@Obliv Let's see, what exactly did I say?

I think you said $f(A) = \{f(a) | a \in A\}$

The image is $f(A)=\{f(a)\mid a\in A\}$

oh

The preimage is $f^{-1}(B)=\{a\mid b=f(a), b\in B\}$

I suppose that's right

isn't my definition correct too?

Uh

I don't get what you said

That's not correct set builder notation

$f(a)$ is not a condition

Read it aloud:

Oh wait I think it would be $f(A) = \{ a \in B | a = f(a)\}$

All elements $a$ in $A$ such that $f(a)$

yeah I realized that

What?

That's the definition of the set of fixed points now

a member of B?

$a=f(a)$ means $a$ is a fixed point

isn't the transformed set a subset of B?

so the members of it should be in B, right?

Yes, but $a=f(a)$ is wrong in general

There are many ways to write things in this notation btw

yeah I noticed

I read it as "such that a is equal to f(a)" so

@Obliv did you mean to write $\{b\in B\mid b=f(a)\}$?

for all a that is

@Obliv uhhh, do you not see the issue with $a=f(a)$?

what is different from $b = f(a)$ and $a = f(a)$

Because $a$ appears twice in the second?!?!?!

The second one really doesn't make sense.

You're literally asking what the difference between $f(x)=y$ and $f(x)=x$ is!

Unless you're talking very specifically about the identity function.

members of $B$ are equal to the members of $a$ transformed by $f$. members of $A$ are equal to transformed members of $a$ through function $f$.

is that how one should read both

Right, so f(a) is a member of set B

f transforms A to B

What you're saying is that $a$ in $A$ gets sent to the SAME thing $a$ in $B$

oh..

I see what you mean

LOL

i'm saying a = b basically

what is $b$???

members of B

oh wait no im not

sweats

Dude what is confusing you so much

i'm saying a is equal to the transformed elements in B. please don't sweat LOL

No book is this terrible

I GET IT

ok

@Obliv So the image is $f(A):=\{f(a)\mid a\in A\}$

yeah

I forgot how to use predicate notation. you were right :p

So anyway @0celo7

For now, no more GR

I'm going all QFT

Predicate notation?

@Slereah WHAT

the $|$

Gonna try to do all of Peskin Schroeder

@Obliv No one actually calls it that.

what do u call it then

I really wanna do the Thesis

But GR thesis seems a bit unlikely

The fucking line in the set notation thingie?

So I wanna buff up on my QFT

Why would you need a name for it?

Set builder notation, @0celo7

"such that"

because there's a difference between listing a set's elements out the long way and using a notation where you can specify a condition for the elements to have to meet to be an element of the set

@Obliv you just read it as "such that"

YEAH I KNOW THAT

some people use $:$ for it

$:$ means equivalent, no?

What

In what notation

Usually $\sim$ means equivalent

You're ultimately invoking the Axiom Schema of Reduction.

That is just the Standard Set Builder Notation

Jeez

If S is a set, the elements of S satisfying a condition C also form a set.

@Obliv why are you making this so hard

@0celo7 no you're right : is equal to $|$ I must have looked in the wrong area

⊢ (𝑥 ∈ {𝑦 ∣ 𝜑} ↔ [𝑥 / 𝑦]𝜑)

if u want an equivalent notation

@0celo7 Patience is a requirement for pedagogy.

@Slereah stop

@Slereah I applaud that notation.

@Slereah Now do: "there exists a statement in ZF that is neither provable nor is its negation provable"

Let me get my Godel book

@Slereah Well that's too easy.

Technically, I should add "OR there exists a statement in ZF that is provable and its negation is also provable"

what's the inverse image of an element though?

$f^{-1}(x)$

the book calls it a fiber

It is a fiber if it's the projector of a bundle, yes

okay I see.

you don't get set notation but are reading about fiber bundles oO

this is just in the preliminaries&basics part of the book

@obliv what book are you reading?

dummit & foote "abstract algebra"

Godel writes it as $(x,y,z)[\{\tilde{} x \mathfrak{B} y \& z \mathfrak{B} \text{Neg}y\}]\rightarrow (x) \tilde{} x \mathfrak{B}\gamma(p,p)$

Yeah, that did seem like a bit of a jump

@0celo7 Sayeth the guy who can be posting differential geometry proofs one hour and whining about PDEs the next.

lol

@dmckee Whining != don't understand

Whining is whining

@Slereah literally satan

Not ZFC, tho

@Slereah Don't kill yourself :)

It's old logic notation from Russell

$(x)$ is $\forall x$

@Slereah For all x, y, z... what is cursive B?

@Obliv let's get a legal copy

Not CURSIVE

third edition?

It's FRAKTUR

@0celo7 yes.

You peasant

Dummit calls the inverse image of an element a fiber

what page

15

@Slereah It worries me that I don't know these things.

@bolbteppa that's so weird

I think

@Obliv did you already do calc 1-3?

