I'm reading a post on BRST by some crazy German guy

ACuriousSocks or something

@ACuriousMind it is on my list, just really far down :)

@0celo7 Ah

I understand why they're not happy about people skipping calculus, I guarantee AP calculus does not prepare you for Hon. ODE.

@ACuriousMind Looks like I'll be flying into Vienna on the 14th...won't be anywhere near you :(

@ACuriousMind $\operatorname{tr}X=\operatorname{tr}X^\dagger$?

sorry did not see

@ACuriousMind you have exactly 3600 close votes

Stalker alert

No, I have 3600 close reviews

what is the difference

I have cast quite a few more close votes, I suspect

@Danu no, I looked at everyone in the queue

I was curious

@0celo7 So you stalk everyone equally.

@ACuriousMind what is the difference

@ACuriousMind sure, does that hurt your feelings?

@0celo7 ...you do know you can cast a close vote any question, even when not reviewing, right?

@ACuriousMind ok, that's what it means

@0celo7 Made me think of... youtube.com/watch?v=0S3foICf5uI

@Danu I don't even read through all of his answers

now that would be stalkerish

Not even :P

I will read them (and usually upvote) if I see them

but I don't seek them out

@ACuriousMind On page 24 of Rel. Quant., I have some questions. Why is the collection of Vectoren zu endlicher Teilchenzahl (English?) dense in $\mathcal{F}$? Is it because the Fock space is the set of Vectoren zu endlicher Teilchenzahl + the "limit points" with infinitely many nonzero? What is $\mathcal{D}$ in the middle of the page? I don't think I quite understand what wesentlich selbstanjungiert means.

Okay, the Vektoren zu endlicher Teilchenzahl are "states of finite particle number". The states of finite particle number are dense in the total Fock space the same way finite sequences are dense in the space $\ell^2$ of square-summable sequences (do you see why this is true?).

ello

badumtiss

@ACuriousMind Is that not by definition dense? Let the finite sequences be $\{x_n\}$, $n<\infty$. Then the $\{x_n\}$ plus the $\lim_{n\to\infty}x_n$ gives $\ell^2$, no?

ello^2

@0celo7 Not really

The $\mathcal{D}$ is just the domain on which $A$ is essentially self-adjoint, it is defined by that sentence.

sigh I need to learn analysis

This is too much

is this supposed to be obvious?

@0celo7 A subspace is dense if you can approximate every element of the total space by a sequence of elements of the subspace. Given a square-summable sequence $\{x_n\}$, it is the limit of the finite sequences $\{y^{(m)}_n\}$ where $y^{(m)}$ agrees with $x$ on the first $m$ entries and is zero thereafter. Therefore, the space of finite square-summable sequences is dense in $\ell^2$.

In the same way, every state of infinite particle number is the limit of states of finite particle number, which are therefore dense

@ACuriousMind Ok, I misunderstood/derped what "finite square summable sequence" is. (Derped because what I said does not make sense in hindsight.)

What I really meant is what you said.

Well, good then, and you see that it is the same for the Fock space and the space of finite particle number?

Yes.

Every time I open up this analysis book I get depressed...I can't do the problems in the "prerequisites chapter".

How on Earth am I supposed to take this next year...

What's with the ello, everyone?

Don't delete it, explain it! :D

@0celo7 do you know the book by Carothers?

No.

@ACuriousMind I was just writing \ell as ello :p

@0celo7 it has a shit ton of easy to medium difficulty (and occasional hard) analysis problems

you could do those in preparation

@FenderLesPaul I literally cannot do any of the "prerequisite" problems in this book.

which book?

Jost, Postmodern Analysis

postmodern?

It's the book that the undergrad analysis sequence here uses

is this an art book?

No, it has nothing do with that

"Now that we have proven Heine-Borel, let us look at its implications on posthuman postmodern neo-classical liberal ideals"

Socratic help?

start by writing down the definition of a limit point in terms of $\epsilon > 0$

I have no clue what that is!

err accumulation point rather

since this book uses that term

"a limit point of a subsequence of a sequence is called an accumulation point"

ok now can you translate that into $\epsilon >0$ language?

it says here a limit point $x$ is one for which $$\forall\epsilon >0\exists N\in\mathbb{N}\forall n\ge N:|x-x_n|<\epsilon$$ if $\{x_n\}$ is a Cauchy sequence

so let $\{a_n\}$ be the subsequence

since it converges (by assumption), it is Cauchy (I know this is true, don't know how to prove it), and thus we have the same stuff above just with $|a-a_n|$?

actually no I do know how to prove that, it's given in the text

@FenderLesPaul ello?

why does everyone say this is easy...

I'm struggling

Meanwhile, y'all need to play this game

@0celo7 It's something like a mindset: You have to get into it initially, but once you do it gets a lot easier.

