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16:01
$X = \{(0, y) : y \in [1, -1]\} \cup \{(x, \sin(1/x)) : x \in (0, 1]\}$ be the topologist's sine curve. Prove that this is connected, but not path-connected. @Remember
This is hard, don't google.
I will try my best....
When you think you've come up with a proof, tell me why you think so, i.e., what's your geometric intuition.
For an inner product, is it true that $\forall x\in X[\langle x,0\rangle=0]$ - I want to say yes because $0=0y$ for some vector $y$, and if we remove that we get $\langle x,0\rangle=0\langle x,y\rangle$ (Not worth posting an answer, just want confirmed)
When the correct geometric intuition is there, there should be no problem in translating that to rigorous math.
@BalarkaSen ah the "topologists sin curve" - I''d be surprised if it were not on the front cover of his text book :P
16:03
yeah
I agree @BalarkaSen
@AlecTeal Yes. Follows from axiom of linearity on each coordinate.
Thanks, confirming stuff (even simple stuff) is important if it is "own stuff"
right, you never know when you're misunderstanding a defn.
@KarimMansour but the translation should be correct too :P
That's what I'm telling myself. I just ... on that wiki project of mine, I never commit a page with my own work on it without marking it as my own or putting a warning saying "unconfirmed" - now I can have a confirmed inner product page!
(as Ted would say)
16:07
@BalarkaSen I have an idea for the proof that it is connected .. Thinking about path connectedness
connectedness is easy.
write down a rigorous proof for that
tell me if you have thought about path connectedness
I have, sorta.
Okhay... Well it is better than having nothing at least :)
@Soham you sure if you haven't googled it? :P
16:09
good
I have a vague picture in my mind, is all.
yeah @BalarkaSen that is is same with me and the problem I was doing yesterday I thought I have already done it but when I was proving it the proof the way I did it didn't allow me to completely proof it
tell me (via mail) when you have a formal picture (i.e., a picture you can translate into math)
that is I was missing some stuff
so when you proof it that is when you know your right or wrong
do you guys keep binders of the stuff you solve ?
or do you solve it on paper and throw it away ?
solve it in a paper, and keep it in my head.
16:11
I have a stack (heh) of exercise books full of stuff.
@BalarkaSen you do forget. like Heine-Borel or whatever it was.
:P
yeah, I keep my notes, kind of.
yeah I think that is what I do too because too much work
heine borel
what about it
@SohamChowdhury I have the Tychonoff proof in mind.
oh, yeah, that.
so no need for me to go through all the laborious details of the B-W proof
16:12
anyway. back to studying chem.
me too
gonna study A-M
@BalarkaSen you are so selling me on studying ch. 5 :P
@BalarkaSen that's just cruel, man
@BalarkaSen can this be a reason for path connectedness failing :
The topologist sine is just a curve which is reverberating infinitely many times near 0?
what's cruel?
A-M vs. redox reactions
@Rememberme wtf
16:13
@Rememberme why does it imply it's not path connected?
@SohamChowdhury hehe
"reverberating"
that's just a silly word
you mean "oscillating"
even then, why should oscillating mean it's not path connected?
wtf... That was cruel @Soham
16:14
well, I think I have an idea.
say you have some point on the y-axis and some point on the main curve of the TSC, @Balarka.
then path-connected implies that the main curve "touches" the y-axis.
Hmm... I am thinking about it
16:15
@SohamChowdhury er, rigorize that
path-connectedness means that there is a path going all the way around the main curve and joins some point on the y-axis
Hmm... Got the reasoning..
Yes got the idea
But just have to write it
just give me the geometric idea
Rigorously
you don't have to write down the whole proof
@BalarkaSen exactly, that's the idea I had.
16:17
nah, if you can't do it geometrically, probably you have some flaw in your algebraic rigorous proof
@SohamChowdhury but how does that contradict anything?
it's just the definition of path connectedness i have written down there
yes, I have to figure that out.
I'm feeling a bit better now.
ohh.
the path has to be infinitely long?
learn the defn. of a path.
cts map $[0,1]\to$ something?
a path is a continuous map $[0, 1] \to X$.
so length is irrelevant
well.
is it that the path becomes discontinuous at the point where it joins the y-axis?
