@SohamChowdhury it's pretty standard, that stuff. if you do identification spaces, you'll get to know about it.

it's true that any 2-manifold can be constructed from identifying edges of polygons.

where did you read linear algebra from @BalarkaSen ?

@BalarkaSen hi!!! I have a question

@Karim Artin.

You free?

almost.

I see

ask away @Rememberme

Well correct me if I am wrong.... Is it that every metric space is a topological space or the other way around ?

The other direction is false.

Learn what a topological space is, then we can talk about it.

How can I think of it...

I mean picturizing

Fine will come back

After few days..... And I am still stuck with your problem....

Which problem?

I can't think of a nice idea for it

Sorry to be pedantic, but I think this is an important distinction. Every metric space can be given the structure of a topological space. But a metric carries nontrivially interesting amounts of information beyond that. eg there's no way to "topologize" the notion of a Cauchy sequence.

A metric space is really a sort of combination of topological space and a uniform space at the same time.

He doesn't know what a topological space is, so it's ok to be less pedantic :P

The compact question @BalarkaSen

@Remember That $\{0\} \cup \{1/n : n \in \Bbb N\}$ is compact?

Yes yes

Well, take an open cover of this space.

There is an open set which contains the point $0$.

consider $\mathbb{R}$ with the metric $d_1(x,y) = |x-y|$, $d_2(x,y) = |\arctan x - \arctan y|$. same topology, but one space is complete, the other isn't

Yes ....

There is an open set which contains the point $0$.

I was saying that for your benefit more than his. It's a common misconception (and I am not claiming you have it) that the only interesting things you can get from a metric space are topological ideas. But uniform ideas like completeness and uniform continuity are important.

I would suggest reading stuff sequentially @Rememberme

thats how I like to do things like climbing a ladder that is if you want to learn topology then read munkrees from first chapter etc

What do you mean Karim well when I was going through some topology questions on the net... I saw this one it felt weird about it

so you don't lose ideas in your head and become confused

I get you, @MikeMiller. I like to think of metric spaces more than topological spaces because of it's extra structure, to emphasise.

But thanks!

well just find a book and go through the chapter and then solve its question

munkrees is excellent choice

Yes munkres it really nice

@Remember D'you agree that there is an open set in your chosen cover that contains $0$?

@BalarkaSen the set you have mentioned isn't it a closed set? I feel it is

Which set?

The union one....

Sorry for being informal

I don't know what you mean.

$\{0\} \cup \{1/n : n \in \Bbb N\}$

Is this set closed??

closed with respect to what space?

I feel it is....

It can be open, closed, both or neither depending on the total space you have in mind.

R

The notion of a set being open/closed is nonsense.

why?

@Remember OK, good. Figure it out!

Nonsense??

@copper.hat We call something open with respect to a topology. "A is closed" is either nonsense or abuse of notation, as it's not specified what the topology is.

true, but language precludes exactness...

writing $A \in \tau_X$ everytime gets tedious...

But @Rememberme didn't make it clear what's the topology he had in mind :)

but i agree, there are many situations in mathematics (mostly learning) where precision in notation would be highly beneficial.

Well @BalarkaSen but it being a closed set does it help me in the proof....?

well close means with respect to topology that its complement is open with respect to the topology

are you tring to prove that the set above is compact in the usual topology?

@Remember It might or might not, but you should prove it directly.

Yes....@copper.hat

I was trying to giving you hints above, which you can take if you want.

you can do it with sequences, or using closed & bounded, or...

using an open cover. take your pick...

... or do it by hand.

:-)

or just by doing it

lol

in fact, starting with an open cover is probably the most direct and illustrative

right

if you have an open cover, then you must have 0 in one of te open sets. then since 1/n \to 0, you have all but a finite number in that set...

I am starting to read about Banach space and Banach algebra interesting stuff

@copper.hat shh

oops...

:P

lol...

@Remember Now that @copper.hat told you the proof, can you see/prove why any open set containing 0 contains all but finitely many of the elements?

in some sense, this example captures the essence of compactness

(for finite dimensions)

Sorry balarka but still having a bit of difficulties...

Ha...understood

OK, explain.

that and $[0,1]$ are by far the most instructive examples, I agree

@TedShifrin Okay, so for part B, I am mapping from $\Bbb{R}^{n+2} \mapsto \Bbb{R}^{n}$? I am unsure what function is supposed to be your $\mathbf{F}$. Is $H$ the analog of $\mathbf{F}$?

@Rememberme: what does it mean for $x_n \to x$ in terms of open sets?

@StanShunpike How do you get that bold font?

