@JasperLoy I know, but now I can start working on a new day on the site. I had capped yesterday. :-)

Twice in a row here: $290+260$ in the last two days.

@Brian Can you go through an argument of mine?

@BrianMScott I got 200 on Sunday, but only 85 on Monday. 215 on Tuesday.

@KannappanSampath Sure, I can take a look.

Suppose $A$ is connected. Claim: $cl(A)$ is connected.

Okay.

Suppose $cl(A)$ has a separation: That is, it is written as union of disjoint non-empty closed sets.

$cl(A)=L \Pi M$.

$A$ is connected and lives in $cl(A)$.

So, we must have that $A \subseteq L$ without loss of generality.

We know that $cl(A) \subseteq cl(L)=L$, which means $M$ is empty, a contradiction.

Looks good.

Except that you want $L\coprod M$.

In general, I claim: Let $E$ be connected. $E \subseteq F \subseteq cl(E)$. then $F$ is connected!

@BrianMScott $\LaTeX$ically speaking, yes!

Is my claim right?

It is.

I'll attempt a proof, here.

Okay.

Let $G$ be a clopen set in $F$. Will I succeed in proving $F$ is either empty or itself.

Say, I started with a non-empty clopen set in $F$. I claim it must be the whole of $F$.

By waiting until I was uncapped, I may have missed the voting on one of my answers. It was a calculated risk.

We'll see :-)

@Brian Am I doing it the right way?

@KannappanSampath You’re fine so far.

I am not seeing a proof though :/

@Brian Stuck, please tell me what to do!

Okay, you have $H$, a non-empty clopen subset of $F$.

Can you show that $H\cap E\ne\varnothing$?

(Sorry, I forgot that you were using $G$ instead of $H$.)

Yes.

$H \cap E \supseteq F \cap E =E$

Oh, did I mess those containments?

How do you get that first containment? $H\subseteq F$, not $H\supseteq F$.

Am I missing something obvious?

@KannappanSampath Yes: there’s no reason to think that $H\cap E\supseteq F\cap E$.

@BrianMScott That's OK. I got that. But, is the proof as well easy?

It’s actually quite a bit like the previous one, for $\operatorname{cl}E$.

Use the fact that $H$ is open to show that $H\cap E\ne\varnothing$.

Then use the fact that $E$ is connected to conclude ... what?

Wait let me think.

No. Only in vain. :/

$E$ is dense in $F$, and $H$ is an open subset of $F$, so $H\cap E\ne\varnothing$.

vain?

You mean in vain.

Veins are blood vessels.

Corrected all!

$E$ is connected, and $H\cap E$ is a non-empty clopen subset of $E$, so $H\cap E=E$.

Therefore $H$ is a closed set containing $E$, so $H\supseteq\operatorname{cl}E\supseteq F$.

But we already know that $H\subseteq F$, so $H=F$.

@BrianMScott :-)

@KannappanSampath That’s not necessarily true.

Why is $E$ dense in $F$?

Draw the right diagram...

:\

Because $F\subseteq\operatorname{cl}E$.

We must have that $cl(E)=F$ right?

No, $\operatorname{cl}E\supseteq F$.

No, our definition was that $cl(E)=F$ iff $E$ is dense in $F$.

It better not be your definition!

$\operatorname{cl}E=F$ iff $E$ is dense in $F$ and $F$ is closed.

mediafire.com/?g9a2cwm4nsy0mvx

Please peruse through section 1 in the pdf I linked, please. @Brian

I am getting too nervous now!

It’s downloading now.

Okay, it’s downloaded, and I’m taking a look.

I think that I see the problem. If $X$ is a space, and $D$ is a subset of $X$, then $D$ is dense in $X$ iff $\operatorname{cl}_X D=X$.

Yes, precisely. I know just this definition.

However, if $A$ is a subset of the space $X$, and $D\subseteq A$, then $D$ is dense in $A$ iff $\operatorname{cl}_X D\supseteq A$; it’s quite possible for $\operatorname{cl}_X D$ to be bigger than $A$.

Oh, I see. This is the way it is defined, right?

Example: $\mathbb{Q}\cap(0,1)$ is a dense subset of $(0,1)$, but $\operatorname{cl}_{\mathbb{R}}(\mathbb{Q}\cap(0,1))=[0,1]$, not $(0,1)$.

@KannappanSampath It could be.

There are other ways to define it.

Alright! :-) A big confusion to have got clarified!

For instance, $D$ is dense in $A$ iff every non-empty open set that intersects $A$ also intersects $D$.

