@DiscipleofBarney for step 5 their is something I don't get here we at first we constructed a number $\mu_{U}(K)$ that depends on the open set U now we do it to make it construct a number on u(K) without requiring specific open set of U however it says at the end we may pick some u $\in$ intersection of C(V)

but how do we actually pick them ?

I am not seeing that in step 5, could you be more specific about where it is? @KarimMansour

theorem 4.3

It can be any $\mu$

, it doesnt matter

as long as its in the intersection

@KarimMansour

I see

@DiscipleofBarney understand when he says in step 6 thinking of elements of X as functions from K to R

I mean for any arbitrarily cartesian product we can think of it as a function from

X to the union

however I don't see why this is a composition of function

I understand that we can think of cartesian product as a function f : X $\rightarrow$ U($X_i$) such that f($X_i$) $\in$ $X_i$ , so here we will have the union will be elements of the real numbers.

I don't understand that step at all.

I mean we showed that $\mu_{U}($K_1$)$ $\leq$ $\mu_U(K_2)$ however why do we even need to do this since this happens for every element u in particular we should have $\mu$($K_1$) $\leq$ $\mu$($K_2$)

?

nvm I understood it

I still don't understand why its composition of maps

@KarimMansour First, these questions seem like you basically want me to explain the whole proof which I don't really feel like going through in its entirety. I am fine with the questions, just be warned that I am just going by the "local" context so an explanation I give you might actually not be what the author is referring to.

I see

It looks like the composition of maps things is saying that since $f$ is continuous, then $f \times f$ is continuous, and minus is continuous so the function $X \to \mathbb{R}$ is the composition of $f \mapsto f \times f \mapsto f - f$

@KarimMansour

I see thank you

(actually the continuity of $f$ is probably irrelevent)

@DiscipleofBarney What's $1+1$?

Just those maps are continuous

I see

@ʇolɐǝzǝɥʇqoq Its $1+1$.

actually we don't even need a map this happens

for each U in the collection of open sets M

$1+1/$

so it must be true that if we pick a point inside C(V)

@KarimMansour Actually continuity of $f$ is important. If the map isn't continuous though you don't know if the inequality holds

$$$

I forgot the last map $f-f \mapsto f-f(K_1,K_2)=f(K_1)-f(K_2)$

don't we have that based on the definition of C(V) and for each point there we have $\mu_U(K_1)$ $\leq$ $\mu_U(K_2)$

so it must be true that $\mu(K_1)$ $\leq$ $\mu(K_2)$?

?

Well it is proven that is the case, but $ \mu$ is in the closures, so it might not be of the form $\mu_U$ so you need continuity to also show that the inequality extends to $\mu$.

ohhh

Thats like having a function that is continuous on the rationals, but I can extend the function in a bunch of non continuous ways on the reals, so we need some restriction on how everything interacts

yeah

(it is sort of like that)

I see

@DiscipleofBarney still don't entirely get the elements of function of X I understand now why we need to see if the map f($K_2$) - f($K_2$) is continuos. But I don't understand the composition itself.

What about it? @KarimMansour

so for each point in X we have a map f and now we send f $\mapsto$ $f \times f$

right ?

and then we do the maps according what you did is my understanding correct?

@KarimMansour Its actually $f \mapsto (f,f)$ now that I think about it, so $$f \mapsto (f,f) \mapsto f - f \mapsto f(K_1) -f(K_2)$$, its just a fancy way of saying that if $f,g$ are continuous real valued functions then $f-g$ is also continuous.

I see

I never encountered this before

It comes up in topology, sometimes analysis, basically giving standard constructions on how to construct continuous functions from other continuous functions

I see

good thanks alot @DiscipleofBarney I fully understand whats going on now

btw in the summer I will study topology from munkrees so we will have long discussions :D

Sounds good @KarimMansour

@DiscipleofBarney still here?

want to ask one last question

Yes @KarimMansour

Did you read the proof? @KarimMansour

yes

the proof of that statement no but what I want to understand why do we need it to contain borel subsets of G?

because I want to understand why do we need this before begging reading that statement

oh I see that is we want to show it is borel measure

That is what it says,

but why don't we do things in gneeral

general

why we restricting only to borel measure

Because it has properties like the space is hausdorff which might not hold in general

I see

@Sayan, you know that $f(x) = 0$ if $x$ is an integer less than $n$.

yes i do

What do you know about it when it's a number between integers less than N?

you mean x>n?

