@PeterTamaroff You found a jet in a router's box? wow!

@robjohn Haaha but I did!

@robjohn, what?

@robjohn, I didn't understand you

OK, this is going to be an alternative when someone here doesn't understand something!

@PeterTamaroff, or when someone is trilling by trying to act "cool" and saying ambiguous things then ignoring questions.

I believe that very similar symptoms are noticed with Facebook Girls

The following is a valid example "OMG! Life is like i donno.. ahhh.. you know.."

@PeterTamaroff I don't understand....

@Adnan Sorry, I had misdirected the comment.

@robjohn, respect! :)

@robjohn It's all there black and white, clear as crystal. You get nothing!

@PeterTamaroff :-)

@robjohn I'm stuck with something on step functions.

@PeterTamaroff such as...

@Adnan lol

Whaaa?! :D

@Adnan What's that site on your profile, you've made it?

@robjohn Given two step functions $s$ and $t$ on $[a,b]$, with partitions $P=\{a=x_0,x_1,\dots,x_{n-1},x_n=b\}$ and $P'=\{a=x_0^\prime,x_1^\prime,\dots,x_{m-1}^\prime,x_m=b\}$ and constants $s_k$ and $t_k$ for $k=1..n$,$k=1..m$, how does one define it's sum explicitly? I need to prove that given two step functions on $[a,b]$, $$\int_a^b (s+t)=\int_a^b s+\int_a^b t$$

@GustavoBandeira, no, that's the website of the company where I work

@Adnan I thought you were a teenager.

@Adnan Just don't ask why.

I AM gonna ask why..

so, why?

:D

@Adnan No reason. =D

Alright! :)

@Adnan Baseless speculation FTW.

So are you from India?

@Adnan Me?

yes

@PeterTamaroff by step functions, you simply mean functions that are constant on intervals?

@Adnan Nope, Brasil.

oh oh right

you said it before

@Adnan Peter Tamaroff and Mariano are my neighbours.

@robjohn Yes. In partiular the partition yields the $k$th interval $(x_{k-1},x_k)$ and there the function is constant with constant $s_k$

On the endpoints, the function is defined arbitrarily.

I'm not good at handling the discrete stuff!

@Adnan clicks on sad face sad face turns happy "What kind of sorcery is that?!"

@PeterTamaroff, oh today I made it even better :D

This morning you weren't able to click to make it sad again :D

@PeterTamaroff Yep. It was only possible to make him happy. =/

@Adnan Do you guys design railways like the one in the front page animation? Not sure I'd want to travel on one of those.

@HenningMakholm, ehehehee.. no we don't make railways

With curve radiuses of less than 2 car lengths, and no cant...

@PeterTamaroff Can't you just show this is so for constant functions and then piece-by-piece on the step functions?

@HenningMakholm, oh you're from Denmark!

@robjohn You mean kind of inducting on the amount of intervals?

Do we like Denmark?

Right. And you're from Finland, I take it?

@Adnan I like.

I guess it is one option. I don't have a "proof plan" yet.

@PeterTamaroff sure, I guess that depends on how detailed you want to get.

Well, I live in Finland, but I'm not originally from here

and in order to "integrate" better I'm trying to hate the things Finland hates :D

like Sweden!

Oh.. here's a Finn!

@JaakkoSeppälä

@Adnan I'm sure I have a finnish course

Is there some kind of lemma that allows you to choose the partitions instead of allowing them to vary arbitrarily (like, uniqueness wrt how one refines the partitions). in order to set the partition associated to one integral the same as the other integral, so that additivity becomes obvious

I read from Finnish discussion forum that one can represent every even number as a sum of two integer n and m where absolute values of them are primes. Is that true or a conjecture?

@Adnan Yes. I'm a Finnish man.

it's called Goldbach's conjecture if we restrict to positive n and m, I'm unsure of this full integer variant you have though

@JaakkoSeppälä, of course you are! One cannot miss such a name :D

@Adnan Found!

@GustavoBandeira, what did you find?

the light

@Adnan Finnish course.

@GustavoBandeira, you're studying Finnish? Niiiice

lovely language

but difficult

@anon I know the name of the positive case but I have never heard a proof of this variant.

@anon I just proved that if one takes a step function $s$ with a partition $P$ and sets up a refinement of $P$ by adding a collection of finite points $\mathscr C\subset [a,b]$ and defining a "new" step function $s^*$ by $s(x)=s^*(x) \text{ if } x\in [a,b]-\mathscr C$ and arbitraily for $x\in \mathscr C$, then $\int_a^b s=\int_a^b s^*$

@JaakkoSeppälä Well, since all numerical evidence points to the Goldbach conjecture being extremely, overwhelmingly true ...

@Adnan Hei olen Gustavo. =D

@GustavoBandeira, niiice :D

No terve Gustavo.

Indeed, barring modular considerations the "all numbers in this very regularly-distributed class are sums of two numbers of this class with so-and-so growth rate" is statistically quite high as we move up the number line, purely as a matter of probability and additive combinatorics / number theory.

