullback the spherical metric on CP^1 by stereoprojection

the 4 is there because the equator of the sphere gets mapped to |z|=1

near that, the scaling factor is 4/(1+1^2)^2 = 1, as it should be, since stereoprojection is an isometry near the equator

Hey guys! I was just hoping that someone here can verify my proof for the method of substitution: math.stackexchange.com/questions/4190612/…

@BalarkaSen ah, yes thats a good way to see it

david it looks good. the first answerer makes valid points.

he also plugs a book. i wonder if it is any good.

Hello

I think I've asked this question on this chat before, but did not receive any input

So here it is again: imgur.com/a/Rm7kYAe

How do we get (9) from (8)?

i don't know what theorem 2.14 is but it probably contains the answer.

Theorem 2.14 is the Riesz Representation Theorem

oh, hrm.

It says that to every linear functional we can associate a Borel measure and a sigma-algebra....

So doesn't really give the answer :(

what happens between 2 and 6? i don't see a definition of Lambda.

Oh, shit. The order got messed up. Let me fix it

sometimes results of this flavor are referred to as 'choquet theory' after a frenchman who represented functionals as integrals over the extreme points of a convex set.

ok, lambda is evaluation at x.

that's how 9 follows from 8.

Lambda f is f(x) by definition.

imgur.com/a/JRfFDOK (corrected link, order fixed)

@leslietownes Oh that wasn't the issue

The problem is, (8) says $f\in C(H)$

(9) says $f\in A$

Why does that happen

i tihnk it's that paragraph between (3) and (4). they are identifying things in some random function algebra with their unique extensions to things that restrict to H.

sorry if my thoughts seem scattered, my daughter has been yelling at me.

she's having quite a morning.

this isn't choquet theory after all but i do like choquet theory. peter lax's functional analysis text has a good chapter on it.

@leslietownes aha no problem. let me read the paragraph between (3) and (4) again

@leslietownes thanks for pointing it out, i'll check it out!

there's some kind of abuse of notation going on.

is it obvious that this P will satisfy the above inequality?

any tips for going about showing it I dont remember strategies for showing this matrix inequalities

@leslietownes you do have her well trained!

my daughter's now yelling at the cat, which is a useful change from me.

i wonder if otimes really means tensor there. i haven't checked the inequality but if they say it, it must be true.

i rhetorically asked my daughter where the cat was. "i think she's under the couch" was the response. this turned out to be correct. she then started yelling the cat's name over and over and throwing toys under the couch.

@leslietownes I think I've got it! Just want to run it by you

famous last words.

First, observe there is a one-one correspondence between $M$ and $A$, hence the author uses the SAME LETTER to denote functions in $M$ and their unique extensions to $K$, which are in $A$.

(8) holds for $f \in C(H)$, so it also holds for $f\in M$

$M$ is a subspace of $C(H)$

Now, the integral is over $H$

So we might as well consider the extensions of the functions in $M$, to $K$, which are in $A$

because it doesn't matter whether you integrate $f\in A$ or $f\in M$, the result is the same as long as the integral is over $H$

Since (8) holds for $f\in M$, it holds for $f\in A$ as well

Does that make sense?

yes.

Am I the only one who disagrees with this answer that says that the function $f(x)=1/x$ is strictly decreasing? The definition of "strictly decreasing" that I'm familiar with is that if $a$ and $b$ are in the domain of $f$, then $a<b \implies f(a)>f(b)$. In this case, it seems that we can take $a=-1$ and $b=2$ to show that the function is in fact not strictly decreasing.

i don't know what the right authorial choice is here. if you don't use the same letter, you're putting funny hats and things on stuff to make these tiny differences. if you do use the same letter, the same letter can mean different things.

@leslietownes Awesome, thank you! I'm surprised how I could figure it out after just talking with you

this is why you are the epsilon emperor.

Ahahah

joe, you are not the only one who disagrees with that answer.

the only right definition of 'strictly decreasing' is the order theoretic one which you state. it does not distinguish between intervals on which a function might or might not be continuous.

points of discontinuity complicate all of the calculus-theoretic characterizations of order properties.

