Mathematics - 2017-06-15 (page 7 of 9)

Akiva Weinberger

19:00

Oh, the curiosity got the best of me

Alex K Chen

No, I don't want @Semiclassical :P

Akiva Weinberger

While looking up this stuff to see if I had the right statement of the problem, I accidentally found the solution

@Semiclassical Yeah

Another way of defining rank is: Define $V_0=\emptyset$, $V_{n+1}=\mathcal P(V_n)$

(So $V_n=\mathcal P^n(\emptyset)$)

(Fun fact: $V_n\subset V_{n+1}$ for all $n$.)

Semiclassical

My thought, I guess, is that there should only be a finite number of rank one sets I can construct using a finite set of rank one seta

Akiva Weinberger

Then $V_n$ is the set of all things of rank less than $n$.

Oh, uh, limit ordinals: $V_\lambda=\bigcup_{\alpha\in\lambda}V_\alpha$

So you can prove that every set is in some $V$ set.

Semiclassical

And in that case it hardly seems plausible that I could have enough rank-one sets to make that work

Akiva Weinberger

19:03

And the rank of a set is the index of the smallest $V$ set it's in.

Wait, that might be off by one

Semiclassical

You said you resolved the puzzle though?

Akiva Weinberger

Oh, the rank of a set is the index of the smallest $V$ set it's a subset of.

@Semiclassical I accidentally read it, yeah

Semiclassical

Ah

Akiva Weinberger

though I'm not sure I quite see why it works?

"Accidentally." Wikipedia has no spoiler tags! :P

Abdullah UYU

ıs there anyone using 3b1b's repo: github.com/3b1b/manim

Akiva Weinberger

19:07

By the way, non-well-founded sets have no rank. Good thing ZFC disallows those!

TheGreatDuck

@Semiclassical sorry for annoying you so much the last few days. I know I have been a bit of a parabola with everything.

Akiva Weinberger

@AbdullahUYU Are you Turkish, by any chance?

Semiclassical

So what is the resolution? @AkivaWeinberger

Akiva Weinberger

I know from the lowercase dotless 'ı' :P @AbdullahUYU

TheGreatDuck

I'm going to head out. This place is just making me unhappy at this point and it is my fault. :-(

Akiva Weinberger

19:08

@Semiclassical You want the answer?

Abdullah UYU

Right, "I" am Turkish :)

Akiva Weinberger

Apparently it's known as the Quine–Rosser definition @Semiclassical

'I', not 'İ'? :P

Semiclassical

Is this the type-level thing?

Akiva Weinberger

In any case, to answer your question: Probably not

@Semiclassical Yeah

Abdullah UYU

@AkivaWeinberger Is it a general habit for Turkish, or you know me from one of my questions on se?

Semiclassical

19:10

Is it obvious that type in this sense is the same as the rank?

Akiva Weinberger

@AbdullahUYU The Turkish alphabet has two 'i' letters: Iı and İi.

I think it's the only language that does that.

The rest of us only have one: Ii

theDoctor

@LeakyNun: which chat are you in: with me or here?

Akiva Weinberger

So, you'll only find lowercase dotless 'ı's or capital dotted 'İ's in Turkish writing

Leaky Nun

@theDoctor both

theDoctor

@what's the deal with the insult about h-bar?

TheGreatDuck

19:14

@theDoctor no clue man.

Abdullah UYU

@AkivaWeinberger thanks for the infos

TheGreatDuck

btw man, your ideas sound very nice. However, I would suggest posting a question about it.

Leaky Nun

bye.

Akiva Weinberger

Hi, Arctin Tern

TheGreatDuck

@LeakyNun who?

Leaky Nun

19:15

@TheGreatDuck everyone

TheGreatDuck

oh you're leaving.

cya

Akiva Weinberger

${B\atop C}Y{E\atop A}$

4

(BYE, CYA)

Abdullah UYU

btw, where do you know this specialness of Turkish, i don't even know :)

arctic tern

@AkivaWeinberger hi

seems I've missed some stuff

Akiva Weinberger

I dunno, I just like languages

One night I tried to learn Turkish. I learned like 3 words, stopped, and then forgot them all :P

Abdullah UYU

19:21

Turkish is hard to learn, really

Akiva Weinberger

What even is the ğ thing

Is it like a silent letter?