@HenningMakholm Maybe this animation is more interesting

@PeterTamaroff, oh that's awesome!

@JaakkoSeppälä I know only Hei and olei. =/

@JaakkoSeppälä Related.

@PeterTamaroff Gah! There's no driver!

@HenningMakholm Safety comes first!

Well, I know they say driverless cars will be safer, but that's not how it looks in the video.

@HenningMakholm Hehe. This one is cooler

And you got test dummies.

@robjohn Maybe I'll just ask for a proof on main...

@robjohn There are fires in CA!

Can I have four paramedics, please?

@J.M. Done!

@J.M. One thing, wait.

He asks for $\int_0^\infty$ not $\int_0^1$

@PeterTamaroff That's why I'm asking for paramedics. :)

@J.M. OH!

Done.

@PeterTamaroff sure...

@PeterTamaroff Not executioners. ;)

@robjohn Here

@PeterTamaroff Of course; it's summer.

This two are nice too.

@robjohn Are they a big deal?

@PeterTamaroff They can be. It depends. The Station Fire was a big deal.

I can't stay for long, so: later, y'all.

@J.M. Later!

@robjohn Did you see the maquette?

@PeterTamaroff I had to look up "maquette" :-) I did see the model.

@J.M. Aw! I didn't see you. Nice new gravatar!

@robjohn Some parts are missing though :/

@JasperLoy =P

@JasperLoy nice =)

@PeterTamaroff =)

@PeterTamaroff What do you expect for free?

@PeterTamaroff maquette is good

@skullpatrol here, in the above example, you get an illustration of attachment to freebies at the cost of paper.

@JayeshBadwaik This was a sarcastic response to the question: " What do you expect for free?" ;-)

: ) @ comments.

@skullpatrol ;-)

Hey that Leonid guy is going to give a talk in my department later this semester...

@skullpatrol a $\text{sarcastic}^2$ response?

@HenryT.Horton Leonid Schifrin?

@robjohn Yep ;-D

Good night everyone.

If (!) = sarcastic response, then does (!!) = odd sarcasm?

(removed)(!!)

@skullpatrol ?

@HenryT.Horton Do you plan to attend the talk?

Good evening mathochists

Is there an element of masochism in studying math?

@yunone no - but there is an element of mathochism :)

I wonder if mathochists stab themselves with axes ...

@GustavoBandeira No, the Leonid Kovalev who posts on math.SE a lot

Hi guys.

hi

Hi all

yo @JonasTeuwen

@OldJohn Which definition of "mathochist" are you using Sir?

@skullpatrol Number 2 would be my definition

@OldJohn Mine too.By the way,Hi everyone!Hi OldJohn!

My ex-students would probably say number 1

@MeAndMath Hi

@OldJohn how are you?

@MeAndMath Great thanks - a bit too much wine with some friends this evening, but otherwise great!

How are things with you?

@OldJohn I'm a little apprehensive...I have a test in Friday...

@OldJohn Yo. @t.b. Yo.

I need to tinkle.

@MeAndMath what is it on? - and is it really important?

@OldJohn topology of metric spaces:limit points,connectedness,complete,compactness,these things.But I'm scared,nervous...

well i think CalTech actually executes people who fail math tests, but the trend hasn't really caught on yet.

Check out the number 2 definition of Math on the Urban dictionary :-D

@MeAndMath why are you scared?

@DavidWheeler I´m scared cause I don´t wanna fail...

@MeAndMath have you got examples of previous tests? past paper questions?

@MeAndMath are you in danger of failing?

@DavidWheeler nope.it´s the first of 5 tests...but I wanna do this right.

@OldJohn yes ,I do.

@MeAndMath If you have examples of the sort of questions, I am sure there are people here who will help or advise .. :)

@OldJohn Thanks ,I appreciate!:-D

@MeAndMath I may not be much help (I usually talk rubbish) but there are more able people here than me who will :)

metric spaces are very "nice spaces" in topology, the condition of being a metric space is quite "strong". for example, we have very strong "separation properties".

On the other hand Horton spaces are very badly behaved. Not even a restraining order will suffice to separate me from you.

@DavidWheeler and they have a distance function that makes them quite concrete to deal with

so...not even $T_{off}$, eh?

Standard example of a space that isn't even $T_{-\infty}$

well, i suggest we consider co-Horton spaces instead, then.

room goes quiet while someone things of something as witty as the previous posts ...

@MeAndMath when in doubt, think of the euclidean plane ($\Bbb R^2$ with the metric induced by the inner product $||x|| = \sqrt{x \cdot x}$). this is the space we are trying to generalize from.

... and fails :)

so $d(x,y) = ||x - y||$

@MeAndMath Is $F$ not necessarily bounded?

@HenryT.Horton No,it doesn't say.

Then the answer is no... for compact set it should be true

Have you proved that for fixed $a \in \mathbb{R}^n$, $d(a,x)$ is a continuous function in $x$?

Oh you said you did the similar one already?

