Mathematics - 2017-01-22 (page 5 of 5)

If I have an not identically holomorphic "around" $0$ function $f$ such that $f(1/n)\le e^{-n}$, what can said about $f$ ? I write $f(z)=z^p g(z)$ but I get $g(1/n)\le n^p/e^{n}$ and it converges to $0,$ so not sure how can I continue.

Socrates

8:25 PM

I would be pleased about questions that I could answer.

Akiva Weinberger

Remind me what "not identically" means?

Socrates

@TedShifrin do your kiddos sometimes ask you something where you first have to think about a while?

JeSuis

Sorry it's poorly written, I mean that $f$ is not zero everywhere

Daminark

OK yeah, I was wondering what "not identically holomorphic" meant.

JeSuis

@AkivaWeinberger Ah, of course I get $g(0)=0$ but when I write $f(z)=z^p g(z)$ I do have $g(0)\ne 0$

thanks

Akiva Weinberger

8:30 PM

I think I made a typo (which is why I deleted it)

What's $p$? @JeSuis

The multiplicity of the zero?

JeSuis

yeah but when I read your comment, I get the point ^^

Akiva Weinberger

Ah, OK

$f(z)/z^p$ goes to $0$ at $0$, but it should be nonzero

Yeah

JeSuis

$p=min\{z: f(z)\ne 0\}$

Balarka Sen

The point is holomorphic functions are like polynomials. They can't decay so fast.

But I see you already got an answer

JeSuis

Yeah, I get the idea when I saw Akiva writes $z^p g(z)$ I don't know why ^^

but nice "intuition" @BalarkaSen

Akiva Weinberger

8:34 PM

Wait, so it's not the multiplicity?

Balarka Sen

@Akiva It is

Akiva Weinberger

You don't mean $\min\{p:z^{-p}f(z)|_{z=0}\ne0\}$?

Balarka Sen

It's just the exponent of the leading term in the Taylor expansion

I had more questions, but Ted is gone. :(

Akiva Weinberger

He's listening to music

Probably not "♫ E B r B C D F G# :|| ♫", but it could be

Theoretically

LaTeX should implement music notation.

That would be so cool.

Balarka Sen

Music is great stuff

Akiva Weinberger

8:45 PM

Googles

Alessandro Codenotti

musescore is the way to go if you need to write music

Simple

1

Q: Show $K_2=f+O(\Delta\,t)=f+\Delta\,t(f_t+ff_u)+O(\Delta\,t^2)$

Consider the single differential equation $u'(t)=f(t,u(t))$ and the numerical method $$\begin{align*}U^{n+1}&=U^n+\frac{\Delta\,t}{2}(K_1+K_2)\\ K_1&=f(t_n,U^n)\\ K_2&=f\left(t_n+\Delta\,t,U^n+\frac{\Delta\,t}{2}(K_1+K_2)\right)\end{align*}$$ Let $u(t)$ denote the solution of the differen...

Akiva Weinberger

$\require{abc}$
\begin{abc}[name=c-dur]
X: 1 % start of header
K: C % scale: C major
"Text"c2 G4 | (3FED c4 G2 |
\end{abc}

Simple

I need some help do this problem

Akiva Weinberger

Aw

Apparently LaTeX can do it, though

Not MathJax

Still, very cool

Socrates

8:56 PM

screw it

Daminark

Lmao @Socrates

Try it in Latex?

Socrates

"Apparently LaTeX can do it" - Akiva

Daminark

No I didn't capitalize the T

Or X

I meant the material\

Probably shouldn't have capitalized the L

Socrates

I understood you before you typed it

Daminark

Oh lol

Socrates

9:00 PM

i mean, after the second line

i fear that everything I experienced is nothing

that everyone I communicated with were very sophisticated cleverbots

Daminark

Dammit guys we failed the Turing test

Alessandro Codenotti

We're definitely not bots, oxygen is awesome, long live oxygen!

Astyx

We ? It's clearly your fault

Daminark

( ._.)

Socrates

what you think about this emoticon? ._O

I told someone to read the forum rules, and it seems he/she didn't understand them

every pc game is trash, heroes of might and magic III, a beloved game by me, is now boring. And every game is boring.