Let's say that $n = 5$ and $x \in (2, 3)$

That's what I mean.

It will be 0?

$f(2.5) = .5$

I'm interested to see how this turns out. My intuition tells me we'll be integrating $g(x) = x$, just "shuffled", but I'm probably wrong about that. Possibly just $n\int_0^1 x dx$

but f(4)=0

oh sorry

ok so now @Axoren

If you understand this behavior, then you can separate the integral into two more easily workable parts.

$\int_0^{n+1}f(x)dx = \int_0^{n}f(x)dx + \int_{n}^{n+1}f(x)dx$

What do you know about $f(x)$ when $x \in (n, n+1)$?

What's, say, $f(5.5)$ when $n = 5$?

It's still the same sort of thing.

$f(x) > 0$

right

@SayanChattopadhyay We're assured that $f(x)$ is positive from it's composition.

ok

But what we can see is that each interval is the same, even between $n$ and $n+1$.

So, really, you have $n$ intervals that act the same as each other as $x$ increases.

$\int_0^{n+1}f(x)dx = n \int_0^{1}f(x)dx$

How can you say that i didnt understand it....

If I told you that $x$ was some number $d.1234567 \le n$

I guarantee that you could tell me what $f(x)$ was equal to

Is that the part you don't follow?

i didnt understand the interval part

I described them as an interval because it's easier to work with a smaller grouping of numbers.

And from the structure of $f(x)$, we can see that it's behavior is easily understood on the integers.

f(x) is always positive right

@Sayan at least never negative.

$f(1) = 0$

Oops

@Axoren i dont understand this statement

Let's go back to the number I mentioned before: $d.1234567 \le n$

$f(d.1234567) = 0.1234567$ regardless of what $d$ is.

Err...

yes

Yes, I didn't make a mistake.

I thought I did, though.

Ok next

Let's say $d = 2$

That number would be greater than 2 and less than 3.

And so would other numbers like it.

you mean the function value at d=2

The function at $x = 2.1234567$, correct.

wont that be 0.1234567

Right, and if $x = 3.1234567$, it'll be 0.1234567

Now, let's say we have any number $d.e$

For any $d \le n$, we have $f(d.e) = 0.e$

Correct.

We want to put all the numbers who have the same $d$ in the same grouping, so we pick an interval that's 1-wide.

Specifically, $(d, d+1)$

fine ....

If we take any number in that interval and add one to it, it's the same as pulling from an interval $(d + 1, d + 2)$

And we know that $f(d.e + 1) = f(d.e)$

ok yes

So if we're integrating over $(d, d+1)$ and $(d+1, d+2)$, instead, we can integrate over $(d, d+1)$ twice.

yes

Where did I lose you, then?

You seem to have been following.

so in that way if we are integrating on (0,n+1) it would mean the same thing as integrating on n(0,1) right?

Right, but only because we know $f(x)$ behaves the same over those intervals.

Like sine functions over intervals of $(2\pi n, 2\pi (n+1))$

Now, the hardest part is finding the integral over $(0, 1)$

thats what i am thinking how do i divide the function

No need.

Then what should i do

Can you think of a function $g(x) = f(x)$ for $x \in (0, 1)$?

One that you know the integral of?

@robjohn Can you delete something from the transcript for me? I put my email address on here(not the one listed on the account) haha

What is $f(0.1234)$?

0.01234

@Incurrence Spam emails hitting you that fast already?

@SayanChattopadhyay off by an order of 10, but yeah.

@Axoren I put it up months ago, but I ran a google search on my gmail and it was one of the only 4 entries

$f(0.543)$?

Wouldnt it go to the trash transcript @Incurrence

@DiscipleofBarney That's how non-moderators get rid of messages

@DiscipleofBarney A moderator from any S.E can delete them fully though

@DiscipleofBarney I could trash things only because I owned the room

Can you think of a function yet, @Sayan?