Abdullah UYU

No, i don't know how to explain it :)

Akiva Weinberger

Haha, OK

TheGreatDuck

brb

Astyx

Would you guys know how I could remove chlorine from my eyes ? I've been crying for the last hour, it's preventing me from reading and thus learning

Semiclassical

19:25

For possible later convenience / present amusement, I'm-a just leave this here: math.ucr.edu/home/baez/crackpot.html

Abdullah UYU

@AkivaWeinberger I found very nice example for it: trttelaffuz.com/kelime/19231

Akiva Weinberger

@Astyx Uhh

Would you be able to read our answers?

Probably try to wash it out or something

Ask Google

Try washing it out first though

with water

Semiclassical

You mean, chlorine from a pool?

Astyx

Already tried that

Akiva Weinberger

Maybe soap? Idk

Astyx

19:26

Yes @Semi

Akiva Weinberger

Is soap in eyes a bad idea?

Astyx

I didn't put chlorine in my eyes on purpose if that's the question :p

3

Akiva Weinberger

Probably

Astyx

Probably

Semiclassical

It seems ill advised

But yeah, flushing with water seems like the right idea

Eg take a shower and rinse your eyes out that way

Astyx

19:28

It didn't do too much good on the long run, even though it does appease the "pain" for a short while

I guess I'll just have to wait

Akiva Weinberger

Look at your eyes

with your eyes

Also touch your right elbow without using your left hand

Astyx

19:43

And see my local shaman for medicinal incantations ?

TheGreatDuck

@Semiclassical Do you think I am a troll?

please be honest

Akiva Weinberger

@Semiclassical With regards to the ordered pair thing

$a\cup b$ is clearly type-level (or flat), but it also clearly doesn't work for this

Is there a way to modify it so that it does work?

Alessandro Codenotti

Is there a reasonable condition on the topological space $X$ to ensure that the diagonal of $X\times X$ is homeomorphic to $X$?

This can fail for relatively nice, for example $T_4$, spaces

Akiva Weinberger

@AlessandroCodenotti Really? Example?

Liad

isn't the diagonal is $\{(x,x) :x\in X \}$ ?

Alessandro Codenotti

19:56

The product of the sorgenfrey line with itself

Yes @Liad

Akiva Weinberger

Why isn't $x\mapsto(x,x)$ a homeomorphism?

Alessandro Codenotti

My question is motivated by the one you had earlier in fact

Akiva Weinberger

@AlessandroCodenotti Hm, let me think

Liad

huh, you wrote the diagonal of , i read "the diagonal $X \times X$ " and thought something is missing.

Akiva Weinberger

Wait, how does it fail for Sorgenfrey?

Alessandro Codenotti

19:59

Ah, no wait, with sorgenfrey it fails on {(x,-x)}

Akiva Weinberger

Yeah

So what fails on the diagonal?

Shouldn't be any

Liad

maybe im wrong but doesn't Hausdorff enough for the diagonal to be homeo' to $X$ ?

Akiva Weinberger

Yeah, I think the diagonal is always homeomorphic to the space

Alessandro Codenotti

Hm, I'm not sure

Akiva Weinberger

It doesn't even fail for Sierpiński

Alessandro Codenotti

20:02

Hausdorff is enough for the diagonal to be closed in $X\times X$ but that's not very useful here

Akiva Weinberger

$x\mapsto(x,x)$ seems to be continuous because the intersection of two open sets is open

Alessandro Codenotti

Oh, wait the projection $X\times X\toX$ restricted to the diagonal is bijective and open, right?

Akiva Weinberger

(A basis on $X\times X$ is sets of the form $A\times B$ for $A,B$ open;

TheGreatDuck

@AkivaWeinberger you don't think I'm a troll do you?

SteamyRoot

It's a homeomorphism.

Akiva Weinberger

20:05

the inverse image of $A\times B$ is $A\cap B$, which is open)

Liad

@AlessandroCodenotti i wanted to write it, wasn't sure about it :P

Alessandro Codenotti

@SteamyRoot yeah, that's where I was going with bijective+open, thanks

(And continuous clearly)

Akiva Weinberger

$(x\times x)\mapsto x$ is continuous because the inverse image of $A$ is in $A\times A$.

