@robjohn If you have $x$ in $M$ then since $f_n \to g$ almost everywhere the almost all of those $x$ will be in one of the $M_m$.

@t.b:I guess Real analysis was taught to me piece-wise in a discrete manner as I know some that are typically covered in the real analysis

@Jason But you've chosen M such that f(0) < f(M)

Or f(0) <= f(M), it makes little difference.

Ah :) now that seems reasonable :P let me think of that for a moment XD

Thanks :)

@DylanMoreland She was confused and all.

@Asaf Right, that is scary.

@MaX Well, measure theory is an advanced branch of real analysis. You usually learn that in a second or third course. It's pretty abstract and the need for it arises and becomes clear only later on, so you usually don't want to expose students that don't have solid basis in real analysis to it. I can very well imagine that you can do a scientific career in CS without knowing the first thing about measure theory, so it's not really surprising that you weren't taught it.

I see this MO question and I'm reminded of a friend's joke: "What's the difference between n-categories and a pile of dead babies? Dead babies are useful."

The only thing funny about it is the level of disdain, really.

@DylanMoreland Quite.

@AsafKaragila but now everything's well again, if I understood correctly?

@DylanMoreland What is a pile of dead babies useful for?

@tb Yeah.

I hope :-)

I hope so, too. Glad to hear that!

Hi.

@t.b.:Our mathematics syllabus mostly have calculus,(both single and multi ), discrete mathematics and abstract algebra (which will be taught next year). Although I am much interested in number theory sadly it's not in our curricula

@Jason So maybe there is a global minimum at an x >= M. But Then you could just take 0 as the location of such a thing instead.

Hi Matt.

Hi Matt.

@AsafKaragila: Do you have another minute? I thought I had solved my homework and then I was going to type it up and while doing that I noticed it was wrong.

@t.b: I forgot to mention differential equation and graph theory

I wonder, should I post an answer to the "algebraic closure" question or not...

@Matt Sure.

@Asaf Very good to hear that.

@DylanMoreland What do you mean by "take 0 as such a thing"?

@MaX Then you need not really worry about measure theory now, and I'd guess not anytime in the near future.

@t.b:I am not worrying, I just feel bad when I don't even recognize something in mathematics.

@AsafKaragila Nice! : ) Ok. So I'm asked to prove that $\bigcup G$ is a set consisting of infinite subsets of $\omega$, i.e. $\bigcup G \subset [\omega]^\omega$. My previous idea was that $\bigcup G$ is the maximal element in $G$ and then argue that all elements in $[\omega]^\omega$ have the same cardinality $\omega$ and so $g := \bigcup G$ will have cardinality $\omega$.

The definitions:

@MaX well, you got now the Wikipedia page and the nice blog post Dylan pointed you to, this should give a first impression. You may also want to check out the Banach-Tarski paradox (even though it isn't measure theory proper).

@t.b:Thank you :) I would check them up! :)

Why do mathematicians hate memorization?

@t.b: Did you checked the Wikipedia page you hyperlinked?

@Jason As in, if I had a global minimum at x >= M, then from f(0) <= f(y) for all y >= M I know that f(0) <= f(x), and since x is a global minimum I also have f(0) >= f(x).

So f(0) is a global minimum too.

My mind is cloudy from yesterday. I don't like this. Every time I drink I lose a day of my life. Life's too short for that.

Well, the terminology is always awkward. "There is a global minimum at 0" might be better.

@skullpatrol: because there are too many subtle thing and variations in mathematics and memorization would take you no where, blind memorization would give you frustration!

@AsafKaragila: The longer I think about it the more elusive it seems.

@Matt Again, are you using Jerusalem notation or "regular" notation? What is $\le$ in terms of "stronger" or "extends"?

@AsafKaragila googling Jerusalem notation

@DylanMoreland Ah Great Thanks :) - I Finally get it XD

@Matt I guess I'm the same. I enjoy drinking quite a bit but I lose too much time on it. I still do it only because that's how a lot of people I know in math socialize.

Uh, I guess people outside of math socialize in this way as well :)

@Matt In the Jerusalem notation $x\le y$ means $y$ is stronger; the rest of the world works the other way around.

But I don't know anyone outside of math anymore. When a friend comes here to give a talk, for example, we go to a Chicago bar; there's no question that this is what we'll do.

@Jason Sure. Glad to help.

