so what is a set

Can't you just use an example on the category of integers :p

are the integers a category?

Well from what I gathered

Sets are a category

So integers should be

No, a set is not a category, what are the morphisms supposed to be?

what is a Potenzmenge?

Power set?

@ACuriousMind identity

Functions on sets?

You can interpret a group as a category with one element, though

I dunno

So what is a category then

@Bass : that definition causes issues. Switch to the Shapiro scenario and the issues go away. Here's some more simple English below:

@Slereah Oh, but the sets are objects in the category $\mathsf{Sets}$. The integers are an object in the category of groups or rings, but that doesn't mean they themselves are a category

@Bass : I have to go. Bye for now.

what is a category??

@0celo7 Yes

Alright

So what is a functor on groups

I think category theory is one big troll

Well it is called "abstract nonsense"

it makes even less sense than PhD level intro set theory

*high school level

ZFC is PhD level, at least

@Slereah E.g.: The forgetful functor $\mathsf{Groups}\to\mathsf{Sets}$ that sends every group to its set of elements, and, well, every group homomorphism to the function on sets that it is.

What we discussed wasn't ZFC

It was basic propositional logic

Oh shit

that's postdoc at least

find me one mathematics PhD student that knows that

So... sending $(\Bbb{Z},+)$ to $\Bbb{Z}$

protip: you can't

Another example is the abelianization functor $\mathsf{Groups}\to\mathsf{Groups}$ that sends every group $G$ to its abelianization $G/[G,G]$.

for instance

@Slereah Yep

what is an abelianization anyway

when you make it abelian

It's like balkanization

But with abelian groups

why is it $\mathsf{Groups}$ and not $\mathsf{Grps}$

Is $(\Bbb{Z},+)$ to $\Bbb{Z}$ a morphism

@Slereah No

and the functor is the abstraction of the function

Hm

What a functor does to morphisms is better seen with the abelianization

@Slereah what

wtf is an abelianization

I guess that you identify the elements of a group that differ if they are not abelian?

Identify $g_1g_2$ with $g_2 g_1$

Or something

There's a theorem that every group homomorphism $\phi: G\to H$ from a group $G$ to an abelian group $H$ factors uniquely through a group homomorphism $\phi^\text{ab} : G/[G,G]\to H$

god

I know no math

Well

Do u know numbers

1

no

2

what are the axioms of the real numbers

3 (in advanced lessons)

Okay, I'll have to restate this to make it coherent:

$\Bbb{R} \in V$

$0 \in \Bbb R$

$1 \in \Bbb R$

$A \in \Bbb R \wedge B \in \Bbb R \rightarrow (A + B) \in \Bbb R$

$A \in \Bbb R \wedge B \in \Bbb R \rightarrow (A . B) \in \Bbb R$

$A \in \Bbb R \wedge B \in \Bbb R \rightarrow (A + B) = (B + A)$

@Ϻ.Λ.Ʀ. you there?

$A \in \Bbb R \wedge B \in \Bbb R \wedge C \in \Bbb R \rightarrow ((A + B) + C) = (A + (B+C))$

$A \in \Bbb R \wedge B \in \Bbb R \wedge C \in \Bbb R \rightarrow ((A . B) . C) = (A . (B.C))$

do you know these off the top of your head?

you're just definining what a field is

@Slereah: Let $\mathsf{Groups}$ be the category of groups and $\mathsf{Ab}$ be the category of abelian groups. Then the abelianization functor $\mathsf{Groups}\to\mathsf{Ab}, G\to G^\text{ab} := G/[G,G]$ sends every group $G$ to its abel. $G^\text{ab}$ and every morphism $\phi: G\to H$ to the morphism $\phi^\text{ab}:G^\text{ab}\to H^\text{ab}$ that is obtained by factoring $G\to H\to H^\text{ab}$ as $\phi^\text{ab}: G^\text{ab}\to H^\text{ab}$.

$A \in \Bbb R \wedge B \in \Bbb R \wedge C \in \Bbb R \rightarrow (A.(B+C)) = (A.B + A.C))$

@0celo7 Real number is a field

just say $(R,+),(R,.)$ are abelian fields

done

Well you asked

then there's some more, right?

