Ah, nifty

@AkivaWeinberger nice

there's a hugely important PDE result that comes up due to Allard though

Or, parsed differently/alternately (a + bsqrt(n))^k + (a - bsqrt(n))^k is an integer so if one is an integer, the other is too. That means (a + bsqrt(n))^k - (a - bsqrt(n))^k is an integer; that's crap because it's an integer multiple of sqrt(n)

Oh speaking of lectures, in the last few days of the REU they have participant lectures and I emailed Peter about doing my Ramsey theory one

very technical proof and all that

But that clearly fails for negative $n$ since $|a+b\sqrt n|=|a-b\sqrt n|$ becomes always true (they're complex conjugates)

@Daminark is such a technical topic appropriate for an REU talk

@BalarkaSen You'd want to show that that last expressing equaling $0$ implies $a$ or $b$ equals $0$

it sounds like not such a good venue for it to me

@AkivaW Ah yeah true

There's where n > 0 comes in of course

I think this is equivalent to—

—oh wait my idea from earlier does work—

I don't think it'll be too much of a problem

—saying that the sixth roots of unity are the only ones with rational real parts

If nothing else, I don't think I have that much time so I'll probably just handwave some of the technical parts

which follows from Niven's theorem

I'm too brain dead to think about it, but that sounds like a fun idea

and I had discovered a cool proof of that a while ago but I forget how it went

Plus like, more of the people in that audience will have already taken or at least seen some algebra

idk man, the REU talks i attended last year were like super handwavy and non-technical, just presenting interesting tidbits. It just doesn't sound like an appropriate place for it to me. Unless you hand wave literally everything

Well, I mean so I had this one juggernaut theorem, so if I just give a sketch of that, the rest of it is gonna be moderate

@BalarkaSen Wait, $(1+\sqrt{-1})^4$ is like $-4$ isn't it

yeah if you're doing just like Van der Waerdan or whatever, that's interesting and stuff to an audience that doesn't know ramsey theory, the other theorems i remember from the section aren't really obviously interesting though

so that gives us another four counterexamples ($\pm1\pm\sqrt{-1}$)

@Akiva Uh, yes

(@LucasHenrique)

Yeah, I mean the whole section was just a means to the end of the Ramsey theory one

@AkivaWeinberger ?

i still don't really think it's that good of an idea on the face of it but i'm sure you can pull it off

well, I should have specified that $n > 0$

@LucasHenrique Yeah I know

I was just curious about the $n<0$ case

I'll see how it goes after Peter May gets back to me on the length of the lecture

@Daminark So, where were we on FA

"mixing" $\Bbb Z[\sqrt{n}], n > 0$ and $\Bbb Z[i]$ seems a bit weird

Okay well, so we talked about bounded linear functionals, and in particular we know that the dual of a normed space is a Banach space

Okay so, let's say we have a Banach space and some bounded linear functional on it

@LucasHenrique (The other one was $(1+\sqrt{-3})^3=-8$)

oh, wait, did I prove the last step in B(X, Y) being a Banach space for Y Banach and X normed?

Oh, uh, I think we did?

It was that $\|T_n - T\| \to 0$ in the operator norm, right? That's the thing I was stuck on maybe

you said that's a consequence of $\|T_n - T_m\| \to 0$

Well, in any event, the way it goes is that if you have $\|T(x) - T_n(x)\|$ for $\|x\| = 1$, you have that this is equal to $\lim_{m\to\infty} \|T_m(x) - T_n(x)\|$, but that's less than $\epsilon$ for the right $m$ and each one beyond by uniform Cauchyness

So then in the limit, that holds as well, so that $\|T - T_n\| \le \epsilon$ for each $\epsilon$, but then yeah

Hey chat!

Hey @Perturbative!

Hey @Dami, how's goes it?

It go k, you?