Alrighty! But the last one was pretty cool.

Or: $D\subseteq A$ is dense in $A$ iff $\operatorname{cl}_A D=A$. (Note where the closure is being taken: not in $X$.)

This last one is natural extension of what my teacher had to say and I must have thought.

One of the things that you have to watch out for with closures is just where they’re being taken: in the whole space, or in some subspace.

Yes, this screwed all my confidence just a few seconds ago!

I think I have now got it right.

@BrianMScott I'll definitely do this. Did you go through some of the rest of the things in .pdf?

No, I’ve been trying to do three other things at once.

Do you want me to take a look through it later when I’ve a bit more time?

@BrianMScott Yes, surely.

Okay; I should get to it later this evening.

Oh, fine. I'll love to hear from you.

Okay; if you’re not around, I’ll leave some comments and a ping.

Yes. I think I should be around once the exam is over only to roll and cry. Lost all the confidence. :/

How long an exam is it?

For $3$ hours and $5$ problems out of $7$ given.

That sounds fairly reasonable, if the problems are chosen sensibly.

My teacher is a bit demanding. A question is only "little" long if he had to ask compact iff sequentially compact together with Bolzano weirstrass.

:-)

And, he asks sometimes extremely hard nuts to crack.

@JasperLoy No, sorry, I am in a depressed state, worried about exams, I type in what I find. So, please do not mistake me!

I do it routinely: $7$ looks better than 7 for some purposes.

Hah.

That countable $\epsilon_0$ guy asked another question. So far I got accepted answers for all the questions he asked.

Wow, I wrote a lot.

@AsafKaragila Good luck on this one :-)

Which one?

The question that was just posted by the $\epsilon_0$ guy.

Ah.

I left him another comment on the nature of his questions...

I may have missed the boat on the $y''=-y$ question.

What scares me is that I wrote that answer in 15 minutes. It seems pretty long.

Pun on Grad?

@robjohn You might like the last link. So will @Asaf!

@AsafKaragila Well, it is stuff that you know extremely well, so that usually flows pretty fast.

@robjohn I guess so. It scares me, though! :-)

@Brian: You're a friend of Kunen, right?

I knew him reasonably well 35 years ago, but I’ve not seen him since.

Oh.

@KannappanSampath ESL (English as a seventh language)?

I figured because you called him "Ken" in a comment once... :-)

@robjohn I don't understand. I meant that punny thing!

It’s how I think of him; as I say, I did know him fairly well at Madison, and he’s not all that much older than I.

I was wondering if you've seen his new edition. If you're not really in touch with him then I'm guessing that the answer is negative.

You’re right: I’ve not.

I'm dying to know what's in there.

Has anyone at all seen that question, I linked to?

Yes. I'm not sure what's so funny about it.

But, is it a Real question?

Third grade of high school and signal processing without limits. But, he knows the word limits. Claim looks not real to me!

Look at the pre-edit version. I should get a prize for that edit.

@KannappanSampath I think they meant third year of high school, which may have somehow gotten translated third "grade". That's why I was blaming the language barrier.

I recall there was one question that was like that, but like six or seven times as long. Someone cleaned the first half of it up and then gave up :)

@AsafKaragila I upvoted your edit ;-)

Huh?

@AsafKaragila It's a joke.

Oh. It's 3:30 in the morning, and I am getting nowhere with these diagrams. I'm not in a very receptive state for bad jokes. :\

@AsafKaragila bad? forced? my jokes these are!

No one said anything about forcing. If I was forced to do forcing I'd be much better.

In fact, I'd likely to be actually working or even done two days ago!

@anon It was I. Like a cut and paste from some web source.

Oh yeah, that's the one

I have to go. I just noticed the time and I need to take Lilly for a walk. bbl

hmmm

with no JM or Mariano around, consider the following scenario

someone posts something considerably offensive / inappropriate, but it is in LaTeX

it gets flagged, and globally mods swarm the room

but all they see is random code and don't understand it (if it's done cleverly)

meh, dunno where I'm taking this

Oh for crying out loud.

I tried to work all day long and I got absolutely nowhere!!

>:(

Holy shiza - this user just put 99.3% of their rep on a single bounty

Yes.

I felt like downvoting the question so he'll end up with just 1 reputation point :-D

@AsafKaragila Why don't you tell your instruictor, that his assignments are hard and should give some hints for you?

Plus, I'm not sure it fits.

@KannappanSampath Well, he gave me some hints. I got nowhere because $\Huge\mathbf{I\ HATE\ DIAGRAM\ CHASING!!!!}$

@AsafKaragila Is it so hard you could not figure out an online source discussing similar thing?