16:20
why should it be that?
that's why you need analysis, @Soham :P
what limits?
I am not going to reply to arbitrary words from now on unless you put them into a sentence.
the limit as you approach the y-axis along the path isn't defined?
precisely.
okhay...
Path connected is just that f(0) has some point on the y axis and f(1) has some point of the y axis..
Now in the case of the topologist sine curve it is oscillating infinitely many times than would mean it is touching the y axis after an.infinitely long time . Now if i try to find a point I wont be able to since we cannot find a point uptill infinity... ( I feel intermediate value theorem has some use but I cannot find an idea for using it )
16:22
$\huge{\text{OHHH MYYY GODD!!!}}$
with apologies to Chris'ssis
Hmm... Yes limits can be used.....
there you go : if a path goes all along the main curve and touches some point on the y-axis, then it must hit the y-axis somewhere around.
but the path along the main curve cannot converge.
that contradicts everything.
@Rememberme that's just silly.
righto. so that was correct?
16:23
Silly ?
It wasn't right :(
So I was wrong ? @BalarkaSen
same thing can be done with the infinite broom
I feel a lot better now. :P
That said, I shall go off for now.
@BalarkaSen let me check that out
@BalarkaSen was i wrong ??
i don't know. what you wrote down is not mathematically sensible.
no path from $(1,1)$ to $(1,0)$ on the broom, amirite?
16:26
Oh okay....
@Rememberme this is not physics, time does not matter. :P
@BalarkaSen was that for me?
oh, good.
Any connected neighbourhood of a point p must contain all a_i's ..... For the info it's broom @BalarkaSen
16:28
what's p?
what's a_i?
Ph let me tell for your picture....
I don't know what you mean.
in any case, I have to leave.
We can show that the infinte broom is not locally connected for a point p on it...@BalarkaSen
Yes we can do that
@Soham do you know about locally connectedness?
Ha I got the idea....
17:03
@Rememberme You can parametrize the curve so that the parameter is finite, but then the function is not continuous at $0$
No matter how small a $\delta$ around $0$ you make, the function will be both $1$ and $-1$ within that $\delta$ of $0$
@SohamChowdhury what was that about?
@robjohn I solved a little problem which Balarka said wasn't too easy, and celebrated just like Chris'ssis does.
@SohamChowdhury okay... I just saw the big exclamation and didn't know what was going on.
haha, sorry.
what are you doing?
well. is the answer small?
don't break my chat, please, if the answer is huge. :P
I know that integral by heart now, I've seen it so many times. :P
:D That one is a modified version, the advanced squared version.
@SohamChowdhury No.
eh, well, your secret is safe then. :P
@SohamChowdhury :D
@robjohn math.stackexchange.com/questions/1351862/… : it seems that is your answer style :P
17:22
Does anybody know how many numbers the Florida lotto has? The question that I'm doing is Finding the number of ways to match third prize in Florida LOTTO. There are so many different Florida lottery games that I don't know which one it is(The mega money, megaball etc).
@Rememberme sure, but I asked for path connectedness.
@Hippalectryon entrain de réviser les oraux ?
17:45
@BalarkaSen isn't the order topology on $\Bbb R^2$ different from the metric topology?
@Soham surely so? order topology is not even metrizable on R^2.
Anotherrrrrrrrrrrrr amazingggggg result!!!
I can't bear them anymore! :-))))))))
@BalarkaSen right, just making sure my intuitive picture is okay.
can you tell me a way to make this rigorous?
not right now, no. am not willing to think about it.
of course not. if I were in your place, I wouldn't either :P
still doing A-M?
17:52
yep
ch. 2.
it's the smallest such book I know.
apart from Serre's NT book, that is
it's super-terse.
but very dense and dry.
it would take you a few months to go through all of it.
you = everyone.
me included.
achha.
also, you have me sold on the importance of point-set stuff:
17:55
prof said upto ch. 4. is ok for Hartshorne ch. 1, though.
yes, of course.
@BalarkaSen you'll study AG now?
I will, later on.
Dunno, I may study it this year. May start on it earlier than I thought, depends on what prof thinks.
But I want to do ch. 2. for now and then move along, think about my problem, do topology.
