@ChrisWhite Found some Bulgarian lecture notes with the proof I was looking for.

Hi everybody.

@DanielSank hi

> sockpuppets

Reporting in.

@DanielSank hello?

Hello friends

@BernardMeurer I had a goal today and I accomplished it.

This is very, very rare.

@DanielSank What was it??

@BernardMeurer Write enough of a python program to parse and upload data to an SQL database that the guy actually doing the measurements can continue on himself, asking questions as necessary.

@DanielSank Sweet! What SQL db? MySQL?

This is really awesome because in the past I tend to try to do everything myself. It's good to learn to train others.

@BernardMeurer Yeah, mysql, via Google cloud.

@DanielSank Which library did you use? MySQL connector?

Which, I'm given to understand, is a crappy version of SQL.

Oh, uh, lemme check.

Meh, MySQL works alright tbh, as far as I can see hating on it is hyped

PostgreSQL is supposedly the king of the hill though

mysql-python

Actually, figuring out all the details of which connector to use, how to configure the cloud sql instance, etc. etc. took a long time.

@DanielSank Use PyMySQL

I did this project as a learning experience.

I did a rather large project that involved MySQL (private repo) and I used pretty much all connectors available and settled with that one

it works like a bliss!

...a bliss?

@BernardMeurer I never even think about the connector.

I use sqlalchemy :-)

The saying is "charm", fyi

ORM for the win.

@DanielSank sqlalchemy? YOU WERE LIKE A BROTHER TO ME

:p

@BernardMeurer Dude... it's awesome.

It's so, so awesome.

You don't like it?

Well, what I needed involved performance, and it was significantly slower than just using a connector and me programming all the rest myself

@BernardMeurer I found a Portugese book on dynamical systems, can you translate

TBH this thing is my magnum opus of Python

I'm surprised. You can get a lot of control over exactly how/when sqlalchemy emits requests to the database.

@0celo7 How large

Don't remember

Less than 500 probably

@0celo7 Pay me

^ Yes

Correct response.

@DanielSank He received payment from my kin already.

Food and shelter

@0celo7 BS

He owes the clan

1. That wasn't payment

@DanielSank Yeah, I did a lot of benchmarks ended up going with a Python-based connector for ease, figured the speed of a C-based one wasn't worth it

@DanielSank He stayed with my family in LA

@0celo7 Let's ask Michelle and Ron that shall we?

2. It doesn't come close to how long it would take to translate 500 pages of technical text.

It's not 500

300 at most

My opinion is unchanged.

That would take a long, long time.

He'd probably have to actually learn some of the material in the book to make sense of it.

@0celo7 I'll translate it for 20 dollars

on google translate :)

Deal.

No deal.

nooooooooooooooo

HA

@DanielSank don't have faith in your product?

Ok I'm outta here. See you all laterz.

I got the both of y'all hahaha

bye

@DanielSank See ya!

@0celo7 Huh? I was "noooooooo"ing because $20 isn't enough.

@DanielSank ...

He owes me $20 anyway

@DanielSank Did you see what I said after?

@0celo7 NO, YOU OWE ME 20 YOU SCUM

I distinctly remember it the other way around

I have your Michelle and Ron on my side

and Kathy

but not Bob

That's just because he doesn't like me

It's because you're

Did you watch the latest Ethan & Hila video

With the tooth paste

Not yet I think

The dumb products one?

yes.

Yeah I watched that

@BernardMeurer Your countryman liked my proof

@0celo7 Guessing it involved nudity then

It involved balls B)

As a non-hausdorff manifold I find that offensive

@NeuroFuzzy hey

did you learn about cohomology in Renteln?

@0celo7 youtube.com/watch?v=QWveXdj6oZU

@BernardMeurer Implying how many times isn't the greatest of all time.

@0celo7 That's actually a really good video

@BernardMeurer Not really.

They censor nigga

And...not anything else.

They don't censor it in audio either.

facepalm

Because they're all white I guess

and that doesn't make it any worse, I mean it's a good musical analysis, I don't give a fuck about that they censor

It makes them suspect

@0celo7 I actually kind of like this youtube.com/watch?v=WUH9NaZI9eA

for the first time

Khalidius album in like two days

@0celo7 :o he said hiiii

@0celo7 Not really, I wanted to get to the good immediately physicsy stuff.

so I only read it lightly. I don't have a good sense of the word "homotopic"

other than it is looser than uh

homeomorphic

@NeuroFuzzy Your mom's homeomorphic

@NeuroFuzzy I thought you said you took topology

@BernardMeurer that's not even wrong

@0celo7 I only do scientifically accurate fake insults

@NeuroFuzzy I think @ACuriousMind would crucify you for saying that

@ACuriousMind what the heck is the last word here

@0celo7 Hungarian?

