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11:00 AM
It's not hard to do the former (which is intuitive), but I'm interested in a proof of the latter (which is un-intuitive at a glance, but can be pictured geometrically, and then translated into rigorous math).
 
@Hippalectryon One can do it with pen and paper easily. I use Mathematica just to ease my work.
 
It seems very long though
 
@Hippalectryon Yeah, but no worry about that. It can be done as I said, with pen and paper.
 
user147690
@BalarkaSen I'll go home now(at uni atm) and I'll check it out. If I google it am I going to get spoilers instantly?
 
user147690
11:03 AM
Okay cya later guys
 
yes, I think you will. don't google!
 
@AlexClark Did you speak?
 
@Chris'ssistheartist How did you start ? I'd be tempted to make some beta appear then do some substitution and play a bit with that, but I doubt it would give that huge result..
 
@Hippalectryon It's related to some of my important research tools, I can't tell you anything at the moment.
 
Ok :-))
 
11:05 AM
@BalarkaSen I can show that topologist curve is connected and not path connected
 
No, you don't understand path-connected well enough. You previously claimed that an interval is not path-connected.
If you can show that, I am betting you googled.
 
I am not... That's there in munkres...
That's the first part of the connectedness chapter....
 
Then you did look at the proof of Munkres.
I am not discussing path connectedness with you.
 
I am really not googling trust me...@BalarkaSen
 
Learn connectedness first.
 
11:09 AM
That's what I am doing@BalarkaSen and I am not cheating... What will i get cheating with you
 
hard to trust a person who already plagiarized a couple math of others
hoya @Soham.
 
I did that ... I don't do anything now. @BalarkaSen
 
it's not easy to believe that a person who says an interval is not path connected can prove that the topologists sine curve is not.
 
Why wouldn't intervals be path-connected?
 
@Hippalectryon I didn't use beta function, but you can tell me if it works.
 
11:11 AM
I don't know.
 
@BalarkaSen hahaha
 
Obviously they are path connected.
 
@BalarkaSen Do you recall if there is a name for "connected via line segments" which tends to show up in complex analysis?
 
nope, what's that?
 
heya, @Alex. long time no see, how's things?
 
11:13 AM
@BalarkaSen Well, like path connected, but the path must be a union of line segments (so one needs enough structure to make sense of that)
 
@BalarkaSen eh, so that we can help you with your analogy finding? The Sen program? :P
 
@BalarkaSen that was a mistake.. Didn't I prove everything else with quantifiers? Please do discuss pathconnectedness with me ... Not now but after few more days....
 
@Rememberme a few*
I'm not happy, I have to let off steam.
 
Yes, yes articles :P
 
@BalarkaSen I'm still waiting to read the motivation for the mathematics you do.
@Rememberme maybe you also share your thoughts.
Here is about love and passion for integrals, series and limits, I see the mathematics I do as an art.
@TobiasKildetoft maybe you also have some motivation for the mathematics you do.
 
11:18 AM
@TobiasKildetoft ok. weird.
 
@BalarkaSen I think it mainly simplifies some proofs (and in most of the interesting cases, one will have this stronger version anyway)
 
@SohamChowdhury nah, it's just that it'd presumably be fun. I studied Galois theory before alg. top., so I have been deprived of that fun.
 
@Chris'ssistheartist Sure, I have plenty of motivation for my research, stemming from a desire to better understand the characters of certain representations of algebraic groups
 
and it's silly to describe what I am doing by sen program. it has already been thought about and done.
 
@BalarkaSen eh, why? can you explain it to me?
@BalarkaSen can you spell "joke"? :P
 
11:21 AM
@Chris'ssistheartist I study algebraic topology because it deals with algebraic invariants of spaces, i.e., it helps to determine which spaces are not homeomorphic by using simple, easy to compute algebraic objects - like groups, modules and rings.
 
@Chris'ssistheartist it makes us happy. that's what all art (because math is an art, not a science) is about! :)
4
 
@SohamChowdhury Very well! (excepting the part with not a science) :-)
 
@Chris'ssistheartist I study topology (and want to study algebraic topology.. an many more) because it is a subject (I feel) which talks about stuff not bounded by the normal geometry....
 
Although what I am studying now is an analogy between galois theory and covering spaces, hoping to get to use some topological facts and analogize it in galois theory to understand the absolute galois group better.
 
@SohamChowdhury Wrong person to tell that to
3
 
11:23 AM
@SohamChowdhury maybe later. e-mail me to remind me about it, I have to run right now.
 
the derivative of the function expressing the number of stars on that comment isn't defined everywhere ;)
@BalarkaSen sure.
 
So I am doing something which doesn't have bounds and allows you to think free ... @Chris'ssistheartist
 
Why do people accept wrong answers?
 