I am not as jumpy as you :P
eh, you already have enough background in a subject to be able to think about a problem.
it's easy to be non-jumpy when you have a, uh, basepoint. (no pun intended)
I have none yet.
17:58
depends on your defn of a basepoint.
cool fact : I learnt Tor functors from Hatcher (a topology book, instead of A-M). getting more excited about cohomology.
something that you know a little bit about. I know absolutely nothing about anything yet. that's why I'm jumpy.
@BalarkaSen nice.
@SohamChowdhury yes, you do. you know quite a bit of algebra.
"quite a bit"
i dunno even what an ideal is
also, if you want to construct a basepoint, fix a basepoint
@SohamChowdhury learn it!
well, I do know what ideals are.
not too well yet.
18:02
ideals are an analogue of normal subgroups.
yes, Aluffi says so.
Maybe you wanna upvote a new nice question
0
Q: Calculating $\int_0^1 \frac{\operatorname{arctanh}\left(\sqrt{1-\frac{u}{2}}\right)\sqrt{\frac{2 \pi \sqrt{1-u}}{u-2}+\pi } }{u\sqrt{1-u}} \, du$

Chris's sis the artistWhat real tools would you employ for calculating the integral below? $$\int_0^1 \frac{\operatorname{arctanh}\left(\sqrt{1-\frac{u}{2}}\right)\sqrt{\frac{2 \pi \sqrt{1-u}}{u-2}+\pi } }{u\sqrt{1-u}} \, du$$

fun definitions : a prime ideal $\wp$ of $R$ is an ideal such that $xy \in \wp$ implies either $x \in \wp$ or $y \in \wp$. a maximal ideal $m$ of $R$ is an ideal such that there is no ideal $i$ such that $m \subset i \subset R$
(inclusions are strict)
fact : if $I$ is a prime ideal of $R$, then $R/I$ is an integral domain. if $I$ is a maximal ideal of $R$, then $R/I$ is a field.
prime ideal def parallels NT
like all of ring theory :P
no, not really all of ring theory.
but yeah, primes in $\Bbb Z$ are a way to think of prime ideals.
18:06
the definitions seem motivated by $\Bbb Z$.
well, basic defns, yes, because $\Bbb Z$ is an ideal (no pun intended) example of a ring.
people look at UFDs because, as I have mentioned before, one wants to do number theory over integral closures of higher number fields.
wwhat?
I suspect, if I finish Aluffi, I'll learn a fair bit of algebra. Have you ever skimmed the contents?
nope. I don't want to.
@SohamChowdhury ?
"integral closures . . ."
ah, I need to learn more.
"higher number fields" = $\Bbb Q(\alpha)$ for algebraics $\alpha$
18:09
oh.
"integral closure of this" = $\Bbb Z[\alpha]$
a group ring?
just a ring.
a ring ring?
oops
you adjoin an algebraic to your ring.
18:10
I thought $\Bbb Z[\Bbb Q(\alpha)]$.
ugh.
what does that even mean.
ring ring lol
= nonsense
Wow what the heck is that?
O_o
@BalarkaSen it's a group ring, leastways.
18:11
a ring ring
no, Q(\alpha) isn't a group
is when you ring at the door
so it ain't a group ring or anything
lol @BenDover
18:11
@BalarkaSen what is it, then?
1 min ago, by Balarka Sen
= nonsense
no, I'm serious.
define $\Bbb Q(\alpha)$.
why would you even associate a ring to a ring
i'm perfectly serious
@SohamChowdhury I just told you that uh, a day ago.
its just all polynomials with Q coefficients evaluated at alpha
son
18:12
oh.
vec. space generated over $\Bbb Q$ with basis $\{1, \alpha, \cdots, \alpha^n\}$
=ring generated by alpha over Q
oops, no inverses?
or Q-algebra generated by alpha
no, there is inverse
18:13
best way to put it
then why isn't it a group?
@Soham you're being silly. verify that there is inverse.
@BalarkaSen soham is trolling man
@SohamChowdhury there is the 0. if you take the multiplicative structure underlying it, then sure, it's a group
no, he's not. he's just confusing everything he ever learnt.