@TedShifrin I wouldn't be surprised at this point if you knew that G&P mentioned 'local coordinates' on pg 4 without even needing to look at the book :p

@Daminark For me it go k too

@Perturbative yeah GP is a bit loose

@Daminark Hahn-Banach?

Not quite, just the fact that for each hyperplane there's a linear functional, unique up to a constant.

oh i mean that's not even an infinite dim thing

Yeah that's my only irritation with G&P so far

I mean yeah but it's a thing that isn't often talked about in linear algebra courses

So might as well bring it up

There's the relation with closed hyperplanes and continuity that's a uniquely infinite dim phenomenon

@SimplyBeautifulArt You should be able to do this math.stackexchange.com/q/2379145/166353

Oh, okay?

I think his scheme will only get him the Veblen ordinals? But I kinda forget what the Veblen ordinals are

xD

@aminliverpool when you see this: To that end I restate what I mentioned earlier, that there's some sense in which generic discussion happens here, including that which can get quite focused, and when people have something of a longer term "Hey let's do a group stufy of blah", that's when they're likely to create a side chat.

Oh wait no

I mean Veblen function

Hahn banach is something you could probably prove right after learning about what a hyperplane is tbh @Daminark

Great theorem

@amin Though I don't think this is the sort of thing that Clay would focus on though, like mathematical research is important to the community, and those are the types of problems that would be huge if solved, so it makes sense that they'll make a big deal about it. Plus, nifty self-studying isn't quite the sort of thing you'd give a million dollar award to, you know?

Yeah Hahn-Banach was nifty

It's also like the result that tells you that the dual space is an interesting object

Yeah, Hahn-Banach is nifty, I intend to get to it soon

Is there an interesting "operations on simple graphs" equivalent to the coldatz conjecture?

A key issue, the multiplications I know of and understand, cartesian and rooted, don't seem to have inverses for most simple graphs, everything is careasian and rooted prime!.

@Daminark, in the stuff G&P do on tangent spaces, there's this one thing that's been bothering me, they say the following "Suppose that $X$ sits in $\mathbb{R}^n$ and the $\phi : U \to X$ is a local parameterization around $x$, where $U$ is an open set in $\mathbb{R}^k$, and assume that $\phi(0) =x$ for convieniance."

Can we actually assume $\phi(0) = x$?

Yeah, let's say $\phi(p) = x$, then let your chart be $\phi(y-p)$

Or sorry, choose $\psi(y) = \phi(y+p)$

That's easier to see, so you'd then have that $\psi(0) = \phi(p) = x$

Ohhh thanks Dami!

Can you guys help me with a beginner-level linear algebra question?

Sure!

yes, it's the sort of question I can definitively say yet about, iYstead of "maybe, perhaps".

Let $\mathbf{v_1}, \mathbf{v_2}, \mathbf{v_3} \in \Bbb R^3$. If each vector is not parallel to another, show that $\mathrm{Ger}(\mathbf{v_1}, \mathbf{v_2}, \mathbf{v_3}) = \Bbb R^3$

What's Ger?

Generated set

@LucasHenrique You mean span?

That's usually called span in English, and this isn't true

I'm using Ted's book in portuguese, I'll google to see if the naming/notations are the same

Take the vertices of a triangle

Centered at 0

That'll generate a plane

But none of them is a scalar multiple of another

So my proposition is false?

the span of just 2 2D vectors yeilds a plane in most cases

Yeah, unless I'm misunderstanding it

they can be complex multiples of each other

Btw, my "generating/generated set" is given by $\mathrm{Ger}(\mathbf{v_1}, \ldots, \mathbf{v_k}) = \{ \mathbf{v} : \mathbf{v} = c_1\mathbf{v_1} + \ldots + c_k \mathbf{v_k}$ for some $c_1, \ldots, c_k\}$

@alan we're doing things in 3D, and R^3 isn't a complex vector space

jesus, where am I destroying notation?

Jesus logs on chat to let you know

I don't understand how is this proposition false.

