@BandeiraGustavo First and foremost category theory and logic. Coming with that, axiomatic set theory (forcing etc.) has been placed on my possible areas of future investigation.

@BandeiraGustavo (Referring to your comment on reading books) Hm, I would never do multiple passes in the same book. I think it would be much more beneficial to do a superficial pass in one book and then delve deeper into a different book on the same subject.

@Lord_Farin What was the name of that comic book you recommended again?

@user1 It's also an alternative.

@skullpatrol I'm almost sure I never did recommend one. But presumably you mean Logicomix, as it's the only one I know of that pertains to mathematics.

@BandeiraGustavo As the answerer says, these things do not apply to everybody, and I just feel that it is better for me to do it the way I described.

@Lord_Farin Yes, that's the one written in a comic book style :-)

Im going to read some stuff. See ya, guys.

@BandeiraGustavo Bye.

In fact, I'd better be going as well. Lunch, then thesis.

Bye!

later

@Lord_Farin I apologize if I offended you in any way by calling Logicomix a comic book, what I meant was a graphic novel, thanks again :-)

hey guys, anyone fluent in graph algorithms?

can you name me an algorithms that gives me a path in a connected graph with all vertices?

basically a hamilton circle without start node == end node

I don't know any graph theory sorry

damn :) np!

@mike I know some stuff about graph algorithms, what did you want to know?

@mike Oops didn't see this, what you're looking for is a Hamiltonian path, one way of looking at the problem is as follows: Add a new vertex in connected to all other vertices, find a Hamiltonian cycle in the new graph (by normal means) then remove the new vertex you added and you will be left with a Hamiltonian path in your old graph.

Hi all, what is the cardinality of the set of Natural numbers, is it 'infinity' or 'undefined'?

@vvavepacket It's $\aleph_0$.

@whats that N shaped thingy

@vvavepacket The aleph \aleph

i see

@vvavepacket But the definition of such concept seems to be kinda deep.

What set theory book are you reading?

How to Think Like a Mathematician - A Companion to Undergraduate Mathematics

@vvavepacket Is it a ST book?

I mean, having a ST chapter in the beginning does not make it a ST book.

Almost every math book has a chapter on ST or Logic.

No.

what does it do

@vvavepacket I'll give you a noobish definiton. Do not trust it fully, I dont trust my mathematical knowledge.

ok..

@vvavepacket The cardinality for infinite sets deals with a connection: Is it possible to enlist every number of an infinite set?

That means that for every number in a set, there is a 1 to 1 correspondence with the natural numbers.

For example:

Is it possible to enlist every number of an infinite set? ->> I think yes. Just list every number

@vvavepacket Yes. But that is only true up to $\mathbb{Q}$.

whats with the matrix

There is a proof called diagonal argument that shows that such a correspondence can not be made for the set of real numbers.

the first row represents the natural numbers

and the second row represents a set of it

@vvavepacket The matrix shows a correspondence between the naturals and the even numbers.

ohhh

ok

@vvavepacket Did you get it?

no

@vvavepacket What are you missing/not understanding?

I'm trying to figure out whats missing

@Bandeira, Ill get back to this later. Thanks

@vvavepacket You're welcome. =)

@vvavepacket One more thing: Cardinality is a different concept for infinite sets and for finite sets.

@mariano Are you around?

Ok noted

What is the easiest way to show that the solution of $1 = x \log x$ is $x = e^{\Omega}$, where $\Omega$ is defined to satisfy $1 = \Omega \cdot e^{\Omega}$.

@AlexJBest Thank you! ...Hamilton path instead of hamilton circle, who could've known...:D

is this $X\Y = {..U(-1,0)U(0,1)U..}$ the same thing as puu.sh/3kVdb.png

notice that I just put brackets, will it mean the same thing?

@vvavepacket For unions use \cup or if need be, \bigcup

@vvavepacket What do you mean?

@Peter is this $X\Y = ..\cup(-1,0)\cup(0,1)\cup ..$ the same thing as $X\Y = $\{..\cup(-1,0)\cup(0,1)\cup..\}$

@vvavepacket If you want to see the { } use \{ and \}

And the answer is no.

If $A$ is a set $A\neq \{A\}$.

The right hand side is a set with one element, the set $A$. The left hand side is the set $A$.

ok

then why would someone not use a {} here puu.sh/3kVdb.png

isn't it that all sets are represented by {then whatever inside here}

it could be {a,b,c,d} or {x:x bla bla}

why is that the braces has been omitted

Why would braces be put there?

its only weird that its the only set representation without braces

might be a correction

i have a book here and every set definition has a pair of braces

even empty set 0 = {}

@vvavepacket Well, but $(0,1)=\{x:0<x<1\}$, for example! =)

And so on for the others.