18:14
ah, thank god.
if $F$ is a field, $F^\times$ in general is a group
I'm just wondering if ring rings exist :P
keep wondering
@BalarkaSen yeah, that's a nice way to show that $(\Bbb Z/p\Bbb Z)^{\times}$ is a group
iirc
anyway, I defined a prime ideal above. collection of all prime ideals of $R$ is called $\mathsf{Spec} \, R$. it can be given a topology (closed sets are sets of all prime ideals containing a certain subset of $R$).
this is called the Zariski topology.
18:18
very useful in AG, right?
all your fonts are wrong, man
it's actually more than a topological space : at every point of $\mathsf{Spec} \, R$ (i.e., a prime ideal $\wp$ of $R$), there is an associated field $R/\wp$. so every point has a field attached to it. you can do the same with every open set (localization).
This is called a locally ringed space, actually. In general, an affine scheme.
hi @Ted.
schemes . . . sheesh.
Hi @Balarka ... How'd your meeting with Sir Prof go?
18:19
I just described you what it is.
@TedShifrin Pretty cool. Got ideas to use in my work, did a couple exercises from chapter 2 A-M (tensor products), talked a lot about topology, learnt what a derived functor is.
Woo, I got a new book which my employer mostly paid for! :D
Also, got to hear a few things about Bass-Serre theory (he's a geometric group theorist after all!)
Have you thought about tensor product and the Möbius strip?
hi @Clarinet
18:21
Morning @Ted
Now you're as bad as Mike, @Clarinet.
I am thinking. I can't make head or tails of it.
@Ted Oh man, what did he do again? I forgot :P
Not even when I know a bit about tensor products now.
How do you define tensor products of bundles anyway? You tensor the fibers. How does that give you a well-defined bundle?
As I said, you want to think about gluing together two copies of $(0,1)\times \Bbb R$, where the two intervals glue to make the circle. How you linearly identify the two different copies of $\Bbb R$ on the overlap of the two intervals tells you what line bundle you're building.
18:23
Hey everyone, does anyone want to hear my (quite possibly stupid) question about the continuum hypothesis?
@BalarkaSen The best way to understand this is to work with local trivializations.
Has anyone read Concrete Mathematics by Graham et al., by the way? That is the book I bought today
:)
Nope, sorry.
I had one volume of that, @Clarinet, but I gave it to someone years ago.
Once I get the money, I may purchase The Art of Computer Programming by Knuth
18:25
don't
@SohamChowdhury Why, out of curiosity?
an updated version is coming out, I hear
@SohamChowdhury Yeah, I heard something about that
with some modernized ASM. that's why.
@TedShifrin Right, sure. If you paste preserving orientations, then you have the trivial bundle. If you flip while pasting, you get the Mobius strip.
I still don't see how tensor product interacts with that.
18:26
it also tends to make people feel inconsequential in the grand scheme of things.
oh, wait.
You have to go back to my question yesterday about the induced map on the tensor products coming from $T\colon V\to W$.
What software does everyone use to open .gz files?
Right. Well, $T : V \to W$ gives you $T^k : \bigotimes_k V \to \bigotimes_k W$ by $v_1 \otimes v_2 \otimes \cdots \otimes v_k \mapsto T(v_1) \otimes T(v_2) \otimes \cdots \otimes T(v_k)$.
18:28
There are more mathematical ways to say that, @Balarka.
is that mathematical enough?
Yes, that's far better. So you should in fact be able, given a matrix representation of $T$ with respect to given bases, give a matrix representation of $T^{(k)}$ with respect to some reasonably natural bases.
Start with $\dim V = 1$, and then try $\dim V = 2$ if you want.
i'll do it, let me eat up quickly. i see what you're getting at.
18:32
Amazing :)
There's lego man.
You all assembled, @MikeM?
No, about halfway in. One of the pieces is too long for me to get a handle on myself.
Waiting for a friend to walk over and help out.
Should be done with the bed by lunch, the bedroom by early afternoon, the sofa by evening.
Better you than I :)
Oh, someone else liked my complements of compact sets answer. Maybe it was worth writing after all... ;)
18:39
I know not whereof thou speakest.
18:52
@Kyth'Py1k What is your question? Generally you should just ask, not ask to ask
ok, done. let me think about that problem, @Ted.
hi @PaulPlummer.
Is anyone familiar with the Tower of Hanoi problem?
Hello @BalarkaSen
how goes it?
i.e., how're you and what kind of math have you been thinking about?