@Incurrence Cool. The voting worked!

f(0.x)=0.0x?@axoren

@DiscipleofBarney Oh nice! I was asleep ahaha

@Sayan, no, it would just be $g(x) = x$

The identity.

f(0.1)=0.1

There was like a 15minute window this time, not sure if it will be that long every time though @Incurrence

@SayanChattopadhyay Right.

@DiscipleofBarney Did you miss a day though?

@DiscipleofBarney I have 5 more votes than you?

@SayanChattopadhyay So, now you know that $\int_{0}^{1} f(x) dx = \int_{0}^{1} \text{___} dx$.

But @Axoren f(0.1)=0.9

No @Incurrence, for some reason my page isn't updating, but I have gotten 320 votes

i am getting that......

math.stackexchange.com/users?tab=Voters&filter=week

@DiscipleofBarney Oh okay, yes that one does always update instantly I have found, good call

@Sayan, oh, I keep mistaking your function for having $|x|$ as a min-term.

Not sure how you got the five extra votes though

Did you do those today

@SayanChattopadhyay But this only changes it slightly.

No, I got them in the 3 hour window instead

Because $f(0.9) = 0.1$

yup

So, either way, the integral would be the same.

Just backwards.

That's where Pizza gets most of the extra votes

In the 3 hour window

Ah cool, tried deleting some closed questions. How many of those are there?

so what will be g(x)

You can probably plot it, given 3 points.

@DiscipleofBarney Thousands of old closed questions

@Sayan This is where it suddenly becomes the easy part, I won't steal that from you.

So to really maximize on Fridays/Saturdays we do the 40 extra vote thing, then vote on as many closed questions, then vote again. If you tried really hard you could probably get like 100votes on those days

Probably

Too bad its saturday morning ahaha

@DiscipleofBarney I've been trying to follow what you guys are talking about, sorry. Exactly what does this have to do with pizza?

Are you guys voting on toppings?

LMAO

HAHAHA

:P

HAHAHA was that an awesome joke or serious hahaha

I'll leave that as an exercise for the reader.

@Axoren what is g(x)=x the identity function

lol

@SayanChattopadhyay $g(x) = x$ is how $f(x)$ acts on every interval except $(0, 1)$.

On $(0, 1)$, it's backwards.

If you worked really hard you could probably get 80 most days, depending on how many closed questions have answers and a lot of upvotes @Incurrence

@DiscipleofBarney Yeah, if you have heaps of time, you could

I need some sleep. All the cancellations earlier this year have left me with a morning class tomorrow.

on (0,1) is it 1-x@Axoren

If your goal was solely to beat Pizza it would be necessary, but it wouldn't be very helpful haha

Yah, Mabe someone needs to come up with a stackexchange site where they votes up all the bad questions, so that they can be easily found on the real site @Incurrence

@DiscipleofBarney Hahaha

@SayanChattopadhyay What do you know about $\int_0^1 (1-x) dx$ and $\int_0^1 x dx$ ?

It would also make it easier to collude on the more resilient questions with answers that need to be downvoted.

I have an exam in a few days, and an assignment on that day :\

It is 1/2@Axoren the first integral

I don't think I will put that much time in to "beat" pizza

That sucks

and the second one is also the same@axoren

@Incurrence pizza is almost the top voter of all time on the site, just today or tomorrow he should top the current one

Tomorrow yep

@SayanChattopadhyay So then you know that $\int_0^1 f(x) dx = \int_0^1 (1 - x) dx = \int_0^1 x dx$

That being said

Which is enough to connect it to all the other intervals other than $(0, 1)$

He has over 30k votes combined when including his other accounts

@Sayan Now, since you've told me what the integral's value was, you can tell me what $\int_0^{n+1} f(x) dx$ is.

You have all the pieces, and I need to sleep.

so the final integral will be $n\int_0^1 x dx$

Good night, everyone.

Good night Axoren

good night and thanks

@Incurrence I assume that your comment has been cleaned up since I don't see it. If not, let me know.