So what you have is that $$\Bbb R\setminus \Bbb Z=\bigcup_{k\in\Bbb Z}\{x:k<x<k-1\}$$

@PeterTamaroff: facepalm

@Peter are u there

so a brace is only use if you want a 1.) list of elements 2.) : such that

@vvavepacket Well, yes, the braces are used in those cases.

no more exceptions

?

@vvavepacket Well, usually it is accompanied with the "set builder" notation. That is $S=\{x:P(x)\}$ where $P$ is some property $x$ should have for it to be in the set. Have you heard about the "Axiom Schema of Specification"?

nope

Braces are read "the set whose members are"

You may use braces { } to show a set of numbers.

@vvavepacket Well, naïvely it is something like this: "To every set $A$ and to every condition $P(x)$ imposed on the elements of $A$, there is a set $B$ such that $x\in B\iff x\in A$ and $P(x)$ holds."

so many ways to write a single definition.

why is that

@Peter regarding your puu.sh/3kWP8.png, isn't it k-1 be k+1

coz if we let k = 6, then we have 6<x<6-1 -> 6<x<5 which makes no sense

how can x be greate than 6 and at the same time, less than 5

@vvavepacket Heh, I meant $k<x<k+1$ =)

i see

=)

@skullpatrol, how is that Peter was able to think of puu.sh/3kWP8.png when I wasn't? Peter was able to get another set definition than what I provided

dunno, maybe experience?

@vvavepacket One can define sets is many, many different ways!

experience?

@skull could be

do \{ for curly brackets

@skull what is the name of that set

its not x right?

x is simply a variable which can assumed a value 0,1,2,3

@Peter is 0,1,2,3 read as 0 and 1 and 2 and 3?

or, 0 or 1 or 2 or 3?

@vvavepacket Sorry I lost my connection ..

Read: "$x$ belongs to the set whose members are 0, 1, 2, and 3."

@JulianKuelshammer Thanks :-)

@vvavepacket No, the name of the set is not x, you could call it the set whose members are 0, 1, 2, and 3.

Given a finite dimensional topological vector space $W$ Which the topology Is considered? The one wich makes it homeomorphic to $\Bbb R^{\dim}$?

@vvavepacket You should read it as "0 [pause] 1 [pause] 2 [pause] 3"

@leo Don't you mean $F^{\dim}$ where $F$ is the underlying field?

Or is $\Bbb R$ the underlying field?

Recall $\in$ is read "belongs to" or "is a member of"

"Resources for learning mathematics for intelligent people?"

@Peter Don't know. It wasn't explicitly stated, so I assumed it was $\Bbb R$. Is that is the case, then my comment can be an answer here

@leo Heh, better stick my foot out of this before I start talking silly.

But the same argument works with any field I think

I am stuck with a problem now.

Gurrdarmn.

@Peter why?

Which one (why to your other comment)?

@leo I just gave the reason.

I see

@leo The problem is the following:

Suppose $\{x_n\}$ is such that $s_n=\sum\limits_{k=1}^n x_k$ neither converges nor diverges properly and such that $x_n\to 0$.

Then $\{s_n:n\geq 1\}$ is dense between $\limsup s_k$ and $\liminf s_k$.

Btw, it seems there is a lot of people learning topology stuff. Me being one of them :-)

@leo Heh, yes.

@Peter isn't that true if you change the series for any sequence with that convergence hipothesis?

I mean, the hipothesis $x_n\to 0$ seems to be there only for the "no diverges properly" thing, that is it is a necessary condition for that

@leo No, no. Look at $1/n$.

When we say "diverges properly" we mean the limit is $+\infty$ or $-\infty$.

Exactly

@leo See my latest answer @leo

@Peter what is your problem

But $1/n$ satisfies that the set $\{1/n:n\geq 1\}$ is dense between $\limsup 1/n$ and $\liminf 1/n$ because between those two there only the 0 which can approximated by sub sequences of 1/n

Oh! You counter example was for the other thing!

Yes that's right, $x_n\to 0$ it's not a necessary condition to do not diverges properly

@PeterTamaroff +1 on your latest.

By the way, the SE Chat Modifications inform me that we have passed 10M chat messages about an hour ago.