Because they think it is right? @robjohn
 
a science makes predictions which may or may not be true.
whatever math says is always true (as long as you don't turn into Kronecker :P).
 
11:24 AM
@Rememberme I accept you do what you do because you like it, right?
 
Because it is not computations and has no fixed path to do it @Chris'ssistheartist
@SohamChowdhury Rightly said
 
@Rememberme topology is full of computations.
 
@TobiasKildetoft I don't see much difference in what you said, you talked about a desire that probably comes from passion.
 
I havent encountered one though@Soham
 
@Chris'ssistheartist we respect you. but it's not very nice to make fun of people who can't do your sums even though they're "high school stuff". just be nice :)
 
11:26 AM
@Chris'ssistheartist The difference is that mine is a desire to understand something broader. Yours is a desire to to calculations. I have no problem with your desire, but it does not constitute mathematical research
 
@Rememberme a lot of AT is about computing $\pi_k(S^n)$ (homotopy groups of spheres')
 
@Chris'ssistheartist So please dont be judgmental about other parts of maths...
I know that @Soham
 
then?
nothing wrong with computations.
 
But I like the proof part of it @Soham
 
@TobiasKildetoft Well, to do calculations, you sometimes need to do research, find new tools, new approaches, otherwise it were pretty difficult to do calculations. Maybe you might like to read, say, the first notebook of Ramanujan and then tell me again he didn't do research.
 
11:29 AM
I also like topology because it is a lot about Visualizations @Soham
 
@TobiasKildetoft you think what you do is more important and broader than what I do, but this is just an illusion, your illusion.
 
@Chris'ssistheartist I am not really that familiar with what he did, apart from his $\tau$ function. As far as I understand it, most of what he did was not really mathematics prior to being "discovered", due to the lack of any sort of proofs
 
@TobiasKildetoft You, like many other, shouldn't talk about integrals and series without reading in advance notebooks of Ramanujan say.
@TobiasKildetoft You see, you don't even know what Ramanujan did. I doubt you understand anything from his notebooks.
 
@Chris'ssistheartist I don't talk about integrals and series anyway. But why would I want to read an outdated book to be able to talk about it?
 
@Chris'ssistheartist For example quotient spaces .....
They are so much filled with visualizations ....
Stuff which we when children used to think were far apart are now being joined by a map... It so nice!!
 
11:32 AM
@Rememberme I never criticized you mathematics. You pushed me into such discussions a lot of times.
I never start such discussions with anyone of you, and I don't make fun of anyone, but let's then respect each other.
 
Its just that your are too judgmental about all other parts of maths and I remember @MikeMiller telling me the first day "never be judgmental about maths"@Chris'ssistheartist
 
@Rememberme I'm convinced that mathematics is amazing in all its branches, but I attend the area I like the most.
I'm totally fascinated by the mathematical connections amongst integrals, series and limits, it's the most fascinating thing I ever attended.
I like it, I love it, it's like a burning passion to me, I wanna learn more about it and develop new tools, strategy to attend harder and harder integrals, series and limits (that have never be calculated so far or not even invented).
 
@Chris'ssistheartist Now this is wrong. You have made fun of, say, Balarka, on many an occasion for not being able to do one of your "elementary" problems.
 
@SohamChowdhury Ask him and see if he ever felt I intended to make fun of him.
 
And Ramanujan did a lot more than just his (amazing) identities.
 
11:39 AM
@SohamChowdhury Definitely.
 
He worked on, say, partitions, number theory, and a lot more.
And it's for these that we remember him now.
His integrals were amazing, but they did little more than show how brilliant he was.
 
You have said to me that I have never entered any of my country's top places with outstanding marks.... @Chris'ssistheartist No offense but isn't that humiliation ?
 
@SohamChowdhury I wouldn't agree, I recollect Ramanujan for his integrals, series and limits.
@Rememberme No, since you were doing fun of me as far as I remember.
 
Yes it is.. you show that you are the most superior here
I was trying to tell you...
Don't be judgmental about maths
 
@Rememberme Really? What I said all the time? I'm just a self-educated, but I was pushed to say more that I didn't plan to ever say.
 
11:42 AM
Who knows someday another Taniyama Shimura might come and link your series integrals to something you never liked
 
that's two people, not one guy.
 
I know that
 
then you don't know English.
just kidding, don't get angry.
 
Yukata Taniyama , Boro Shimura
 
Why are people so agressive q_q
 
11:44 AM
No worries ... I am not angry
 
@Rememberme found the Riemann Hypothesis crank. :P
Jan 29 at 10:04, by Sayan Chattopadhyay
Well I want to prove the Riemann hypothesis that is why I am learning allghis
:P
 
Oh.. Yes thats one thing I would love to prove...
But,
 
Look who wants to become celebre here!!! :-)
Good to you!
 