\mathbf{}

Well correct me if I am wrong when I was going open sets I always saw that inside an open set there were infinitely many points because when I was doing the problem that Alex Clark gave me I saw that even if I make the ball arbitrarily small there will be infinitely many elements in it . noe in this case we are talking finitely many subsets of the given set (0,1) which contains infinitely many elements ts and hence cover the full set... Sorry if this is nonsense @BalarkaSen

And with the 0 we complete the finitely may sets

@remember you're on the right track. but why would an open set around 0 contain infinitely many points?

(of the sequence 1/n)

For example the set of all {1,1/n} where $n \in N-{0}$ is an open cover for the interval (0,1) and now with the 0 we have the whole set covered with finitely many sets

Brb

@SMF: it is not really clear what you are looking for...

i have to go, whoops

@Rememberme: slow down. what does it mean when we write 1/n \to 0?

Hey, does anyone know how to prove that for $f$ an isomorphism (in a category), $(f^{-1})^{-1} = f$? I mean, is there any kind of cancellation law so that $ab = cb = 1_A \implies a = c$ for isomorphisms?

I could then just compose on the right by $f^{-1}$ so that $$(f^{-1})^{-1}f^{-1} = ff^{-1} = 1_A$$.

Hello @user218931

@evinda: did you sort out your mst algorithm?

it reminded me how much i have forgotten :-(

what is a good book to study operator theory and functional analysis?

:(

just introduction

@copper.hat do you know?

What is your profession? @copper.hat

hmm, not sure i am the right one for that. i like kantorovich & akilov's "functional analysis", my backround is continuous optimization (as opposed to discrete)

so much of my analysis background has an engineering (as opposed to mathematics) focus.

@evinda: nominally an electrical engineer, semi-retired...

oh

Aha! @copper.hat And maths is your hobby?

@KarimMansour: but k&a is a very solid book, much in the soviet style

cool

I like russian books

@evinda: for want of a life, yes :-)

Hi @TedShifrin

i find mathematics a bit depressing, yet i cannot stay away :-)

lol

No, @Stan: What are all the equations we have? You told me this right before.

a bit like chess, i will always be at 1700, never a grand master :-)

And do you live in California? @copper.hat

near berkeley

Howdy @TedShifrin

a little enclave called albany

That's my old home, @copper.hat!

@copper.hat How is the weather there now?

13 yrs in Kensington, 5 in Berkeley ...

perfect for me (cloudy * cool) my wife wants a warmer climate

@TedShifrin Sind Sie umgezogen?

Hi @Remember

we're near kensington circle

ted, where r u now?

Noch nicht, @evinda

du kanst deutsch sprechen @TedShifrin !

in Athens, GA, @copper.hat, but moving to San Diego this summer.

wu hast du deutsch gelernt ?

haven't been to ga yet.

@Karim you know French??

no

I know 3 languages though

Aha.. @copper.hat Then you should maybe go on vacations so that your wife has a warmer climate... :) Will you go now in summer?

I was just there last July, @copper.hat

@KarimMansour Which languages?

@evinda English,german,little arabic

Hello @evinda

i am from ireland, so will go back to visit family, will take a short break from there to barcelona (so my wife is happied, irish weather is no improvment for her on albany)

Hi @Rememberme

at berkeley,ted?

And Arlington Circle ...

I want to learn greek and french later

Irish.... I love English and irish people @copper.hat such open minded people

@copper.hat Oh nice... :)

hmm, i think people are much the same the workd over :-)

Plan to drive up to the Bay Area at leadt once a year ...

ohhj, long drive!

And I love the irish accent....

thats the amazing thing about america for me, you can just drive & drive...

There are distances, yes ...

Rightly said....and in India in the blink of your eye you see a traffic jam....

and there is so much to see (i like the outdoors).

howrah bridge at rush hour

my gf is of irish descent

How do you know about Howrah bridge @copper.hat did you build it :p

was in calcutta/kolkata a long time ago. loved the place. maybe my favorite (non rural) place in india...

@TedShifrin What you said before that the (optimal) solution $(\mathbf x,\lambda)$ varies smoothly with the budget $c$. I take that to mean that, as I vary $c$, $\mathbf{a} = \psi(c)$ varies smoothly as a solution to $\nabla f(\mathbf{x})=\lambda \nabla g(\mathbf{x})$.

You were in Calcutta what on earth were you doing in Calcutta?

@SohamChowdhury You look young.. How old are you?

on my way home from darjeeling :-)

Biriyani hunting

@evinda 16

But I have a beard . . .

Okay, not a very nice one. But still . . .