It certainly ever appears in other places

Maybe it's an algebra thing?

calc 1-3 = calc -2

no i still have to take 3

It's actually really clever, tells you there's more to fiber bundles than you think

LOL barry

$x \mathfrak{B} y$ means that x is a proof of y

Yay, someone got it.

Where x is a proof and y is a formula

@Obliv you should probably do that first.

In fact right there is the secret to abstract algebra, but I'l scribble that in a margin to small to contain it

@bolbteppa what?

@Slereah You realize I was joking when I asked you to write that, right?

Well I did, whatcha gonna do about it :V

@3075 my motivation for studying abstract algebra is that I want to know the architecture of algebra so that linear algebra comes more intuitively.

@Obliv normally you learn linear algebra first.

you're a physicist right? so you're wasting your time I think.

@3075 I'm a physics student. nothing is for sure. I like math a lot too and I have downtime so I don't think it's a waste of time.

@Obliv what linear algebra books you having trouble with?/

@bolbteppa I was studying axler's "linear algebra done right" and found that linear algebra is a study of like 2 algebraic structures and I want to learn all of the structures before I delve into the specifics of each one. thats my reasoning anyway

@Obliv no I mean you could be learning linear algebra, calc 3 in your free time instead, which will allow you to learn more advanced physics.

@Obliv You're better than physics. Join the Math side.

wow.

@barrycarter What did you say? ::gets pitchfork::

@ACuriousMind I meant... uh, physics sure is great for physicists!

@barrycarter Note to self: keep peasants happy

@DanielSank You probably know John Martinis then, right? I'm gonna go listen to his talk today at Cal!

Probably a simple question, but I couldn't find it/the right search term. Say you have two qubits, a and b, with lifetimes and decoherence times T1a,b and T2a,b. Now say that these two qubits (or any two level system) are in a symmetric superposition. How does one then determine the decoherence time (and life time) of the symmetric superposition? I imagine it might be the lowest of the two constituent qubits, but it would be nice to quantify

Isn't a function injective by definition?

No

@Obliv No.

Let $f(x)=x^2$.

for the map $f: A \to B $ $ f$ is injective if whenever $a_1 \ne a_2$, $f(a_1) \ne f(a_2)$

this is the definition right?

@Obliv Yes. It means two different values of a can't map to the same value of b.

That's sort of the "foil" of the definition of function: one value of a can't map to two values of b.

isn't the definition of a function: a relation where for each of its domain's, there exists only 1 co-domain. or something along those lines?

@Obliv Function: for every one a, there is one b. Injective: for every one b, there is at most one a.

It's like the difference between polyandry and polygamy.

oh i was reading that wrong I see now

thx @all

@Anthony Great. He was talking about that yesterday in the office.

Don't be afraid to ask questions.

o/

@Obliv I think that's the wrong reasoning.

Would you really like to learn about sheaves and topoi before mastering linear algebra?

@Obliv Answer carefully. You must choose between the paths of righteousness and physics.

@Obliv Think about a particle moving on a circle, round and round. Define $f:\Bbb R\to \Bbb R^3$ by $f(t)=x(t)$ where $x(t)$ is the position of the particle.

This is not injective if the particle goes around the circle at least once, or if it stands still at any point

Oh, I see your reasoning was already corrected. Feel free to ignore my example.

@Danu That's the simplest example you could come up with?

and it wouldn't be a function if the particle was in two positions at some time t right? @danu

No i think that's a great example honestly.

@barrycarter It's physical.

@Obliv It wouldn't even be a particle if it was in two positions at the same time.

@Obliv Yeah, then it'd be ill-defined (not single-valued).

@Obliv Thanks.

Physical examples help because you intuitively feel that it makes sense, more so than with abstract functions :P

Perhaps he is a physicist after all.

Of course, a much more simple example is $f:\Bbb R \to \Bbb R; f(t)=0,\ \forall t\in\Bbb R$.

Yes, or f(x) = x^2 someone noted earlier.

{0,1} -> {0}

That's probably the simplest possible example.

Interesting little exercise: $f:X\to Y$ is injective if and only if $f^{-1}\circ f=\operatorname{id}_X$

Isn't that pretty much the definition

Similar flavor, $f$ is surjective iff $f\circ f^{-1}=\operatorname{id}_Y$

@Danu Lies! That doesn't work if f isn't surjective, since f^-1 wouldn't be defined.

@Slereah Na

@barrycarter You can always take preimages.

A function must be defined on all elements of its domain.

You don't need an inverse function for that.

$f^{-1}$ is defined even if it has no inverse

Then it's a different function!

It just happens to be a set

Not a function

$f^{-1}(y)$ is a subset of $X$, and so is more generally $f^{-1}(K)$ for any $K\subset Y$

Oh geez, so f^-1 could be the empty set?

@barrycarter $f^{-1}(y)$ is empty iff $f$ does not "reach" $y$.

I'm going to be whiny and say f^-1 to me means the function that is the inverse of f.

Well then you're going to have a bad time reading any elementary math book.