@CuriousOne Yes, but it is for the Higgs particle. I am talking about the Higgs field. Its value is the same everywhere. This is its Lagrangian density: theory.sinp.msu.ru/comphep_old/tutorial/img472.gif Can you see the deltas on the left side? These are the so-named kinetical terms.

@CuriousOne These kinetical terms are added to the well-known mexican hat potencial (right side of the formula). On my little understand, it effectively means, that the space derivate of the Higgs field has an energy density, this energy density wants to be minimized, and this is why the Higgs field is the same nearly in the whole visible Universe.

@CuriousOne Now consider if we could perturb the Higgs field in many points in a volume. What would the kinetical terms in this case do?

@CuriousOne This is why I said, that monocrystallous material of large, asymmetric nuclei (with asymmetric isospin/hypercharge distribution) in strong magnetic field could maybe produce interesting effects.

@CuriousOne For example, the mass of the electron would be a little bit different in it.

@CuriousOne And it would be a permanent, low-energy effect.

@0celo7 Copy-paste the word, or the sentence, I can translate it to my super-level English :-)

(sorry I don't have youtube access here)

Sanity check: A rotation in 3-d is specified by 2-parameters, or two basis-vectors right? Thus in considering the action of a representation of a rotation on a QM wave function, when you find it's infinitesimal form to derive the (orbital) angular momentum operators it involves rotating the coordinate vector, but this is tantamount to rotating the coordinate vector using only one of the rotational parameters, so you have not fully represented an entire rotation in considering just this right?

There is a whole extra degree of freedom that is basically the origin of spin angular momentum, right? You can sidestep this by using a fixed axis generator AND the total angular momentum casimir, but if you rotate around any other axis the wave function has to become a linear combination which indicates the existence of spin.

So the double-cover nonsense is really just saying if you rotate one of the basis vectors by 360', the other basis vector could have been oriented in one of two directions, hence the double cover!

@peterh There is no difference between the particle and the field. The particle is just the way we see the field in our accelerators in a high momentum frame. More importantly, a Higgs field doesn't just happen by itself any more than an electromagnetic field. If you want it, you got to put energy in.

If you write it in terms of wave functions $\psi(x) = <x|\psi>$ then you represent a rotation $R = R(\theta,\phi)$ by $U(R(\theta,\phi))$ and want to analyze $<x|U(R(\theta,\phi))|\psi>$. Then, expanding $U$ infinitesimally and exponentiating the end result allows you to write $<x|U(R)|\psi> = <x|R*S(R)|\psi> = <Rx|S(R)|\psi> = [S(R)(\psi)](Rx)$ or something like that.

@0celo7 I think I read you quit physics over spin, maybe the above explanation will help bring you back to the dark side ;)

^$R* = R^* = R^{-1}$

@CuriousOne Ok, here are coming the things I know probably not enough precise. I asked some questions on the site into this direction, unfortunately I didn't get really clear answers. So, first logical step: 1) the weak field couples to the Higgs field. 2) the electric charge, the weak isospin and the hypercharge mix to produce the weak and the EM interaction as we know.

@CuriousOne | Does it mean, that a collection of particles [i.e. nuclei] with a non-zero isospin/hypercharge have actually a small, permanent perturbation in the Higgs field around them?

This is some incomprehensible shit I wrote in the past that is supposed to be a sentence encoded into something that is made to look like a mathematical formula

because of that, I have no idea what me a few years ago is trying to say

@CuriousOne Btw, I feel a little bit scared on the "no-no-no" style what I get in this line everywhere. If the 60GeV gammas wouldn't be actually produced in the LHC, I would have an impression as if the Higgs field is only a purely mathematical construction without any physical meaning. But they actually produced. I suspect, Higgs constructed his theory really as a purely mathematical trick and even he didn't think that he found a key segment of the reality.

@Secret I tried to play with QFT math a little bit on a very layman level and produced similar things, although $d^\frac{x}{y}$ -like terms weren't in it.

@peterh: On some level it feels as if you are trying to construct some effective field theory here that allows for some static contributions that could, in principle, be differentiated from the entire model. While this may be mathematically possible, I can't assign much physical reality to it. Take the effective electron mass in semiconductors, it can differ by over an order of magnitude from the free field mass and heavy fermions, I believe can be thousands of times heavier.

