Mathematics - 2019-06-25 (page 3 of 3)

Leaky Nun

7:00 PM

@Thorgott I didn't get it until I thought of the same joke

Ted Shifrin

There was a joke? I still don't get it.

Alessandro Codenotti

I never use unnecessary curly braces actually, which annoys the people I write exercises with on overleaf

Thorgott

lol

Leaky Nun

$e = 3 = \pi$

Alessandro Codenotti

And I have no unnecessary whitespaces either

Balarka Sen

7:01 PM

bad

Ted Shifrin

No, $\pi=22/7$.

Leaky Nun

we all know $\pi = 3.2$

Tobias Kildetoft

@AlessandroCodenotti I hope you will never write code I have to read :)

Balarka Sen

Hah!

Alessandro Codenotti

@TobiasKildetoft my code is a little better

Balarka Sen

7:02 PM

So I had a first course in C in first semester, I never used proper indentation and braces to format

It's just too much work

Leaky Nun

surely there's a way to solve it by using the fact that $\Bbb N^\Bbb N$ is homeomorphic to some subset of the irrationals

Ted Shifrin

Were you fired, Balarka?

Leaky Nun

I forgot which subset

Thorgott

If you're going for efficiency, [ln2,e] would be even faster

Alessandro Codenotti

I'm mostly writing Haskell now so at least some indentation is enforced (and it annoys me)

Leaky Nun

7:03 PM

probably $[0,1] \setminus \Bbb Q$

Tobias Kildetoft

I let my IDE do all the formatting for me

Alessandro Codenotti

@LeakyNun to the whole of the irrationals

Balarka Sen

@TedShifrin Nah just had some points taken off in all my assignments

Leaky Nun

@AlessandroCodenotti they're probably homeomorphic anyway

$[0,1] \setminus \Bbb Q$ and $\Bbb R \setminus \Bbb Q$

Ted Shifrin

@Alessandro: I guess I don't see it.

Alessandro Codenotti

7:05 PM

@LeakyNun seems like it

Ted Shifrin

I see infinite decimals with no zeroes ...

Balarka Sen

the first is (0, 1)\Q so if you map (0, 1) to R homeomorphically you get a ctble dense subset of R

countable dense subsets of R are homeomorphic. extend that to a homeo R -> R and restrict to the complements

Ted Shifrin

Oh, no, I don't. Never mind.

Alessandro Codenotti

@TedShifrin the idea is to use continued fractions to build an homeomorphism $\mathcal{N}\to(1,\infty)\setminus\Bbb Q$, compose with $x\mapsto 1/x$ to get those in (0,1) and with $x\mapsto -x$ to get the negative ones

Ted Shifrin

@Balarka: I don't see that either, now, but I'm being dumb.

Agh.

Alessandro Codenotti

7:07 PM

So that's an homeo from the disjoint union of four copies of $\mathcal{N}$, but this is homeomorphic to $\mathcal{N}$

Ted Shifrin

Continued fractions is too esoteric.

Alessandro Codenotti

More weirdly $\mathcal{N}$ continuously surjects onto $\Bbb R$

Balarka Sen

N^N is just a cantor set

C surjects into every compact metric space

Alessandro Codenotti

Not quite, it's not compact

Ted Shifrin

I'm used to doing $\{0,1\}^\omega$.

Balarka Sen

7:08 PM

Oh ok

ya

Thorgott

(0,1)\Q is uncountable though. I feel like I'm missing something.

Alessandro Codenotti

N^N surjects onto every Polish space

Ted Shifrin

Is $\mathcal N$ otherwise $\Bbb N$ or what?

Leaky Nun

N=N^N

Ted Shifrin

Ah.

Balarka Sen

7:09 PM

Boom

Alessandro Codenotti

$\mathcal{N}$ is N^N

Balarka Sen

gotcha

Ted Shifrin

Now I'm slightly less confuzled.

Alessandro Codenotti

That's common notation for the Baire space (N^N)

Ted Shifrin

I've never seen it.

Alessandro Codenotti

7:10 PM

Well it's common in descriptive set theory :P

Leaky Nun

is the Baire space a Baire space?

Thorgott

Clearly, it's a normal distribution in this context

Ted Shifrin

Right.

Thank goodness, friends are visiting from out of town and I'll be missing a few days of this garbage.

Balarka Sen

We'll be missing you though :3

Alessandro Codenotti

@LeakyNun sure it's Polish, so in particular completely metrizable

Thorgott

7:13 PM

Have fun!

Ted Shifrin

I'll try.

Semiclassical

talking to a prof about the teaching evals from his latest course, and got this gem

in the 'how to improve the course bit' : "Less information, more details"

MatheinBoulomenos

I wrote a blog post on how to think of natural transformations as "homotopies", among other things, in case anyone is interested: wlou.blog/2019/06/25/… (sorry for the blatant self-advertising, but I don't make any money or internet points from this blog)

Semiclassical

I'm sure there's some way to interpret that meaningfully, but...

