Mathematics - 2012-07-04 (page 1 of 4)

Peter Tamaroff

00:00

@MarianoSuárezAlvarez Hola/Hello

=P

@BenjaLim Are you around?

I'm dealing with $Y \subset B$ then $ Y \supset f(f^{-1}(Y))$ now.

I think words might be better to prove both.

Let $f: A \to B$, $X \subset A$. We have that $f(X) = \{f(x) \in B : x \in X\}$.

Since $f$ is not assumed to be one-one, then $f^{-1}(f(X))$ is composed by precisely those $x \in X$, and other $x \in A$ but $\notin X$ such that $f(x) =y$, so that at least one element of $f^{-1}(f(X))$ is not in $X$, so that $X \subset f^{-1}(f(X))$

@JasperLoy Am I making sense there?

@anon ?

user19161

@PeterTamaroff I did not follow just now, but why do you say that f is not injective?

Peter Tamaroff

@JasperLoy I'm not saing it is not. I'm saying it can be, or not.

Actually, I should be writing $X \subseteq f^{-1}(f(X))$

user19161

@PeterTamaroff Oh, then you don't have to mention it at all.

Peter Tamaroff

@JasperLoy Well, I guess, but my point is the preimage operation adds points that weren't there.

Is the proof viable?¿

@JasperLoy Listen to this, bro

►

user19161

00:19

@PeterTamaroff Your sentence here is difficult to parse. First you have not defined y. And you can't be sure there is "at least one element" there.

user19161

@PeterTamaroff I think you get the idea but cna't express yourself properly.

Peter Tamaroff

@JasperLoy Let me tweak it.

Jonas Teuwen

"It is not even false!"

Peter Tamaroff

Assume $f$ is not one-one, then $f^{-1}(f(X))$ is composed by precisely those $x∈X$, and other $x∈A$ but $∉X$ such that $f(x)=y$ for some $y \in B$, so that there is at least one element of$f^{-1}(f(X))$ is not in $X$, so that $X⊂f^{-1}(f(X))$

@JonasTeuwen ?¿

Jonas Teuwen

@PeterTamaroff What is that for gibberish? If it is not one-to-one we have distinct $x, y$ such that $f(x) = f(y)$. Done.

Peter Tamaroff

00:25

@JonasTeuwen Just remember I'm learning.

Jonas Teuwen

Yes, I know. Don't see it as a statement with any emotional value.

I am writing down how I see it.

As I want you to know that you should think about what it means, not just about the literal definition.

And this is what it means.

Your statement is so full of symbols, but I think it is correct. I wonder if you will have an easy time manipulating it.

Peter Tamaroff

@JonasTeuwen OK. Yes, I understand what it means. The preimage operations "adds" extra elements.

Jonas Teuwen

Remember, the definitions are usually just to encode some reasonable simple mathematical concept rigorously.

Which can look quite scary but means nothing.

Peter Tamaroff

Whilst if $f$ is one-one the inverse image returns to the original set $X$.

Jonas Teuwen

Yes, but that is still "gibberish".

It means... That if you have two distinct points, they will not get mapped to the same element.

Remember, I am using implicitly "for all distinct elements".

So that statement for something not being injective is that there just exists two distinct elements that map to the same element.

Peter Tamaroff

00:29

@JonasTeuwen How would you say it? I'm talking about a subset $X$ of $A$, its image $f(X)$ and the preimage of this. I'm saying that if $f$ is not $1-1$ the preimage of the image adds elements that weren't in $X$ originally.

How is that gibberish?

Jonas Teuwen

The existence of those two elements refute the injectivity as this is for all.

user19161

@JonasTeuwen Actually his statement is still wrong.

Jonas Teuwen

@JasperLoy Perhaps, I did not completely read it because it is way too complicated for something so simple.

@PeterTamaroff It is gibberish because you use jargon where it not necessary.