It's a nice answer by David Mitra.

@DylanMoreland Technically you could order a soft drink, no? That's what I'll do from now on.

@MaX So memorization is avoided by mathematicians because it "would take you no where"???

@AsafKaragila I'm not sure what this means. I thought "is stronger" is a synonym for "\leq". For comparison, can you tell me what the rest of the world does?

@MaX Here's the link that should work, sorry.

@Matt If $(P,\le)$ is a notion of forcing we say that $x$ extends $y$ if $x\le y$. This the common notation, Jerusalem notation is that $y$ extends $x$ if $x\le y$.

By the way (To everyone) , I don't know any of you as I use this site mainly to ask questions which I come to during my studies at university (I am still a newbie :P) but from what I can tell: Each and every time I had a question you guys came and gave me excellent answers and stuck with me until I got It. So thanks a-lot for all the time and help - You are all just awesome!

@Matt Technically.

@AsafKaragila Yes but either way it makes no difference: it's just words. I think it's defined above, where I wrote that $[x] \leq [y]$ means $y \setminus x$.

@Matt Yes, but it helps me understand whether or not the generic filter goes in one direction or the other :-)

@Jason You're welcomed! :-)

Goodbye everyone have a good day/night

@AsafKaragila The filter goes down I think because it's defined that if $p$ in $G$ and $q \leq p$ then $q$ is also in $G$.

Bye, Jason!

@Jason Bye Jason and thanks for the nice words!

@Matt This is exactly why I prefer to work with "stronger" and "weaker" in forcing posets. Filters are "closed to weaker conditions", if someone is weaker than $p$ and $p\in G$ then it is in $G$. This is ambiguous when talking about general posets and whatnot.

Suppose $x=\{2k\mid k\in\omega\}$ and $y=\omega$ (or any tail segment). Do we have $x\le y$?

No because the odd numbers aren't finite.

Does anybody think there would be any interest in a question threat entitled "Why mathematicians hate memorization?"

You have $x \leq y$ if $y \setminus x$ is finite.

@Skullpatrol No.

So what is $\{y\mid y\le\omega\}$?

Anybody else have an opinion?

@Skullpatrol: Same opinion here: no.

@AsafKaragila The set of all natural numbers? Union $\omega$, so $\omega \union \{ \omega \}$

@Matt No... which are the sets that below $\omega$ in this relation?

I got a little trigger happy...

@AsafKaragila Did I save myself there? : )

@Skullpatrol: I have an opinion

@TheChaz care to elaborate?

@TheChaz Do you agree that mathematicians hate memorization?

My opinion is that math.stackexchange.com/posts/11267/revisions is a poor revision.

@Matt No :-P

@The Chaz: That's off-topic here :P

@TheChaz I wondered about the "trigger happy" :)

@tb The limit question math.stackexchange.com/questions/89720/…

@JonasTeuwen okay, so we are assuming that $f_n\to g$ pointwise a.e., so all of $\mathbb{R}^n$ is in $\bigcup_nM_n$ except a set of measure $0$.

Is that not what we were talking about?

@tb He's Texan. Probably killed someone today... :-P

@robjohn Wait. Did I mess up?

With my beltbuckle

@Jonas: However, how does that relate to $f$?

@AsafKaragila Noooo : D I did it again. I keep on reading the bloody \leq as \in -- Grrrrr

@robjohn Err.

I'm claiming that it is almost equal to $M$. Probably messed up.

@TheChaz you still haven't learned to stay away from $\frac{1}{x}$ have you? :)

@Matt Oops you did it again, you played with the \leq and lost in the game? If you start singing Britney Spears I'll force you to stop :-)

Please elaborate, impressing me with your memory in the process, if possible!

@AsafKaragila That's about the last thing I'd sing (or listen to).

@Matt Either way, if $x\le\omega$ then $\omega\setminus x$ is finite, therefore $x\sim\omega$.

@TheChaz I respect your opinion about my revision.

@robjohn Let me reconsider the problem.

@TheChaz I was talking about this :)

You assume that $f_n\to g$ pointwise, but how does this relate to $f$? Are you showing that $f_n\to f$ pointwise as well and I missed it?

Bed time for me see you tomorrow fellows :)

See you, MaX

Bye, MaX!

@tb Ah yes. Henning's comment to Eric's answer (to the new question) is what I should have written as my answer to the earlier question!