$\mathrm{Ger}((1,0,0), (0,1,0), (1,1,0))$ is a plane but none of the vectors are parallel to each other @Lucas

Since when on $\Bbb R^3$, $\mathrm{Ger}(\mathbf{v_1}, \mathbf{v_2})$ is a plane.

Did you understand my counterexample?

@EricSilva Oh.

Yeah, since they're not parallel. But the set of vectors parallel to either is just two lines in the plane, so find a third line

So $\mathrm{Ger}(\mathbf{v_1}, \mathbf{v_2}, \mathbf{v_3}) = \mathrm{Ger}(\mathbf{v_1}, \mathbf{v_2})$ does not necessarily mean that $\mathbf{v_1} // \mathbf{v_3}$ or $\mathbf{v_2} // \mathbf{v_3}$

Precisely

Schlag voice

nope, it just means that $\mathbf{v_{3}}$ could be written as $a\mathbf{v_{1}} + b\mathbf{v_{2}}$ for some numbers $a, b$.

just that $\mathbf{v_3} \in \mathrm{Ger}(\mathbf{v_1}, \mathbf{v_2})$

@EricSilva yeah. :)

how can I deduce this?

btw @Lucas, generally portuguese terminology is faaiiirly close but it definitely diverges in some places

and it basically diverges in the same place french terminology diverges i think

So, imagine a vector which is a linear combination both of all 3 and of just 2

math.meta.stackexchange.com/questions/26726/… .

@Lucas: In English we write that Span. What's Ger in Portuguese?

That doesn't sound like a correct statement of anything I wrote :P

Ger is actually not a word. More like an abbreviation (and personally I've never seen that in common language)

How do you say subspace spanned (generated) by ... ?

What page are you on, @Lucas?

@TedShifrin Nope! You divided cases where either $\mathrm{Ger}(\mathbf{v_1}, \mathbf{v_2}, \mathbf{v_3}) = \mathrm{Ger}(\mathbf{v_1}, \mathbf{v_2})$ or $\mathrm{Ger}(\mathbf{v_1}, \mathbf{v_2}, \mathbf{v_3}) = \Bbb R^3$, and I (erroneously) assumed that the first case is where $\mathbf{v_3}$ is parallel to some of the other vectors

@Ted Gerar, to generate is what it's going for

Aha, @EricS, that makes sense.

Maybe I can learn Portuguese if Lucas asks enough questions. What page, @Lucas?

Page 14, example 11

Ah, OK ... Wow, this is weird. Good thing I sort of know what I'm reading :)

What do you mean, prof?

At the end of section 2, @Lucas, are all sorts of interesting geometry theorems to prove using vectors and dot products.

I mean that I don't know Portuguese, so it's weird to try to figure it out. :)

by the way, Ger was defined as "conjunto gerado" (conjunto = set, gerado = generated; Ger probably came from Gerado)

OK, that makes sense :)

Yes, just as I usually write $\text{Span}(\mathbf v,\mathbf w)$ to mean the subspace spanned by ...

I hope you enjoy the pictures, too, @Lucas :)

I am. The geometrical intuition makes it easier.

Which of Ted's books is this?

linear algebra, translated into Portuguese recently

It looks pretty good (paperback).

Ah cool cool

They never asked us for permission to get the translation done, but I hope it gets used places in Brazil/Portugal.

It's so weird: I've already had vectors, but the geometric aspect is making it hard, which implies that my knowledge was extremely limited, since the geometric aspect is very important.

@TedShifrin wait, what? hahah

By the way, Eric, the OP might appreciate your writing a non-moving frames answer to this one. :) @EricSilva

@Lucas: Lots of mathematicians who teach linear algebra do no geometric stuff at all. :(

I think I'll get to do a bunch of geometric stuff with vectors in this high school precalculus class I teach this year ... That'll be fun.

Oh yeah I could write an answer when I get to a computer

I could do it in the "usual" notation, but why bother? I hate it. :)

I actually don't like the question so much. I think you should understand that you get that $2\times 2$ minor of the second fundamental form even if $\Sigma$ isn't invariant under the shape operator.