Peter Tamaroff

@JonasTeuwen "Communication using various noises, illegitimate words, nonsense or sounds. "

Jonas Teuwen

Yes, it is an exaggeration.

A hyperbole, so to speak :-).

Peter Tamaroff

00:31

@JonasTeuwen How would you say it? I give up.

user19161

OK @peter I am confused now. You are trying to prove that the preimage of the image possibly adds extra stuff right?

Jonas Teuwen

@PeterTamaroff I have already done so.

Okay, right so you have $f: X \to f[X]$ with some slight abuse of notation (perhaps it can be made okay).

Peter Tamaroff

@JasperLoy I'm trying to prove that $X \subset f^{-1}(f(X))$ for any $f:A\to B$, $X \subset A$.

Jonas Teuwen

So, if we have a point $y$ in $f[X]$ this means that there is an $x$ such that $f(x) = y$.

Peter Tamaroff

@JonasTeuwen Yes.

Jonas Teuwen

00:34

And that $x \in X$. Perhaps there are multiple ones. If we have injectivity we don't.

Peter Tamaroff

@JonasTeuwen Right,

That is what I'm saying.

Go on, please.

Jonas Teuwen

So, I didn't think about any definition, just what it means.

So, what do you want to prove?

Peter Tamaroff

@JonasTeuwen See what I wrote to Jasper there.

Jonas Teuwen

Okay.

So $f[X]$ are all the points mapped from $X$ to the codomain.

user19161

@PeterTamaroff OK. This is what I will write. Let $x\in X$. Then $f(x)\in f(X), so that x\in f^{-1}(f(X))$. Done.

Jonas Teuwen

00:35

Excellent, so basically we don't care about surjectivity so we can redefine our function so that it is: $f: X \to f[X]$.

Now, we know that if we take those points back from the "range" (by surjectivity) we should get the whole of $X$ if $f$ is injective.

Right? As the preimage would be unique (injectivity).

So, what if it isn't unique? So, the function is not injective?

That means that it is possible that the preimage actually maps back to several elements.

Peter Tamaroff

@JonasTeuwen Then more elements are added.

@JasperLoy Why does the last follow?

Jonas Teuwen

Yes, but in $f[X]$ there are only elements that come from $X$ itself, but if the function is not injective it is actually possible that there are also points outside.

But if $X$ is the complete domain then of course not.

(deleted because it was not even wrong what I said 8-))

user19161

@PeterTamaroff From the definition of the preimage.

Jonas Teuwen

@PeterTamaroff My point is: think intuitively, not just manipulate symbols. Or just try to, I know it is hard for new things.

Peter Tamaroff

@JonasTeuwen I knew why it was true. I couldn't explain it well it seems , I don't know.

Jonas Teuwen

00:40

Note that I made too many steps as to restrict the codomain.

This is quite common: try to make it easier by removing irrelevant things.

user19161

@PeterTamaroff Could it be because of your English as well?

Jonas Teuwen

@PeterTamaroff Yes, new concept. It is normal.

Peter Tamaroff

@JasperLoy I don't think so. My english is OK.

Jonas Teuwen

But don't fall into the "rigorousity trap".

user19161

@PeterTamaroff You just need to read more then. For a first year student you are doing fine.

Jonas Teuwen

00:41

Mathematics is not about being rigorous.

Peter Tamaroff

@JasperLoy I'm not in first year yet.

user19161

@JonasTeuwen This should be starred.

Jonas Teuwen

(But you should strive for a good level of rigor in the result).

@JasperLoy Go ahead 8-).

Peter Tamaroff

@JonasTeuwen Yes, you're right.

user19161

I think reading good mathematical exposition helps @pet.

Jonas Teuwen

00:43

Good :-). So this means, just mess with things, don't worry too much about the definitions as long as what you write down does have these properties!

Peter Tamaroff

@JasperLoy The thing is now I have to prove that if $\{ X_\alpha \}_{\alpha \in I}$ is an indexed family of subsets of $A$, analogous relations hold.