I'm reading the same proof in five different books.

It's just the details... oh lord. The details...

I hate details. I hate them even more when I feel like a retard moron moving my finger across a diagram.

"We go from here to there to there to here to there to here to there to here to here, and this is well defined because this and that and this and that and this and that and this and that and that and that and this and this". Only replace each instance with letters on top of more letters and every argument with symbols and more symbols on top of other symbols.

Well, last suggestion: Catch hold of another moron who has done all of this. Ask him to give you a lecture on HOW he did this. Repeat the same steps. Note that to do sth, one need not understand what he has to do but still do rightly successfully @AsafKaragila

I cannot do things that I don't understand.

I cannot, for the life of me, repeat an argument correctly without understanding it fully first.

Sure you can. Just memorize a very short syllogism in a language you don't know.

But, still asking him means he will tell you something he does not understand and in the process you may understand!

Nah.

It's more complicated than that.

I need to understand the big picture.

Then the finer details.

Mariano once told me that understanding comes later after familiarizing oneself with math, after I complained I wasn't understanding something.

I'm most certain that I'm an exceptional person when it comes to most things. In particular the way I understand is very different.

Well, bang the doors of your instructor, tell him he must repeat everything but not more than what he covered in class and learn it over again!

@KannappanSampath Things don't work this way. You're living in a fantasy.

Sure, The last idea was not meant to be of any help, after you rejected some of the previous realistic versions of how to succeed in alg.top.

Yeah, algebraic topology in low-dimensional spaces is the least likely thing for me to understand.

The easiest way to succeed in algebraic topology is to avoid it like the plague.

I'd avoid it even more.

If you have the plague then you don't need to study algebraic topology!

One last attempt to get you to finish this: Tell yourself you're never ever going to meet this guy in life. The one last time, just wrap things around your head, as if it was set theory and finish this assignnment and hence the course (presumably)

And, You'll be a happy man.

You clearly don't know me at all.

This approach wouldn't work even if the assignment was in set theory.

Why, the one last time approach, you mean?

Indeed.

I tell myself every time, that was the last time I am going to injected for a disease and bare the pain.

(As if it was multiplying a 5 digit number by 15 digit number)

You really need to work on your English, or go to sleep.

@AsafKaragila An edit would be welcome although I am typing it all in Semi-Conscious mode)

@KannappanSampath ‘bear’

Yes, I noted that and I missed a get before injected. I noticed!

But it was already too late to edit. :-(

I suppose that you mean inoculated for a disease, not "injected for a disease" too.

receiving an injection?

@KannappanSampath Just work in base $10^5$: then you’re multiplying a three-digit number by a one-digit number!

@KannappanSampath That’s fine.

That is not the part I find weird, rather "for a disease" is strange sounding.

Brian, you're the native speaker... your call?

It sounds a little odd, yes.

receiving an injection for a disease is odd? : (

I am an Indian, and this is all the English I know!

One can be inoculated against a disease, or one can be given an injection as part of the cure for a disease.

I never cared about injections.

@KannappanSampath I shouldn’t worry about it too much: you usually make yourself understood pretty easily.

@BrianMScott Oh, if that's the case, then the language serves its purpose and I am happy!

@BrianMScott I have now learnt the right usage for sth today.

And the colloquial term for an injection is a shot.

Then, you take shots?

No, you get them.

So $|A|\le|B|$ iff $A$ can be shot into $B$?

Someone might say, ‘I have to get a tetanus shot, and I’m not looking forward to it.’

@BrianMScott Yes, I get that now.

You're getting a shot now?

@AsafKaragila Interesting idea. ‘And $f$ is a shooting function, so $|A|\le|B|$.’

@AsafKaragila No, not any time in the near future, will I have to get shots because I am hale and hearty.

Well we cannot really use "shooting" because we are "shooting a club" with forcing.

@KannappanSampath Come again?

I was very healthy when I got the largest number of shots that I’ve ever had at one time. (It was in army basic training.)

I am so erroneous, I made the same mistake another time!

This is amusing: a question that has 15 upvotes, 2 downvotes, and 8 answers with a cumulative total of 63 upvotes and 3 downvotes also has four votes to close.

Okay. Screw that. I am going to sleep.

If I am not working, at least I'll get some rest.

Probably a good idea; isn’t it almost 0430 your time?

I voted to close although I held a opposite view point a day ago.

I may vote to reopen: the question has not in fact stirred up any unpleasant debate, and several of the answers have clearly been considered worthwhile.

