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1:03 AM
2
Q: A line OA of length $ r$ starts from its initial position OX and traces an angle...

user541396A line OA of length $ r$ starts from its initial position OX and traces an angle AOB=$\alpha$ in the anticlockwise direction.It then traces back in the clockwise direction an angle BOC=3 $\theta $(where alpha is greater than 3 $\theta$).L is the foot of the perpendicular form C on OA. Also $\f...

 
@BalarkaSen Are you a professor, Sir?
just curious
What did you specialize in?
 
1:43 AM
@SubhasisBiswas No, I'm a first year undergraduate. I don't specialize in anything, but I like topology.
 
well, which college are you in?
 
ISI Bangalore
 
Well, it will take me at least 5+ years of work to reach at your current level
any advice on how to improve my mathematical abilities ?
 
Don't think about reaching any certain "level" too much, just be interested and keep doing math. What kind of math do you like?
 
analysis, mostly. I am still a big noob
 
1:53 AM
Nice, analysis is a good think to be interested in. If you want to learn it thoroughly I'd recommend you to go through the first seven chapters of Rudin, exercises included.
 
How much did it take you?
 
I didn't do it! But you should
 
I will. But I am very very slow paced and it takes way too much time to grasp on concepts.
I went through the first 4 chapters.
 
I honestly think that's all it takes to learn most of real analysis that an undergraduate can learn (multivariable calculus aside)
@SubhasisBiswas Oh, that's interesting. Did you do the exercises?
 
Quite a few. The problems takes sometimes 1-2 days to solve
you can go through my profile and see (how much my mathematics lacks rigour).
I am still very basic at it.
 
2:00 AM
Alright, then, I have a question for you.
 
ask away
 
Given any two bounded sequences $\{a_n\}$ and $\{b_n\}$ of real numbers, there's a way to measure how far they are apart. Namely, consider $\sup_{n \geq 1} |a_n - b_n|$. Do you know how to make this idea of "distance between two sequences" precise, and put it in a formal context?
(I suppose this will be a series of questions than a single specific question, but that's ok)
 
"distance between two sequences" : do you mean term by term distances?
 
Supremum of those, like I defined.
 
I need a bit more clarification on "formal context" (don't answer the question right away, let me try)
you'll need to be patient, I am a slow learner
 
2:10 AM
I don't seem to get this, what is the reason that when I have to define the degree of a smooth map I have to take the residue class modulo 2 of the number of elements in the preimage of a regular value? What's special in mod 2. Like I get that the number of points in the preimage of a regular value is locally constant and that motivates some definition out of it but why mod 2
 
odd or even maybe?
 
@SubhasisBiswas What's the precise formal meaning of "distance" in mathematics?
 
In a certain metric space, the distance is the output of a distance measuring function
it measures the "gap"
 
@Albas The number of points in the preimage of a regular value depends on the regular value, but it's residue class mod 2 doesn't, is the key reason. It's a topological invariant of the smooth map.
You can remove this mod 2 reduction at the cost of keeping track of certain signs at the preimage points coming from orientation reasons. This is the "oriented degree", and is independent of the regular value once again.
Look at Milnor's little book for some pictures and intuitions regarding this
 
Yea but like why? I don't seem to have any intuition for it.
Oh okay
 
2:16 AM
You can easily construct a map $f : S^1 \to \Bbb R$ with regular values $x$ and $y$ such that $f^{-1}(x)$ and $f^{-1}(y)$ have different cardinalities but mod 2 they are the same. Try it out
 
"The distance between two points can be regarded as the shortest length of the path possible joining the two"
 
@SubhasisBiswas Precisely. So consider the collection $X$ of all bounded real sequences and define the function $d : X \times X \to \Bbb R$ such that for any two points $\{a_n\}, \{b_n\} \in X$, $d(\{a_n\}, \{b_n\}) = \sup |a_n - b_n|$.
Is $(X, d)$ a metric space?
 