Sylent Nyte

9:11 PM

hey guys, need some help with normal distribution

I have $Y~N(100,10)$ and I need to find $P(Y<95)$, so I try to convert to $Z~N(0,1)$ and that changed my $P(Y<95)$ to $P(Z<-5/\sqrt{10})$ How can i find the value for this without using the table

Akiva Weinberger

Use \sim and \sqrt

$\sim$, $\sqrt{\phantom{Hello!}}$

@SylentNyte

Sylent Nyte

there we go

sorry, forgot the syntax

anybody?

Alessandro Codenotti

@Balarka are you still here?

Balarka Sen

yep

Alessandro Codenotti

Is $\Lambda^k(V)^*$ spanned by stuff of the form $dx_1\wedge\cdots\wedge dx_k$ with $dx_i$ in $\Lambda^1(V)^*$? (also isn't $\Lambda^1(V)^*$ just $V^*$?)

Balarka Sen

9:26 PM

Well, $dx_{i_1} \wedge \cdots \wedge dx_{i_k}$. But yes, and yes.

Alessandro Codenotti

yeah that's what I meant, annoying indexes :P

Balarka Sen

Formally, it's called an exterior algebra. But the algebraic story isn't something I find very important.

Let me know when you want to actually get to differential forms.

Alessandro Codenotti

Wikipedia says the exterior algebra is the direct sum of the $\Lambda^i(V)^*$, which makes sense to me, or a quotient of the tensor algebra, which doesn't

whenever you want (and can)

Socrates

can you explain me your question? :34946651

Balarka Sen

tensor algebra is just the same story, but with multilinear forms with no alternating condition.

Sylent Nyte

9:33 PM

@Socrates ah its okay, I had to find $P(Y<6.957010852370434)$ and as the table didnt cover it I wanted to know if there was an equation, but then I realised that itd be 1

Balarka Sen

@Alessandro Let's do it

Socrates

@SylentNyte one decimal place is more then enough for almost anything

Sylent Nyte

@Socrates felt like being extra

Socrates

@SylentNyte i appreciate

can you explain what you searched @SylentNyte

Alessandro Codenotti

Ok, I have no idea what the tensor product is (and I don't want to find out right now), but I guess that $\omega\wedge\omega=0$ because of that quotient?

Sylent Nyte

9:38 PM

@Socrates what like on google?

Balarka Sen

$\omega_1 \wedge \omega_2 + \omega_2 \wedge \omega_1 = 0$ happens because of the quotient in general.

But yes.

Socrates

@SylentNyte no, what is $Y\sim N(100,10)$?

Sylent Nyte

@Socrates Is just a normal distribution my teacher gave me as part of some prep for next lesson

Alessandro Codenotti

Ok, I'll add the tensor product to the never ending list of things I should find out more about

so, what's a differential form?

Sylent Nyte

however I got stuck at the very last bit at where I had to do $P(Y<6.957)$ so I started to google and then went onto here before realising its just $1$

Akiva Weinberger

9:41 PM

Tensor $V\otimes W$ is the set of things of the form $k_1v_1\otimes w_1+k_2v_2\otimes w_2+\dotsb+k_nv_n\otimes w_n$

Mary Star

Does it hold that the cardinality of $\mathbb{C}$ is the same as the cardinality of $\mathbb{R}$, $\frak c$ ?

Akiva Weinberger

with $(a+b)\otimes c=a\otimes c+b\otimes c$, and similarly for $a\otimes(b+c)$

@MaryStar Yep

Mary Star

@AkivaWeinberger Ah ok. Thanks!!

Balarka Sen

Nah, it's ok. You're not working basis-wise.

@Alessandro Given an $n$-manifold $M$ a differential $k$-form $\omega$ on $M$ is a smoothly varying alternating multilinear $k$-form $\omega(p)$ on $T_pM$ for each point $p \in M$. Smoothly varying in what sense? Well, just that given $k$ (tangential) vector fields $X_1, \cdots, X_k$ on $M$, $\omega(p)(X_1, \cdots, X_k)$ is a smooth function $M \to \Bbb R$ of $p$.

Akiva Weinberger

Also $(ka)\otimes b=a\otimes(kb)=k(a\otimes b)$

OK, so I don't need the constants in the general form thing

Balarka Sen

9:47 PM

Start with an example. Let $M = \Bbb R^n$, the simplest example of an $n$-manifold. $T_p \Bbb R^n$ can actually be identified with $\Bbb R^n$ itself; tangent vectors at $p$ are vector in $\Bbb R^n$ with the "tail" at $p$.