I think that "generally involv[ing] facts, references, or specific expertise" is not mutually exclusive with "solicit[ing] opinion, debate, arguments, polling, or extended discussion"

I think that there was no good reason to close it. In fact, I have now voted to reopen it.

I also voted to reopen.

If people wish to reopen, I'll vote to do so. But, the problem is, it will solicit more and more opinion.

It hasn’t been a problem so far.

@Foool Do you have reason to think that it can be integrated by standard techniques?

"I'd love to know who the person is who keeps downvoting me for no reason. Just love to know." lol'd @ this comment dripping with attitude

@JasperLoy I am taking a nap before $2$ hours for the exam. Ping me like this after an hour to ensure I wake up, if you can .; -)

I believe I am only semiconscious 90% of my waking life.

@Kanna: I use this as my alarm clock because I'm too lazy to plug my actual one in

@JasperLoy Yes, but I was getting some arguments clear. Now it's just time pass.

@anon I’m afraid that he brings it upon himself, though I didn’t think that this was one of his really bad answers.

Also, getting less than a full cycle of sleep can actually put you in a worse position cognitively IIRC, so unless you can fit a full 90 min of napping in you may want to forgo it. If you find it hard to stay awake now I'd go ahead with the short nap though.

@anon Boy, I should try sleeping a full cycle some day :-p

Oh, you are talking about the 45 min cycles, or 90 for 2. :-)

"In humans, each sleep cycle lasts from 90 to 110 minutes on average" Deprivation has also been linked to psychosis.

Also, REM (the most important part for cognitive function) is the very last stage in the cycle.

Of course, if you're deprived regularly, your body will go into the REM stage sooner than natural to compensate. This allows one to condition oneself into tighter sleep schedules, extreme example being the ubermensch or however you spell that word. This schedule is highly volatile, though, in that slight deviations can wreck your system.

I had heard 45 minutes, but I don't remember where. It was probably my wife when she was studying psychology as an undergrad (31 years ago).

I heard 90 minutes in my psych class, which was 3 years ago.

What do you call a category such that if $f:A\to B$ and $g:B\to A$ and $f\circ g$ is an isomorphism then $g\circ f$ is too? I can't think of a counterexample off the top of my head but I see no reason why it has to be true categorically.

so $f\circ g:B\to B$ is an isomorphism, but not necessarily the identity?

Yes.

I think this applies to $\mathbf{Vect}_{\mathbb{C}}$ anyway so my comment here is useful

However, it is still 1-1 and onto.

and bicontinuous

in category theory you're not supposed to think about the things inside the objects, I thought? though that reasoning suffices for my purposes

however, it could be factored into two discontinuous parts

where are we getting the notion of continuity from? in an arbitrary category I don't think we speak of continuity, and in Set we need only speak of bijections for what I want

bah, this is what happens when I try and learn cat.thry on the interbutts

Hi, quick question: What's the name for the matrix operation $ P^{-1}DP $?

i.e. multiplying a matrix on both sides by some given matrix and it's inverse

similarity transformation

Aha, thanks a lot. The name slipped my mind :(

no problemo

hi

looks like all are asleep

more or less

I am reading the chapter Functions of several variables from the book Pric. of Math. Analysis by Rudin

@anon

BTW its morning here

okay

guess you are asleep or tryin to...so won't disturb you

nah jus half asleep

if I didn't want to be disturbed I wouldn't have responded

okay

but you say disturbance is disturbance ?

whether liked or not

what

I mean you want to be disturbed now is it ?

I just don't care

@anon how apathetic :)

Morning.

@anon : )

well, technically after midnight is morning for where Im at ...

@KannappanSampath Good luck! I'm keeping my fingers crossed, although I think you won't need that.

@BrianMScott You got 2 stars for that. Wot!? 0_o : )

@KannappanSampath Next time you sleep more before the exam. I hear enough sleep helps with thinking...

Morning Brian : ) I see you helped Kannappan with topology last night.

A little. I’m not sure how much I helped.

Time to look at Fourier transform and chapter 0 of Pseudo-differential operators.

@robjohn Are you there?

@MattN yes

@robjohn Great : ) Do you mind talking about the Fourier transform?

I'd like to continue where we left off... i.e. talk about why it's continuous.

No (I don't mind) Sure (whatever)

okay, what is continuous?

$ \mathcal{F}(u(x)) := u(\xi) = \int_{\mathbb{R}^n} e^{-ix \xi} u(x) d x$ is linear and continuous. Linear follows from the definition because the integral operator is linear.

