Mathematics - 2016-12-28 (page 4 of 4)

Null

22:01

@KajHansen you where just joking or?

Kaj Hansen

About what @Null >

?

Null

impostering

Balarka Sen

@Kaj nice hat lol

Kaj Hansen

100% serious. en.wikipedia.org/wiki/Impostor_syndrome.

ikr? I set trends. It's what I do.

alan2here

Q: Is {{1, 2}, {1}, {2}} a power set? Call this set 'x'.

A:
Step 1: {1, 2} is the member of 'x' that iteself contains the most elements.
Step 2: The power set of {1, 2} is {{1, 2}, {1}, {2}}, call this 'y'.
Step 3: Is x = y? Yes, therefore 'x' is a power set, if not then 'x' is not a power set.

It seems to me that this should work.

Null

22:03

@KajHansen there is even a bigger syndrome. the fantasy that everything is exchanged. truly horrible stuff

Balarka Sen

You're missing the nullset.

Mike Miller

wikipedia page is using the dumb spelling

alan2here

How can "power" have an incorect spelling?

Kaj Hansen

w/ an "o" ?

Mike Miller

@Null That's not relevant. Almost every grad student suffers from imposter syndrome. Almost none suffer from Capgras.

alan2here

22:05

@GFauxPas Oh, yes :¬P I wrote some steps for that too below.

Null

@MikeMiller right, i mixed those up

@KajHansen i can nothing and still have this snydrome. so when i finally can something i am cured. I fail to see the flaw in my logic :)

GFauxPas

forgot the empty set

but otherwise good

the empty set is a subset of every set

alan2here

ohh, TY, true :)

GFauxPas

giveaway that your set wasnt a powerset: if a finite set has $n$ elements, its power set has $2^n$ elements

and I believe that's true for cardinals as well but I dont know cardinal arithmetic

Alessandro Codenotti

Yeah, the powerset of $A$ has always cardinality $2^{|A|}$

Semiclassical

22:19

@MikeMiller While I entirely agree that impostor syndrome != Capgras claim, I think there's at least a vague sense in which the earlier statement is true---namely, that one may go from believing oneself to be an impostor, to believing that the entire institution is pretending to be something that it's not.

(i'm feeling tired and somewhat philosophical, so take it with a grain of salt)

@BalarkaSen It's a sign of the mood I'm in when I find TS Eliot lines spontaneously coming to mind

Balarka Sen

heh.

Semiclassical

'unreal city'. not sure why that one phrase is in my head right now. (from "The Waste Land")

alan2here

Thought I may as well :¬P
http://math.stackexchange.com/questions/2075349/how-to-determine-if-a-set-is-a-power-set

Akiva Weinberger

A function can't be in its own domain in ZFC, can it?

Null

22:34

@AkivaWeinberger how you mean?

Akiva Weinberger

Hm. I'm sure there are ways to go around it. Give the domain a surjection into the set of functions you want to study, or something

@Null Like, I want to have $f(f)$

Null

i have to pass here ;)

Akiva Weinberger

In any case, let the set $A$ consist of functions whose domains include themselves, and let $f$ be the function whose domain is $A$ defined by $\forall a\in A,f(a)=a(a)$

Null

@AkivaWeinberger that would be a mapping of mappings, im not sure

GFauxPas

Oh, well, let's see if it contradicts anything Akiva

let's start with general relations

domain $\text{Dom}$ codomain $\text{Cod}$

$D$ and $C$ for ease of typing

this is exploratory research Akiva I hope youre excited, I don't know the asnwer

A relation is a tuple, a triple in this case, of $(D, C, S)$ where $S \subset D \times C$.

Call this $R$

Null

22:41

call of duty=cod=codomain=coinscidence?=i think not

GFauxPas

that's model theory Null, beyond my grasp

so for $R$ to be in $D$

Akiva Weinberger

Well I don't know what $f(f)$ would be. 'Cause the definition would just give you $f(f)$. Maybe you could define it to be anything without causing contradictions. $f(a)$ would be defined to be $a(a)$ for $a\ne f$ and whatever you want for $a=f$.