That's beautiful Darjeeling ....

So you go to school, right? @SohamChowdhury

@evinda I do, yes.

What is your favorite subject? @SohamChowdhury

Well you should look at Calcutta at winters early morning its the most beautiful place I have ever seen

@evinda Guess.

@evinda obvious maths

Aye.

@Rememberme: i was there in winter. it has so much stuff for a tourist. i only scratched the surface (and this was pre-kids travel).

and very easy to interact with people.

CS or physics might be close seconds. Although "Hey the determinant is $ad-bc$ now do the sums for homework" is painful.

@copper.hat which year? Were you in india

@evinda Why the curiosity all of a sudden? :P

oh my, i think '96, it was the time of one of the everest disasters..

@SohamChowdhury Just for the heck of it... :)

when we have some hilbert space H

Ahh I wasn't even born then @copper.hat

@evinda: are you ingermany?

what is this notation

Lol

B(H)

@copper.hat No, but my mother is from there.. Have you been in Germany?

indeed, a few times over the years.

Well I am thinking what is knot theory about connectedness of different objects ?

@evinda I was in Germany last year.

@copper.hat Do you like it there? @copper.hat

St. Peter Ording. Wunderschoen. :)

i do. i found people very helpful in unexpected ways.

@SohamChowdhury in Germany? People are so lucky...

@SohamChowdhury As an exchange student?

@copper.hat Nice.. Have you been in Greece? I live there..

not yet, sadly, even though one of my best friends is from greece...

@TedShifrin If the BH isn't equal to zero, then what function is? I mean, I have $f(\mathbf{a})-\lambda g(\mathbf{a})$. I know that equals zero. That also includes $\lambda$, $c$ and $\mathbf{x}$. I suppose I could define $F(\lambda, \mathbf{x}) = f(\mathbf{x})-\lambda g(\mathbf{x})$. Then at $\mathbf{a}$ this would equal zero.

@copper.hat From which region?

my friend? patras

Where on earth do you study...

guys anyone knows what is this notation here ?

B(H)

@SohamChowdhury A nice... :)

Basis of a Hilbert space @KarimMansour :p...

@copper.hat Aha.. I haven't been there..

hm

Morning, @Ted

@Rememberme A lot of schools in Kolkata offer this. It's not like I go to some "elite" school or something. Look at the website, you'll feel sick. :P

@KarimMansour: bounded linear operatorson a hilbert space

@Ted timezone?

oke that makes sense thank you @copper.hat

@Soham He's in EST for now. 1PM.

So I was related to linear algebra...@KarimMansour ;p

@TedShifrin That seems like nonsense. That's a function onto $\Bbb{R}$ not onto $\Bbb{R}^n$, so that cannot be right. hmmmm.....

Mm.

lol @Rememberme

later folks, i need to attend to some reality :-(

oke cya copper hat

@MikeMiller 1PM is morning? (This came off the wrong way)

We are virtual?? @copper.hat cya peace

Oh, I see your question.

Sorry.

@TedShifrin I only have two functions that map to $\Bbb{R}^n$. The BH and $\psi$

Repeating a question that got swallowed up.

No, @Stan. Write down again the LM equations.

Seems easy, but . . .

An isomorphism is an invertible morphism: an $f$ such that there's a $g$ with $fg=gf = \text{id}$. Suppose there was an $h$ with these properties. Then $h = hfg = g$.

Inverses are unique.

@TedShifrin $L = f(\mathbf{x}) - \lambda g(\mathbf{x})$

$\nabla f(\mathbf{a}) = \lambda g(\mathbf{a})$

$g(\mathbf{a}) = c$

All I see is scalar fields @TedShifrin

But I thought $\mathbf{F}$ is a vector-valued function...

Sorry! $\nabla f(\mathbf{a}) = \lambda \nabla g(\mathbf{a})$

Typo

Okay, so we are mapping from $\Bbb{R}^{n +2} \mapsto \Bbb{R}^{n+1}$

@Stan: $g=c$ is one scalar function, $\nabla f \dots$ is $n$ equations. All of these give you $\mathbf F$.

@StanShunpike Ah, right, so you got it. Now get to work :)

@copper.hat wow!! nice to know that :D I'm from Kolkata!

hi @Ted

@TedShifrin please is the dual space of $W_0^{1,p}$ is the same ? i.e (W_0^{1,p})^*=W^{1,p}_0$ ?

please

Good Afternoon Professor @Ted :)

@r9m Never asked you that. Do you regularly practice some kind of sport? I mainly do a crazy amount of cardiovascular fitness, but it's great! :-)

hello @KajHansen

Hey @BalarkaSen

Hello.