Alessandro Codenotti

hmmm I'm not sure I follow the vector fields part

Balarka Sen

Don't worry about it, it'll be clear once we get a good handle with $\Bbb R^n$.

I just threw you the abstract definition to see how you react to it.

Just think of "smoothly-varying" hand-wavingly

In any case, a $k$-form on $\Bbb R^n$ is an expression of the form $\sum f_I dx_I$ where $I$ runs through ordered $k$-tuples, and $f_I$ are smooth functions on $\Bbb R^n$.

So instead of the constant term in front of the $dx_I$ in the alternating multilinear theory, you have a smooth function. Does that make sense?

Alessandro Codenotti

it does. So in the simplest case a $1$-form is just $fdx_{i_1}$ with $f$ smooth right?

Balarka Sen

write $i$ instead of $i_1$ for simplicity. Yep

Alessandro Codenotti

yeah I never know how to write those indices

Balarka Sen

9:55 PM

Eg $dx$, $dy$, $-ydx + xdy$ are all 1-forms on $\Bbb R^2$

$dx \wedge dy$ is a 2-form on $\Bbb R^2$

@Alessandro why does this match up with the abstract definition? If $\sum f_I dx_I$ is a $k$-form on $\Bbb R^n$, then "evaluating at a point $p$" gives an alternating k-multilinear form $\sum f_I(p) dx_I$ at $T_p \Bbb R^n$ :)

since $T_p \Bbb R^n$ is as a vector space same as $\Bbb R^n$ you can also see it as a alternating k-multilinear form on $\Bbb R^n$ (which is a vector space this time, not a manifold). but technically it's on $T_p\Bbb R^n$.

Daminark

Lol @Balarka when we talked about forms, that was one of our main examples

$-ydx + xdy$

Balarka Sen

it's an important example of a 1-form

Daminark

The main one was $\frac{-ydx + xdy}{x^2 + y^2}$

Mike Miller

You probably had a stretch factor on it.

Balarka Sen

@Daminark That's a 1-form on $\Bbb R^2 - {(0, 0)}$, @Alessandro.

We're already getting examples of forms on more general manifolds than R^2

Daminark

10:01 PM

Since we defined winding number about the origin using it, as well as when we did that problem about de Rham cohomology of the punctured plane

Alessandro Codenotti

hmm, ok, I see how those are forms, I'm not sure how do we get an alternating k-multinear form on $T_p\Bbb R^n$

Daminark

Actually we ended up using forms to prove the fundamental theorem of algebra

It was a nice proof

Alessandro Codenotti

Ah, no, wait, I do

Balarka Sen

@Daminark Right, winding numbers are key to FTA

@Alessandro :) Eg, $xdx$ is a 1-form on $\Bbb R$. evaluate that at $2$, you get a 1-form $2dx$ on $T_2 \Bbb R$

evaluate that at $0$ you get the trivial 1-form

so it's really a family of alternating k-multilinear forms at each point, or rather, at tangent spaces at each point

Daminark

We had two other proofs, both of which had the same idea of saying that some positive minimum for the polynomial is achieved, but then you can't have that geometry of the image of a polynomial being tangent to a disk

Alessandro Codenotti

10:06 PM

@BalarkaSen that $1$-form actually is just $x\mapsto 2x$ at the end of the day since $T_2\Bbb R=\Bbb R$ and $1$-forms are linear functionals, right?

@BalarkaSen ok that makes sense now

Balarka Sen

@AlessandroCodenotti Exactly!

Daminark

The proof we did in class just manufactured the term, the other proof was on the homework and used another exercise to show that a polynomial is an open mapping

Alessandro Codenotti

So, sanity check $dx\wedge dy+dx$ is a meaningless expression, $dx\wedge dy+dz\wedge dx$ is a $2$-form on $\Bbb R^3$

Balarka Sen

That's right.

You have a vector space of $k$-forms on $\Bbb R^n$, but this is a vector space over the space of smooth functions $C^\infty(\Bbb R^n)$ (funny... this is not a field)

Because given a $k$-form $\sum f_I dx_I$, scalar multiplication is really multiplication by a smooth function $g$ on $\Bbb R^n$. Addition works exactly the same way as before.