Completeness

Completeness is part of the definition of a field?

trichotomy?

ordernedness

Otherwise it's just a ring, no?

@Slereah definition of the reals

completeness is that an upper bounded set as a supremum

@ACuriousMind Oh so it just maps the functions on groups to the functions on abelian groups

Yes (but you have to show that $\phi^\text{ab}$ actually exists and is unique.

Which one can either do by foot, or use a theorem about the existence of adjoint functors that tells us there has to be an adjoint to the inclusion functor $\mathsf{Ab}\to\mathsf{Groups}$.

mind = melted

Which is precisely the abelianization.

@0celo7 : $((A \subseteq \Bbb R \wedge A \neq \varnothing \wedge \exists x \in \Bbb R \forall y \in A y < x) \rightarrow \exists x \in \Bbb R (\forall y \in A \neg x < y \wedge \forall y \in \Bbb R (y < x \rightarrow \exists z \in A y < z))$

Or something

wtf

@Slereah Jesus, enough with the obfuscatory notation :P

oh that's the archimedean principle

Do you mean STANDARD NOTATION

I think

"A non-empty, bounded-above set of reals has a supremum."

why is ^ not true for Q

@0celo7 What?

or Z

@0celo7 the reals are equivalence classes of cauchy sequences of rationals...not enough?

It's supposed to be a damn shorthand, not the way you actually reason about things

@DanielSank You control 95% of the people on this site.

@0celo7 I do? That's awesome!

Are you one of them?

@0celo7 make me a sandwich.

@DanielSank i wish... there are some things about @0celo7 i want to control!

lol

@ACuriousMind Well I read a lot of Russell and Bourbaki back a few years ago :p

@0celo7 sudo make me a sandwich.

No problem with the notation

can someone give me an example of a subset of $\mathbb{Q}$ that shows it fails completeness?

It's not working...

@ACuriousMind make me a sandwich.

@DanielSank Yes, master.

!!!

@DanielSank he is a poor innocent guy.. pls spare him!

@TanMath I'm not harming him, just getting sandwiches.

@0celo7 The subset that converges to $\sqrt{2}$?

@Slereah proof?

@HDE226868 lololololol

@0celo7 You will regret asking for it

@DanielSank I cannot disobey, for I am one of your sockpuppets according to that deranged user

@ACuriousMind Did not even notice that

@ACuriousMind I see.

@ACuriousMind wait..who?

Retrospectively: no

@TanMath MEEEEEEEE

@0celo7 you are a sock too!

I am one of the 5%

I fight freedom at all costs

Wait, that's wrong

@DanielSank It's spilled onto meta now, in case you care

" To give you a concrete feel for why the supremum axiom fails for rationals, consider the infinite set of numbers 3, 3.1, 3.14, 3.141, 3.1415... Each of these numbers is a rational number. But the supremum is pi, which is not a rational number. "

I couldn't find the symbol version quickly :p

bah

that's one example

one example does not prove something

well you only need one example

no

It does

Thanks to

you need infinite examples

THE AXIOM OF CONTRADICTION

Oh, wait. The guy on Meta misinterpreted the original claims, and now thinks that DanielSank is a sock of a mod (probably dmckee).

lol

you all are sheeple

literally puppets

The theorem is that not all subsets of rationals have a supremum

@HDE226868 he is weird...

is @DanielSank insane?

Not that all sets have no supremum

He talks to himself all the time

Take the set {0}, it has a supremum

(0)

@0celo7 no.. you are! XD

@TanMath ?

Baseless accusations...

@Slereah exactly

so the integers are complete

thus $R\cong Z$

@ChrisWhite um.. I have been to much more active chats.. I think the answer would be, people over here do not have lives!

The integers do have a supremum for every subset

But not the rationals

@0celo7 XD

so are the integers complete?

@0celo7 Playing stupid is not as funny as you think :P

what is the difference between the integers and the reals then?

I think it also needs to be dense?

oh the reals are a field over .

Or somesuch

I dunno

Z is not a multiplicative field

What about Z*

@ACuriousMind baseless accusations

Who said I'm trying to be funny

@Slereah what be that

Z - 0

still not

@Slereah Nope

the inverse of an integer is not an integer in general

@Slereah $Z^\times = \{1,-1\}$.