AS Balarka quotes "Interests Change' @Soham
 
BBL (I have to finish some proofs)
 
11:46 AM
@Chris'ssistheartist Being famous is just a momentary gain ... If I had to pursue that I would have started doing series and integrals though :)
 
@Rememberme llloooooooollllllllllllllllll, you made my day with it!!!! :-))))))))
 
@Soham I presume you doing connectedness?
I am doing that though....
@Balarka Can I ask you a question about connectedness?
 
go on
but I have to leave in a second.
 
@Balarka I want an example of a Hausdorff without a point connected space
Does $R^2-(0,0)$ work
 
I don't know what you mean. You want a Hausdorff space such that removing a point gets you a connected space?
Sure, why won't R^2 - {(0, 0)} work? There are a lot of other examples.
 
11:59 AM
No a space without a point which is hausdorff and connected
 
define "space without a point"
it doesn't make sense
 
Punctured plane...
R^2-(0,0)
 
you want a punctured plane which is hausdorff and connected?
 
Yes
 
that's a tautologically trivial problem
because a punctured plane by definition is R^2 - (0, 0)
:P
 
12:01 PM
Oh okay.. haha
I got lost in my own words :p
Thanks for solving my dilemma .... :)
 
12:13 PM
@BalarkaSen I don't see why explanations above would be better great than mine, that I also addressed to @TobiasKildetoft, I wanna learn more about this part of analysis and develop new tools, strategy to attend harder and harder integrals, series and limits. Besides that, I wanna develop tools for solving elementarily most of the classical problems about integrals, series and limits that usually require tough approaches to be solved.
I can even talk a lot about my stuff once I arrange my thoughts in English. I don't see anything more fascinating, more motivating than what I have, attend, but I respect what you do.
 
@Chris'ssistheartist But you never do talk about your stuff. You only talk about specific integrals and series. Whenever the talk turns towards you research, you don't want to say anything
 
@TobiasKildetoft Well, just partial true. I might show you paper with my research but I prefer to remain anonymous (I have a lot unpublished yet).
But I don't feel the need to justify each of my word.
 
@Tobias I have a kind of proof to show you ....
 
@Rememberme What do you mean "a kind of"?
 
I am not sure it is a proof or not
 
12:18 PM
@M.N.C.E. nice approach to my problem! :-)
 
@Rememberme Ok, let me see it
 
So I got to prove that $\{A_\alpha\}\cup A$ is connected where $\{A_\alpha\}$ is the set of connected spaces of$X$ and $A$ is a connected subspace of $A$ such that $A\cap \{A_\alpha\}\neq \emptyset$
So I have though of a separation of $\{A_\alpha\}\cup A$ $Y=K\cup M$ where $K$ and $M$ are disjoint
 
@Rememberme Your notation confuses me. Do you really mean $\{A_{\alpha}\}$?
 
Yes thats what is given in munkres
 
@Rememberme Are you sure you do not want to take a different union then?
that union gives a set containing some subsets of $X$ together with the elements of $A$
 
12:24 PM
Wait for a sec.....
 
@Hippalectryon
 
It should be....
$\{\bigcup\{A_\alpha\}\}\cup A$ @TobiasKildetoft
 
@Chris'ssistheartist
 
I think it is fine now @Tobias
 
@Rememberme Right
 
12:27 PM
@Chris'ssistheartist That's fun !
 
@Hippalectryon :D
 
@Rememberme But what about the condition $A\cap \{A_{\alpha}\}\neq \emptyset$?
 
@Chris'ssistheartist Thanks.
 
Wow, traffic on my blog exploded because of the post on p-adics.
 
heh
 
12:28 PM
Okay so I take that separation... and say that $\{\bigcup A_\alpha\} \subset K\cup M$
 
@BalarkaSen thanks, man :P
 
@Rememberme You still need to clarify what that condition is meant to be
 
@Hippalectryon I think it's good to add it to my book. :-)
 
Yep :D
 
no problem. looking forward to seeing you putting a bit about short exact sequence in there.
 
12:29 PM
@BalarkaSen diagrams are hard to put up. but, sure!
also, doesn't every surjection have (in general) many right-inverses ("sections" afaik) by default?
 
i forget what is right and what is left, but yes.
 
"sections"
 
That means that If I take any set in the $\{A_\alpha\}$ and intersect with A it will always contain an element....
 
no, then that's not true
 
@Soham Can you link me to your blog?
 