The formal word is that the space of differential $k$-forms on $\Bbb R^n$, denoted as $\Lambda^k(\Bbb R^n)^*$, is a "module" over the ring $C^\infty(\Bbb R^n)$ of smooth functions. It's a "vector space over a ring"

I guess you know what a module is though.

Alessandro Codenotti

I know the definition and a couple of basic facts, but not much

Balarka Sen

10:16 PM

That's fine, I don't either. None of the algebra, or the abstract formalism I told you, matters much

Alessandro Codenotti

but that makes a lot of sense to me

Balarka Sen

Great

Alessandro Codenotti

begin{rant} why do mathematicians always use the same symbol ($\Lambda^k(\Bbb R^n)^*$) for $2$ different things? end{rant}

Balarka Sen

In one case R^n is a vector space. In the other R^n is a manifold :)

it's rather confusing, but such is life: R^n has way too many structures

I think Ted uses $\mathcal{A}^k(\Bbb R^n)$ for differential forms though. I forget.

Also you should remove the star. I mistyped

Alessandro Codenotti

do you mean everywhere or just for differential forms?

Balarka Sen

10:20 PM

Just for forms.

Mike Miller

forms should have the star

Balarka Sen

hmm, I thought that's for vector space levels only

actually aren't forms $\Omega^k(\Bbb R^n)$

@Alessandro Note that wedge product of $k$-forms work the exact same way too. $dx_I \wedge dx_J = dx_{(I, J)}$, and then extend to all of $\Omega^k(\Bbb R^n)$ (that's the notation I am going to use for now) as a $C^\infty(\Bbb R^n)$-module.

and it satisfies anticommutativity $\omega \wedge \eta = (-1)^{k\ell} \eta \wedge \omega$ where $\eta, \omega$ are $k$- and $\ell$-forms respectively. just like alternating multilinear forms.

Fargle

Hello chat.

Alessandro Codenotti

Ok, that was expected, I'll check it though. (As far as notation is concerned Ted uses the star for vector spaces and $\mathcal{A}^k(\Bbb R^n)$ for forms)

Balarka Sen

Right, but I am going to use $\Omega^k$ for what Ted calls $\mathcal{A}^k$

Alessandro Codenotti

10:30 PM

sure, as long as we agree on what to use

Balarka Sen

Right.

The real deal with differential forms is that we can differentiate and integrate them.

Alessandro Codenotti

I was starting to wonder what's the point

Balarka Sen

A 0-form is just a smooth function $f$ on $\Bbb R^n$. We can "differentiate" that to get a 1-form: $df$, the derivative of $f$.

Generalize: given a $k$-form $\sum f_I dx_I$, we can "differentiate" that to get a $(k+1)$-form $\sum df_I \wedge dx_I$.

Alessandro Codenotti

ok, the relationship with alternating k-multilinear forms kinda breaks down at 0-forms

Balarka Sen

an alternating 0-multilinear form is... a scalar. a differential 0-form = a smoothly varying 0-multilinear form = a smoothly varying scalar = a smooth function :)

nothing really breaks down

Alessandro Codenotti

10:36 PM

hmm, well, maybe I'm just annoyed by the fact that scalars look like functions with zero arguments there

Akiva Weinberger

$\pi$ is a function with zero arguments. $\pi()$

And now I look like a programmer.

Balarka Sen

@Alessandro The point is not that a scalar is a function with zero argument. The point is that you're assigning to each point of $\Bbb R^n$ a scalar. Done smoothly, it's a smooth function.

$p \mapsto f(p)$

Alessandro Codenotti

ok, I'm convinced I was just being stupid there. So back to differentiation

Balarka Sen

OK. So you agree with me that you can "differentiate" a $k$-form to get a $(k+1)$-form like I said above?

This operation is known as the exterior derivative, denoted as $d : \Omega^k \to \Omega^{k+1}$, BTW.

Alessandro Codenotti

well I'm sure that's a well defined operation turning $k$-forms into $(k+1)$-forms and it agrees with my notion of differentiation for (smooth) functions, I'm not sure yet that's the right generalization

Balarka Sen

10:43 PM

It's funny how none of these definitions actually make sense at a glance. You wanna differentiate the $1$-form $fdx + gdy$ (on $\Bbb R^2$) for me, see what you get?