Oh right

nvm

I am being dum

SEE

everyone is dumb

some more than others

We don't even believe Einstein and the evidence

although people like @ACuriousMind are smart all the time, so there goes that theorem

@Slereah we should try to get that Skyrim mod to work later

I'm sure we can trick @ACuriousMind into being dumb

Hm

What's a thing he doesn't know too much about

he knows more about everything

not kidding

he could probably knock out HE in an afternoon and do GR for the rest of his life

it's unfair

He probably couldn't because HE is missing a lot of proofs!

then he reads the literature

or does them himself

I think he'd just as soon do the latter, probably quicker for him

Hey @ACuriousMind, do you know how to prove Turing completeness of the game of life!

@Slereah I don't know much at all about computation theory

See? Told you!

I stopped taking the theoretical computer science course because it just bored the hell out of me

There is a very weird intersection of general relativity and computer science, by the way

Two, even

1) Super turing computing 2) closed timelike curve computing

Very small fields but not as small as you'd think

he's just saying that so I don't feel even worse

@Slereah I'd think the field is 1 crazy prof + his PhD students

ask him about some random GR thing

he claims to know little to no GR but he seems to know a whole lot of it

Oh there's also the proof on the computational capacity of the universe

which is based on GR and QM

@Slereah probably not enough to run Witcher 3 on ultra

IIRC it's about $\approx 10^{100}$ bits

Give or take a few OoM

that's not even that good

my computer has like 1 flop

maybe

The difference between $\approx 10^{100}$ and a flop is $\approx 10^{100}$

your computer is pretty shit

You'd better get a universe

what's a random GR proof that ACM wouldn't know

Well

You're not going to believe me that I don't know it anyway, so why bother? :P

there was one he didn't know

The proof of submaximal symmetry :V

that sentence was a mess

@ACuriousMind no I will

how about the proof that light is a null geodesic

not many people know that one

@ACuriousMind ?

But light is an EM field, not a point particle :O

@Slereah a photon

@Slereah isn't it both (wave-particle duality)?

whatever

that the wave vector is auto parallel

A photon isn't a point particle either >:\

@Slereah it is in GR

@ACuriousMind do you know the proof?

By the way

@0celo7 Perhaps

@Slereah really?

There is a neat proof that C isn't Turing complete!

@ACuriousMind what does that mean?

yes or no

be honest -- could you derive it?

Since the function sizeof() has to return an integer, the size of a pointer is always finite

Hence you can't have arbitrarily huge memories

just take the infinite integer

No infinite integers tho

There's an infinite float in C

uh, $\infty$ is pretty infinite

Two, even

It's not an integer

of course it is

There are infinite ordinals, on the other hand

But this is

PHD LEVEL

just take $1,2,3,...$

it converges to $\infty$

and is an integer

You can totally add infinites to real numbers btw

you know what bothers me

$\Bbb R \cup \{\infty\} \cup \{-\infty\}$

"light rays travel on null geodesics" is an approximation

:(

but if you add infinites to the real, then you break down some properties of addition and product

dude

light traveling on null goedesics is an approximation

is it

Wald page 71

Straumann page 40

but is the solution derived from an approximation, or is it not actually on a null geodesic

I'm...not sure

Pretty sure it's on a geodesic

You know what I'm gonna ask

candy?

Proof?

Hm

Let's see

Best way to find the motion of "light" is to take a wavepacket, I guess

Apparently the standard arguement is just to invoke the Eq principle

hmm

maybe one needs an approximation in the Maxwell theory because light is not always a plane wave in GR?

so it's a wave packet

but one would have to solve the Maxwell euqations

Actually the Einstein Maxwell equations

to find the recoil on the geometry

$A = e^{(x - x_0(t))^2}$

Or something

Too lazy to do it tho

is that just a gaussian

Yes

Wavepackets are gaussians

wonder if @ACuriousMind the boy genius knows

@Slereah are they

Well coherent states are, anyway

I'd like a proof of this

I suspect that a gaussian EM field would have its center travelling on a null geodesic

@ACuriousMind ^ pls prove

if you prove it

I will give you $10 in an account of your choosing

@0celo7 Stop pinging me, please.