So I guess I could use some symbols, maybe? Or just explain it with words. I'm starting to like words more.

user19161

@PeterTamaroff Whether you use words or symbols, the important thing is to write it correctly. What you initially wrote, I can point out various mistakes in it.

Jonas Teuwen

@PeterTamaroff Pretty much the same, just more messy.

(Try if you can apply the thing you just proved! Don't immediately try to copy the argument)

Peter Tamaroff

@JasperLoy Please do.

user19161

Paul Halmos is a master of using words.

Peter Tamaroff

00:46

@JasperLoy I'm reading Bert Mendelson's book. It is good!

@JasperLoy Could you tell me about the mistakes?

Jonas Teuwen

I give away the information it took me years to collect (concerning good practices) 8-)).

Peter Tamaroff

@JonasTeuwen That is the essence of investigation, isn't it?

Jonas Teuwen

Yes, but nobody told me these things I've told you. Or maybe they did and I didn't listen. I wish I did!

Peter Tamaroff

@JonasTeuwen I will remember them, then!

Jonas Teuwen

I didn't try to give you the argument "for free", but I hope you get the idea how I got it?

Peter Tamaroff

00:49

@JonasTeuwen Yeah, I do. I'm good at overly complicating things, however.

Jonas Teuwen

Not by manipulating the definitions but by manipulating what they mean.

Yes, but that is common, I mean, you don't yet fully grasp the definition so the only thing you can do is manipulate it.

user19161

@PeterTamaroff One mistake here is that you should use $y\in f(X)$ and not $y\in B$. Another mistake is that even if $f$ were not injective as you assumed, the choice of $X$ may not give you the at least one extra element you hope to find. Anyway, that proof is convoluted and now you know what to write as I have shown.

Peter Tamaroff

@JasperLoy And a similar short proof would work for $Y \supset f(f^{-1}(Y))$ right?

user19161

@JonasTeuwen Yes, in the same way the reflection of the moon in the water is not the moon itself. Symbols are just a representation of the actual idea.

Jonas Teuwen

Yes, it is like a No Shit Sherlock remark, but it has to be made...

Captain Obvious, or whatever.

Peter Tamaroff

00:53

@JonasTeuwen lol

@JasperLoy I guess I should learn how to produce proofs of inclusion in general.

user19161

@PeterTamaroff It should be short too as well. You can try it.

Peter Tamaroff

@JasperLoy For example, inclusion proofs usually start with "pick an $x$ in $X$",

and they prove a statement about that element in particular

user19161

@PeterTamaroff Right. To prove that a set is a subset of another, we prove that any element of the first is an element of the second.

Peter Tamaroff

@JasperLoy Listen to this song youtube.com/watch?v=bWTuKd2lTo4

I think you'll like it. I think I linked to the full album before, but I like that song in particular.

@JasperLoy This is so trivial I find no sense in proving it.

user19161

@PeterTamaroff Do you listen to music when you are doing math? For me, I don't unless it is say background music in a cafe.

Peter Tamaroff

01:07

@JasperLoy I usually do. It helps me forget about my surroundings.

user19161

Also there are some things that seem trivial but are not trivial or true at all. For example, if f is continuous on A and continuous on B, it need not be continuous on the union of A and B @peter.

Peter Tamaroff

@JasperLoy Because cusps could appear?

user19161

@PeterTamaroff Actually I omitted some detail there. Just think of the function with a point of discontinuity on an interval. Then consider the restriction of the function on each of the two sides of the interval. The restrictions are continuous in this case.

Peter Tamaroff

@JasperLoy And what union are you considering?

user19161

@PeterTamaroff Union of the two sides of the interval to make the whole interval.

Peter Tamaroff

01:17

@JasperLoy You're not being coherent. Say the function is defind on $(a,b)$ and there is a $ c \in (a,b)$ where $f$ is discontinuous.