Right. (^ to add some The Chaz style)

=)

@AsafKaragila So assuming $y \in \omega$ I fail to see what these $y$ look like. What natural number is infinite? None.

@Skullpatrol meta.math.stackexchange.com/questions/3304/… This meta thread is related, especially as it seems motivated by your revision (see the comments)

What you wrote above ready as $y \in \omega$ such that $|\omega \setminus y|$ is finite, I think.

Of course, you'd have to ask the author, if he were ever around...

What? No... $y\in[\omega]^\omega$...

That makes more sense. : )

@robjohn I'm assuming that $f_n \to f$ in measure and $f_n \to g$ pointwise a.e.. I'll rethink that argument.

The Cesaro mean of the Fourier series of $f$ converges to $f$ where $f$ is continuous.

@TheChaz pssst he's absorbed with the holy grail of set theory. Don't dare disturbing the König.

I'll just wait till he goes to visit his girlfriend ('s parents?) again...

The set of discontinuities has measure 0 if $f$ is Riemann integrable.

@AsafKaragila So $y$ such that $y \leq \omega$ are infinite subsets of $\omega$ whose complement is finite.

So the Cesaro mean converges to $f$ a.e.

@Matt But that means that $y\sim\omega$.

@AsafKaragila facepalm

Damn I'm too tired :(.

@robjohn Yes, that works too.

Actually, that is better.

@AsafKaragila: Thank you so much!

Sure. What did I do now?

It is also correct for any demention

@AsafKaragila You pointed out to me the obvious that I had missed.

@tb I am partially demented :-D

@tb Hah.

For any dementia

Those are the things that protect Hogwarts right

(^ I just realized that chat is a more appropriate place for all my offtopic comments. Who knew ?!?!)

@DylanMoreland dementors :-)

Dylan, to quote tb: LOL

@TheChaz I did.

@robjohn Will you post your argument? :-). Then I'll delete my post, I'm too tired to fix it.

I guess someone has to be privy to social norms, etc.

@TheChaz Thank you for bringing my attention to those comments.

Where the social norm is defined by

||.||_{FB}

FB?

FaceBook

It would have been much funnier if I had the FB logo there...

Or maybe it would have been just funny.

I guess I'm getting dimented...

@AsafKaragila But actually, that doesn't tell me anything about $\bigcup G$ which was my previous question.

@Skullpatrol No problem. Sometimes users aren't aware that even miniscule edits resurrect questions.

Hello

Hello, VVV!

Is there a moderator here?

No. What's up?

I put too many questions in one question.

I want split it up

@TheChaz I'd write that as $\|\cdot\|_{\text{FB}}$.

@Jonas It's Cesàro, not Cesáro =)

I know that. Did I mistype?

Yes. I was hoping I can catch you before the edit window closes.

@Matt If $G$ is generic that means that $[\omega]\in G$.

@TheChaz I was just trying to bring more attention to that particular quotation.

Oh I did it one time correct and one time wrong.

@VVV Then copy the text to a file and post the questions separately. A moderator can't help you there.

@JonasTeuwen Nice.

@Skullpatrol Well, you succeeded!

@tb Ok, is there no way to split and keep my upvotes?

@AsafKaragila I think I'm going to copy this homework off someone. I can't think straight today. Thanks a lot and sorry for the bother.

@VVV Can you link me to the question? Your latest one?

sorry for spam

Looks like I succeeded. You're welcome, @Jonas. =)

@Srivatsan: Don't get yourself a puppy, they're too much work. We might have to take ours back because she now decided that the cats don't deserve any attention...

@VVV If you do not want the preview, include at least one word before the link. Like this: math.stackexchange.com/questions/89468/….

@Matt Oh not a problem. Do let me know what you came up with. I had a busy and tiresome day so I can't really think it through either :-)

@Srivatsan test math.stackexchange.com/questions/89468/if-l-is-a-subgroup-of-mathbbz3-linearly-independent-linear-equations

@Matt Thanks for the heads up. I might not be able to.

@Srivatsan merci

@Srivatsan Good for you ; )

@VVV Okay, that looks a bit difficult to split up, but if you have an idea, just keep the first part of your question in that post. And post the other parts separately.

You'll keep the votes on that one post.

@Matt =) What do you call her anyway?

@TheChaz And may I add that the revision actually got an answer to the question after more than a year of being posted ;-)

@tb merci

@VVV you're welcome, VVV.