Ehh.. I have been cleaning up and throughing out a bunch of stuff, making some decisions on what to keep and what I could bring when I move, looked around a bit for places.. Not much math. I think I am going to read some stuff on hyperbolic groups or algebraic topology in a bit
18:56
ok, cool.
I just got to know about Bass-Serre theory from prof.
Not so cool... (well I guess the math is)
yeah, I don't guess the cleaning would be cool.
Sweet, I don't know much about it. Did you figure out your whole van Kampen Galois stuff
I've'nt thought about van Kampen much. But yeah, figured out a lot of analogs between galois theory and covering spaces. I think I am even getting at a formal analogy between covering spaces and field extensions.
The contractability of $S^\infty$ was pretty cool.
19:00
yeah, how did you do it?
Basically contracting the "top" of $S^n$ for each level, in shorter and shorter intervals, and the top of $S^{n+1}$ contains all the lower dimentions, so those get contracted to a point. Exploit the topology that is given to $S^\infty$ to prove that it is continuous
Exactly. All you have to do is to contract $S^n$ in $S^{n+1}$ for each $n$.
And composing all the contractions.
question : does linking a sharelatex project to someone automatically gives him the ability to edit?
The problem would have been a bit more difficult without the previous problem (describe the subcomplexes)
I don't know if it does
19:08
no problem, it was a question to anyone who knows the answer in general
what did you learn about Bass-Serre theory? @BalarkaSen Hmm I wonder if there are any videos floating around on the subject
infinitesimal amount. it's all about universal cover of spaces where $\pi_1$ is amalgated product of groups.
@Paul For example.
Well any cool theorems or applications in particular.
Or just the broad brushstrokes of the ideas
When $X \vee Y$ is your space, i.e., it's fundamental group is free product of $\pi_1(X)$ and $\pi_1(Y)$ with zero amalgation, the universal cover is the tree obtained from attaching a copy of $X$ at each lift of a loop in $Y$, and attaching a copy of $Y$ at each vertex.
For amalgation, an analogous picture holds. The thing's called a Bass-Serre tree, and is some kind of a graph of groups.
@PaulPlummer Just the ideas, I don't know what use it can be of. Prof said it's used in geometric group theory a lot.
Cool. How did it come up? Is he trying to "convert" you :P
19:18
@PaulPlummer Okay, I was wondering why, very generally, the continuum hypotheses is independent of ZFC. math.stackexchange.com/questions/903536/… doesn't seem to help. So first off, in ZFC, the continuum hypothesis implies that there exists a bijective function between R and the set of countable ordinals.
Since the continuum hypothesis is unprovable in ZFC, there must be both a model fulfilling the axioms of ZFC that makes such a function exist and one that doesn't. But why is that? Is it that there is uncertainty as to what objects are in R, uncertainty as to what objects are in the set of all countible ordinals, uncertainty as to what functions exist, or am I just fundamentally mistaken?
no, no. not at all.
I was helping him working out some fundamental group exercises for his students.
He's going to do exercises from basic alg top for his first lecture in the geometric topology class this sem.
Also, I know PaulPlummer said one generally shouldn't ask to ask, but is it okay if I ask the above question on SE? I'm afraid it would get closed as it being unclear what I am asking.
It came up while we were discussing about universal covering of wedged spaces, @Paul.
I believe the basic idea is that bijection does not have to exist but all the axioms still hold. @Kyth'Py1k
You could ask the question but not sure how it would be received.
Basically you just don't have that much control over certain behaviors in ZFC. And in some sense you don't really have much control over what the objects are, if there are any models of ZFC, there are also countable models of ZFC, even though in those models we can still define ordinals and the reals, which will seem uncountable in the model. @Kyth'Py1k
19:43
How to simplify $P_d(n,k)=H_d(n,k)/k!$ with $H_d(n,k)$ defined as
$$H_d(n,k+1) = k! \sum_{j=0}^k \frac{ (-1)^j }{(k-j)!} \sum H_d(n/d, k-j)$$
where the inside sum is taken over all $d$ such that $d|n$ and for $d\geq 2$, is a $(j+1)$-st power ?
I have no idea how to deal with the recursion...
@Gato Oui (je suis à Supélec en ce moment)

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