@KajHansen you where just joking or?

About what @Null >

?

impostering

@Kaj nice hat lol

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

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

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

You're missing the nullset.

wikipedia page is using the dumb spelling

How can "power" have an incorect spelling?

w/ an "o" ?

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

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

@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 :)

forgot the empty set

but otherwise good

the empty set is a subset of every set

ohh, TY, true :)

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

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

@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

heh.

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

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

@AkivaWeinberger how you mean?

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 $f(f)$

i have to pass here ;)

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 $\forall a\in A,f(a)=a(a)$

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

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 $R$

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

that's model theory Null, beyond my grasp

so for $R$ to be in $D$

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 $a=f$.

you'd need $S \subset D$

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

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

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

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

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.

why never mind

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

$\sin\notin A$.

iteration?

Hm, maybe

Let's allow that, why not

@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?

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

@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)

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

axiom of foundation

@Akiva Axiom of regularity, yes.

No idea what happens if you ignore it

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

I got a secret hat?

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

and it would be the same thing essentially

vin diesel has such a pitched down voice lol

if $i$ is a bijection

I will also post another Nice question :)

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

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

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

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

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

@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

@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 $f_{-1}(-1):=17$.

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

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

When?

just set up a little personal website

@AkivaWeinberger A'right.

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

@ZachHauk haha

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

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 $V_\omega$?

Just redefine $f_0$ to be like that.

@ZachHauk if it works ;)

Gonna work on my question bye

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

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.

Bye

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

@Null 69.113.147.32

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

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

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

@ZachHauk cant see

mobile?

no, not mobile

so umm

Wait, no, now I'm confused

what do you mean can't see?

does it not load?

nothing loads

wait a bit

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

it may be my internet OR your internet being shitty

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

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

Yikes.

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

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

It's a countable set

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

Oh yeah

Huh

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

@Alessandro Subset, or an element?

Surely not the latter

Subset.

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

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

THAT'S SO WEIRD.

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

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

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

god

oh

actually

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

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

^yep

That's why inclusion maps are better than subsets

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

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

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?

Yes

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

hrm, I see

The subscripts are ordinal numbers

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

Got it

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

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?

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 $(a,b)$.

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

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

We're doing set theory. Everything is sets.

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

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

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 $f_a(f_b(c))$.

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

And also that you can't have AxA=A

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 $g$.

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

@BalarkaSen Wait, no, sorry.

Depends on $g$.

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

Ah

I see now

@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.

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

But yeah I see what you want to do

Neat idea

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.)

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

The first joke or the second

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

Both

Sure, but atheists are allowed to smack

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

What?

@Mahmoud Neither Balarka nor I are mathematicians

NeverMind, words have let me down.

We just like math, just like you.

@AkivaWeinberger What ?

Never mind.

No, if you are not, who is ?

Maybe Ted?

Definitely,

lol^

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

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"

Okay, that is much better.

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?"

Mathematicians can't define themselves.

Or can they ?

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

maybe

I was gonna say something like that

And the rest are : Math-enthusiasts.

I think there must be another class,

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

Math. Noun. What mathematicians do.