12:30 PM
@SohamChowdhury yes, surjective maps will generally have many right inverses
 
@Rememberme link
 
but only as sets, not usually in more generality
 
Thanks
 
@TobiasKildetoft oh. i forgot that bit.
 
oh, yep, sorry.
 
12:31 PM
@Rememberme ok, so that should have been a "for all $\alpha$"
 
Yes its to tedious to write in latex when you have set theoretic notations
 
@Rememberme try commutative diagrams sometime :P
 
every surjective map has a right inverse. but a SES might not have a section (which is a group hom) : an example is $\Bbb Z/2 \to \Bbb Z/4 \to \Bbb Z/2$
 
Thats what I am saying its too tedious in latex
 
@SohamChowdhury Of course, in more generality, in categorical terms, "surjective" should be replaced by "epimorphism" which need not imply surjective, even in concrete categories
 
12:33 PM
@BalarkaSen right. yours was "show that $0\to A\to B\to \Bbb Z\to 0$" splits?
 
@Hippalectryon the idea is to do it elegantly without using error function in any way. :D
 
yes.
 
@TobiasKildetoft of course.
 
in AbGrp.
splitting is wildly different in other categories
 
@Chris'ssistheartist error function ? I only know that for economy-linked maths.. what do you mean here ?
 
12:34 PM
@BalarkaSen I prefer having the zeros. They make the fact that the $A\to B$ map is injective and the other one is surjective clear.
 
sure, you can have zeros.
 
@BalarkaSen right, I'll try.
 
Ooh that. I remember that from physics :D
 
in general, you can defined exact sequences $\cdots \to A_{k+1} \to A_k \to A_{k-1} \to \cdots$ similarly.
when these continue at the left or at the right infinitely, we say it's a "long exact sequence", but you needn't care about those right now.
 
12:35 PM
@BalarkaSen I skimmed the rings/modules chapter, so I know :P
 
@Tobias If I can show every element in $\{A_\alpha\}$ has an intersection with $A$ is connected then will it imply what I want to show?
 
okay, I have to go for now. toodles.
 
@Rememberme I can't parse that sentence
 
Even I cant :p
 
@Hippalectryon now I'm extracting a very nice integral ... wait
 
12:40 PM
What I mean to say is that if I can show that for all $\alpha_i$
$A\cup A_{\alpha_{i}}$ where $A_{\alpha_{i}} \in \{A_\alpha\}$ is connected then will this imply what I wanted to show @Tobias
 
@Rememberme Have you shown that the union of two non-disjoint connected subsets is itself connected?
 
Yes I have done that
 
@Rememberme this is really just an expansion of that to many sets
 
Yes and that is why I am asking will my method work @Tobias
 
@Rememberme You need to be a bit careful, extending from two to infinitely many sets
 
12:45 PM
okhay.... But anyways it will work right(If I do everything properly)?@Tobias
 
@Rememberme Well, that each of those unions are connected is clear. I am not quite seeing the precise argument why it implies what you want
 
Okay.........
 
@Hippalectryon and finally my research reveals the closed form to $$\int_0^{\pi/2} e^{-\csc ^2(x)} \csc ^2(x) \left(\text{Chi}\left(\cot ^2(x)\right)+\text{Shi}\left(\cot ^2(x)\right)\right) \, dx$$
 
That's crazy :P
 
So I have to go the normal way then.... taking a separation and then showing one of the sets in the separations in the set is empty@Tobias
 
12:51 PM
@Hippalectryon It's $0$. :-))))))
 
Yes @TobiasKildetoft The normal method works
 
@Rememberme Good
 
Though do check it once
 
@Hippalectryon It's weird Mathematica calculated the closed form below $$\int_0^{\pi/2} e^{-\csc ^2(x)} \csc ^2(x) \text{Chi}\left(\cot ^2(x)\right) \, dx=\frac{\sqrt{\pi } \log \left(17-12 \sqrt{2}\right)}{8 e}$$
 
12:54 PM
$\bigcup\{A_\alpha\}$ is in the separation $(K,M)$ then its either in K or in M . Now if $A_{\alpha_{i}}$ is in K then inductively $A_\alpha$ is in K . This shows M is nonempty...@Tobias
Is it fine? @Tobias
 
@Chris'ssistheartist What is that numerically ? (I mean mathematica's result)
 
@Hippalectryon I can calculate such integrals by solving clever systems of equations.
@Hippalectryon $-0.28735$
 
@Rememberme Don't do anything inductively
 
Okay so what should I do @Tobias
Whats wrong anyways with induction
 
@Chris'ssistheartist Why is it weird ?
 
12:58 PM
@Rememberme you need a well-founded set to do induction. What well-founded relation are you putting on the indexing set?
 
@Hippalectryon Well, in general I didn't see Mathematica calculating such tough integrals.
 
Oh..
 

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