GFauxPas

you'd need $S \subset D$

Akiva Weinberger

In any case, $f(g(f))$ would be $g(f(g(f)))$ for any $g:A\to A$

GFauxPas

I'm starting with relations because we dont have to worry about left total or one-to-many

so I suggest making a new relation that builds off of the old relation

Akiva Weinberger

22:43

Defining $h$ to be $f(g(f)))$, we have $h=g(h)$ for every $g$.

GFauxPas

building on $\mathcal{R} = (D, C, S)$

let's do $\mathcal R_2 = (S, S, S_2)$ where $S_2 \subset S^2$

so give me a function or relation and we can see if it works

Akiva Weinberger

Wait, I think I have my contradiction already.

So, starting over: $A$ is a set of functions whose domains include themselves

Oh, no, never mind.

GFauxPas

why never mind

What would $\sin(\sin)$ mean?

Akiva Weinberger

$\sin\notin A$.

GFauxPas

iteration?

Akiva Weinberger

22:48

Hm, maybe

Let's allow that, why not

Balarka Sen

@AkivaWeinberger If f : A --> B is a function, it's technically a subset of A x B. So you're saying, that subset of A x B is an element of A?

GFauxPas

so fine, define $\sin^{\text{WOW!}}: f \mapsto \sin(f)$

Akiva Weinberger

@BalarkaSen Yeah, that's why it can't be done. But I wonder if you really need that axiom (I forget what it's called, the axiom that there's no infinite descending chain of elements)

mick

Maybe this is a dumb question but ...math.stackexchange.com/questions/2075374/…

GFauxPas

axiom of foundation

Balarka Sen

22:51

@Akiva Axiom of regularity, yes.

No idea what happens if you ignore it

Akiva Weinberger

@BalarkaSen Alternatively, supposing you have an index set $I$ and a mapping $i:I\to A$ for your set of functions $A$

alan2here

I got a secret hat?

Akiva Weinberger

you could just have all of the functions in $A$ have domain $I$

and it would be the same thing essentially

Null

vin diesel has such a pitched down voice lol

Akiva Weinberger

if $i$ is a bijection

mick

22:52

I will also post another Nice question :)

GFauxPas

Are we bothered by sets defined by $A = \{ A \}$?

Akiva Weinberger

@BalarkaSen For example, any countable set of functions with domain the natural numbers $f_n$ can be thought of as acting on themselves.

GFauxPas

anyhow Akiva you have the composition operation which gets you what you're looking for

Akiva Weinberger

Instead of applying $f_a$ to $f_b$, you just do $f_a(b)$.

Balarka Sen

@AkivaWeinberger Right. So you want to define $f_a(f_b) = f_a(b)$?

Alessandro Codenotti

22:55

@BalarkaSen you can have stuff like $A\times A\subseteq A$, but I don't know what happens if you want some subsets of the Cartesian product as elements of A

Akiva Weinberger

@BalarkaSen Yeah, but formally that'd be an abuse of notation.

But, sure, lets go with this. Suppose $f_n$ all have domain $\Bbb N_0$. Define $f_{-1}$ on $\Bbb N\cup\{-1\}$ such that $f_{-1}(n)=f_n(n)$ for natural $n$, and $f_{-1}(-1):=17$.

Balarka Sen

@Alessandro I am pretty sure it'd break ZFC big time

Null

@AkivaWeinberger do you want f(f) or f(f(x))?

Akiva Weinberger

When?

Zach Hauk

just set up a little personal website

Balarka Sen

22:57

@AkivaWeinberger A'right.

Zach Hauk

using an old laptop from 2001 that i just but linux server on as a webserver

Null

@ZachHauk haha

Akiva Weinberger

No, actually, let me take that back. @BalarkaSen

Alessandro Codenotti

Hm, it's late so I'm not sure, but aren't there subsets of $V_\omega\times V_\omega$ which are also elements of $V_\omega$?

Akiva Weinberger

Just redefine $f_0$ to be like that.

Null

22:58

@ZachHauk if it works ;)

mick

Gonna work on my question bye

Zach Hauk

@Null since it's a giant thing, i just use SSH via ethernet to code and stuff

Akiva Weinberger

That is, replace whatever $f_0$ was before with the function $f_0(n)=f_n(n)$, and let $f_0(0)=17$ because it can be anything I want.