@HenryT.Horton Yes.$K\subset M$ ,$a\notin K$,and $K$ is compact.Prove there's a $p\in K$ such that $dist(a,K)=d(a,p)$.

@MeAndMath So you proved that one? Using continuity of $d(a,\cdot): K \longrightarrow \Bbb R$?

@HenryT.Horton I used nested intervals...

???

Sounds complicated...

@HenryT.Horton do you have a better way?easier?

Yes

A continuous function attains its maximum and minimum on any compact set

$d(a,-) : K \longrightarrow \Bbb R$ is continuous on the compact set $K$

So it attains its minimum at some $p \in K$

But by definition $\mathrm{dist}(a,K) = \inf_{x \in K} d(a,x) = d(a,p)$

@HenryT.Horton neat and straightforward (I like it)

@HenryT.Horton that is lucid...but, it hides the difficulty in the first line.

@DavidWheeler but doesn't it use a property that is really useful to know?

@HenryT.Horton I liked!

Not only useful to know but also almost definitely already covered at this point

It's less complicated.

no argument there...but if this is MeAnd Math's first exposure to metric spaces, perhaps it hasn't been covered. if it has, then great.

@DavidWheeler so maybe the instructor is expecting them to be able to construct a solution from (more) basic principles ... agreed

@OldJohn That's what I did...

@MeAndMath in that case maybe you now have 2 ways to tackle the problem?

@OldJohn I think I do.

@JonasTeuwen a bit late, but yo!

Hi :-).

@JasperLoy Nice, can you link me?

I wrote something on my development page for my new project.

@JasperLoy Purrfect.

@MeAndMath By the way the one you are trying to do now with $F \subset \Bbb R^n$ should actually be true... I was thinking about the distance between two closed sets (where neither is necessarily a single point)

@JasperLoy For now.

@HenryT.Horton great! - I can stop racking my brains worrying about that! - I was also half-remembering an example from years ago with distance between a hyperbola and an axis

@OldJohn Yes when the two sets are just closed you can take $\{(t,0) : t \in \Bbb R\}$ and $\{(t,1/t) : t \in \Bbb R\}$ in $\Bbb R^2$ for the counterexample

@HenryT.Horton yep - and when you mentioned there might be a problem with the question asked, I was getting my old memories of these things confused - and couldn't see why my mental argument failed

... and it seems it doesn't fail :)

I was doing some statistics exercises,but I couldn't stop thinking about this...

But if one set is just a single point I think you can just create a sequence $\{x_n\}$ of points in $F$ from the definition of the infimum (each $x_n$ within a distance $1/n$ of $a$) that are all within a distance of $1$ from $a$ so that they are bounded in $\Bbb R^n$ and hence have a convergent subsequence... since $F$ is closed the limit of the subsequence is in $F$, this is the desired $p$

@HenryT.Horton as an aside, do you know Fulton's book on Algebraic Topology?

@OldJohn I am not familiar with it, I looked at it one time maybe

@HenryT.Horton yep - that was my mental argument - and I suddenly started panicking when you mentioned bounded, and thought I must be wrong

@JasperLoy @HenryT.Horton I think it is now the only book on AT I now own - looked at it again last night, and wondered if it was "exotic"

@MeAndMath what do you assume about $M$?

I think you should assume it to be locally compact or something like that in order to be able to prove this.

@t.b. The exercise only states that M is a metric space...

@JasperLoy That sounds interesting - although I am not sure I really have time for AT nowadays

Hmm I don't even have an ebook of Fulton

My first brush with AT was many years ago from an old book by Hilton and Wylie

The table of contents looks strange

Fulton

@HenryT.Horton yep - I picked it up years ago in a yellow sale

and never really read it

I don't read a lot of AT books but I haven't seen (co)homology done before fundamental groups before

@HenryT.Horton yeah - De Rham cohomology very early in the book

@MeAndMath A good bounded example where this fails without compactness of $K$ is if you take $e_n$ to be the standard basis in $\ell^2$ and take $F = \{(1+1/n)e_n\,:\,n \in \mathbb N\}$ and $a = 0$. Then $d(a,F) = 1$, but there's no $x \in F$ such that $d(a,x) = 1$.

Of course, $F$ is closed.

@t.b. hmmm good.

@JonasTeuwen what was the reason for your asking about the topological characterization of the line?

If I may ask

@tb I was trying to understand Polish spaces, why do we define them as we define them...

What is your definition? Second countable, completely metrizable?

Homeomorphic to a completely metrizable separable metric space.

Why homeomorphic? That's redundant.

Anyway. I think the main reason really is that you have a natural way of encoding them using the Baire space $\mathbb{N}^\mathbb{N}$.

(which is sort of the mother of all trees).

@t.b. Sorry metric, not metrizable.

I see.

@t.b. Perhaps I must postpone thinking about this until a bit further in descriptive set theory.

@OldJohn Ahh. Joshua Bell!

But no Scotch yet.

@t.b. is the homeomorphic bit really redundant?