@leslietownes: I thought I was losing my sanity for a moment there. Thanks for clarifying. I'm tempted to write up my own answer, even though that post is almost two years old

i'm trying to think of other instances where this comes up. changes of variable in definite integrals, maybe. the formula in calculus books tends only to work in complete generality if the change is a monotone, one-to-one function, although it's usually used beyond that and god help you if there are discontinuities.

it would be fair to say that 1/x is locally decreasing or something like that. that is all you get from the derivative being negative everywhere.

computer algebra packages often f--- up inverse trigonometric functions and definite integration because of the discontinuities that are associated with that.

@leslietownes: It seems that things are stickier than I thought. I just looked at Spivak and Wolfram MathWorld, and they only define increasing/decreasing on intervals

oh, if you implicitly require the domain to be connected, that's another story.

but i don't see this as inherent in the definition of 'decreasing.' it is an order theoretic concept.

i'm still on your side, joe.

@leslietownes: Haha, thanks. Unfortunately, I just checked Rudin, and even he (who generally offers very general definitions) only defines increasing/decreasing on intervals :(

@joe whyd you care either way, I think everyone agrees on its behaviour

well he's dead, so he doesn't get a vote. it's 2-0 in favor of you right now.

the answer to your question is 'however you want define decreasing'

flows, you're ignoring the fun of battling with other people over distinctions.

seems most useful to define the property over intervals

i do think the natural home of the definition is a function from one linearly ordered space to another. this would not incorporate concepts like connectedness.

but if RUDIN disagrees with me maybe i am wrong about this.

a bit rudein of him though isnt it

is that pun made repeatedly on here, if so then sorry

puns are always welcome.

@leslietownes: He's not exactly transparent in his writing though, is he...

rudin did a lot of things extremely well, and for some reason, people rely on him for other things too. i've insulted rudin's treatment of various things many times.

i love rudin

even though his material can be hard to read,

well why don't you marry him.

i can say without a shred of doubt that i have learnt the most math from his books. his books are very effective if you're self studying and like the painful approach of figuring most things out on your own

imo trains you to be a better researcher probably

@leslietownes i would if i could XD

@epsilon-emperor: Pain is essential to self-studying mathematics, I see

i think his treatments of multivariable calculus and measure theory in principles of mathematical analysis are a disaster. and the orgy of point-set topology in chapter 2 is disgusting. thankfully nobody ever gets too far into the book.

@Joe i think pain is essential to studying mathematics. if it's coming easy to you, you're doing something wrong

@leslietownes too far? i'm on chapter 6 of rca right now; hahaha

he's a functional analyst and i am too so i do like a lot of his approach.

@epsilon-emperor

What is the definiton for a first order theory to be complete?

@Prithubiswas i have forgotten logic but just google soundness and completeness, you should be on your way

Wiki :="In mathematical logic, a theory is complete if, for every closed formula in the theory's language, that formula or its negation is demonstrable" can both formula and its negation allowed to be demonstrable?

@Prithubiswas in general, i'd rather not comment. completeness alone is probably not sufficient to make a comment. but if you also have soundness, then: math.stackexchange.com/questions/105575/…

Hello. I'm quite sure my post, which I have deleted is not correct. Could someone please point out the mistake?

Hello. I'm a little surprised why unconditional convergence (en.wikipedia.org/wiki/Unconditional_convergence) is defined only for COUNTABLE series?

Why is that? Why can't we extend to uncountable indexing sets?

I'm surprised that you're surprised that people don't consider uncountable series

@Thorgott It doesn't seem like a bad idea, what can go wrong?

There's probably some trivial reason why people don't talk of uncountable series

there are no absolutely convergent uncountable series (assuming that they're not cocountably zero)

granted, I do not know about unconditionally convergent ones, but unless you manage to produce an example, I will doubt it's a good notion

@leslietownes: I have reluctantly written up an answer about increasing vs. decreasing that reflects the (wrong) definition that is given in textbooks. Any feedback?