I hope we get to moving frames in the bootcamp kind of early

There's so much stuff in Clelland's book that I think the bootcamp people could be ready for and gives nice applications of the method

Well, if everyone knows forms, you can bypass all the yucky stuff for holonomy and Gauss Bonnet.

The problem is, I don't know how well they all know forms

Still, don't get too carried away. There's a lot of intuition to develop along the way that isn't in her treatment.

I don't think they did it as thoroughly in their analysis class as my year did so I don't know the baseline amount of knowledge they have

No, Demonark hardly learned anything about forms ... barely stuff in the plane.

Although he did read some of my G&P notes on them, I think.

But the rest of 'em? ...

@Lucas: BTW, if you find typos or errors, please let me know. I have an errata page for the English edition, but I can add corrections for the Portuguese if you and Eric help me :)

Yeah Im quite fond of her book but I've had a very easy time reading it because I already knew what was going on mostly, so that might influence that

@Ted, since you're here and I'm a shameless person, how does $\mathbf{v_1}, \mathbf{v_2}, \mathbf{v_3}$ being linearly independent vectors imply that $\mathrm{Span}(\mathbf{v_1}, \mathbf{v_2}, \mathbf{v_3}) = \Bbb R^3$?

Heya! @TedShifrin

Hi @Perturbative :)

yo @Perturbative, welcome back :p

I do hope they like the diff Geo

I think it's such beautiful material

@Lucas: We probably don't quite know enough in the very first section to prove that, but we will soon, @Lucas. Once you get to matrices, it will follow by looking at the matrix whose columns are the $\mathbf v_i$. You can think geometrically by saying two of the vectors span a plane and then you use the third vector to move to parallel planes, which will fill up space.

Just remember, Eric, that they are not necessarily all as enamored of geometry as you, nor as quick at it.

@Ted, do you remember problem 2.6 in G&P? :p

Yeah I'm gonna have to try to hold back a little lol

Hum a few bars, @Perturbative :)

So I'm done with the first section. I'm gonna get this exercises :P

Oh, I looked it up. I never assigned that one. What's your question, @Perturbative?

@Lucas: Interesting that so many American students complain that we don't do any examples in the text. :P

@Ted I hope you can read that, I'll type it out if you can't

That's not 2.6 in my book, @Perturbative!!

Whoops sorry, Chapter 1, part 2 (on Derivatives and Tangents), problem 6

That's not #6 for me.

Really? Hmmm, could it be we have different editions of G&P?

Ohhhh, sorry. I was in section 3. Dopey.

I'm gonna go crazy if I don't understand how to construct the reals

The question was. "The tangent space to $S^1$ at a point $(a, b)$ is a one-dim subspace of $\mathbb{R}^2$. Explicitly calculate the subspace in terms of $a$ and $b$"

Your write-up is flawed when you choose $a,b>0$ at the beginning. Maybe you're assuming that the reader knows $(a,b)\in S^1$, but I didn't when I read it.

Rather than the yucky parametrization you used, could we use trigonometry instead and have fewer cases to worry about?

@Lucas, really??!! That was one of my least favorite lectures in real analysis when I took it.

Actually, it went on for weeks. :(

That wasn't my write-up it was a solution I found online :p

But you can read about it in Spivak or Rudin or ...

@Daminark Sorry, I was half following, you'd need the span of 3 3D vectors then indeed, and I'm not even sure how quaternions work, besides which there not (real number) multiples of vectors.

Ugh, @Perturbative, it's not a great solution. How are you doing it?

Quotient of $10^{\Bbb N}$ FTW

Since I started my olympiad classes, I became more rigorous about math. When I started to think about the concept of "existence", the reals do not look real.

DogAteMy!