I simply don't know how you want to find perturbations due to the Higgs contribution in there. Maybe it can be done, but I don't see it.

@CuriousOne The W/Z field couples to 3 components of the Higgs-field. Thus, if you perturb the W/Z, you also perturb the H. Or not?

@CuriousOne And, because of the mexican hat potencial, if you perturb H, it will effectively mean a rotation (i.e. the VeV remains the same, but with different components)

The Higgs degrees of freedom don't get really excited until you collide stuff at the 100GeV/TeV level, up to there is't just a perturbative contribution. Necessary to make the theory self consistent (which is probably mostly a failure of the theory), but not something that can be separated from other perturbative contributions. The standard model, at the eV level, can only be seen in the electron mass, the proton, neutron and effective nucleon masses and the neutrino mass.

I don't know what you mean by "if you perturb the W/Z". The W/Z/H are excited states of fields. You can excite them, then they interact. You can use their energy levels for your intermediate states in perturbation theory (that's what they call "virtual particles" in high energy physics), but it's no different from summing over all atomic states when calculating an optical transition in atomic physics.

Can you populate the electromagnetic field with tons of photons? Yes. Does that change the field? No.

Can you populate the Higgs field? Yes. Does that change the field? Not in the interpretation of quantum field theory that I am used to. Maybe I don't know better, but I don't know what you mean. To me these are tones of a bell and all we are trying to do is to hit the bell really hard to determine its shape. I do not believe that we can change the shape of the bell.

Ok, but the Higgs field interacts with the particles, even on the eV scale, which is far below the required 100GeV excitations. Or the particles wouldn't have any mass. Why can't this interaction work also in the opposite direction?

(more accurately: wouldn't have mass from the Higgs)

Yes, that's what effective mass is. But like I said, the effective mass of an electron in Germanium is 0.041m_0. What does that make your Higgs field?

@CuriousOne It would be an essentially different thing. This effective mass change of the electron is coming from its EM interaction with the Ge electron shells. Out of the semiconductor, the mass of the electron is normal again. Actually, it doesn't change even in the semiconductor. On my little understand, it is only a mathematical description to simplify to calculate the interaction with the Ge electron shells.

@CuriousOne I tried to write: if you perturb the isospin/hypercharge fields.

@CuriousOne As I understand, their excited states (+EM field) are the W/Z bosons.

@CuriousOne Well, here is something what is new to me: "Can you populate the electromagnetic field with tons of photons? Yes. Does that change the field? No." <-- if you have a tons of photons, it is a change of the EM field, or not?

Is this sentence correct (grammatically and choice of vocabulary)? "Bodies are divided by two parts: rigid bodies and non-rigid (deformable) bodies."

@peterh : you should look into the Higgs mechanism some more.

@JohnDuffield what do you think of my spin explanation above, and why do you think it's related to Einstein de-Hass (I'll give you a hint, magnetism is linked to rotations)

@Peterh : the above image is from A Zeptospace Odyssey by CERN physicist Gian Francesco Giudice

@bolbteppa : sorry, I have to go now. I'll have a look and get back to you later.

@JohnDuffield No problem, basic point is orbital angular momentum arises by varying one angle in a rotation, but a rotation is fully represented by two angles, so rotating your reference frame to vary both angles ends up giving you spin. Since lorentz transformations are just rotations, spin is intimately linked to QFT and the Dirac equation exploits that directly.

That children's point took so f*cking long to see :(

@JohnDuffield Thanks! But I know this. This is why I came up with the electron mass in my previous chat messages.

@JohnDuffield But, what I am not sure in: here is the lagrange density of the Higgs: theory.sinp.msu.ru/comphep_old/tutorial/img472.gif . As I know, the first part of the field is the kinetical term, and the second is the well-known mexican hat potential. On my little understand I interpret this formula, that the space-derivate of the Higgs field has also an energy density (from the kinetical term), and this is added to the mexican hat potential.

@JohnDuffield This energy density wants to be minimized, and this is the reason why the Higgs field is the same everywhere in the Universe. Unfortunately, every source I've found was either too trivial, or far over my knowledge level. Is it correct?

@peterh: "It would be an essentially different thing." Why? Silicon is made from the same field as the vacuum. "This effective mass change of the electron is coming from its EM interaction with the Ge electron shells." And the effective mass in vacuum comes from? "Out of the semiconductor, the mass of the electron is normal again." Normal relative to what? Why is the electron in silicon abnormal? It's all the same universe, you know, it's just a different energy range.