@Srivatsan "Aya".

@Matt Oh. It's a different name. =)

@Srivatsan "Shredder" would have been more suitable.

Shredder, like sofas and cushions? Isn't that more the cats' occupation, normally?

Yes, I thoughts dogs are well-behaved and all that.

@tb No, unfortunately like anything she can get her teeth on: sofa, cushion, chairs, tables, shoes, feet, arms, catty toys, cables and even two of my books so far and some of my notes/homework.

Holy crap monkies, why is my bloody tire leak like every bloody week?!

I am not a fan of any cub of any animal.

@Matt Oy! So Shredder seems indeed appropriate :)

Ok, bed time here, good night folks and have nice day or night!

Good night, Matt.

Ciao.

@Matt No drink today?

I fixed my tire. I need a drink now.

@Matt Good night.

@VVV, "Introduction to Algebra" -- can you mention the author as well in your post?

@Srivatsan yes

Cool, thanks.

Damn, there is another leak. Now I definitely need a drink.

@VVV Have you solved your earlier question?

@DylanMoreland no… I tried but I dont know the solution

Ah, okay. Just curious. I made a note of it and was going to come back, but I saw that you posted a follow-up.

I wouldn't write L = (l_1, l_2, l_3). The thing on the right is supposed to be an element of L, right?

@Sivaraman I see. I am ignoring them I think.

Dupe

He's no Liouville.

@tb That's the lesser of my concerns. He seems to be back to unmotivated questions. =)

@Srivatsan Ok see you at 7 then.

@Sivaraman You're not going with them?

@Srivatsan what do you expect, that it's going to start raining frogs today?

(and now he deleted it, oh, well)

@Srivatsan Prior commitment :P

@tb Forgive my optimism. =)

@tb Executed :-).

Chat hit squad

TF?

@Jonas Wow.

@JonasTeuwen Not constructive, I presume. It's sad...

Yes.

But I found your comment curious, Jonas. What's good enough got to do with it? =)

@JonasTeuwen Looks pretty close to what was being asked here.

@Srivatsan Well, the idea of the amphetamines is to be a better mathematician I presume.

@JonasTeuwen Yes, you can be better even if you are good already, right?

There are people I have suspicions about.

Well, that's why I say: If you need it.

But I know little less than nothing about these things =)

@JonasTeuwen Aw, I see. Nice.

I don't like amphetamines.

Hah, tagged under meta-math... If anything meth-a-math would be a better fit!

My favorite comments are those of the form May you use a solution not using induction?

@tb Is it me (sleepy) or is that really incoherent?

It looks like a PE question or something

@AsafKaragila Looks like one. I have left a comment: let's see if he replies.

Yeah. I'm going to sleep now.

So long, suckers! Also, so long those which are not suckers.

@Jonas, why the rollback?

Because titles should be short. You can roll it back again if you want to.

@Srivatsan It is a justified rollback, if we could unbump it - it was even better!

@AsafKaragila No, I don't have anything against the rollback per se (and I don't think one title is much worse than the other) but it's the re-bump that I was thinking about.

Well. I am crashing out now. Ciao.

Bye, Asaf.

Victor gave the source of the problem btw. Apparently not PE.

"I'm afraid I don't understand the method of characteristics." has now become "Smith's fear of the method of characteristics". =)

It seems that my answer will never climb to the heights of its neighbors...

@AsafKaragila turn in your chimps...

Dude.

change of variables.

How did you ever get up to the point that you can ask the question of computing that integral without knowing how to do this? I quite often see these kind of things. I'm curious.

@Victor $x=r\cos(\theta)$ and $y=r\sin(\theta)$

@Victor: can you compute the Jacobian of that change of variables?

@robjohn - i still not sure and i can't compute the jocobian.

Santa Claus is Coming to Town...

Where would I find a list of conferences on say harmonic/stochastic analysis? I feel like going out 8-).

@JonasTeuwen COMS for example.

You should modify your script that it gives an inline preview, robjohn 8-).

@tb Thank you!

Too bad. Nothing useful for me.

Mariano has a pen-name? math.stackexchange.com/a/88832/13425. Apparently, Fernando Q. Gouvea.

@robjohn - what is jocobian?

