hi @peoplepower

@skullpatrol We meet again.

@OldJohn I have seen plenty such proofs. You have never added $0$?

a+0=0+a=a

@JonasTeuwen is this^ what you're talking about?

@skullpatrol Perhaps something like $f(x+h)g(x+h)-f(x)g(x)=f(x+h)g(x+h)-f(x+h)g(x)+f(x+h)g(x)-f(x)g(x)$ which simplifies to $f(x+h)\Delta g(x)+\Delta f(x)g(x)$

@peoplepower ? which simplifies to 0_o

@skullpatrol $-f(x+h)g(x)+f(x+h)g(x)$ is zero, so we're adding zero and expressing it in a compatible way.

@JasperLoy Yep

@anon Hey

Did you see the rest of what I typed yesterday?

yo

yeah

Is it ok for you?

yeah

@anon I also edited my answer completely here

@BenjaLim Hey, Aussie.

EXTERIOR ALGEBRA

@PeterTamaroff I edited my answer in the link there

@BenjaLim BLACK HOLES.

@anon Actually now I think determinants are very natural

@anon Did my explanation to your thing yesterday help

showing the map was balanced helped

yeah because then you get a group homomorphism out of the tensor product

but I think the key point now is that you have some guy being a $kH - k(H \cap K^g) $ bimodule

so the tensor product has the structure of a $kH$ module

and your group homomorphism is now a homomorphism of left $kH$ - modules @anon

@anon Hey I felt because you helped me so much in the past that this time I should try to help you too

@JasperLoy I was just curious, I didn't read it. I'm with analysis now, and I think I'll move on with algebra using my uni's book.

@PeterTamaroff This is the way to think of the determinant

@BenjaLim Are you evangelizing us now?

@PeterTamaroff You should say: Are you evangelizing ?

I don't think you can "evangelize someone"

@BenjaLim You do.

but anyway I wanted to say determinants are beautiful

@BenjaLim Awwwwwwwwwww

Did you propose?

yeah if you think about them in terms of exterior powers

@JasperLoy determinants

Now I disagree with axler

Analysis now?

But if you have read all the books you suggest. You would be an A+ researcher already. I think...

@JonasTeuwen math.stackexchange.com/q/211481/38268

@BenjaLim Can yuo help mewith some¿

@PeterTamaroff yes?

@BenjaLim Well, let $A\subset \mathbb R$ be infinite and bounded.

ok

I have proven the following:

yes?

$(1)$ $x$ is an acumulation point of $A$ iff for every $\epsilon>0$ there exist infinitely many $a\in A$ such that $|x-a|<\epsilon.$

ok

$(2)$

@BenjaLim Hmm, I would argue through what those matrices mean.

$\overline{\lim}A$ and $\underline{\lim}A$ are acumulation points of $A$.

@JonasTeuwen what matrices?

Expanding it seems so unimaginative to me.

@BenjaLim The ones having determinant $1$.

@PeterTamaroff I don't understand your notation

@JonasTeuwen I don't understand what you're saying

@BenjaLim $\limsup A$ and $\liminf A$

ok

Now, my definition of those is:

you normally define those for sequences

but your $A$ could be uncountable

@BenjaLim Yes, of course.

@JonasTeuwen What are you saying to me about matrices?

The problem, of course, is that there is no limit after taking supremum.

It's constant.

But sup $A$ could be some isolated point.

I see. $\sup$

@BenjaLim We say $a$ is an almost upper bound of $A$ if there exist only finite amount of $y\in A$ with $y\geq a$ and we say $b$ is an almost lower bound of $A$ if there exist only finite amount of $y\in A$ such that $y\leq a$. We then define $$\limsup A=\inf\{x:x\text{ is an almost upper bound of} A \}$$ $$\liminf A=\sup\{x:x\text{ is an almost lower bound of}A\}$$

ok

Now I want to prove thatif $x$ is an accumulation point of $A$, $\liminf A\leq x\leq \limsup A$.

Ok. What have you tried?

@BenjaLim Well, I was thinking about arguiing by contradiction, but I had to leave.

Let me see what I wrote.

Wait da fuq?

@BenjaLim Huh?

Isn't that inequality you wrote down obvious from the definition of sup and inf?

oh no sorry

@BenjaLim That is not a set!

But wait now I am confused what is $x$ then?