@Lord_Farin Yeah! =)

@PeterTamaroff Luckily, that SE network-wide, and not this room alone. :)

@Lord_Farin and soon we will have 1M removed chat messages ^^

@Lord_Farin We should announce this in meta: "Look, we have no fucking life at all!" =)

@PeterTamaroff innocent child's voice What is a "fucking life"?

@PeterTamaroff "who has no fucking life at all" can test the features of MO 2.0

@Peter neat answer

Must go

Bye

@leo Bye.

later

@leo Thanks, byes.

hi all

Hi.

Hm.., I finally understand how to copy pdf content to a webpage, without losing it's vector quality : dl.dropboxusercontent.com/u/58922976/pdftosvg2.svg

nice

@MarianoSuárez-Alvarez Drats.

I haven't figured how to fix this

If the sequence weren't of continuous functions one could make it nowhere convergent (just use what I made, but with indicator functions instead of triangles).

@PeterTamaroff I think you can fix it by letting the sequence run over the other sawtooths , as the comment suggests.

To finish it off, you can use two half sawtooths on the end as another term.

Can you imagine it?

hi @Lord_Farin did you get my apology?

@skullpatrol Yes. Don't worry. :)

That wasn't what I meant.

@skullpatrol Yes I saw it.

@Lord_Farin what did you mean?

@skullpatrol That I read cq. know so few comic books that it struck me as unlikely that I recommended one to anybody.

@Lord_Farin What does "cq" stand for?

@skullpatrol Casu quo.

A cq. B broadly means "A, or, if not A, B".

@Lord_Farin which is?

@skullpatrol GIYF.

Lit.: "in which case". But it is understood as "in the other case", as can be inferred from my previous comment.

@Lord_Farin I tried and it came up in the Dutch Wiki...

...pardon my lack of familiarity with other languages :-)

@skullpatrol Hm, consider it a quirk. :)

I always assume that Latin phrases are used in other languages too. Usually, it works.

@Lord_Farin thanks for explaining that to me :-)

@skullpatrol My pleasure. The expression is misused vastly more than it is used correctly. Ever since I learned that I always used it wrong, I try to use it correctly wherever I sensibly can. :)

@Lord_Farin How so?

@PeterTamaroff You want to construct an example that does not converge pointwise for any $x$, right?

@Lord_Farin Aye, but for which $\displaystyle\int_0^1 f_n\to 0$.

@PeterTamaroff Yes. So we conjure the sawtooths, one at a time (passing to finer tooths once we've run over them all).

how can i find all x_i for which a fixed reducible four degree polynomial satisfies |f(x_i)| is a prime. any advice. the polynomial is not provided.

@Lord_Farin Indeed.

But this leaves us with the points $k2^{-n}$.

@Lord_Farin Yes.

@Lord_Farin @PeterTamaroff ANY ADVICE GUYS.

So what we do is, we run over them again, but now we shift the tooths to the right by a distance $2^{-n+1}$.

@Shobhit No idea.

@Shobhit My advice: Less shouting, more patience.

@Shobhit If it is to be prime, all but one of the reducible factors have to be equal to $\pm 1$.

That should leave you with finitely many possibilities.

@Lord_Farin but the polynomial is not given, how can i put it equal to +-1.

@Shobhit You said it was fixed, so I assumed you knew what it was.

I should have read better.

@Lord_Farin Hmm... for $n$ fixed or variable?

@Shobhit But you can still do it algebraically, I think.

@PeterTamaroff Intuitively, you move the peaks to the valleys.

@Lord_Farin how i have been trying it for 2 days.

@Lord_Farin Yes, I tried that.

And then we should take $g_1,f_1,g_2,f_2,\ldots$, yes?

I don't know why I thought it wouldn't work...

I was in a rush, so I didn't had the time to think about it thoroughly.

@Shobhit There are four cases to consider. Those with linear factors should be easiest, but indeed, it may be hard. You can e.g. try to employ modular arithmetic under the assumption that one of these linear factors is prime.

So for $n2^{-k}$ the sequence should go $1,0,1,0,1,0,1,0,\ldots$?

@PeterTamaroff Yes, or first you take the "even" sawtooths for fixed $n$, and then the odd, but that's essentially just a permutation of your sequence.

@Lord_Farin Aha.

@PeterTamaroff Eventually, they will exhibit that behaviour (but not initially).

@Lord_Farin ok i will try that. thanks.

@Shobhit My pleasure. :)

@Lord_Farin Here

@PeterTamaroff Mh. You take one extra $g_i$ per $n$, which instigates a (conceptual) mismatch; you can "fix" this by putting the two half triangles in one $g_i$.