Alessandro Codenotti

@MatheinBoulomenos HOTT intensifies

MatheinBoulomenos

7:16 PM

@Alessandro I'm not going down that rabbit hole

Alessandro Codenotti

Pff

Balarka Sen

Don't categories with morphisms between categories being functors and natural transformations between functors as 2-morphisms make a lax 2-category or whatever

MatheinBoulomenos

it's a strict 2-category even

Leaky Nun

What do you wish to unlearn so that you can learn again?

Balarka Sen

@MatheinBoulomenos Mmk

"(Exercise for the stray logician or set theorist: The above proof doesn’t work in ZFC, why?)"

smh

Alessandro Codenotti

7:18 PM

@BalarkaSen now I'm interested

Balarka Sen

lmao

Alessandro Codenotti

@Mathei they're offering an "introduction to higher categories" course next term here

Leaky Nun

@MatheinBoulomenos well to do your exercise you would have to redefine everything in ZFC

are we only looking at small categories?

Balarka Sen

@MatheinBoulomenos Prop C3.8 is cool! I didn't know this perspective

It's clear once you say it but I never thought about it

Semiclassical

hmm, going back to something I asked about earlier

MatheinBoulomenos

7:20 PM

@LeakyNun that's the crucial point, I allowed a class of objects

Leaky Nun

then certainly you can't choose

MatheinBoulomenos

@AlessandroCodenotti I'd take that, but I guess that's obvious

Leaky Nun

what is a functor then

Alessandro Codenotti

I'll also take it I think

Semiclassical

If I've got a set of $n$ points in a vector space $X$, i.e. $A=\{x_k\}_{k=1}^n$, then the convex hull of $A$ is $C = \{\sum_{k=1}^n p_k x_k|\forall k (p_k\geq 0) \wedge \sum_{k=1}^n p_k =1\}$

MatheinBoulomenos

7:23 PM

@LeakyNun I think you know what a functor is

a lot of interesting categories are not small

but I didn't want to go into Grothendieck universes

Leaky Nun

I'm just thinking whether it is choice that fails since you have a class of objects

Balarka Sen

Locally small or it's a myth

Leaky Nun

but can we cook up an explicit counter-example

Semiclassical

If I have a linear transformation of vector spaces $T:X\to Y$, then for any point $y\in C$ I have $T(y) = T(\sum_{k=1}^n p_k x_k) = \sum_{k=1}^n p_k T(x_k)$ where the same conditions on the p_k apply

MatheinBoulomenos

@BalarkaSen I defined categories as locally small

Semiclassical

7:24 PM

Hence $T(C)$ is nothing other than the convex hull of $T(A)$.

However: That proof is for a set of points.

Balarka Sen

@Mathein Hahah!

MatheinBoulomenos

but I want to relax that a bit to be able to talk about functor categories freely later

Semiclassical

Is it obvious how to make that work for an arbitrary $A\subseteq X$?

seems like it should be obvious

MatheinBoulomenos

not in a necessary way, but yeah

Leaky Nun

@Semiclassical surely just make it have finite support

and you'll get the formula for convex hull

proof: universal property nonsense

therefore the proof goes through

Adam

7:27 PM

When people speak of land ownership they only mean surface of land ownership, about 30 centimetres below the surface isn't owned by anyone but the government probably made up a victimless crime law forbidding you to dig below a certain depth. Everything is so retarded when you look at it closely

Akiva Weinberger

Cuius est solum, eius est usque ad coelum et ad inferos

Cuius est solum, eius est usque ad coelum et ad inferos (Latin for "whoever's is the soil, it is theirs all the way to Heaven and all the way to Hell") is a principle of property law, stating that property holders have rights not only to the plot of land itself, but also to the air above and (in the broader formulation) the ground below. The principle is often referred to in its abbreviated form as the ad coelum doctrine.In modern law, this principle is still accepted in limited form, and the rights are divided into air rights above and subsurface rights below. Property title includes to the space...

Adam

haha good luck spreading that view though I will bet you a 20 I am right about there being victimless crime laws in your locality forbidding you to dig without an excavation license or some similar garbage reason

Leaky Nun

in minecraft there's a bukkit plugin that allows land ownership and it follows the ad coelum doctrine; surely it implies that the ad coelum doctrine is true in the real world

Semiclassical

in a real world, anyways

Akiva Weinberger

@Adam I mean you'd probably run into infrastructure and stuff

Pipes

Sewage and stuff

Adam

7:37 PM

Well I remember being lectured about it before the age I went to school for digging with intent to build underground, but I was in trouble for a lot of crap and have been since I guess it might be a false memory. I mean when my brother was on board we were allowed to make tunnels and did, but yeah I do remember a lecture about it being illegal when I tried myself the first time

Akiva Weinberger

I think your parents (or whoever) was probably just worried about you hurting yourself

although looking online I am seeing stuff about becoming a Certified Excavator

Adam

Well yeah I know that but it seems fairly elaborate

Akiva Weinberger

digsafelynewyork.com/cep

> State law requires you to place a location request with Dig Safely New York prior to digging or excavating.