@robjohn Here > chat.stackexchange.com/transcript/36?m=20584359#20584359

Thanks

@Incurrence deleted.

@robjohn Thanks

@Incurrence how did those 14 upvotes sneak in there?

@robjohn xD

@robjohn Must have been good questions/answers

@WillHunting You changed your name again, noooo

@WillHunting and it's so... green...?

I have wondered about that too, maybe accidentlaly upvoted and it timed out so couldn't take the votes back, must happen every once in a while with that many downvotes

Or never noticed

@DiscipleofBarney No, he did say they were intentional

@DiscipleofBarney But they are to trigger something that allows him to delete I think he said

How could that work?

@DiscipleofBarney Don't remember, but he explained it ages ago somewhere

How many old usernames recently of his do you remember?

Seems like an interesting puzzle.

The only one I can't remember where to place was Fundamental

Was that directly after Behavior and directly before Blue raincoat?

Pizza, Woodface, Blue raincoat, Behavior Carebear, Thursday, 900 push-ups a day,... I don't remember Fundamental

I think there was element, Don't you have some trick that could get you those? @Incurrence

I have a few tricks, but they don't have 100% effectiveness

user98130, user111742, Post No Bills, Post No Bulls, user127096, cheap effective diet pills, user147263, wordsthatendinGRY, This is much healthier, 900-sit ups a day, Thursday, Care bear, Weapon of Choice, babi ji, just a brick in the wall, Rafflesia arnoldii, Raff, Warm Fuzzies, Behaviour, Fundamental, Famous Blue Raincoat, Woodface, Pizza

That's my current rendering

Fundamental I have from the 6th to the 22nd ish from memory

I think this is starting to make the chat look like a fan club... What is your different trick?

Oh yah it was sit-ups

Oh I just get obsessive about some things

I like completeness for some reason

Oh I know, but what is your different trick?

There is a third, that you can probably think of :)

Ah use the chat room instead of the comment pings,

Well I better get to study

Talk later @Disc

@Incurrence Later

@pizza You may already have heard about wkhtmltopdf.org It does quite a good job at converting html to pdf, even for MSE

hey @KarimMansour

hey @BalarkaSen

anyone out there know matlab

Hey @BalarkaSen ever heard of a Grothendieck group? sbseminar.wordpress.com/2011/04/18/…

yeah, heard of 'em.

encyclopediaofmath.org/index.php/Grothendieck_group

@robjohn did you manage to calculate the integral I showed you yesterday? The one with $14\zeta(3)$?

Greetings

what got you interested in them, @bolbteppa? you are not studying K-theory, are you, by any chance?

@Chris'ssis I started and had an idea, but I got taken away by other things. I should take it back up.

@robjohn No need to finish it. I was just asking.

@Chris'ssis I want to see if my way works.

@robjohn Yesterday I discovered a new double integral representation of Catalan's constant. I checked more sources and no one says anything about that representation. I think I'll add it to my book, no publishing before.

@robjohn OK

@BalarkaSen I'm revising Hall algebras, which I've recently seen called a grothendieck ring!

@Chris'ssis I think that $\frac12\int_0^{\pi/2}\frac{x\,\mathrm{d}x}{\sin(x)}$ is my favorite.

@robjohn I like that too. :-)

Is there any savior angel here to answer my question below?

@bolbteppa no idea what hall algebras are.

=I

ughh this code is frustrating

@Incurrence there is a general expression for arcsin(z) as log(something). try fiddling with the Euler's formula for a while, you'll be able to derive it.

Alright, I'll give it a shot

well at least I sort of got it but it's ugly

@Incurrence Solve $\frac{e^{iz}-e^{-iz}}{2i}=6$ using the quadratic equation

@robjohn Okay I'll give it a go(that looks like a hyperbolic with complex components)

afk 1 min while doing this on whiteboard

@Incurrence It's just $\sin(z)=6$

equivalently, $e^{iz} = \sin(z) + i \sqrt{1+\sin^2(z)}$, thus $z = -i\log \left (\sin(z) + i\sqrt{1+\sin^2(z)} \right )$...

quality of MSE questions are heading quite low, it seems.