@0celo7 sorry groupmate came by for a question

yes that's fine

Also, it helps me a lot to try to rephrase things in natural language osmetimes.

o.O help

how is this the prerequisite material

I need to find the professor...if this is for real I can't take it next year

For instance, for convergent implies Cauchy, I'd say: If it eventually gets arbitrarily close to a single point, then eventually all points have to be arbitrarily close to each other because they're all getting close to that one point.

I can figure that out too

but how to put it in symbols?

Now just rephrase that in $\epsilon-\delta$ and you're done.

Jost uses the triangle inequality

Don't worry too much about the formal language if you really think you understand the intuition well. Make sure you do, though.

Yeah, of course

well that would have never occurred to me

In a week it'll be second nature to you

and 2) I have even less of a clue

Triangle inequality is literally all you have for arbitrary metric spaces

@0celo7 Well, start by computing $\lvert b_n - a \rvert$!

@ACuriousMind I have no clue??

I tried that, got nada!

Term by term

:(

what

Try to be concrete. Think about the first term, for instance.

how do you enable mathjax on chat kind people?

@FenderLesPaul ...you're just now doing that?

$d(\frac{1}{n} a_1, a)$

It's...for a friend...

link, top right

What happens when $n\to \infty$

goes to zero, maybe

or not, I have no idea!

@0celo7 Well, use the triangle inequality to get terms of the form $\lvert a_i - a \rvert$. Observe that there is a bunch of terms you can make arbitrarily small by making $n$ larger, and another bunch of terms which gets small by assumption on $a_i$.

or is it just |a|

@ACuriousMind high school did not prepare me for that at all

how are Freshmen supposed to do this

@ACuriousMind You're spoiling the learning process.

@0celo7 lol, high school didn't prepare me for that, either.

@0celo7 What are the properties of the metric?

omg magic

@Danu tringle ineq., positive definite

nondegenerate

Okay, before we consider separate terms it may be better to first talk about why we can split into separate terms

What property of the metric does that follow from?

triangle inequality?

Right.

Also, intuitively, do you see why the new sequence should still converge?

nope

Okay.

What is it doing? It's equally weighing the first $n$ terms of the original, convergent, sequence.

You know that, eventually, the original sequence gets close to $a$. So eventually, you're averaging over more and more points near $a$

So the average slowly shifts towards the "tail" of the original sequence, you see? That's of course towards the original limit

yes

Here we're working with the usual metric on $\mathbb{R}$ right?

well I played around with the triangle inequality and got $$|b_n-a|\le |a|+\sum_{i=1}^n|a_i/n|$$

@Danu yes

@0celo7 Is that a useful form? What do you want $\lvert b_n - a \rvert$ to become?

@ACuriousMind 0, I think

as n to infinity

Right. Does $\lvert a \rvert$ look as if it is going to become zero?

nope

I don't see any other way of applying the triangle inequality

Well, think about the givens. What kind of term do you know something about?

$|a_i-a|<\epsilon$

for some $i$ and any $\epsilon$

Exactly. So your goal should be to massage $\lvert b_n - a \rvert$ in such a way that terms of this form appear.

@ACuriousMind I get that, but I don't see how

Hm. Perhaps writing $a = \sum_{i = 1}^n \frac{a}{n}$ helps?

oh

god damn it why are you so smart

@ACuriousMind Not true for just any $n$

in fact, only in the limit

@Danu wat?

@Danu the sum just gives a factor $n$

wow, lol I thought isaw a sub_n on that second a

I was like... lolwatwatwatwatwat

$$b_n-a=\sum_{i=1}^n\frac{a_i-a}{n}$$

And now I guess you can use the nice properties of this special metric

then $$|b_n-a|\le \sum_{i=1}^n\frac{|a_i-a|}{n}\le \sum_{i=1}^n\frac{\epsilon_i}{n}\to 0?$$

actually, it's probably a generic property of all metrics

@0celo7 Halt!

How does the second inequality follow?

by me having to leave for something

srs, I gotta eat and then head out

will be on phone now

...and what is $\epsilon_i$ :P

Me running out the door because I'm late

@Danu Well, before he defines it it's at least not wrong ;)

I don't see what the issue is

Well, there are $n$ terms, each of which is divided by $n$. You have to show that this really does go to zero.

Does each term not go to zero?

Sure but you keep adding terms

as you probably know, $\sum 1/n$ does not converge

I did know that

But I'm not summing 1/n, am I?

Let's hope not ;)

No clue how to continue :/

Given any $\epsilon>0$, you want to show that this is eventually smaller than it

Hi @DavidZ

How's the QCD?

as normal

As you see, we've descended into the depths of epsilon-deltonics

and personally I'm starting to get sick of open-closed-clopen-or-neither

(reading some topology)

haha

@Danu Aha, the fine craft of epsilondeltamanship