Then $f$ is certainly continous on $(a,c)$ and $(c,b)$, and in $(a,c)\cup (c,b)$ too.

Dylan Moreland

Where are people learning to write generators of ideals like $\langle x, y\rangle$ instead of $(x, y)$?

Peter Tamaroff

@DylanMoreland I write spanning sets (systems of generators) as $<x_1,\dots,x_n>$

Same for a basis.

anon

well, wait, that still doesn't make it sensible

user19161

@PeterTamaroff Sorry I mean for instance, if we define $f(x)=0$ if $x<0$ and $g(x)=1$ if $x\geq 0$, then f and g are continous on their domains. But then if we take the value of $h(x)$ to be $f(x)$ for $x<0$ and $g(x)$ for $x\geq 0$, then h is not continuous on $R$.

Peter Tamaroff

@JasperLoy For $Y \supset f(f^{-1}(Y))$ I should start with "Let $y \in f(f^{-1}(Y))$" right?

@JasperLoy Oh, sure.

user19161

01:25

@PeterTamaroff That's the general principle but sometimes general principles fail. You may try and see if you can continue.

Peter Tamaroff

@JasperLoy I'm letting myself down today. Fuck me.

user19161

@PeterTamaroff Then never mind, go to sleep first.

Peter Tamaroff

@JasperLoy Maybe I should.

user19161

@PeterTamaroff Yes, good night!

Peter Tamaroff

But I'm a little upset this is troubling me.

user19161

01:29

@PeterTamaroff OK let me write it out for you now.

user19161

Let $y\in f(f^{-1}(Y))$. Then $y=f(x)$ for some $x\in f^{-1}(Y)$, so that $y\in Y$. Done.

user19161

Sometimes while trying to prove these silly things, one gets confused.

Peter Tamaroff

@JasperLoy "...so that $y \in Y$" follows from the fact that $y=f(x)$ and $x \in f^{-1}(Y)$ right?

user19161

@PeterTamaroff Yes, again using the definition of the preimage.

user19161

The proof looks quite silly actually!

Peter Tamaroff

01:37

@JasperLoy Yes I know!!!!!!!!!!!!!!

user19161

@PeterTamaroff OK now you may go to sleep!

Peter Tamaroff

@JasperLoy It is not even midnight,

@JasperLoy I have a lot of work to do. =D

user19161

@PeterTamaroff OK then off you go to do your stuff.

Peter Tamaroff

@JasperLoy I'm on it. I come and ask for help sometimes.

As @anon made me see some days ago, I have to work on my logic, and set theory.

user19161

@PeterTamaroff I keep thinking pronouncing his name like an Indian name: Ah-non. Haha!

Peter Tamaroff

01:41

@JasperLoy I'm was convinced "Anon" was a name. It is, actually.

Do you like your name?

user19161

There are names Anand and Ananda as well.

user19161

@PeterTamaroff Yes, I do.

user19161

@PeterTamaroff Do you like yours?

Jonas Teuwen

Good night guys.

user19161

@JonasTeuwen Good night bro.

Peter Tamaroff

01:51

@JasperLoy I'm indifferent. I think any name I'd get I'd find one I like better.

Though my middle name is cool in Russian "Nikolai"

user19161

Oh you look very cute in that photo @peter!

Peter Tamaroff

@JasperLoy In what photo?

(Oh that one?)

The Jimmi Hendrix one?

user19161

@PeterTamaroff Yes, that's the only one I saw...

Peter Tamaroff

@JasperLoy Hahaha but I thought you had seen it already!

user19161

@PeterTamaroff Yes but I had not remarked on it, and I also love to make random remarks sometimes.

Peter Tamaroff

01:56

@JasperLoy Hahaha, well, thanks.

It is funny you thought the sculpture was real.

Though that one is particularily well achieved.

And the pic is not very good.