TheGreatDuck

what are you guys discussing?

Akiva Weinberger

So it's a homeomorphism.

@TheGreatDuck You're annoying sometimes, but only sometimes

:(

SteamyRoot

20:07

@TheGreatDuck If $(X,\tau)$ is a topological space, then it's homeomorphic to the diagonal (with subspace topology) of $X \times X$ (with product topology)

TheGreatDuck

uuh

sounds nice

Akiva Weinberger

@TheGreatDuck Ale said it's possible for the diagonal of $X\times X$ not to be homeomorphic to $X$. This turned out to be false.

Astyx

@SteamyRoot Is that always true ?

TheGreatDuck

umm...

ok

Akiva Weinberger

@Astyx Yeah

TheGreatDuck

20:08

(I have no clue what any of that just meant)

Astyx

(Sorry if I don't follow the conversation too well, I can't see from time to time)

TheGreatDuck

@AkivaWeinberger here's a weird concept. Noneuclidean matrices.

grids of numbers on surfaces other than flat planes

like a number cube but only the surface

or a number sphere

Alessandro Codenotti

@AkivaWeinberger that's what happens when you check only one example and it's wrong :P

Astyx

@TheGreatDuck You can already do this in some sense with Manifolds (things that locally look like $\Bbb R^n$)

user379685

There are three bus companies A, B and C that transport 2000 people from X to Y. Company A takes 30% of them, B takes 35% and C the rest. How many routes does the company A need to do so that the chance of rejecting the client is less than 1%. Buses of company A can hold 40 people.

Astyx

20:12

If that's what you mean

TheGreatDuck

no i meant matrices

the same algebra as matrices

but shaped like spheres or cubes or surface geometries.

Astyx

I'm not sure I follow

TheGreatDuck

you have a matrix

SteamyRoot

That sounds like restriction a tensor to one of its indices being either $0$ or maximal

TheGreatDuck

now imagine one that was a sphere of numbers

Astyx

20:14

Ah ...

So instead of a table of numbers you have a dice or something else ?

TheGreatDuck

well dice with several numbers on each side

but yeas

@SteamyRoot what about a circle of numbers?

Akiva Weinberger

You'd need a good way to spread your numbers onto the surface of the sphere

Maybe vertices of Platonic solids or some such

Astyx

That would be an map from a partition of your object to the natural numbers, I don't know of anything like it though

TheGreatDuck

hmmm

SteamyRoot

A circle of numbers just sounds like a vector with indices mod $n$

TheGreatDuck

20:16

@SteamyRoot not a ring of numbers. A filled circle.

Akiva Weinberger

Or a function from $\Bbb Z_n$ to $\Bbb N$ I guess?

SteamyRoot

A disk, you mean.

TheGreatDuck

yeah

Akiva Weinberger

In the same sense that a usual vector is a function from $[n]$ to $\Bbb N$

user379685

hello, anyone?:c

Akiva Weinberger

20:16

@user379685 What have you tried?

Astyx

Or $D \to \Bbb N$ or $\text{anything} \to \Bbb N$ more genrally

SteamyRoot

Dunno, you'd have to specify how these numbers lie with respect to eachother

Akiva Weinberger

I'd probably start by seeing how much "30% of them" is @user379685

TheGreatDuck

@user379685 you have provided no context. Therefore, we will ignore you.

Akiva Weinberger

Well, not context per se, but at least tell us what you've done so far.

user379685

20:18

Well, I tried using the Central Limit Theorem but i got 1600 which is waay to many

Akiva Weinberger

Ohh

TheGreatDuck

you have a bad history of no context posts

why should we help you?

Akiva Weinberger

Oh, you know what, I'm gonna duck out of this one

I don't actually know statistics

Sorry!

TheGreatDuck

i probably could help but i refuse to help on principle.

sorry!

Akiva Weinberger

You might as well type out what you've tried with the Central Limit Theorem

Incidentally, do you know how to turn on MathJax?