If you want to be very precise and pedantic, you can take care to always "apply to a set" when indicating preimages, e.g. writing $f\circ f^{-1}(U)=U$ for every $U\subset Y$ instead of what I wrote.

Then, no confusion can arise.

I have to prove that $f:A \to B$ is bijective if there exists a map $g: B \to A$ where $f \circ g $ maps $B \to B$ and $g \circ f$ maps $A \to A$. gl me lol.

@Danu Er, I was using the math definition. f^-1 doesn't even exist if f is not both surjective and injective.

@barrycarter "the math definition"

Yes.

I'm just going to tell you that $f\circ f^{-1}=\operatorname{id}_Y$ is a perfectly normal thing that you'll see in many math books.

@Danu as opposed to the "dirty physicist" definition

If you choose to shut your ears, eyes and yell LALALA that's wrong then go ahead.

@Danu: I have never seen someone write $f^{-1}\circ f = \mathrm{id}$ where $f^{-1}$ is not a function.

@Danu Provided that f is surjective and injective, sure.

@ACuriousMind I can show you an example :P

@Danu In a MATH book?

Sigh, I have to admit I don't have the example I saw most recently at hand---it was in a problem set for a course one of my tutees was taking.

Suuuuure

@Danu Is it preceded by some sort of explicit statement that $f,f^{-1}$ are considered as functions on the powerset instead of functions on the spaces themselves?

@0celo7 Hahaha :D

@ACuriousMind I don't remember.

Because I'd say it's fine if one says that, but without that, that's confusing notation.

I don't think I goofed, and as I indicated before, I'm willing to concede that more precise notation would be

$f\circ f^{-1}(U)=U$

Which is definitely 100% correct, and I think the confusion is minimal in any case.

@Danu Suuuuure

@0celo7 Suuuuuure what?

@Danu Whoa calm down

@0celo7 I promise you I'm writing this with a smile on my face :)

But Suuuuure what?

@Danu What does that have to do with anything

I'm sure e.g. Bundy was smiling when he tortured and killed his victims!

Smiling means nothing!

If anything, it means you're a killer!

He was probably calm.

It means that I'm calm ;)

Brah, calm down

@0celo7 Meh, you're just being an ass. That's okay @0celo7, I can understand it ;)

Oh really?

Is it because I'm bored?

Perhaps your mother didn't hold you enough.

lol

@0celo7 If I'm re-learning gr what book should I read?

I'm reading your gr book answer.

all of them there, probably

but I'm not interested in gr.

this is only to be well rounded.

oh right.

the lecture notes I had.

I'll read those.

@Danu No really, I'm interested in your analysis

Did my parents not love me enough

?

hey isn't $\sqrt{x}$ the inverse of $x^2$ making $x^2$ injective?

No

@0celo7 You don't have any parents. Some non-orientable surface just spat you out at some point

@Obliv No, sqrt x means the positive square root

@Danu What the fuck

:D

sqrt(x^2) is abs(x)

What'd you expect me to tell you?

@Danu whoah lol

@Danu you've been reading too much Riemann surface stuff

@0celo7 If it annoys/offends you then I'll delete it

@Danu It triggers me

...in what way?

Best putdown ever!

OK, enough weird shit for today

xD

that was way weirder than it was meant to be

Bye.

@Danu <3

@barrycarter oh and $\pm \sqrt{x}$ isn't a function so $x^2$ doesn't have an inverse. I gotcha

@Obliv Correct. At least, it's not a function from R to R. Some of these "physicists" would argue it's a function from R to R^2.

i don't want to know why that is

2complicated4me

@Obliv I'd ignore him.

He doesn't understand basic physics and comes here and insults physicists?

@0celo7 On the offchance you're not blocking me and are actually offended, I should point out that I'm pretty much just joking.

@barrycarter I've unblocked you, and am not offended. Doesn't mean I want to read shit, y'know?

@0celo7 It's a friendly rivalry between physicists and mathematicians. It's the subject of a great deal of humor.

@barrycarter is your enter key beside the 0 key or c key?

lol

@3075 Yeah, I know, two times in a row.

xD

Oh wow, when I type, "@0" and then tab, I sometimes just tab through items on this page.

tpo.net/humor/Ultra/Canon/scijokes.html

@barrycarter What kind of mathematician are you, anyway?

@ACuriousMind The best kind.

I meant: Algebraist, analyst, graph theorist,...?

@ACuriousMind Yes, I know, Abstract Algebra.

@ACuriousMind seriously?

Graph theory is before topology and geometry in your esteem list?

What? That's not my "esteem list"

Just three subfields that popped into my head

The current division is Abstract Algebra, Real Analysis and Complex Analysis.

Those are the three "main" "super fields"

I should be able to even source that.

@barrycarter And the logicians and topologists wept.

@ACuriousMind That's all under Abstract Algebra.

Well, pure topology anyway.

Once you start metricizing, maybe not.

mathematics.stanford.edu/academics/graduate/phd-program/… ... well, there were three when I took them.

math.stanford.edu/~white/quals/qualinfo.htm <- historical evidence that there were 3 tests required at one point