@peterh: "if you have a tons of photons, it is a change of the EM field, or not?" The field is the same, it just gets excited. Honestly, to me it's all one and the same, just with different degrees of population. Some of these are metastable with very long lifetime. I am not even sure that photons are stable, we probably just won't see their decay channels, ever, we will never get to that energy and time scale because angular momentum conservation is so strongly conserved.

I can't help feeling this discussion has got a bit silly. The Higgs field is an essential component of the Standard Model, and the fact that our universe is (apparently) described by a broken symmetry is quite remarkable when you sit down to think about it.

OTOH the Higgs boson is an accidental byproduct of the symmetry breaking and uninteresting except for what it can tell us about the Higgs field.

@0celo7 I'm giving myself a break, there was a mathematically correct idea there xD

But no, I never took topology. All the point set topology in Rudin? Yeah. Some abstract defs as in Munkres? Yep. But no formal class.

I was just totally unable to intuition-ify the definition of "homotopic" and decided to move on to lower hanging fruit for physics.

@CuriousOne For example, if this monocrystall would have a hole in it, the masses of the particles would change even in this hole. Maybe the processes would be quite similar, but the chain of effect would happen this time by changing the parameters of the vacuum, and not by the "simple", EM interaction with the Ge electron shells.

Definitely plan on revisiting it soon though.

@NeuroFuzzy What's the problem with the definition of a homotopy of curves? Two curves are homotopic if you can continuously deform them into one another. To define that formally you just want to define a continuous map $H$ such that $H(a)$ is $f(x)$ and $H(b)$ is $g(x)$ so that $H : [a,b] \times \mathbb{R} \rightarrow \mathbb{R} | (t,x) \mapsto H(t,x)$ satisfies $H(a,x) = f(x), H(b,x) = g(x)$.

I bet you can define the homotopy linking the top half of a circle to the x-axis contained in the unit circle explicitly ;)

@JohnDuffield Another remark: a change of the Higgs field would change the 4-8MeV rest mass of the u, d quarks, which would probably cause a change in the gluon fields around them, too. Which means, that the hadron masses would be probably also changed, although probably not so significantly. I suspect, if a permanent, static change of the Higgs field would be possible, its main visible result would be caused probably by the reordering of the electron shells in the normal baryonic matter.

otherwise we will have made many of these already

P.S. I don't remember if the Higgs field actually interact with gravity, and I don't remember how its energy momentum (if any) is expressed in terms of a stress energy tensor

@Secret Matt Strassler denies any direct relation which corresponds that its energy density is actually zero.

Ok that's interesting, so a Higgs field cannot actually bend spacetime?

(Given that using Einstein's equation, if something hasa stress energy tensor of zero, then the riemann culrvature tensor must vanish also)

@Secret In the inflation period, its energy density weren't zero, probably at the time it was possible. There are a lot of models calculating back into this era.

@Secret I think, the masses (energy density) still influence the curvature, and these masses can come also from the Higgs coupling.

So basically (at least according to the standard model and only restricting our discussion to the elementary particles) the Higgs field at the present day will not induce a spacetime curvature except when it couples to said particles via the mass it give rise?

correction: couples to said particles and the mass that arises will then induce curvature. I.e. the field itself in present day will not be able to induce spactime curvature directly because its energy density is zero?

@peterh I think you are running into a conceptual problem where you perceive the hole and the stuff around it at different, whereas in reality they are the same quantum fields, just with different local states. You can easily verify that by pushing the stuff around. It doesn't mind where it is. Give it a shove and it well keep floating trough the vacuum forever. Even Newtonian mechanics tells us that something and nothing are the same thing.

@0celo7 I'm more partial to pitchforks and torches

@0celo7 What last word? Are you thinking I speak Hungarian? oO

@ACuriousMind and burning at the stake? ;)

@bolbteppa : I'm not quite clear on what you're saying, but it sounds reasonable. A Dirac spinor is a bispinor. There's two orthogonal rotations in a Möbius strip which is likened to a spinor. It isn't some strip of course.

@peterh : I'm sorry Peter, I just don't know enough about the Higgs mechanism to answer your questions. But do look up the mass deficit.