(for constructing the reals—decimal expansions)

(and also joking)

Have fun defining addition, multiplication, etc., DogAteMy :P

There are only two sets that do not make me uncomfortable: $\Bbb Z$ and $\Bbb Q$

Well, if you're really interested, @Lucas, you can read about Dedekind cuts. (You haven't learned Cauchy sequences yet ...)

Any construction of the reals honestly makes me bored to tears

Eric, me too. But you and I have different mathematical taste from most people in here.

@AkivaWeinberger It's not a joke, in a sense (?), since the infinite decimal expansion of an irrational can be written as a Cauchy sequence

@TedShifrin I've seen someone do it

@TedShifrin I was doing something similar to the yucky parameterization :(, but my real problem was with the part where they say $\phi(0) = (a, b)$

@TedShifrin I know a bit about both. A bit.

and, to be honest, isn't that done in grade schools, in a sense?

@Perturbative: Go on.

The infinity of it makes it harder, but not too much harder

I thjnk

DogAteMy: Stuff arbitrarily far down the decimal expansion can affect stuff right near us here.

@EricSilva: The OP accepted my answer without any questions about the forms.

I guess they know about forms or something

And his answer was yes

Sure, @Perturbative: If $U\subset\Bbb R^k$ and $p\in U$ maps to $x$, just slide everything over and consider $U' = \{x-p| x\in U\}$.

Eric, that's the most frustrating thing about this site. One never knows what people know and what they don't.

@Perturbative: You'll also see repeatedly that if $df_x\colon\Bbb R^k\to\Bbb R^k$ is invertible we may assume (by linear change of coordinates) that $df_x$ is the identity matrix.

Right yeah I see that

@AkivaWeinberger Ironically, I can take infinity pretty much well. Since it's an integer infinity (yeah, I know that such a thing does not exist, I mean that I can see infinity as an uncountable number of things)

Okay thanks for that other explanation! @TedShifrin

Don't you mean countable rather than uncountable there?

@Perturbative: Can you do it with the $\phi(t)=(\cos t,\sin t)$ parametrization? What do we do if we want $\phi(0)=(a,b)$?

Countable infinity is the easiest one to deal with, since, in a sense, it's just $1+1+1+\dotsb$

Ok, I don't have theorical knowledge about this, but I'm using a "common sense" language

I've felt the desire to prowl for diff geo questions to answer but I'm honestly not super fond of the way this site works for that reason @Ted, chat is less stifled I think

If you could count infinity, then infinity would be finite

I don't know, it's weird...

Well, it's good practice for you to try to explain things using different techniques, too, Eric. Plus, occasionally, I suggest questions for you :P

Sometimes, I ask what they know and wait for them to respond. This time I didn't do that. I figured I might have to put in more stuff, but ...

@Lucas: You're going down the set theory rabbit hole.

Oh, so it kind of makes sense when using this concept of countability

Yeah that's true @Ted, I should have some practice explaining things generally. I think I'm bad at gauging the level my responses should be

BTW, @EricSilva, that Frobenius integrability question took me another two hours to get. Yuck. Who the hell would put such a thing on a qualifying exam.

@TedShifrin I'm already seeing that I'll have to forget everything about mathematics to build it from "concrete" postulates/axioms on set theory :P

Eric, in general, if you want to run things by me for comments I'm happy to help.

What frobenius question is this?

@Lucas: When I was about your age (a bit into college) I started going backwards and backwards trying to make sure all the foundations were OK, and then I realized I'd never do that again :P

Ok thanks Ted, I'll probably take you up on that

@LucasHenrique Oh, yeah, there's a formal definition of "countable" in math that's kind of nonintuitive

I used to be really concerned with foundations, but I think Schlag and André knocked that out of me

@Ted: in fact, are the foundations OK?

@TedShifrin I think Secret's been doing something like that, but with a sledgehammer

@EricSilva: This one.

@Lucas: I realized that just wasn't interesting to me.

BTW, Eric, I thought this one was interesting.

