I just misread what you typed

Hi

Does the fact that the identity, I, of G is in U and V automatically mean it will be the identity in UV = {uv | u in U, v in V}? It seems so intuitively...

@Jim_CS: Yes, the identity is unique

if any subset A of G contains the identity I and if there is an element E which is "like the identity" in A, ie. EX = XE = X for all X in A, then E = I

so in particular a subgroup cannot have a different identity

if U is a subgroup of G then there is an injective group homomorphism phi : U --> G defined by phi(u) = u.

so phi(1) = 1 and nothing else gets mapped to 1

hi @Chris'ssister

@spernerslemma hi

This question is really cool math.stackexchange.com/questions/221018/…

I wonder if it will be solved or not

@JonasTeuwen I too asked a question!!

@m.k. Cheers mate

@JayeshBadwaik Nice.

@Jim_CS: no problem

@Jonas I am faster than you!

@JayeshBadwaik Wow.

Hi mr.@JonasTeuwen

Hi.

@JonasTeuwen How are you,Jonas?

Could be better. How are you?

@JonasTeuwen My weather widget really claims there will be snow at the weekend.

That would be too nice.

@MattN. Wooooohoooo!!!

But no teddy to have an argument about it : (

@JonasTeuwen I'm good!

Good.

@MattN. Is there no argument clinic in Zurich?

Hmm, hmm. I can show that $\mu(\pi_i(x_1, \dots, x_d))$ is smooth (to a certain degree) for each $i$ in $[1, d]$. I need to get the complete one.

The tensor product will surely be dense, but yeah, but whatever.

Hi,I have a question: how do we prove that an equivalence relation is well defined?

Does anyone know the answer?

@Pilot What do you mean by "well defined"?

i myself cant get it. I was trying to solve some problem,and in solution it is said that relation is well defined thus we can say....

"The rationale here is that (6) imposes severe restrictions on the possible awesomeness of $(L, \mu)$."

@JonasTeuwen Is that in one of your papers?

@Pilot What is the relation in question?

@PeterTamaroff Yes!

@JonasTeuwen I love the frivolity!

(But you should get serious sometimes too.)

@PeterTamaroff The seriousness is above that.

@JonasTeuwen Did you hear Mariano had a conference these days?

No?

Plenty of people have conferences these days, what is so special about his?

@JonasTeuwen I don't know. I'm not saying it is special. Just small talking, jibbity jabbing.

Oh yes.

What is it about?

@JonasTeuwen "Linear representations of finite groups."

(Representaciones Lineales de Grupos Finitos)

Holy cow, I feel sorry for the bro.

@JonasTeuwen Why?

It's a conference in Spanish? How rare.

@JonasTeuwen Yes. It is in a University in Patagonia.

Yes, but I have been to one in Madrid and it was in English.

@JonasTeuwen This one doesn't seem international.

Well, Argentina is large enough I suppose and many other Spanish speaking countries in the neighborhood. So it is not such a strange choice. For Spain it might be.

I need to go there to habla Espagnol.

@JonasTeuwen You should! =)

relation is x_n is equivalent to y_n (x_n and y_n are Cauchy sequences in some set) iff $d(x_n,y_n)=0$ (i.e 2 sequences are equivalent if the distance d between them is 0)

@Pilot Hmm, I don't think that is what you want.

Don't you want $$\lim \;d(x_n,y_n)=0$$?

@Pilot Don't we want two Cauchy sequences to be equivalent if they have the same poit $\xi$ as a limit? It seems logical.

You want to define a completion I suppose.

yes

but I am not sure about Peter's opinion

Okay, easy. When do you want Cauchy sequences to be the "same"?

When they give rise to the same bloody limit, rite.

mhm

@JonasTeuwen That was what I was saying. Don't scoop me!

=D

Peter has something that looks like it, but I suppose you need something stronger (perhaps this also okay).

Like they are $0$ for some large $n$ on.

@PeterTamaroff Got scooped by Bourgain I've noticed today, so now I'll scoop you son of a crack!

yep

@JonasTeuwen Well, what about $\{n^{-1}\}$ and $\{2^{-n}\}$? They both go to zero, but they aren't $0$ away for some large $n$.

@PeterTamaroff You son of a...

But they are not the same sequence.

@JonasTeuwen But they define the same number....?

For sequences you care about tails. But you care about convergence here.

So perhaps that is better, yes.

@Pilot Suppose that $\lim x_n=\mu=\lim y_n$. Then $$d(x_n,y_n)\leq d(\mu,x_n)+d(\mu,y_n)<2\epsilon$$

As I do not like to read books or many papers, I get scooped all the time.

So $\lim\; d(x_n,y_n)=0$

Conversely assume $x_n,y_n$ are Cauchy and $\lim d(x_n,y_n)=0$ and show they converge to the same thing.

We have $\lim \;x_n=x$ and $\lim\;y_n=y$

This means $d(x_n,x)<\epsilon$ and $d(y_n,y)<\epsilon$.

But also $d(x_n,x_m)<\epsilon$ and $d(y_n,y_m)<\epsilon$.

Sorry, GTG

@JasperLoy Na, then I would never have proven a theorem by Bourgain when I knew he had already done that.

If I only read stuff people have done I will also not get anywhere.

Also, my result surely is new and I bet I understand much better the problems involved in proving the theorem then when I would have been reading the paper.

Opening the door for more proofs under roofs!

Books put a significant strain upon my creativity.

I will not write books.

Hardly anybody reads books. They usually just collect dust on the book shelf. Perhaps you should make a nice drawing for the cover, that is about as much as most people will see. Leave the rest blank. "Content left blank as an exercise for the interested reader".

If I were to write one -I doubt it-, it will be really short, under 100 pages and very specific things.

@PeterTamaroff but the mimit of x_n is not in the set,because the set is given to be not complete,so we cant consider d(x_n,x),or am I doing mistake?

Well, what it means for it to be "well-defined" is that the operation on your equivalence class may not depend on the representative (as equivalence classes cut up your space real good into cosets).

Jonas Teuwen, cutting up spaces real good since 1986.