That way I don't have to deal with the $\Bbb N\cup\{-1\}$ stuff.

mick

Bye

Balarka Sen

@Alessandro I dunno set theory; what's V_omega?

Zach Hauk

22:59

@Null 69.113.147.32

Balarka Sen

@AkivaWeinberger Ok. (I assume your codomain is N when you write 17 but you can literally make that anything)

Akiva Weinberger

Then what's $f_0(f_g(0))$? @BalarkaSen

Zach Hauk

wrote that html and css in a few hours, no templates :P

Null

@ZachHauk cant see

Zach Hauk

mobile?

Null

23:00

no, not mobile

Zach Hauk

so umm

Akiva Weinberger

Wait, no, now I'm confused

Zach Hauk

what do you mean can't see?

does it not load?

Null

nothing loads

Zach Hauk

wait a bit

Balarka Sen

23:01

@AkivaWeinberger What you wrote is $f_{f_g(0)}(f_g(0))$, yep?

Zach Hauk

it may be my internet OR your internet being shitty

Alessandro Codenotti

It's the set of all hereditarily finite sets @Balarka (or the $\omega$-th stage in the Von Neumann hierarchy)

Akiva Weinberger

@BalarkaSen Right. I was hoping it would be $f_g(f_0(f_g(0)))$ though :(

Balarka Sen

Yikes.

@Alessandro shit man, i don't know a thing about that

Akiva Weinberger

It's the set of all hereditarily finite sets. You can write them with finitely many curly braces @BalarkaSen

It's a countable set

Alessandro Codenotti

23:04

Actually I'm pretty sure that $V_\omega\times V_\omega$ is a subset of $V_\omega$

Akiva Weinberger

Oh yeah

Huh

Balarka Sen

Ah, ok, I can parse what wiki said better now.

@Alessandro Subset, or an element?

Surely not the latter

Akiva Weinberger

Subset.

Oh, god, is $(V_\omega)^3$ even smaller?!

Balarka Sen

I thought we were looking for subsets of V x V which are elements of V

Akiva Weinberger

THAT'S SO WEIRD.

Null

23:05

technically $\mathbb{R}\times\{0\}\not\subset\mathbb{R}^2$

Akiva Weinberger

I'm tired and my neck hurts for no reason.

Alessandro Codenotti

Yeah, but finite subsets of VxV will be elements of V

Balarka Sen

god

Null

oh

actually

$\mathbb{R}\times\{0\}\subset\mathbb{R}^2$, but $\mathbb{R}^2\times\{0\}\not\subset\mathbb{R}^3$

Akiva Weinberger

Oh, yeah, 'cause $((a,b),c)\ne(a,(b,c))$ and $(a,b,c)$ is arbitrarily defined to be the latter.

Null

23:08

^yep

Akiva Weinberger

That's why inclusion maps are better than subsets

Alessandro Codenotti

Hm, that last bit about subsets of VxV being elements of V looks suspicious though (where suspicious means headache inducing), but it's to late to think about it, I'll just go to sleep instead

Null

math.stackexchange.com/questions/2075349/… strange but mmh

Balarka Sen

So V_0 = {null}, V_1 = {null, {null}}, V_2 is {null, {null}, {null, {null}}} etc?

and we take union over those to construct V_omega?

Alessandro Codenotti

Yes

Akiva Weinberger

23:15

Yup. And $V_{\omega+1}$ is the next step, the power set of $V_\omega$.

Balarka Sen

hrm, I see

Akiva Weinberger

The subscripts are ordinal numbers

Alessandro Codenotti

Yeah, you keep taking powersets for successor ordinals and unions for limit ordinals

Balarka Sen

Got it

Akiva Weinberger

I wonder if there's an easy way to see if a set $X$ is $V_\alpha$ for some $\alpha$

Balarka Sen

23:16

and by V_omega x V_omega being a subset of V_omega, you mean it's in bijection with a subset of V_omega, surely?

Akiva Weinberger

No

Literally $V_\omega\times V_\omega\subset V_\omega$.

Every element of the first is in the second.