Now I need to see why it's continuous (as function on the Schwartz space).

You had mentioned that if two functions are nearby then so are their Fourier transforms.

You mean $\hat{u}(\xi)$

No, I mean $\mathcal{F}$.

Unless I'm confused.

you wrote $u(\xi) = \int_{\mathbb{R}^n} e^{-ix \xi} u(x) d x$

it should be $\hat{u}(\xi) = \int_{\mathbb{R}^n} e^{-ix\cdot\xi} u(x) d x$

Right. Thanks!

You need the $x\cdot\xi$ in the exponential

True, we're not in $\mathbb{R}$ so we need a dot product. Thanks!

Now, we have to define our range and domain.

If the domain is $L^1$, the range is $L^\infty$

Okay, if we want to simply start with $\mathscr{S}$, then do you know Plancharel?

Not yet. But isn't $\mathcal{S} \subset L^\infty$? (Does that lead anywhere?)

or Parseval is more applicable

They are essentially the same thing, but Parseval is polarized

That is $\int f\bar{\hat{g}}\mathrm{d}x=\int\hat{f}\bar{g}\mathrm{d}x$

If $g=\hat{f}$, we get the non-polarized version which is Plancharel

That the $L^2$ norm is the same for $f$ and $\hat{f}$

But I don't have a norm on $\mathcal{S}$, I only have a bunch of seminorms. How do I deal with that? I need to have some idea of distance to apply the epsilon delta definition of continuity to $\mathcal{F}$.

Is the topology determined by the family of seminorms?

Yes, I think so.

Okay; then continuity isn’t too bad.

You can’t talk about a simple $\epsilon$-ball, but you have something very similar.

Essentially, for each finite subset of the family of seminorms you have an $\epsilon$-ball.

Aha: as it’s done here.

Awesome, thank you, this is exactly what I needed to know.

If we let $\|f\|_{\alpha,\beta}=\sup_{\mathbb{R}^N}x^\alpha \partial^\beta f$

and set $\|f\|_n=\max\limits_{|\alpha| +|\beta|=n}\|f\|_{\alpha,\beta}$

Then $\|f\|=\sum2^{-n}\|f\|_n$

can't we make a norm like that? It may not be as easy to work with though

It looks like a norm. Assuming it is: does it induce the same topology as the semi-norms?

I think it does, but it is harder to work with. However, it gives an idea of how the seminorm topology looks.

I believe that Stein mentioned this.

I have Stein. Let me have a look.

If the family of seminorms is countable and separates points, it can be replaced by a metric.

Hello guys, aced through!

Except a very minor glitch!

@KannappanSampath Hehe, I told you so : D Congratulations!

@MattN I was talking about Stein in lecture. It is probably in one of his books.

@KannappanSampath Good job!

The glitch was with part $ii$ of the second question here.

isibang.ac.in/~creraja/Ana2_12/mex.pdf

@MattN Thank You @Matt And, @robjohn Thank you as well.

@KannappanSampath Crap : ( That's the one you were trying to do when I left yesterday.

@BrianMScott and these seminorms are indexed by $\mathbb{N}^n\times\mathbb{N}^n$

In general I must thank every one in the Chat room for their little helps!

@robjohn It's $\mathbb{N}^n \times \mathbb{N}^n$.

@MattN But, did you read the hypothesis, for one I could show that intersection has atmost one point. To show non-empty intersection, I could not spin that argument using supremums. :-(

@MattN you need the exponents on $\mathbb{R}^n$ for the multi indices, I believe

But now the argument is so simple after I got hints from teacher before leaving the hall. :-(

@MattN but I was mainly implying that the semi-norms were countable.

@robjohn Then indeed you should be in good shape.

@KannappanSampath Not yet, I'll look at it later. I have to hurry and get some stuff done about PDOs.

@MattN I think I am OK even if we discuss this much later!

@BrianMScott thank goodness :-)

@BrianMScott Thank you for the help, professor. I got some arguments so quickly!

@KannappanSampath I’m glad to hear it!

@robjohn Which exponents?

@Matt Also deserves this praise: With your help, some arguments came out of my arsenal naturally!

@KannappanSampath Nice : ) I'm glad.

But I can show that it's continuous if I show that $\mathcal{F}^{-1}(U_\varepsilon) = U^\prime_{\varepsilon^\prime}$ where the $U$-sets are in the basis of the Schwartz space. Right?

Well, up next is probability. I am not too good with this stuff. I'll have to do a lot and may not come here as often although I'll be logged in!