@Lord_Farin Come again?

@PeterTamaroff There are $2^n$ $f_i$s with width $2^{-n}$, but $2^n+1$ $g_i$s.

@Lord_Farin Aha. So?

@Lord_Farin Oh, you mean it is unclear?

@PeterTamaroff At some points, the $g_i$s will be wider than the $f_i$s they are in between. I think this is a bit ugly. But it does not fault the example, of course.

It's perfectly clear, but not as aesthetically appealing as possible. :)

@Lord_Farin But the $g_i$s have the same width as the $f_i$s, well, except the first and last.

I see your point.

@PeterTamaroff Yes, but since you take one $g_i$ more, there will be this mismatch (the $f_i$s having gone one level of width down, but the $g_i$s not yet).

@Lord_Farin Ah.

@Lord_Farin OK; I changed it.

You're right.

@PeterTamaroff This seems to happen surprisingly often. :)

Unfortunately, I can't upvote you for a second time.

@PeterTamaroff :)

One can avoid excessive starring by "highlighting" the conversation!

@Lord_Farin See?

@PeterTamaroff Nice. NB. I didn't star.

@Lord_Farin Now I have to leave.

Cheers!

@PeterTamaroff Bye! Have a nice day!

yo yo yo wassup

I feel dumb today

confused

@Jay hi

hi

@amWhy , hi :)

@MathsLover Hello there!

@amWhy , it's the first time i find you on chat !

or maybe , it is the case as i don't enter the chat room much !

I pop in throughout the day...always on "standby" ;-)

cool !

it takes me a months to think to enter the chat !

hi all

Did you ever have the chance to follow up with the math overflow link I posted in an answer awhile back?

hi vvavepack

how does one make a sine function given that it must pass points (-1,5) and (3,-2)

which link ? , when you add a link for maths overfollow on my questions , i open it and try to understand .

It was a question you haven't accepted an answer to...I was a little later than other posts.

@vvavepacket , really don't know ! but if it's the case for a polynomial then this is easy using solving lin equation systems .

tnx maths lover

oh , this one about finding a a group such that $G \cong G\times G$ ?

Yes, indeed: this one

i forgot to revise it ! that day was so busy and i was so tired so i decided to look at it the next day but i forgot ! , i will check it today.

There was a similar question on math overflow, and the answers were interesting...a couple of references, too, at MO

yes , i have opened the link now , but first i have to revise the definition and some proppeities which was mentioned in exercises in D&F

as i think it needs those definition and properties for constructing such group G .

@amWhy

@MathsLover Fine, fine...I was just hoping for an accepted answer ;-), but by all means, update, if you'd like!

@vvavepacket , your questions attracted me ! i hope i can know the answer ! , one idea is that if we can find a polynomial whose domaiin is [-1,1] and satisfy those conditions , then we can easily put the sin(t) instead of the variable x and we are done !

@MathsLover yeah it really boggles both our minds

@amWhy , we're done :) i will check the theard of maths overfolw today

flow*

@MathsLover You might enjoy visiting the site and seeing what's up there! (and thanks ;-)

@amWhy , i had an account on it from months , it was the first time to visit an english site which is concerned with maths ! so i discovered that this site was for graduate students and they there tell me about maths.SE so i joined the site here !

Good! Lots of people visit both. You'll see usernames there that are familiar to you.

gr8 !

@vvavepacket , i don't think that my method can work

as if it works then it reuqire that sin(t)=3 for some t which is impossible as the range of sin function is [-1,1]

@MarianoSuárez-Alvarez Hey did you see my ping yesterday?

@vvavepacket , it may require methods which i don't know , but i will try !

@BenjaLim hm, not really. Here in the chat?

The OP's answer is easy in the case $Y$ and $Z$ are just projective spaces

/me looks

because we know what $I(Y \times Z)$ is explicitly

in the general case

u have skype math_lover

@MarianoSuárez-Alvarez basically I ask this because I want to know why the hilbert polynomial of a product of projective varieties is the product of the hilbert polynomial of the individual varieties

@vvavepacket , is there anybody don't have skype nowdays ?!

this means , yes, !

@MarianoSuárez-Alvarez ping me when you have had a look

I talked to my advisor yestterday and the only way he knows how to prove it is a method involving terms like "lines bundles, Kunneth formula and taking global sections"

@vvavepacket , if you mean that my skype name is "Maths_lover" then this is not true !