Peter Tamaroff

02:09

Ok, now I have to prove that if $f$ is one one then $X=f^{-1}(f(X))$ which means I have to prove inclusion both ways,

I have one already.

So I need to show that $X \supset f^{-1}(f(X))$

anon

How about this. Suppose there is an $x\in f^{-1}(f(X))$ that is $\not\in X$. Derive a contradiction.

Peter Tamaroff

@anon Then $f(x) = f(y)$ but $ x \neq y$?

anon

mmhmm

Peter Tamaroff

(for some $x,y$ in there)

@anon I understand that is the case, but I can't prove it.

anon

for some $y$. $x$ is given.

Peter Tamaroff

02:15

That is what is going on today with me.

anon

NEIGHBORHOOD DOGS. STOP BARKING AT THE FIREWORKS.

brb

Peter Tamaroff

@anon Don't do anything crazy!!!

Eugene

@anon i'm not sure they participate in MSE.chat

@PeterTamaroff i see you're learning set theory

Peter Tamaroff

@Eugene Well, not really. I'm studying from Rudin, and using an introductory book of topology as a complement, and before relations come functions.

And I found myself unable to prove some seemingly trivial stuff.

Which was a little upsetting.

anon

lol, book of topology as a complement

topology without tears is nice

(nice and free)

Peter Tamaroff

02:31

@anon It is introductory. I'm using it just for metric spaces, open sets, closed sets, limit points, and stuff.

anon

I was loling at the pun.

Eugene

@PeterTamaroff yes this is set theory.

anon

man, weird stuff happens without choice

Eugene

@PeterTamaroff trivial stuff is always harder to prove

Peter Tamaroff

@anon Hahhaha now I follow you.

Though Rudin uses $A^c$ and Mendelson uses $C(A)$

Dylan Moreland

02:33

@anon It's a rough week for dogs. I don't really blame them for being alarmed by loud explosions.

Peter Tamaroff

@DylanMoreland Just sedate them for a while.

Eugene

@anon weird stuff happens with choice too banach-tarski cough cough

Dylan Moreland

@PeterTamaroff The book I taught out of last term used a prime $'$ for complements. It was horrible.

anon

yes that's true

user19161

@Eugene You are so funny sir.

Peter Tamaroff

02:34

@DylanMoreland The horror.

@JasperLoy PLEASE, DON'T.

Dylan Moreland

@PeterTamaroff Well, everything's relative.

Eugene

i guess without choice we would lose a lot though

Peter Tamaroff

@DylanMoreland Though it is cool to use $A^c$ since it gives out that the complement operation is idempotent.

user19161

@DylanMoreland Isn't that fairly standard?

Eugene

i like my maximal ideals.

Dylan Moreland

02:36

@JasperLoy The first I'd seen of it.

anon

are quotients of number fields' rings of integers by ideals supposed to be cyclic?

Eugene

@JasperLoy definitely not

Dylan Moreland

@PeterTamaroff Does it? How so?

user19161

@Eugene Oh I guess that's because I have read over 9000 books.

Peter Tamaroff

:5219366$ (A^c)^c=A$?

anon

02:37

@PeterTamaroff you could say the same for ' notation: $A''=A$

Eugene

@JasperLoy ?

Dylan Moreland

@PeterTamaroff I guess I don't see how the notation helps you remember that. You are, of course, correct.

Mark Dominus

$(A^c)^c$ ought to be $A^{2c}$.

anon

@Eugene over 9000

Mark Dominus

So you just have to remember that $2c = 0$.

Peter Tamaroff

02:38

@MarkDominus I makes no sense to write that.

Dylan Moreland

@JasperLoy Showing off your power level. Damn.

Eugene

@anon this meme again...

Dylan Moreland

@anon Not always, I don't think.

anon

wait, you know about dbz, surely you know this meme

user19161

@anon I learnt the meme from the ELU room a few months ago.

Eugene

02:39

@anon yes i do. i just didn't connect the two. plus he wasn't crushing a scouter when he said it so it's harder to tell.