@BenjaLim $x$ is an accumulation point of $A$.

@BenjaLim One's a free variable, the other is chosen to be an accumulation point.

Ok.

@BenjaLim What shouldn't be hard to show is that if $x$ is an accumulation point of $A$ then $\inf A\leq x\leq \sup A$.

You see I am trying to do a proof by contradiction now and say that if $x < \lim \inf A$

@PeterTamaroff If $a$ is an almost upper bound and an accumulation point, then for small enough $\delta$, where will the $y\in A$ come from such that $|a-y|<\delta$.

i.e. which direction?

@peoplepower What is your native language?

@PeterTamaroff English.

@peoplepower I'm having a hard time understanding "come from such that"

@PeterTamaroff Which side of $a$ (left or right)?

@BenjaLim Such operators can be represented by matrices...

what operators

@BenjaLim Yes, me too.

Hence, you do that. Think about what it says and you are like "ah, easy."

@BenjaLim The thing you've linked.

what what says?

@JonasTeuwen I gave the proof as to why that is true in terms of exterior powers

Yes, and I feel that it is killing a mosquito with a hammer.

I feel more for seeing your exterior powers as a (kinda) natural generalization of the "easy stuff" than the other way around as a specialisation.

@peoplepower I think I got it.

@JonasTeuwen If you like algebra that is the way to think about it. What mosquito did I kill with a hammer?

@PeterTamaroff What is your proof?

I beg to differ that is.

suppose that $x < \lim \inf A$

@JonasTeuwen Why is it nuking a mosquito?

I am pretty sure the algebraists (I know some) have an intuitive understanding based on easier concepts.

[\bigwedge\nolimits^{\!k}V]

And exploit that intuition to solve harder problems.

Nevermind.

@JonasTeuwen we are having a debate

but I'm willing to listen to what you have to say.

because it may be useful

I don't feel like having a debate. You figure it out by yourself when you need to do research!

I had the same thing when I was younger.

Now I figured how people come up with these things... the other way around!

but I don't understand $x$ could be neither of these

@JonasTeuwen I mentioned reduced singular homology in my fourier analysis talk :D

@BenjaLim Why not?

@BenjaLim To prove what theorem?

$\limsup A$ and $\liminf A$ are accumulation points of $A$, always.

@JonasTeuwen Jordan curve theorem

Oh yes.

@PeterTamaroff ok.

@JonasTeuwen Btw Thierry Coulhon was listening too

goddam he was asking a fuck ton of questions I was shivering as fuck

The Brouwer proof?

But dimension two is sufficient, isn't it for Fourier analysis?

yeah but I don't know of any other except singular homology

anyway I just mentioned the general ideas used to prove it

not the full proof

@JonasTeuwen You never answered how a+0=a and a=a are the same?

Yea? So what?

You said they were.

@BenjaLim The one in $\mathbb R^2$ can be proven quite directly.

you need van kampen even

Mm. Just homotopy.

@skullpatrol What is your question?

@PeterTamaroff have you ever seen a proof where the step "a=a" is a vital step?

@JonasTeuwen still algebraic topology

@PeterTamaroff Jonas says that when you use a+0=a you are using a=a.

@skullpatrol How would that be a step? It is an axiom.

@PeterTamaroff Use the axiom in a step.

@skullpatrol We always use that. I don't understand what you mean.

@PeterTamaroff Example please.

@skullpatrol I mean, $=$ is a relation defined so that $a=a$, always.

@PeterTamaroff Please provide a specific example of how you would use that relation in a proof. @JonasTeuwen

No. Please shut up or stop pinging me!!!

@JonasTeuwen LOL

You pinged Old John saying a+0=a is the same as a=a.

@JasperLoy NEVER.

good answer^

@skullpatrol You need to be clear about what you're talking about.

$a+0=a\iff a=a$

@JasperLoy Err... Yeah well.

See you guys.

later

@JonasTeuwen Good luck.

@JonasTeuwen GUSFRAVA!

@JasperLoy Well, one lasts longer, so no =)

@PeterTamaroff One could last longer, so maybe ;-)

@JasperLoy Damn you, Loy.

@JasperLoy Well, you aren't black, so J is out of the question.

And you don't have a neuralizer.

mib

@PeterTamaroff J.J. was a character in the Good Times.