@Thorgott That's fair, but I will think more about it

let me know if you find that such a thing exists, that would be a mild curiosity

The veronese surface has the property that any curve on it can be obtained by the intersection of a hypersurface with the veronese surface. Is there a name for surfaces which have this property?

Sounds like projective normality, although I never thought of it this way.

@TedShifrin if $\nabla f$ is linear, is $f$ necessarily quadratic?

@TedShifrin Oh yeah, projective normality might work.

This is what I do in my spare time now. Whenever I get tired of GIT, I look at random examples of varieties. They are really cool.

Apparently it is true...

Neat.

@SayanChattopadhyay what is GIT?

Geometric Invariant Theory

Geometric Invariant Theory should get a new abbreviation, GIT is already Geometric Integration Theory

Is that actually a thing, @Thorgott?

Ugh, no @Thorgott

I know of GMT, but not a geometric integration theory.

it is a thing, albeit perhaps not a well-defined one

Krantz & Parks have a book called Geometric Integration Theory, which is actually just a GMT book. Whitney has a book called Geometric Integration Theory, which is Topology/Geometry with a heavy focus on integration, measurable forms and stuff like that.

@Thorgott I should get that book. I think Krantz is excellent.

@Thorgott so how does GIntT differ from GMT?

@anakhro How could this not be?

@TedShifrin it possibly could be if someone had a lapse in judgement.

But it's probable that someone who did have a lapse in judgement has resolved it since asking.

;)

But not certain.

Ted, what have you been up to lately?

Other than helping me with foolish math problems

Watching Wimbledon, having a cold, cooking. Nothing deep.

Are you especially a tennis fan?

Yes, have been forever.

@robjohn I've read little of it, but I liked what I've read. It has a very nice exposition on the area and coarea formulae.

@anakhro read Whitney and find out

@TedShifrin a cold on the Fourth of July (observed)? that's un-American!

Perhaps so. It sucks.

@Thorgott I am not sure I am initiated to GMT enough to notice a difference.

Hey. I need some hints to proceed with this sum :

Is it more on the level of the scope of problems that it seeks to solve? Kind of like a topology vs. geometry sort of deal?

@Thorgott I had Krantz for Graduate Real Analysis and Complex Analysis and an independent study in Fourier Series (ala Katznelson). Those were some of the best classes I had at UCLA.

@TedShifrin I am sorry.

I cooked for 30 people (BBQ) yesterday. I am resting today.

show that open interval (3,4) is complete wrt the metric d = mod(x-y) + mod(f(x) - f(y)) where f(x) = 1/min{mod(x-3),mod(x-4)}.

I made yummy cole slaw and potato salad and grilled hamburgers. Not sufficient division of labor, although a few others contributed good stuff.

will only the sequences that converge to interior points be cauchy here?

Of course.

@TedShifrin I only cooked the main courses people brought: chicken, steak, burgers, hot dogs, brats, etc. Luckily, no fish this year. The salads were shared and the drinks and dessert was provided by the host

Why use 3 and 4 to make things difficult …

just to make things difficult.

How do I proceed?

Prove what you said.

Okay and I have checked that {3+1/n} and {4-1/n} are not even Cauchy. Did I do it right?

*w.r.t this metric

Well, you need a more general argument, but that idea will do it.

On the note of cooking, I will be making spaghetti and meatballs tomorrow, and a bunch of penicillins. So that should be interesting

Take any sequence converging to 3.

Okay okay, yes

Convert to a question about sequences converging to 0 with the analogous metric.

You mean in (0,1) ?

Sure

@SayanChattopadhyay A bunch of penicillins?

@Ted It's a ginger based whiskey cocktail

Does that go with spaghetti and meatballs?

Well I just had some whiskey on hand, and people asked to drink something. So it was either this or a whiskey sour, people chose this.

I'd do red wine and have the whiskey before and after.

Thanks!