Peter Tamaroff

@anon English is the only language in which they said 9000

In Spanish they said 8000

user19161

@PeterTamaroff That is so weird.

anon

I know. I can't stand the Japanese dub, though. In the English one they all have fittingly gravelly, hypermasculine voices :P

Eugene

@anon i can't stand the cartoon. it takes like 5 episodes to charge up...

and all the different angles they have to show it from...

anon

took many an episode for namek to blow up in 5 minutes like freiza said it would..

Eugene

02:41

@anon lol! i remember that!!

anon

it came out in the US when I was a kid so nothing really bothered me. 'cept maybe aliens all speaking English.

Eugene

i think i'm starting to like my elliptic curves to be over $\Bbb Q$ now...

@anon well they were bilingual...

i'm working to learn what an elliptic curve looks like over a ring and it sure don't look pretty

@anon piccolo still spoke namek once in awhile

user19161

@Eugene Hahaha, an elliptic curve looks like a curve and a ring looks like a wedding ring silly!

Eugene

@JasperLoy oh wow. all that time i wasted studying

user19161

@Eugene What text are you using? Milne? Knapp?

Eugene

02:51

@JasperLoy for what?

user19161

@Eugene Elliptic curves.

Eugene

@JasperLoy milne doesn't even talk about them over general number fields though right? his "book" is tailored towards proving FLT

@JasperLoy i'm using silverman

@JasperLoy his book with cornell talks about them over schemes

@JasperLoy schemes look like the coyote trying to kill the roadrunner

user19161

@Eugene Yeah, you are full of schemes too I see.

Eugene

hold on. brb. steaming a char siew pao.

Peter Tamaroff

03:06

@JasperLoy Have you seen how anon suggested to prove that $X=f^{-1}(f(X))$?

user19161

@PeterTamaroff Yes I saw it.

Peter Tamaroff

I understand that means that $f(x)=f(y)$ for some $y$ but $x \neq y$ . But I can't prove it, again.

I'm trying now.

In fact, I'm lying in saing I can't.

I haven't tried.

I'll give it a shot and get back to you.

anon

$u\in f(W)\iff \exists w\in W: f(w)=u$

user19161

@PeterTamaroff Yeah, you should not do math when tired, not productive. I am going to sleep now.

anon

$u\in f^{-1}(M)\iff f(u)\in M$

Peter Tamaroff

03:14

@anon Thaks. I like symbols much better.

Words sometimes help but they tend to confuse me.

I usually write $f(X) = \{ f(x) \in B : x \in X\}$ but your writing is clearer.

anon

Thus $x\in f^{-1}(f(X))\iff f(x)\in f(X)\iff \exists y\in X:f(y)=f(x)$. But $f$ is injective, so $y=x$. But this implies $x\in X$; hence $f^{-1}(f(X))\subseteq X$.

Peter Tamaroff

@anon Yeah, that's better.

One thing.

I hope you don't "kill me" because of this

anon

sharpens knife

Peter Tamaroff

$\subseteq$ means "a subset or equal to"?

anon

yes

or just "a subset" if you understand that subsets can be improper (ie equal)

Peter Tamaroff

03:17

@anon Then it suffices to go the other direction now.

@anon Yeah, that's why I asked.

@anon Have you seen this?

anon

nope

Peter Tamaroff

@anon I just can't understand why would one publish that without checking the formula like Robert Israel did.

anon

Um, is that supposed to be the multiplicative or functional inverse of $\zeta$ in the paper?

Peter Tamaroff

@anon Multiplicative inverse.

anon

seriously, who writes $\zeta^{-1}(s)$ for $\zeta(s)^{-1}$? Also lol @ the referee comments.

Peter Tamaroff

03:31

Wait. Maybe the OP wrote it wrong.

@anon Yeah, I will edit the question accordingly.