I am trying.. wait a while
other than the triangle inequality, everything is trivial
 
That is true. What about the triangle inequality?
 
Now, to prove the triangle ineq:

$|a_n -b_n|+|a_n-c_n| \geq |b_n-c_n|$. So, $\sup(|a_n -b_n|+|a_n-c_n|)\geq \sup |b_n-c_n| $ (we are taking the supremum of the sets of values).
supremum of $|a_n -b_n|$ and $|a_n-c_n|$ might not occur at the same $n$.
 
2:32 AM
I don't follow your last statement. The supremum need not occur at any specific $n$ in the first place.
 
so, $\sup |a_n-b_n|+\sup|a_n-c_n| \geq \sup (|a_n-b_n|+|a_n-c_n|) \geq \sup |b_n-c_n|$
 
That's correct.
Can you say $\sup |a_n - b_n| + \sup |a_n - c_n| = \sup (|a_n - b_n| + |a_n - c_n|)$?
 
No. Not always.
 
Think again.
 
(I guess)
wait wait.
 
2:38 AM
Oh yes it does. It does seem kinda beautiful. Though S^1 is a relatively easy manifold to deal with and the maps from S^1 to R are easy to visualise.
 
@Albas Easy examples do the trick, not the convoluted ones. :)
 
Lol yeah.
 
You should be able to now figure out from your example, whatever it is, the appropriate definition of oriented degree.
 
Yea as you said the "sign" needs to be taken into consideration
 
Correct.
So what is the (mod 2 or oriented) degree of a map $S^1 \to \Bbb R$? (I can say this without having to fix some specific regular value now, by fact)
 
2:43 AM
In few cases I got a 0... So 0?
 
That's right.
Consider the simplest map $f : S^1 \to \Bbb R$, just the height function on the standard circle $x^2 + y^2 = 1$ in $\Bbb R^2$. The preimage of a regular value (let's take something which is in the range of $f$) contains two points in general.
 
Cool.
 
These will have opposite orientations, $+1$ and $-1$ (think of tracing the circle by your finger eg clockwise, then note that the direction you are crossing one is opposite the direction you are crossing the other)
 
Yea
 
So the oriented degree will be $(+1) + (-1) = 0$.
 
2:45 AM
Thats the example I took
 
Yes, the equality does not always hold.

for example... $7, 3, 8, 10, 0,0,... (b_n)$ $5, 6, 3, 2, 0,0,0...(a_n)$, $13, 2, 9, 4,0,0,...(c_n)$

$\sup (|a_n-b_n|+|a_n-c_n|)=11$, $\sup |a_n-b_n|=8$, $\sup|a_n-c_n|=8$
 
So for my summers these people have asked me to do degree of maps and stuff from Milnors book and then do some symplectic geometry later in the project. Lots of stuff to do.
 
@SubhasisBiswas Good example. But what if $\sup_{n, m} (|a_n - b_n| + |a_m - c_m|)$ was taken?
 
Then the equality holds good
 
@Albas Cool, keep me updated about it.
@SubhasisBiswas Can you prove that?
 
2:51 AM
Yea, how good a book do you thing Berndt is for Symplectic topology. I am not really a fond of it but I can't McDuff's book because it is too difficult for me at the moment.
 
I don't know any symplectic geometry to be honest. There's a good summary of results/crash course outline in Eliashberg-Mishachev's book on h-principles that I scroll through occasionally.
 
I see. Will have a look.
 
@SubhasisBiswas That side question aside, let's take a look at what you have proved so far: $(X, d)$ is indeed a metric space. This is known as the space of bounded real sequences with the "sup metric".
 
fix $k=3 $ (say). $\sup_{n,k}(|a_n-b_n|+|a_k-c_k|)=(\sup |a_n-b_n|)+|a_k-c_k|$.

Let $\sup|a_n-b_n|=\alpha$

Now, we unfix $k$ and take sup again
 
What do you mean by fixing $k = 3$? The equality you wrote is not correct.
 