I'm studying the field equations and ran into a problem. If the energy-stress tensor fulfill that Tij = 0 for all i,j, with the exception T00 = E at a single point in space, then solving the equation should give me the metric at that point. Although, assuming that T00 = 0 at other points, solving the same thing at those points would give the metric g fullfilling gij = 0. This doesn't make any sense, considering that spacetime should be bent at a distance as well. Does anyone understand why?

Just to be clear, I do see how it works out when both the metric and the stress-energy tensor are functions of position, but not when focusing on a single point

I...don't think that's a valid stress-energy tensor

oh, that's why it doesn't work?

Okay, maybe it's valid, but you need to be a bit more tricky: Have you thought about solving Maxwell's $\nabla\cdot E = \rho$ in that manner for a point charge?

Just putting $\rho = Q$ at one point and $\rho= 0$ at the others doesn't give you the correct answer either

hmm, yeah

I get what you mean, but my head won't understand what exactly makes it impossible

Both $T$ and $\rho$ are densities

they give energy/momentum/charge when integrated over a subset of space(time)

But integrating a function that's only non-zero at a single point gives you zero, always

oh yes

The remedy for the charge density is to use a delta function, $\rho(x) = Q\delta(x)$.

I guess you can do the same for the stress-energy, but I don't guarantee that that's a good thing to do

You can have the stress energy tensor of a point particle but then it's gonna be $T = u \otimes u m \delta(x - x(\lambda))$

or something like that

Of course if you want to do actual GR with it then things get complicated

I'm trying to enable latex so I can see the equations

If you want to do it properly you have to use mollified functions

just a quick side-note question - is it possible to ahve latex parsed immediately? I see all equations as "$\rho..." etc

nevermind, found a parser, now I can see the tensor

@Max Look in the upper right corner of the chatroom

oh, thanks!

okay, so the real "at a point" energy-stress tensor is more complicated I see

Well to find it you just use the standard method

Find the variation of the action

And remember that $\frac{\delta f(x)}{\delta f(y)} = \delta(x-y)$

LinkedIn recommended I join its theoretical physics group. I did so. Worst mistake of my life. Nothing but "Is gravity EM?" style questions from the uninitiated. Thank, LinkedIn, for once again proving we should never expect anything good from you

Since the action for a point particle is $\int d\lambda g_{\mu \nu}(x) \dot x^\mu(\lambda)\dot x^\nu(\lambda)$ it's not too complex

oh, I see, I see

processing everything :P

SPOILER : The metric of a point particle of mass $m$ is just Schwarzschild

but since it involves the delta function, won't the same problem appear?

Same problem as what

that there's a distinction between "at point" and "not at point"

Delta isn't a function, it's a distribution

I'm realizing a bunch of things while writing these messages, while realizing that I've been thinking incorrectly about other things, so it may sound confusing, haha

You can have such distinctions that make differences in integrals with distributions

oh, so I'm guessing that I should study the delta function/distribution a bit more in-depth then?

Always a good idea

I'm feeling that it may be a key reason for my misunderstandin

alright, I'll definitely go a head and do that! Also, thanks for all your help! And yours too, ACuriousMind

@ACuriousMind Hungarian? That was German.

@0celo7 Ah, I didn't catch that right at the beginning

The last word is "meschugge"

Which means "crazy", and is probably Yiddish originally

Right, I thought it sounded Yiddish.

Does anyone use that word?

Can I be forgiven for not knowing it?

Oy vey

Shut up goyim

@bolbteppa No, I quit it over group theory in general.

Or representation theory.

@0celo7 It's not ubiquitious, but it should be in the passive vocabulary of most native speakers, I think.

@ACuriousMind Hmm, it doesn't even sound familiar.

So, I found a proof of the Poincare thingie

I'm assuming you'd like to live on without seeing it?

@ACuriousMind There is one detail I'm unsure about. Why does a co-dense set have measure zero?

"co-dense" means its complement is dense?

I'd guess so

Same as cofinite

"Quantum mechanics makes extensive use of group theory, e.g. the irreducible representations of the rotation group are at the heart of the theory of spin and angular momentum. Many famous people have related that they also experienced a feeling of alienation when they tried to make sense of group representation theory.

then just use $\int_X = \int_A + \int_{X-A}$.

Abdus Salam is quoted to have said in Trieste in 1962 that he remembered having thought after a lecture of Racah’s on Lie groups in 1951 ‘I can never learn all this. All this is too damned hard and unphysical’"

@ACuriousMind No, it means interior is empty.

You're not alone in feeling that way apparently

@0celo7 That's equivalent.