Oh I saw this problem when it was posted and tried to think it through but gave up after half an hour

Which? The first or the second?

@TedShifrin Yep we should be able to, define $\phi : (0, 2\pi) \to S^1$ by $\phi(t) = (\cos t, \sin t)$, this parameterizes $S^1 - \{(1, 0)\}$. Then suppose $\phi(\alpha) = (a,b)$,define $\psi : (0-\alpha, 2\pi -\alpha)$ by $\psi(t) = \phi(t+\alpha)$ then $\psi(0) = (a, b)$

The first

What does it mean by the "existence of a mathematical structure"?

Yeah, Eric, it felt like the usual Cartan game, but then it wasn't. I don't know what the qual writer was thinking.

OK, great, @Perturbative. And do you see that the calculus is easier to get the tangent space?

I've never thought about the second question but it's the kind of question I would normally want an answer to, pretty cool, I'm gonna think about it a bit before I read any answers

BTW, @Perturbative, in concrete problems, there's no need to make it be $0$. :)

@EricSilva: I did send you my diff geo exams, right? I have this recollection that you asked me to.

Indeed you did

@Lucas: What does who mean?

DogAteMy: I think the definition of countable is totally intuitive? I can enumerate the elements by the counting numbers. ...

@TedShifrin Let me try to compute the tangent space now from that :), it should just take me a few minutes

@Akiva, @Daminark: Any ideas about this one?

Man, varifold stuff seems to have some nasty computations sometimes

@TedShifrin Yeah, but if you didn't know that, you'd assume "countable" meant "you can count them", which means "finite" (and perhaps even fairly small)

(Small, as in, I'm not going to count something that takes days to count)

It depends how you interpret countable :P

But, whatever, once you know the definition it makes sense

You could also see it as "any element can be reached eventually by counting"

@TedShifrin Is this a rhetorical question?

Who's talking about existence of what mathematical structures, @Lucas?

I always felt like enumerable was a better word than countable

To me, that's a total synonym, Eric, just using Latin.

@TedShifrin well... I'm confused, but I think I am.

I mean it is a synonym, but in my head it has a "put it into a list" connotation

So what particular mathematical structure, Lucas?

Every single mathematical structure. I mean, more generally.

I see, Eric. OK.

I mean so does countable, but less so

So, Lucas, you're asking for an example of every structure (group, vector space, field, differentiable function, ...) that you come across?

No. I mean that the concept of existence is unclear to me.

How do we know that something exists? What does it mean to such a thing exist?

@TedShifrin So when trying to compute the tangent space I got the following

I mean if you know an example of something you know it exists @Lucas

For example, I feel it's clear that the integers exist. Those are concepts we created to count things.

Would you accept the existence of geometric objects?

Yes

Oh oh. Lucas is turning into one of those people who won't allow proofs by contradiction. Another damn constructivist.

All is Mathematics is Geometry.

d0n3.

Are you fixing that, @Perturbative?

//

Oh well that's not rendering :(, I'll post it again just now

@TedShifrin Yep, sorry about that

Pls don't take away my law of excluded middle

@TedShifrin Here it is

I've heard that there's some constructivist differential geometry or something, I wonder what that's about

The final step would be to say that vector is $(-b,a)$, @Perturbative. I would also use $\alpha$ or $t_0$ or some letter other than $p$ for an element of $\Bbb R$, but I'm being picky.

Yeah, this is way easier, agreed?

Agreed! :)

Thanks so much!! @Ted

Back later.

@Waiting Did you see Olivier shared a solution to the problem here? :)

Evening all. Is anyone here reasonably familiar with finite simple groups and GAP?

What do you mean, @AlwaysNeedHelp?

@r9m No, never saw it before. I'm amazed to see it there.

@Waiting well he spoiled the eggs ,, -_-

@r9m Ah, I know understand the point. I thought it was an older post!!! Now I see it's a new one. I would have felt bad otherwise.