That's the sort of thing that turns $a^b$ into actual exponents for us

(There's a link in the room info on the top-right corner)

tinyurl.com/cfqcvpc

user379685

20:28

$P(S_n > 600) < 0.01$
$P((S_n - n*0.3)/(sqrt9(n)*sqrt((3/10)*(7/10)))>((600-n*(3/10))/((sqrt(n)*sqrt((3/10)*(7/10))) < 0.01$
$P(Z>x_o) < 0.01$
$1-P(Z<x_o) < 0.01$
$0.99 < P(Z < x_o)$
$0.99 < Φ(x_o)$
$2.33 < x_o$
$((600-n*(3/10))/((sqrt(n)*sqrt((3/10)*(7/10))) > 2.33$
$n=1849$

TheGreatDuck

use English

Akiva Weinberger

Let me fix the formatting

TheGreatDuck

@user379685 where is your explanation of those lines, hmm?

Akiva Weinberger

What's sqrt9(n)?

user379685

whops that was a mistae

Akiva Weinberger

20:29

$\sqrt{9n}$ or $\sqrt[9]{n}$?

user379685

theres no 9

Akiva Weinberger

Oh, OK

TheGreatDuck

please revise your post before posting....

:p

Akiva Weinberger

$P(S_n > 600) < 0.01$
$P\left(\dfrac{S_n - 0.3n}{\sqrt{n}\sqrt{(3/10)*(7/10)}}>\dfrac{(600-(3/10)n}{\sqrt{n}\sqrt{(3/10)*(7/10)}}\right) < 0.01$
$P(Z>x_o) < 0.01$
$1-P(Z<x_o) < 0.01$
$0.99 < P(Z < x_o)$
$0.99 < Φ(x_o)$
$2.33 < x_o$
$\frac{(600-n*(3/10)}{(\sqrt n*\sqrt{(3/10)*(7/10)}} > 2.33$
$n=1849$

Alessandro Codenotti

@Akiva take an $n$ dimensional topological vector space $V$, are all $n$ dimensional subspaces of $V\times V$ homeomorphic to $V$? It sounds reasonable

user379685

20:32

the second line the probabilty is of the whole thing not just the numerator

yeah that's correct

Akiva Weinberger

$P(S_n > 600) < 0.01$
$P\left(\dfrac{S_n - 0.3n}{\sqrt{n}\sqrt{(3/10)\cdot (7/10)}}>\dfrac{600-(3/10)n}{\sqrt{n}\sqrt{(3/10)\cdot(7/10)}}\right) < 0.01$
$P(Z>x_0) < 0.01$
$1-P(Z<x_0) < 0.01$
$0.99 < P(Z < x_0)$
$0.99 < \phi(x_0)$
$2.33 < x_0$
$\dfrac{600-(3/10)n}{\sqrt n\sqrt{(3/10)\cdot(7/10)}} > 2.33$
$n=1849$

user379685

well obviously that's wrong, could you help?

Akiva Weinberger

Where did 2.33 come from?

@TheGreatDuck (I hope your moral qualms are quelled since they posted their work) Do you know this stuff?

TheGreatDuck

@AkivaWeinberger i don't actually know statistics at this level. I'm in high school!

user379685

from tables for the N(0,1)

TheGreatDuck

20:38

:-P

Akiva Weinberger

Yeah, OK, I don't really know this stuff either

Sorry!

TheGreatDuck

sometimes I read about things and talk about them to sound smart, but tbh I don't know much advanced math.

Astyx

Why is probability so ... tricky ?

Alessandro Codenotti

I'll have a probability exam in July... definitely not looking forward to that one

Akiva Weinberger

@Astyx Provability > probability :P

(Almost every time I try to write "provably" on my phone, it autocorrects to "probably")

TheGreatDuck

20:53

@Astyx because it is evil incarnate along with calculus and geometry.

Akiva Weinberger

Hey, what's wrong with calculus?

TheGreatDuck

idk

i just don't like much math really

Astyx

Great choice of chat then :p

I'll go to bed now, and hope a night's sleep will solve my eye problem

Cya

Akiva Weinberger

What time is it where you are, 10pm?