Sure thing, umm …

Yeah I would totally love that but for that I have to go out, and that seems risky

Gotcha.

Are there any results about uniqueness of $f$ as a sum of quadratic and linear functions when $\nabla f$ is affine?

Huh?

I am trying to figure out if there is anything that can be said when $\nabla f = Ax + b$ for a matrix $A$, vector $b$.

About $f$, that is.

If $\nabla f = Ax$, this means $f$ is quadratic, and so $A$ must be symmetric.

Didn’t we already cover this?

I guess not.

You are suggesting the $b$ part makes no difference?

Derivative of sum is …

sum of derivatives.

So what is your issue?

no worries

@Joe that gets at another issue, decreasing, strictly decreasing, nonincreasing. i am very fine with 'decreasing' to mean a > b implies f(a) \leq f(b) because it reflects a lot of conventions from functional analysis (where 'positive' means 'nonnegative').

@leslietownes: Indeed, I did mention that in the answer because I wanted to reduce confusion, not enhance it. I was surprised that the OP actually saw my answer and accepted it, but I'm glad that they did, because that question had over $1000$ views, and I didn't want any more people to be misled...

Also I am glad because I am addicted to rep

haha

maybe it is a leniently decreasing function?

a captiously decreasing function

What is meant by ties possible or not possible here please?

that means that two or more horses can finish at the same time

the outcome of the race is not necessarily a linearly ordered set of participants.

A and B can finish first and C finishes third

A, B, C can finish at the same time. A and B can be first and then C can finish. etc.

robjohn is stealing my thoughts.

eww. what's this slime all over these thoughts?

he's also concealing his tomfoolery by going back in time and saying them before i can say them.

@leslietownes. Hahhaha

@leslietownes. @robjohn. You are both awesome. Thanks.

lego time

i count 13 ways assuming vaporisation is not allowed.

@copper.hat No vaporization? where's the fun in that?

whoah, no vaporization?

just horsing around

suspect peta will show up outside my door shortly

my great grandfather litigated an insurance case about a horse being set on fire by a bolt of lightning. it made it to the supreme court. (long pause) of new hampshire.

he lost.

@copper.hat A pair of german shepherds in dark glasses.

i'm not having total recall right now, the movie name escapes me

it wasn't "total recall," presumably.

:-)

and my daughter's yelling at the cat again.

i was thinking mib

my daughter's really yelling at the cat. i have this issue every day. at what point does this become my problem?

i never know.

earlier the cat handled the situation by escaping to a nearby closet. now, for some inexplicable reason, she's sitting there and frowning at this verbal abuse.

dunno, only my wife yells at me.

@leslietownes. If she is not busy with the cat, she has to be busy w/t else. Kids!

my wife wants to have another one of these. i think one is enough.

cat?

kids.

one was more than enough.

it's fine with me either way, i don't have to do any of the work. but i don't think the world needs more human beings.

definite gruner veltliner discussion material

did you get any?

@leslietownes. Agree. Law restriction for 2 kids could be amazing thing!

not yet, might pop down today after my 'run'.

paul marcus at the rockridge bart station had a number of good offerings. anything from $10-60 in price range.

i hit my limit with 2 kids in the middle of an intense startup. i definitely lost my edge from overwhelming tiredness.

i tended to sample from the cheaper portion of the spectrum.

the problem is i do not like drinking on my own and my usual accomplice is away for a few wrrks

i find little real correlation with price (within reason)

i can drive up. i could even locate the wine.

@copper.hat Sorry, we're here for you.

a ton of wine is overpriced. i was born in napa and i know what i'm talking about.

at some point there was a wine glut and tjs two buck chuck was excellent wine. i would buy a bottle, open & taste and then go back in and buy a box

@robjohn yours?

@copper.hat No. I do have a shepherd mix, but you wouldn't probably see that much in her.

tjs has some really good stuff at reasonable prices.

TJs gruner is horrible. they get this hungarian stuff.

hungary is a great winemaking region but not for gruner.

my recent import was worth the hassle

4 bottles left

@copper.hat of what?