@KajHansen been thinking about anything interesting?

Surely Junior Soares' worry here isn't valid? If $K$ is a finite extension of $\mathbb{Q}$, then there is a Galois extension $K \subset K'$ of $K$ where $K'$ has finitely many subfields of degree $2$, which follows from the fundamental theorem. Hence, $K$ itself can have only finitely many subfields of degree $2$? math.stackexchange.com/questions/1285655/…

lol at Junior Soares' worry

Up to isomorphism that is.

Junior Soares?

let me read that

Wait

He deleted his comment @BalarkaSen

I guess he figured it out lol

k.

I don't think so, @Vrouvrou, but I haven't done this stuff in 35 years.

Hi @r9m.

Hi @Kaj

I greeted you hours ago, Balarka.

waah, @Ted

@BalarkaSen, in case you were curious, he was claiming that there might exist finite extensions of $\mathbb{Q}$ with infinitely many non-isomorphic subfields. I was like "That's definitely wrong!", but then wanted to be careful before embarrassing myself

Hey @TedShifrin

It's definitely wrong, @Kaj.

It is wrong.

mhmm, since you can fit finite extensions inside of a Galois extension whose Galois group is finite, and so the Galois group has finitely many subgroups, which implies what we want due to the one-to-one correspondence.

If you mean field isomorphism, that is.

Indeed

It's true if you replace that with vector space isomorphism, say.

No, no. Just sticking to field isomorphisms.

Eh, my internet is fluctuating.

@TedShifrin Oh, I guess I didn't see that.

There's still finitely many subfields, even if several are isomorphic!

How is an isomorphism between vector spaces defined? A bijective linear map @TedShifrin ?

Yup.

a K-linear isomorphism, yeah

that reminded me of the question i was pondering on years ago : given galois extensions $K, K'$ over $\Bbb C(z)$ with isomorphic galois groups, is it necessary that $K \cong K'$?

not sure if i recall it correctly, though. if it feels trivial, then it's probably not the question i had. if it's not, most likely it is.

i guess this one is the correct question.

note how this is false when $\Bbb C(z)$ is replaced by $\Bbb Q$.

@BalarkaSen and in fact very false

By $\mathbb{C}(z)$, are you referring to the field of rational functions over $\mathbb{C}$ ?

for $\Bbb Q$? yes, obviously.

@BalarkaSen Is it obvious that all quadratic extensions are isomorphic?

@KajHansen Yes.

I didn't say 'all'.

@BalarkaSen All quadratic extensions have the same Galois group

@Chris'ssis sports? no .. I stay in a hostel where food sucks! I wouldn't dare expend calories that'd force me to ingest more of that shitty food :P .. I used to swim at one point in time regularly (but that was a long time ago) :)

Yes, true. But to contradict my statement for $\Bbb Q$, you just have to come up with one pair of fields $K, K'$ for which this is false.

@BalarkaSen Right, I meant quadratic extensions of $\mathbb{C}(z)$

@r9m Ah, sorry to hear that. Yes, swimming is great too.

Oh.

I don't know how to prove that all quadratic extension of C(z) are isomorphic.

At least, not classically.

I can prove this with a bit of Riemann surface theory, I guess.

@Chris'ssis ya! I was good at it :) I used to swim like a wild beast :P

@r9m :D

@TedShifrin: No hi for me?

@TedShifrin hello Professor :)

@KajHansen I guess you can think about this one if you're interested. It's a cool problem I came up with when I was doing Galois theory.

Probably nontrivial too.

@BalarkaSen: $\Bbb Q(\sqrt{3})$ is not isomorphic to $\Bbb Q(\sqrt{2})$.

It's false for $\Bbb Q$, like I said.

But if you can come up with a $\Bbb C(z)$-example, I'd be veeery surprised.

Misread your question.

@r9m did you see the series I created above?

I guess this came up as the last link to resolve the "quintic-mysteries", which was about proving that the modular equation for X(5) (or X_0(5), don't recall) and the quintic z^5 - z - a are related by some rational transformation (and indeed they are! it's what Kronecker did) by purely using Galois theoretic tools. (they generate Galois extensions with the same Galois group).

$$\sum _{n=1}^{\infty } \frac{H_{2 n+1}-H_n}{(2 n+1)n}$$

@r9m ^^^ :D

@Chris'ssis oh! yea I saw it! :-)

@r9m Easy?

@Chris'ssis very ;)

@r9m :D

@robjohn I think you also like very much this kind of series.

@r9m The point is to hit it deadly in one line (or two). :-)