Astyx

11

Akiva Weinberger

21:05

Oh

Astyx

I wish I could go to bed at 10 :(

Akiva Weinberger

Yeah sleep

TheGreatDuck

@AkivaWeinberger i mostly come here because I have nothing better to do.

Akiva Weinberger

Eat, drink, piss

TheGreatDuck

that's not better

Daminark

21:26

Hai everyone!

5

Fargle

Sup nerd

Semiclassical

@LeakyNun this turned out to be a prophetic comment

Akiva Weinberger

@Semiclassical Want me to tell you the flat ordered pair thing? Or do you still want to think about it

Daminark

How goes it @Fargle?

Semiclassical

Not right now

Fargle

21:29

@Daminark Alright. Listening to the Hotline Miami soundtrack so I feel lucid and yet strangely violent. You?

Daminark

I just woke up like, basically now

Semiclassical

Actually I was curious

Daminark

shrugs This sleeps schedule is like, not sustainable for long

Semiclassical

How sure are we that type is equivalent to rank in the senses described?

Fargle

yeah I woke up about an hour ago with some serious sleep paralysis

Daminark

21:31

Kek

Semiclassical

I need to avoid taking naps during the day

I tend to end up feeling really tense/anxious after

Akiva Weinberger

How is type defined?

Daminark

Eek

Initiating homotopy type theory

Semiclassical

I dunno. But that's the word used in the Wikipedia bit on Quine-Rosser ordered pairs

Akiva Weinberger

Oh, so you saw how it's defined?

Semiclassical

21:35

I saw it described as being type-level

I didn't really follow the definition though

I'm mostly paranoid about whether type-level implies flat in your sense

Akiva Weinberger

What's your definition of type-level though

Daminark

This sounds odd but, I've been running wild and free

Can I ransack someone's city?

Fargle

@Daminark Yeah sure, I've been needing an excuse to move

Akiva Weinberger

Wat

Semiclassical

Again, I don't know. I am literally saying that, since the portion on Wikipedia is not in terms of rank, I can't assess whether or not the Quine-Rosser definition is flat

Daminark

21:41

Thx

Semiclassical

All I can tell from Wikipedia is that it's type-level. That sounds like flatness, but I can't draw such a conclusion just from that

Daminark

Fargle, being ransacked

Fargle

Temba, his arms wide.

Akiva Weinberger

Bill Wurtz?

@Semiclassical It's hard to draw conclusions from words you don't know their meaning

Semiclassical

ST:TNG if in not mistaken

Daminark

21:44

Yup @Akiva

Fargle

@Semiclassical In my case, yes.

Darmok and Jalad at Tanagra.

Semiclassical

Shaka, when the walls fell.

Daminark

Rip walls

Faust7

blarg

Semiclassical

Blargle

Fargle

21:50

Gargle

Faust7

not a redvs blue fan?

Semiclassical

That's a pretty vague reference

Faust7

not for anyone in my generation

Semiclassical

Given that I've watched all of RVB and have kept up the latest season, yes it is

Faust7

when Kaboose finds the elite on the planet

it says Blarg and he as asks it if Blarg means yes?

Semiclassical

21:54

Yes, and that's all he can say

Faust7

it says Blarg and Kaboose decided that it means yes

Semiclassical

I know the joke, but not the specific phrase

Faust7

:s

k

Semiclassical

Because, again, that specific word is not a familiar reference

Faust7

anyway hows ur morning?

Semiclassical

21:55

"You ever wonder why we're here?", by contrast...

Faust7

^^

Semiclassical

Eh, late afternoon here.

Not very productive

Faust7

i feel like a kid again im drinking grape juice from a measuring cup

Semiclassical

Could be worse. Could be a sippy cup

Faust7

Onle piece of clean glassware i could find

ive done 4 loads of laundry today and i probably have 5 more to go...

Night shift makes the fiance retarded

Semiclassical

21:58

I should admit, though, as far as internet refs go I'm more of an abridged series fan

Faust7

the abridged DBZ was awesome

Kirill

Hi! I have a partial derivative $\partial_{rr}F(r, \varphi)$ for some function $F$, where $r \in (0, \infty)$. What these $rr$-s actually mean?

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$Mathematics$ Mathematics