Remember that $df = \sum_i \partial f/\partial x_i \cdot dx_i$, by definition if you wish.

Mike Miller

@Balarka The dual means they eat vectors, which is what forms do. Now R^n is canonically identified with its dual, but you shouldn't use that.

Alessandro Codenotti

@BalarkaSen $df\wedge dx+dg\wedge dy$?

Balarka Sen

@Alessandro Yes. Expand using what I wrote? Or rather, write in terms of the basis $dx, dy$ in $\Omega^1(\Bbb R^2)$

@MikeMiller I agree, I just wasn't sure if that's what one uses for manifold-level forms too. I was aware of that star only at vector space-level.

But we agreed on a better terminology

Mike Miller

Sure you're aware of it on the manifolds level. You use the cotangent bundle!

Balarka Sen

Yeah good point

Alessandro Codenotti

10:52 PM

modulo mistakes I got $\big(\frac{\partial f}{\partial y}-\frac{\partial g}{\partial x}\big)dy\wedge dx$

Balarka Sen

That's it. $(g_x - f_y) dx \wedge dy$.

Does $\partial g/\partial x - \partial f/\partial y$ remind you of anything? Eg, from vector calculus?

Alessandro Codenotti

not really? But I might be missing something obvious

Akiva Weinberger

Threen's guiorem

Or something

Balarka Sen

lol

Alessandro Codenotti

we'll see Green's theorem in the next semester :/

Balarka Sen

10:59 PM

Ahh, I see.

Eh, what the hell. The discussion has gotten nonlinear enough that I can tell you about it

Call $\omega = fdx + gdy$. Then you just proved $d\omega = (g_x - f_y) dx \wedge dy$

Alessandro Codenotti

lol maybe we should get a separate chatroom. I'm copy/pasting everything in a LaTeX file at the moment to look back if I need to.

right

Balarka Sen

Let's spam the geometry/topology chatroom with this garbage

the room owner won't ban us because i'm the room owner

Daminark

There's a geometry/topology chatroom?

Alessandro Codenotti

that's one way to keep it active I guess

Daminark

Oh lol

Balarka Sen

11:04 PM

@Daminark This is the one

Daminark

Ah, cool, thanks!

Jasch1

11:32 PM

Can someone help explain this to me

Well-order

In mathematics, a well-order (or well-ordering or well-order relation) on a set S is a total order on S with the property that every non-empty subset of S has a least element in this ordering. The set S together with the well-order relation is then called a well-ordered set. In some academic articles and textbooks these terms are instead written as wellorder, wellordered, and wellordering or well order, well ordered, and well ordering. Every non-empty well-ordered set has a least element. Every element s of a well-ordered set, except a possible greatest element, has a unique successor (next element...

Alessandro Codenotti

what exactly are you having troubles understanding?

Jasch1

Just the general concept

I am familiar with sets and the concepts with sets and infinite sets

and I need to understand this better for a paper I'm writing

arctic tern

which part don't you understand?

Jasch1

I'm having trouble understaning the concept of being well-ordered as a whole

I was hoping someone could attempt to help me by explaining it in more basic terms

Alessandro Codenotti

do you understand what an ordering on a set is?

Jasch1

11:37 PM

I think I'm looking at something in the wrong area, I was looking for stuff more related to Zermelo and Cantor but this seems to be a little different

I'm not familiar with ordering on sets no

arctic tern

a total order is a way of arranging the elements of a set in some order so that every two elements are comparable (given any two elements, one of them comes before the other). there are orderings which are not total, e.g. the collection of subsets of {1,2,3} can be ordered by "A<B whenever A is a proper subset of B," in which case e.g. {1} and {2,3} are not comparable to each other.

Jasch1

Thats seems relatively easy to understant

arctic tern

a well-order goes further: every subset must have a unique minimal element. in general, subsets of total ordered sets may not have minimal elements (e.g. the even integers as a subset of the integers has no minimum).

Jasch1

Could you give some examples to make it a bit cleared?

Thank you for helping me btw

arctic tern

examples of what specifically?

Jasch1

11:43 PM

Well-ordered sets vs non-well-ordered sets

Akiva Weinberger

The positive integers $\{1,2,\dots\}$ are well-ordered. The integers $\{\dots,-1,0,1,\dots\}$ are not.

It's not well-ordered because you can take the entire set as the subset; there's no smallest integer.