Château Lynch-Moussas 2016

that makes it sound like i am some sort of connoisseur, i am not.

good, I take it?

you kind of are, though

but if i find something nice, it remains in my memory

if you like Bordeaux, then a shouting yes

not a sweet wine

i bought my advisor some bordeaux and i heard from people who were able to make it that they served it at his funeral. that wasn't paul marcus, it was "vino!" by the safeway.

i am working on making sure the possibility of that is minimal

i mean for myself

we're going to drink four loko at your funeral. we will also toss it upon the coffin.

the irish epitome of bad wine (:-)) was either blue nun or black tower

or you can do that at my funeral.

night train and thunderbird are technically wines. i don't know if they sell them in ireland.

funerals are of course sad, but there should be laughing as well.

the reality is that just the first few sips matter, in general. after that it is whether or not it induces a headache.

i've given one funeral speech, it was a series of insults. everybody is complicated.

i do not subscribe to spitting the wine out

@copper.hat Ireland is such a wonderful climate for grapes :-)

my funeral speech was me insulting the decedent in as many ways as i could think of. he would have wanted that.

@robjohn many folks would take the ferry to france for a summer holiday and bring back a supply

i choke up when called upon in such circumstances

@copper.hat wine, surrendered for a fee.

he had written some books and i said that thanks to him, everyone had discovered their love of reading . . . amazon's return policy. that was the nicest joke.

:-)

it's very hard to say anything at those kinds of events, so if you have the opportunity, you need to lean into being as offensive as possible.

please don't listen to me.

my brother made a slightly negative comment at my mom's service about god and choice of pouring rain on the day. amusing, but slightly awkward reception.

For any constant $c$ and a variable $x$, does it hold that $o(cx)=o(x)$?

$o()$ is the little oh, with its argument tending to $0$.

if $c \neq 0$.

Yes :)

generally i prefer to avoid doing calculations with little o stuff.

So $$\left \{f(cx) : \lim_{cx\to 0}\frac{f(cx)}{cx}=0 \right\}=\left \{ f(x) : \lim_{x\to 0}\frac{f(x)}{x}=0 \right \}.$$

I don't see it quite yet, to be honest.

The equality, that is.

Is it true that any open set in the Sorgenfrey line is of the form $\bigcup\limits_{k\in K} [a_k,b_k)$ ?

@copper.hat is it also true that $o(cx)=o(x)=co(x)$?

if $c\neq 0$

Actually, $o(cx)=o(x)$ is quite straighforward.

$cx\to0 \iff x\to0$ and thus $\frac{f(cx)}{cx}=0$ as $cx\to 0$ or, equivalently, $\frac{f(cx)}{x}=0$ as $x\to 0$. So $o(cx)=o(x)$.

$o(cx)=co(x)$ follows a similar reasoning.

Suppose $\int_{0}^{1}f(x)*e^{n*x}dx=0$ for all natural numbers and 0. f must equal 0. I've tried proving it with Weierstrass theorem. But I am out of ideas. Anyone have any?

you need some hypotheses on $f$ for this to be true

but yes, I think Stone-Weierstrass should suffice to prove this

Im trying proof by contradiction with f does not equal 0. But im struggling with the contradiction

@Thorgott any ideas?

you've said the idea yourself already, use Stone-Weierstrass

no need for a contradiction btw

I tried that already and I havent had much luck @Thorgott

Wouldnt a proof by contradiction be needed, I struggle how to do it without as well

ok, what can you say about $\int_0^1f(x)p(x)dx$, where $p$ is a trigonometric polynomial?

@Thorgott Im not sure about anything worthwhile. If f(x) = 0 for all x is would be 0. But idk about some f which does not have that

then reflect on your hypothesis

@Thorgott im still not sure

@JakeFreeman If $f$ is the characteristic function of a non empty set of measure zero then the integral is zero but $f$ is not. There is no point in continuing without some hypotheses.