$\dfrac{\partial(x,y)}{\partial(r,\theta)}=\begin{bmatrix}\cos(\theta) &\sin(\theta)\\-r\sin(\theta) &r\cos(\theta)\end{bmatrix}$

I know a guy that went to a conference like two times a month... I wonder how he did that if I see how many related conferences there are for me (0).

@Victor Have you heard of polar coordinates?

@JonasTeuwen Did he go to only the related conferences or unrelated ones also? =)

What area is they interested in?

@Victor That is the Jacobian of the change of variables. Its determinant is $r$.

@Srivatsan - yes, i did

Probably also unrelated ones. I want to learn to know more people in analysis.

It seems to me (I might be mistaken) that these conferences are more like social events to learn to know other people in the same field.

@Srivatsan that's a joke (I don't get it) but definitely a joke

I've seen a few talks and even if I should be able to understand the stuff the talk was too bad for me to understand what they said. 8-).

@tb Ya, it's kind of weird given that one of the two (same? =)) people is not in MSE.

Perhaps they know each other too well or something.

@Victor Thus, $\;\mathrm{d}x\;\mathrm{d}y=r\;\mathrm{d}r\;\mathrm{d}\theta$

@Srivatsan I've given up on understanding Will's "jokes" a long time ago.

@Victor: do you have the MathJax bookmark installed?

@robjohn @Victor, You can find some pictures in the wikipedia page.

[The "Calculus" section is relevant.]

@robjohn - that matrix is definitely too complicated, and how do i install the bookmark?

@tb That's a good thing to know, Thanks. =)

@Srivatsan Especially the Generalization.

@Victor Use this link

All all the conferences on the AMS calender?

I can find 0 on harmonic/Fourier analysis. I should switch 8-).

@robjohn - how to use it?

You have this habit of asking before reading, don't you, Victor?

@QED Why did you delete your question? =)

@Victor Do you know how to install a bookmark?

no

@tb :-)

@Victor what browser are you using?

Can't we just make a <a href="all that crap">click bookmark!</a>?

@robjohn - IE

Is anybody here using IE who can help install a bookmark?

@robjohn Good luck with finding someone using IE =)

Maybe you should make a .html page somewhere with that javascript-thingie as a link. Then people can just rightclick and press "add bookmark" or something like that?

@Srivatsan Thanks.

@JonasTeuwen the bookmark would affect that page, but not the chat page.

[At least in this gang, none uses IE afair.]

@robjohn Yes, but that way you can add it is a bookmark in your bookmark bar, no? Then you can press it when you're on this page.

Final vote needed: math.stackexchange.com/questions/89737

@Srivatsan That felt good.

@Srivatsan 8-).

(I stopped wearing contact lenses, what a misery!)

that didn't come out that correctly, scratch that. =)

Could someone please flag this answer and ask for merging the question account into the answer account + moving the answer as a comment to Ryan's answer? I accidentally flagged it simply as "not an answer".

@tb Done. "Not an answer" is also a valid flag in this case. =)

@JonasTeuwen I tried and it doesn't seem to work. When I drag the link to my bookmark bar, it tries to prepend http://... to the bookmark.

@Srivatsan Thanks! Sure, but I prefer to flag more specifically to make the moderators' job easier.

@robjohn Oh :(. I'm sorry.

@tb But not an answer is specific, no?

Do I also need to flag it?

@JonasTeuwen No, one flag should be enough :) thanks.

@Srivatsan yes, but less helpful

@JonasTeuwen no need to be sorry. I thought it was a good idea, but it doesn't seem to work the way I tried it.

There may yet be a way.

@robjohn And if you right-click and press "add bookmark"?

Oh, right, that is not an option here.

was that retag really necessary? ...

And how did Erdos die?

@tb I will press rollback. We don't need that tag.

@Srivatsan nothing too spectacular, as far as I remember. He was old and had a heart attack. Let me find a source.

@tb Did you read the 'gician's comment?

Lol, he rolled it back again.

"And considering how Erdos died,was it worth it"

He died in the saddle.

Ok, I am off, guys.

Actually which math conjecture you people want to focus to prove or disprove if you people have severe brain cancer?

@Victor Excuse me? Out of curiosity, are you always this pessimistic? =)

@Srivatsan Bye bye.

@Victor :D :D :D LOL.

@Victor Thanks. You've just made me horrible day better.

@Srivatsan See here for example.

@JonasTeuwen I added another tag and left a comment.