If $a$ and $b$ are both hereditarily finite, so is $(a,b)$.

Balarka Sen

I'm confused. Elements of $V_\omega \times V_\omega$ are ordered pairs, not sets. Whereas elements of $V_\omega$ are sets.

Akiva Weinberger

$(a,b)=\{\{a\},\{a,b\}\}$, I think.

We're doing set theory. Everything is sets.

Balarka Sen

Ah, ok, so you're defining ordered pair like that.

Sorry, I'm a little unfamiliar with this territory. Yeah, so I get it.

Akiva Weinberger

Relevant

Hm. An annoying thing about having functions in their own domains is that you don't necessarily have $(a(b))(c)=a(b(c))$

Stuff isn't "associative".

In our strange encoding thing, it's saying $f_{f_a(b)}(c)$ need not be $f_a(f_b(c))$.

Alessandro Codenotti

23:24

This question confirms that VxV is a subset of V (where V is V_omega)

And also that you can't have AxA=A

Akiva Weinberger

Hm. Suppose there's this crazy function $f:A\to A$ such that $f\circ g=g\circ g$ for all $g:A\to A$ (for some set $A$). Show that there exists a function $h$ such that $g\circ h=h$ for all $g$.

Balarka Sen

Is h universal, or depends on the choice of g?

Akiva Weinberger

@BalarkaSen Wait, no, sorry.

Depends on $g$.

Balarka Sen

Hrm

I'm going to chicken out because now I am getting a headache too. Maybe I'll try to get another nap or something

Akiva Weinberger

23:42

@BalarkaSen In any case, this all has to do with Gödel's construction of a sentence that says "This sentence is unprovable". $g$ is "The statement with Gödel number $x$ is unprovable", and $f$ is "Replace all instances of the free variable in the formula with Gödel number $x$ with the number $x$."

Balarka Sen

Ah

I see now

Akiva Weinberger

@BalarkaSen So part of the proof of incompleteness involves showing that $f$ and $g$ can be written in PA, using Gödel numbers.

And the final bit is figuring out how to make a sentence refer to itself ("This statement is unprovable")

And $h$ would be the solution.

Balarka Sen

Hmm. The proof I read went through T-schemas though.

But yeah I see what you want to do

Neat idea

Akiva Weinberger

I have no idea what those are.

But presumably they're the twentieth of a series, after A-schema, B-schema, C-schema…

(Similarly, Jesus must have been God's seventh child.)

Balarka Sen

I wonder what Ted's reaction to that would have been

Akiva Weinberger

23:49

The first joke or the second

He's an atheist, I doubt he'd care.

Balarka Sen

Both

Sure, but atheists are allowed to smack

Mahmoud

I should stop getting in here to pretend I'm a questionably accurate parallel image of a kind-of-a-mathematician.

Akiva Weinberger

What?

@Mahmoud Neither Balarka nor I are mathematicians

Mahmoud

NeverMind, words have let me down.

Akiva Weinberger

We just like math, just like you.

Mahmoud

23:53

@AkivaWeinberger What ?

Akiva Weinberger

Never mind.

Mahmoud

No, if you are not, who is ?

Akiva Weinberger

Maybe Ted?

Mahmoud

Definitely,

Eric

lol^

Mahmoud

23:54

If he isn't, I'm leaving this.

Akiva Weinberger

In any case, try not to self-deprecate.

@Mahmoud He wrote several textbooks and taught at a university so that makes him a mathematician

I shouldn't have said "maybe"

Mahmoud

Okay, that is much better.

Balarka Sen

So I have this single page proof of Godel's incompleteness TeXed up in my folder, which I presumably wrote a year and a half ago when I causally went through a proof of it... I have no idea what all this means now

"Every math you've learnt, you've forgotten?"

Mahmoud

Mathematicians can't define themselves.

Or can they ?

Akiva Weinberger

Mathematicians are people who do math; math is what mathematicians do

maybe

Balarka Sen

23:58

I was gonna say something like that

Mahmoud

And the rest are : Math-enthusiasts.

I think there must be another class,

Akiva Weinberger

Maybe only the second part of that definition works, then.

Math. Noun. What mathematicians do.

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$Mathematics$ Mathematics