Adam

yeah I mean naturally if I really want to do it tomorrow I'm going to do it regardless, its not like I take the government seriously when they forbid stuff lol I mean who would really

Akiva Weinberger

Dunno what counts as "excavating" though

@Adam If you hit a utility though you have yourself to blame

Outside of an urban area it's probably fine

I am not a lawyer though

(The Bar Association isn't returning my calls)

Adam

7:47 PM

no I wont hit a utility lol there is a "dial before you dig" hotline here so ill just have to call up for everywhere but and get the depths for everything and take a guess when I need to be careful and when I should worry about a visit from the idiots

The Bar Association? haha yeah I will speak to my attorney also which is definitely not exactly the same type of person as what is referred to as an attorney in the film "Fear and Loathing in Las Vegas"

user193319

8:33 PM

Can there exist 2x2 matrices which are nonsimilar but have the same characteristic and minimal polynomials? I've tried to search for a pair, but I haven't had any luck so far.

MatheinBoulomenos

@user193319 consider two cases: diagonizable and non-diagonizable. (we can read off diagonizability from the minimal polynomial, so that's not an issue)
In the diagonizable case, clearly the characteristic polynomial tells us the eigenvalues with multiplicity. In the non-diagonizable case, the minimal polynomial is either irreducible or of the form $(x-a)^2$, now use Jordan canonical form and Frobenius normal form (or generalized Jordan, same thing here), to conclude that the matrices are similar

user193319

Hmm...so they can't exist...

MatheinBoulomenos

there's an easier argument using modules, though :)

user193319

What about the 3x3 case?

MatheinBoulomenos

the question is equivalent to the following thing: if two $k[X]$-modules having dimension two over $k$ have the same annihilator (minimal polynomial) and Fitting ideal (characteristic polynomial, then are they isomorphic? From the classification, all modules are of the form $k[x]/(f)$ for $f$ of degree $2$ or $k[x]/(x-a) \oplus k[x]/(x-a)$, which can be distinguished by Fitting ideal or annihilator.
In the 3x3 case, we have modules of the form $k[x]/(f)$ for $f$ of degree $3$ or $k[x]/(x-a)(x-b) \oplus k[x]/(x-a)$ or $k[x]/(x-a)^2 \oplus k[x]/(x-a)$ or $(k[x]/(x-a))^3$, all of which can be …

For the $n=4$ case, though you have $k[x]/(x^2) \oplus k[x]/(x^2)$ and $k[x]/(x^2) \oplus k[x]/(x) \oplus k[x]/(x)$ for example, both have annihilator $x^2$ and fitting ideal $x^4$

These correspond to matrices $\begin{pmatrix}0&0&0&1\\0&0&0&0\\0&1&0&0\\0&0&0&0 \end{pmatrix}$ and $\begin{pmatrix}0&0&0&1\\0&0&0&0\\0&0&0&0\\0&0&0&0 \end{pmatrix}$

user193319

8:51 PM

Wow, very interesting.

Adam

The worst part about the world I can think of is the number of highly intelligent people that see laws, rules, and social for pars as absolute, and don't see how easy it is to bend and break them when it's necessary, I really think I could be speaking to you if you are reading, they are just idiots in respect costumes, you don't have to do what they say

MatheinBoulomenos

Oops, these should be $\begin{pmatrix}0&1&0&0\\0&0&0&0\\0&0&0&1\\0&0&0&0 \end{pmatrix}$ and $\begin{pmatrix}0&1&0&0\\0&0&0&0\\0&0&0&0\\0&0&0&0 \end{pmatrix}$

GFauxPas

As part of a longer proof, I'd like to show $N^N>(N+1)^{N-1}$ for $N\ge3$ odd (probably true for even $N$ but it di

I don't need it to be). I couldn't find an inductive step that made things easier. What else could I do?

Or what inductive step would work

Akiva Weinberger

So equivalently $N+1>(1+\frac1N)^N$?

(multiplying both sides by $\frac{N+1}{N^N}$)

GFauxPas

Oh I found a question on the site

17

Q: Prove by induction that for all $n \geq 3$: $n^{n+1} > (n+1)^n$

I am currently helping a friend of mine with his preperations for his next exam. A big topic of the exam will be induction, thus I told him he should practice this a lot. As at the beginning he had no idea how induction worked, I showed him some typical examples. Now he showed me an exercise he ...

Akiva Weinberger

9:05 PM

You have $e>(1+\frac1N)^N$ for all positive $N$ I'm pretty sure

GFauxPas

Yea

There we go

Thanks

So its true for $N=2$ also but as I said I only need odd

Leaky Nun

taking the stalk at x is a functor from the category of presheafs on X to the category of all sets

Adam

9:37 PM

The Chasers War on Everything - Middle East TV

►

lol how, how is this possible youtube.com/watch?v=8SBzuG_-O2o

Leaky Nun

10:29 PM

J. S. Bach - Toccata et Fuga in Fa maggiore BWV 540

►

this looks impossible to play

3:18

Adam

10:54 PM

I played the baritone saxophone when I studied music as a kid but I feel sorry for piano players in that they have so much competition, like being the pianist in the school orchestra is basically a super human feat, I don't want to be racist and elaborate but lets just say there is at least one kid in every high school whose parents refused to let them out of their room until they are a musical prodigy or a doctor

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$Mathematics$ Mathematics