:-)

Say if you use GAP, you can look at things such as the conjugacy classes of a group and there is a command "MathieuGroup(11)" which'll give you standard generators. However, that has nothing to do with generating by conjugacy classes which satisfy those conditions.

I'm sorry, it was not about this. I was talking about the "all math is geometry" part :p

;-;

@r9m there are some very amazing identities with the alternating series, and highly advanced which can be proved without using extraordinary tools.

problem: if $\frac{a^2 + b^2 + 1}{ab} = k \in \Bbb Z$, show that $k=3$

Ah, I was just citing Plutarch who said that Plato uttered: "God eternally geometrizes".

It was very much just an offhand remark.

:p

@Waiting :) Cool!

Who knows, maybe Witten will show that some 11-dimensional SUGRA is in fact just number theory. For then, it's just a nonsensical comment. :p

Ah, time to try to find some stuff on rigid groups.

G'night.

Gnite.

@LucasHenrique Do you know any set theory? Do you know, say, the difference between {} (the empty set) and {{}} (the set containing the empty set)?

And if so, would you say those "exist"?

Not the "rigorous" set theory (ZF)

But yes, I do.

@AkivaWeinberger yes, I would. "an empty, infinite box", "an empty, infinite box containg an empty, infinite box"

(ugh, I'm "almost" a physicist using the arguments with real life examples xD)

It's not we have infinite-sized boxes, but we do have boxes. Or collections of things. :p

@LucasHenrique From a formalist point of view, you could think of $\Bbb N$ just as a symbol, for which the collection of symbols $“1\in\Bbb N”$ is labeled "true"

yes, I kind of understand that

and $“2\in\Bbb N”$ as well, and in general any collection of digits followed by an $\in$ symbol followed by the $\Bbb N$ symbol

I'm actually reading the construction of reals

I have in mind that functions are quite interesting for this concept of existence

It's easy to accept that $\exists i$ such that $i^2 = -1$ if you think of multiplication as a function

under the usual definition of multiplication, given any real number, that would not make sense. but since we can arbitrarily define a new element, we can define that $i \times i = -1$

I find the construction of the reals via Cauchy a bit weird. how can you work with limits without reals?

You can define the limits using rationals

So the $\delta$-$\epsilon$ definition becomes, for all $\epsilon \in \mathbb{Q}$ positive, blah blah

A lot of you in here might be interested in this newly made meta post:

Just advertising, don't mind me.

@LucasHenrique Oh, fun.

Does there always exist a surjection $\Bbb R\to\omega_1$? I think the answer is yes with choice, but without choice I'm not sure

What's $\omega_1$ ?

$\omega^{\omega^{\ldots}}$

@Daminark No, that's $\epsilon_0$

@Astyx The first uncountable ordinal. Its cardinality is $\aleph_1$.

Right

Is that not how you generate the first uncountable ordinal?

Oh duh math.stackexchange.com/q/2379316/166353

@Daminark No that's still countable

Well, that got answers fast

Proof by absurd ?

Wait, no

The answer is 'yes'

Huh

If you have two sets $A$ and $B$, does there always exist an injection $A\to B$ or $B\to A$ ?

I don't think that's true in the absence of choice

With choice, yes

Well-order them.

Yeah, with choice it's certainly true

@Daminark If there is a single notation that can name every ordinal less than that and show that for any named ordinal $a$ and $b$ whether or not $a>b$, $a<b$, or $a=b$, then that ordinal is computable < uncountable.

Huh

Particularly, every ordinal less than the supremum of $\{\omega,\omega^\omega, \omega^{\omega^\omega},\dots\}$ can be described using Cantor normal form in base $\omega$.

i.e. it should be relatively obvious that you can name every such ordinal as a linear combination of finitely many smaller ordinals of the form $\omega^{\gamma_n}\delta_n$

And its not too hard seeing whether or not $\omega^2$ is larger than, less than, or equal to $\omega\cdot50$, for example.

> Lemma 0. You're beuatiful.