@anon Hhaha yes

anon

@PeterTamaroff note that $\zeta^{-1}(s)$ is the one in the paper

(so, not Jose's fault)

Peter Tamaroff

@anon Yeah. I think that should be then the functional inverse.

Not the multplicative inverse as José wrote.

Frank Science

@anon $\zeta^{-1}(z)$ means $(\zeta(z))^{-1}$?

anon

@FrankScience it shouldn't, but I believe that's what the author intended by it

@PeterTamaroff I don't think so. Take a guess as to why $0<Re(s)<1/2$ is stipulated.

Frank Science

@anon I saw $f^m(x)=(f(x))^m$ only when $f=\sin,\cos,\tan,\sinh,\cosh,\tanh$.

Peter Tamaroff

03:37

@anon I see.

anon

@FrankScience not when $m=-1$ :)

Peter Tamaroff

@anon Hehehe true.

Eugene

@JasperLoy ok i'm back

Frank Science

@anon $\tan^{-1}(x)$ is somewhat ambiguous with $\arctan$, but in math it seems that $\arctan$ is more conventional.

anon

somewhat ambiguous? in my experience it's almost universally used as arctan. unless you mean ambiguous to someone not already familiar with the conventions (in which case the conventions would naturally seem counterintuitive)

Peter Tamaroff

03:40

@anon I guess I can procede by contradiction to prove $Y=f(f^{-1}(Y))$ when $f$ is onto, too.

Frank Science

Well, in my high-school, when we are working on physics, we can use $\tan^{-1}(x)$, but in math, the notation shall be avoided.

Peter Tamaroff

@FrankScience I preferer $\arctan$ for aesthetics motives only.

Frank Science

and some notation like $\DeclareMathOperator{\tg}{tg}\tg$ should be avoided.

Peter Tamaroff

@FrankScience why?

You can use \operatorname{tg}

Frank Science

@PeterTamaroff Otherwise we would get $0$ score.

Peter Tamaroff

03:43

@FrankScience I'm mean why you should avoid it, not what the punishment is for not doing so.

Frank Science

@PeterTamaroff For the wrong notation.

Peter Tamaroff

@FrankScience $\operatorname{tg}$ also means "tangent".

In fact it is used in Argentina for example.

Frank Science

@PeterTamaroff Many years ago it was commonly used here, but the revolution of education changed it.

Peter Tamaroff

@FrankScience I think it is good to have conventions, but the measure seems drastic.

@anon I don't know if I should start with "suppose $y \in Y$ but $y \notin f(f^{-1}(Y))$" or "suppose $y \notin Y$ but $y \in f(f^{-1}(Y))$"

Frank Science

@PeterTamaroff Oh, I cannot make sure. It might be a caution from teacher, not the measure.

anon

03:52

@PeterTamaroff find out which containment is trivial, then pick the one that contradicts the nontrivial contaiment

Frank Science

@PeterTamaroff Is there any condition constrainting $f$?

Peter Tamaroff

@anon The trivial containment was $Y \supset f(f^{-1}(Y))$

anon

@FrankScience surjectivity

Peter Tamaroff

@FrankScience It is onto.

@anon I should go with the first one.

anon

yes

Peter Tamaroff

03:55

@anon In fact I started writing it out with that one and flinched.

Frank Science

$f^{-1}(y)=\{x:f(x)=y\}$

Peter Tamaroff

@FrankScience Yes.

Frank Science

$f(f^{-1}(y))=\{f(x):f(x)=y\}$

Peter Tamaroff

@FrankScience You should be explicit about the underlying sets.

Frank Science

$f^{-1}(Y)=\{x:f(x)\in Y\}$

Peter Tamaroff

03:58

Oh, you were using values, sorry.

Seems short: "Suppose $y \in Y$ but $y \notin f(f^{-1}(Y))$. Then $ \not \exists x \in f^{-1}(Y) :f(x)=y$.

Frank Science

$f(X)=\{f(x):x\in X\}$?

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