@BalarkaSen Is that inclusive of mine above?

no, no, i was merely commenting on MSE questions.

Well I mean to say, was it my question above that partially triggered you to say that

i have been looking for answering a few questions for a while, but the algebraic topology tag barely gets 3 or 4 questions in the whole day

it was nothing related to your question. we ask trivial questions all the time.

Oh okay haha good :)

I try not to ask trivial questions on main

yeah, and i don't post questions on main at all [whispers everything i ask are trivial questions]

Hahaha you are really really advanced

I find it weird that I feel so shit at math, and I seem to know more than all of my classmates

Hi @balarka @Incurrence

@SayanChattopadhyay Hey Sayan, how are you?

I am trying to catch up my crap complex analysis knowledge right now

@Incurrence well, if you're not feeling shit enough, you're not learning enough. :P

@BalarkaSen Yeah probably xD

@BalarkaSen I wish I had more time to do math each day

how the hell do you make those graphs.

Like that

@BalarkaSen: When do you sleep?

it's 12:52 PM in here, @Huy

@BalarkaSen: Just wondering, because you seem to be awake always.

the correct conclusion would be that i am almost always awake, except when you're not here.

lol

i am just spending a lot of time in chat because my school's off. gotta change this habit soon.

Your schools off @Balarka

Lucky guy.............

@MikeMiller i see, thank you. how do u know that we can bound the cardinality of the set of maps between a pair of separable smooth manifolds?

=/

yeah, it's been off for a week.

balarka

@Balarka tell me something if all the elements in set B are mapped onto by some elements in set A leaving some elements in set A this is a bijective function right?

remember this exercise

@SayanChattopadhyay it is not bijective then.

sorry sorry i mean surjective

it is surjective right

it is not surjective either. there are elements in A with null preimage by construction, right?

@iwriteonbananas which exercise?

Yup there are

so that contradicts surjectivity.

yes!

it's a cool exercise.

that's the space from an example in that section

$\mathbb{R}^3 \setminus (A\cup B)$ where A,B are linked circles

right?

Okay fine thanks..................my professor is wrong again

@iwriteonbananas no, that's not it.

why not?

$\alpha$ and $\beta$ are not circles.

oh yeah, the top and bottom of the cylinder have holes, we cant contract them to a point

huh?

the cylinder is $D^2 \times [0, 1]$

i meant the cylinder after we removed $\alpha \cup \beta$

yes, right.

What do 'branches' refer? Is this referring to, for example with a hyperbola, we have two branches, left and right(if x^2-y^2) and top and bottom branches with (y^2-x^2)?

there are a lot of empty spaces, @iwriteonbananas. you can easily deformation retract a lot to get a easy-looking space.

@BalarkaSen what space should we deformation retract to?

@iwriteonbananas A separable manifold has cardinality at most $\Bbb R$. Prove this.

well, the space deformation retracts to two cylinders, linked transversely like the $\alpha \cup \beta$ link and wedged at a point.

yeah right

i got that too

and that in turn deformation retracts to the center circle of the two cylinders wedged to a point right?

i.e., $S^1 \vee S^1$.

where does $\gamma$ go to after these two deformation retracts?

@BalarkaSen hmm yeah i suppose so

i can draw it for you if you're unsure, but i guess you see it

@BalarkaSen it walks both circles once

yeah. prove that you can't nullhomotope this loop (this is easy).

of course you cant nullhomotope it

you still have to prove it, but ok.

it represents a nontrivial element in $\Bbb{Z} * \Bbb{Z}$

fair enough.

@BalarkaSen i kinda get it, but i would appreciate a picture regardless

then i can see how u think of it

wait a couple of seconds. this is a cool problem. i didn't come up with a solution when i was doing chapter 1 form Hatcher.

okay

this is a solution of someone from here or the alg top prof.

i don't recall who gave me the hint.

which alg top prof?

i took a class on alg top in the uni i often go to. only recently gave up as they started cohomology. thought i'd rather learn it in a De Rham-ish way.

okok