@ACuriousMind ...o.o

I don't believe it

@yuggib It'll take me ages to figure out where those papers really address my question, but thanks nevertheless ;)

@ACuriousMind Let's assume I believe this crazy theorem.

What then?

@ACuriousMind My notion of measure has nothing to do with integration...

that's not relevant, but I realized that doesn't work anyway

Not relevant?

Hm

Googling says the statement is not even true

NØ!!

What is true is that measure-zero sets have empty interior

But the converse fails

Damn, there goes that proof

@ACuriousMind I guess you've never seen the proof of "physics" Poincare?

Nope

It's apparently highly nontrivial :/

@ACuriousMind What is the measure of the irrationals on $[0,1]$?

Still not closed :/

Done.

@0celo7 Pretty sure its 1

But you have to explicitly show that, it doesn't follow from density

Ah nice one

@ACuriousMind So...isn't that an example of a codense set without measure zero?

Unless it's not codense?

It's...both dense and codense!

So why the example with the balls

Good question

Why not use the irrationals

you're completely right, the irrationals appear to be a much better example

@ACuriousMind I mean, $1=\mu([0,1])=\mu(\Bbb Q\cap[0,1]\cup\Bbb I\cap[0,1])=\mu(\Bbb Q\cap[0,1])+\mu(\Bbb I\cap[0,1])=\mu(\Bbb I\cap[0,1])$.

Because $\Bbb Q\cap[0,1]$ is countable.

And the sets are disjoint.

@Acuriousmind My profile said I can see closed votes of a question, but where exactly I can view that information in a PSE question. Is there some kind of number that I should be finding near some buttons such as flag, close?

@ACuriousMind Codense + X = measure zero. Solve for X?

@0celo7 I got nothing

@Secret If there is a close vote pending, you'll see it as a number in brackets next to close. You'll only be able to see this on your own questions until you reack 3k, however

i see

Hi!

I came across thepimanifesto.com and across tauday.com/tau-manifesto

I was wondering.. what is the opinion among you guys?

Who cares

^that

@ACuriousMind ...wow

Maybe I should get a rubber duck

@0celo7 What? You think the distinction between pi and tau is in any way interesting or meaningful?

It's a fundamental issue

It's as "fundamental" as choosing the sign of the electron charge.

Which is also fundamental...your point?

And in both case, switching it by this point would be unnecessarily confusing

@0celo7 The point is that the choice is arbitrary

If you switched the definition of $\pi$ in 1500 that would be fine

Nowadays it basically can't be done

A choice needs to be made, but which one is utterly irrelevant

Irrelevant?

I don't think so

$\tau$ requires two strokes to write. $2\pi$ needs four.

hahaha

I like to minimize my action.

But $\tau / 2$ requires 5

Don't write $\tau/2$ then. $\tau/2\cong\pi$.

And well.. the unit of charge got changed when it was convenient (natural units).

Why not a symbol for every fraction of $\pi$

Too confusing.

Why not a symbol for every real number

Symbols are countable

@Slereah We have such a symbol. It's called $x$ ;)

Oh snap

@0celo7 the argument you're perhaps thinking of has some surprising subtleties, cf. this MO thread

what

wtf is a "definable real number"

lol.

Every single number can be defined in just a few words. =).

Assume there exists a number $x$ which cannot be definable in just a few words. Thus, one can define $x$ as the number not definable in just a few words. Contradiction. This concludes the proof.

@Physicist137 If there's more than one such x that doesn't work

But I guess it is the basic idea behind why the concept is tricky

What

This problem can be easily solvable.

Set theory is so disinteresting.

Anti-interesting, in fact.

@ACuriousMind You know lots of PDE stuff, right

Said the man who doesn't understand set theory

@0celo7 I would for the life of me not want to work in it, but I love all its glorious weirdness

Because he's a big thicky bobo

@ACuriousMind Assume there exists a set of numbers $S$ which is not definable in just a few words. Take the smallest of the numbers in this set. Let it be $x$. One can define $x$ as the smallest number not definable in just a few words. This concludes the proof. =D.

What does that even mean @Slereah

@Physicist137 Nope, not every set of reals has a minimum

@Physicist137 You're assuming that a lower bound exists

True. Hm... Does it works for natural mumbers?

Its a paradox apaprently: en.wikipedia.org/wiki/Berry_paradox

Well...with choice you could well-order the reals, then it should work

Well for defining definable numbers

You have to do it with respect to a mathematical system