On the other hand, the nonnegative rationals are not well-ordered

since the subset of the positive rationals has no minimum

even though the entire set does.

Jasch1

Essentially there needs to be a defined min for it to be described as well-ordered?

arctic tern

suppose L is an infinite well-ordered set. then it must have a minimum element, call it 0. then L-{0} must have a minimum element, call it 1. then L-{0,1} must have an element, call it 2. and so on. we get {0,1,2,3...} inside L. but then if L-{0,1,2,...} still has elements in it, it must have a smallest, call it ω. if L-{0,1,...,ω} still has elements, call its minimum ω+1. we can continue in this way to get L={0,1,..,ω,ω+1,...,2ω,2ω+1,...} (although it has to stop eventually)

Akiva Weinberger

$S=\{0\}\cup\{1,1/2,1/3,\dots\}$ is another example; it has a minimum, but the subset of nonzero elements doesn't, so it's not well-ordered.

On the other hand, $T=\{0\}\cup\{-1,-1/2,-1/3,\dots\}$ is well-ordered.

$\{\infty\}\cup\{1,2,3,\dots\}$ (with $\infty$ defined to be greater than any other element) is also well-ordered, and is in fact order-isomorphic to $T$.

Jasch1

So essentially the set and all subsets must have a defined min?

Akiva Weinberger

11:48 PM

Yup.

All finite sets are well-ordered, also, clearly

arctic tern

@AkivaWeinberger ....

Jasch1

LMAO I got that far on my own

Akiva Weinberger

…?

arctic tern

all totally ordered finite sets are well ordered

Akiva Weinberger

We're assuming everything here is totally ordered, I think

@Jasch1 Do you know about countability and uncountability?

It turns out that there is an uncountable well-ordered set

Jasch1

11:50 PM

Yes I understand countability vs uncountability

Akiva Weinberger

though it's hard to construct, kind of

Jasch1

But I am a bit more confused now, because although some sets are not well-ordered, the Well-ordering theorem says "In mathematics, the well-ordering theorem states that every set can be well-ordered"

from wikipidia

Akiva Weinberger

By changing the order.

Jasch1

but in the set that you mentioned before

Akiva Weinberger

Like, the negative integers are not well-ordered,

arctic tern

11:52 PM

@Jasch1 that means if you start with a set that has no ordering on it, it is possible to find a way to well-order it. it doesn't mean if you start with an already ordered set that it's well-ordered.

2

Akiva Weinberger

but if you define a new order $\prec$ such that $-1\prec-2\prec-3\prec\dotsb$, then it becomes well-ordered.

It's just not well-ordered with the usual order $<$.

Jasch1

$S=\{0\}\cup\{1,1/2,1/3,\dots\}$ how would this be rearranged or have a new order defined such that it is well-ordered?

Akiva Weinberger

Use the opposite order

2

Meaning, define a new order $\prec$ such that $a\prec b$ iff $a>b$ (where $>$ means the old order)

2

Jasch1

ok

Akiva Weinberger

So we'd have $1\prec1/2\prec1/3\prec\dots\prec0$

2

This set (with $\prec$) is isomorphic to the set $T$ from before (with the old ordering $<$)

2

8 mins ago, by Akiva Weinberger

On the other hand, $T=\{0\}\cup\{-1,-1/2,-1/3,\dots\}$ is well-ordered.

That one. ^

2

Jasch1

11:55 PM

ok, i seem to have a decent understanding

Thank you guys!

Akiva Weinberger

Alternatively, you could define a new order $0\prec1\prec1/2\prec1/3\prec\dotsb$, making it isomorphic to the nonnegative integers @Jasch1

Alessandro Codenotti

@AkivaWeinberger fun fact: a set is finite iff it has a well order whose reverse order is also a well order

Akiva Weinberger

There's really an uncountable way to order any countable set. EDIT: …even if you don't count isomorphic orderings as different.

@AlessandroCodenotti This is true.

@Jasch1 You should learn about ordinals, which essentially measure how complicated a well-ordering is.

That's way too many stars

3

The ordinal corresponding to any finite set is just the number of elements in it, so $\{0\}$ is the ordinal $1$, $\{0,1\}$ is the ordinal $2$, etc

We define $\omega$ to be the order type of $\{0,1,2,3,\dots\}$.

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