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7:00 PM
@Thorgott I didn't get it until I thought of the same joke
 
There was a joke? I still don't get it.
 
I never use unnecessary curly braces actually, which annoys the people I write exercises with on overleaf
 
lol
 
$e = 3 = \pi$
 
And I have no unnecessary whitespaces either
 
7:01 PM
bad
 
No, $\pi=22/7$.
 
we all know $\pi = 3.2$
 
@AlessandroCodenotti I hope you will never write code I have to read :)
 
Hah!
 
@TobiasKildetoft my code is a little better
 
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
 
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
 
Were you fired, Balarka?
 
I forgot which subset
 
If you're going for efficiency, [ln2,e] would be even faster
 
I'm mostly writing Haskell now so at least some indentation is enforced (and it annoys me)
 
7:03 PM
probably $[0,1] \setminus \Bbb Q$
 
I let my IDE do all the formatting for me
 
@LeakyNun to the whole of the irrationals
 
@TedShifrin Nah just had some points taken off in all my assignments
 
@AlessandroCodenotti they're probably homeomorphic anyway
$[0,1] \setminus \Bbb Q$ and $\Bbb R \setminus \Bbb Q$
 
@Alessandro: I guess I don't see it.
 
7:05 PM
@LeakyNun seems like it
 
I see infinite decimals with no zeroes ...
 
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
 
Oh, no, I don't. Never mind.
 
@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
 
@Balarka: I don't see that either, now, but I'm being dumb.
Agh.
 
7:07 PM
So that's an homeo from the disjoint union of four copies of $\mathcal{N}$, but this is homeomorphic to $\mathcal{N}$
 
Continued fractions is too esoteric.
 
More weirdly $\mathcal{N}$ continuously surjects onto $\Bbb R$
 
N^N is just a cantor set
C surjects into every compact metric space
 
Not quite, it's not compact
 
I'm used to doing $\{0,1\}^\omega$.
 
7:08 PM
Oh ok
ya
 
(0,1)\Q is uncountable though. I feel like I'm missing something.
 
N^N surjects onto every Polish space
 
Is $\mathcal N$ otherwise $\Bbb N$ or what?
 
N=N^N
 
Ah.
 
7:09 PM
Boom
 
$\mathcal{N}$ is N^N
 
gotcha
 
Now I'm slightly less confuzled.
 
That's common notation for the Baire space (N^N)
 
I've never seen it.
 
7:10 PM
Well it's common in descriptive set theory :P
 
is the Baire space a Baire space?
 
Clearly, it's a normal distribution in this context
 
Right.
Thank goodness, friends are visiting from out of town and I'll be missing a few days of this garbage.
 
We'll be missing you though :3
 
@LeakyNun sure it's Polish, so in particular completely metrizable
 
7:13 PM
Have fun!
 
I'll try.
 
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"
 
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)
 
I'm sure there's some way to interpret that meaningfully, but...
 
@MatheinBoulomenos HOTT intensifies
 
7:16 PM
@Alessandro I'm not going down that rabbit hole
 
Don't categories with morphisms between categories being functors and natural transformations between functors as 2-morphisms make a lax 2-category or whatever
 
it's a strict 2-category even
 
What do you wish to unlearn so that you can learn again?
 
@MatheinBoulomenos Mmk
"(Exercise for the stray logician or set theorist: The above proof doesn’t work in ZFC, why?)"
smh
 
7:18 PM
@BalarkaSen now I'm interested
 
lmao
 
@Mathei they're offering an "introduction to higher categories" course next term here
 
@MatheinBoulomenos well to do your exercise you would have to redefine everything in ZFC
are we only looking at small categories?
 
@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
 
hmm, going back to something I asked about earlier
 
7:20 PM
@LeakyNun that's the crucial point, I allowed a class of objects
 
then certainly you can't choose
 
@AlessandroCodenotti I'd take that, but I guess that's obvious
 
what is a functor then
 
I'll also take it I think
 
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 $C = \{\sum_{k=1}^n p_k x_k|\forall k (p_k\geq 0) \wedge \sum_{k=1}^n p_k =1\}$
 
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
 
I'm just thinking whether it is choice that fails since you have a class of objects
 
Locally small or it's a myth
 
but can we cook up an explicit counter-example
 
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
 
@BalarkaSen I defined categories as locally small
 
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.
 
@Mathein Hahah!
 
but I want to relax that a bit to be able to talk about functor categories freely later
 
Is it obvious how to make that work for an arbitrary $A\subseteq X$?
seems like it should be obvious
 
not in a necessary way, but yeah
 
@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
 
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
 
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...
 
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
 
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
 
in a real world, anyways
 
@Adam I mean you'd probably run into infrastructure and stuff
Pipes
Sewage and stuff
 
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
 
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
 
Well yeah I know that but it seems fairly elaborate
 
> State law requires you to place a location request with Dig Safely New York prior to digging or excavating.
 
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
 
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)
 
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"
 
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.
 
@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
 
Hmm...so they can't exist...
 
there's an easier argument using modules, though :)
 
What about the 3x3 case?
 
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 $\begin{pmatrix}0&0&0&1\\0&0&0&0\\0&0&0&0\\0&0&0&0 \end{pmatrix}$
 
8:51 PM
Wow, very interesting.
 
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
 
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 $\begin{pmatrix}0&1&0&0\\0&0&0&0\\0&0&0&0\\0&0&0&0 \end{pmatrix}$
 
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
 
So equivalently $N+1>(1+\frac1N)^N$?
(multiplying both sides by $\frac{N+1}{N^N}$)
 
Oh I found a question on the site
17
Q: Prove by induction that for all $n \geq 3$: $n^{n+1} > (n+1)^n$

HuyI 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 ...

 
9:05 PM
You have $e>(1+\frac1N)^N$ for all positive $N$ I'm pretty sure
 
Yea
There we go
Thanks
So its true for $N=2$ also but as I said I only need odd
 
taking the stalk at x is a functor from the category of presheafs on X to the category of all sets
 
10:29 PM
this looks impossible to play
3:18
 
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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