3:00 AM
I meant:

$\sup\ {f(x)+3\}=\sup f(x)+3$
similar to this
 
Ah, $\sup_n(|a_n - b_n| + |a_k - c_k|) = \sup_n |a_n - b_n| + |a_k - c_k|$, then. The subscript $k$ in the first sup shouldn't have been there.
 
fix $k=3 $ (say). $\sup_{n}(|a_n-b_n|+|a_3-c_3|)=(\sup |a_n-b_n|)+|a_k-c_k|$.

Let $\sup|a_n-b_n|=\alpha$

Now, we unfix $k$ and take sup agai
yeah. typo
 
That does the job, very nice.
In general your proof shows that for any two subsets $A, B \subset \Bbb R$, $\sup(A + B) = \sup(A) + \sup(B)$, which was my punchline in this digression.
 
:

equivalent to the case when we vary $n$ and $m$ of the sequence simultaneously and take the sup?
 
Well, in case $A$ and $B$ are the absolute termwise differences as above, sure.
 
3:06 AM
okay, now how did you get so good?
You cracked an exam where Phd s and Masters compete
 
Next question: Consider the subset $Y \subset X$ consisting of convergent real sequences. Is it a closed subset?
@SubhasisBiswas lmao I don't know what exam you are referring to
I certainly didn't take one in my recent recollections
 
ATMW differential geometry
 
You don't need to take any exam to go there.
Those are random workshops, happens all the time throughout the country. I just knew people who can give recommendations
 
you are more than qualified
I am trying your question
wait
well, every sequence in $Y$ converges to some $l \in \mathbb{R}$ . Now, we consider the constant sequence $(l)$
bounded & convergent. So, in Y
 
I agree. Then?
 
3:20 AM
so, the limit of a sequence can be considered as a constant sequence
 
That's not what convergence in the metric space $(X, d)$ means
 
okay
let me try again
 
Write down precisely what it means for a sequence of sequences $\{a_{n, 1}\}, \{a_{n, 2}\}, \cdots$ to converge to a sequence $\{a_n\}$ means in the sup metric.
 
I forgot to ask you Balarka, how was the Probabilistic methods in Negative Curvature thing in ICTS?
 
Oh quite nice. I only attended the first week though.
 
3:27 AM
I see. A friend of mine was supposed to go there but wasn't able to for various reasons.
 
There wasn't any hardcore probability involved, surprisingly. They prove lots and lots with techniques that are quite standard in probability
@Albas Ah, I see.
 
Hmm interesting. What kind of stuff can you do with probability in things like differential geometry roughly? Whenever I hear probability and stuff in diff geometry my mind straight away goes to dynamical systems
I just took a pretty basic course in probability last semester, hope to strengthen it with concepts from measure theory in my next semester.
 
@Albas Mostly these people do probability & geometry in the discrete setup, eg studying random walks on group.
 
A sequence of sequences $\{\{a_n\}\}$ is said to converge to a sequence $\{x_n\}$ if,

$\forall \epsilon>0$, $\exists k \in \mathbb{N}$ such that $d(\{a_n\}_n, \{x_n\})< \epsilon$ $\forall n \geq k$
 
The geometric side of the story is hyperbolic groups
I don't understand the differential geometric side of the story that much, but there you do Brownian motion on (negatively curved) manifolds
@SubhasisBiswas Should use $n$ and $m$, different scripts, for avoiding confusion. Yes, that's correct, where $d$ is the sup metric, remember.
 
3:34 AM
Oh interesting.
 
4:21 AM
can we visualize/regard these sequence of sequences as ordered tuples in $\mathbb{R^\infty}$?
 
You're certainly going to have hard time visualizing anything in $\Bbb R^\infty$.
Just do the calculation if you don't have a good mental model for what's happening.
 
Lmao the top of the starboard
Also yo
 
wait.... correct me if I am wrong. Here is my intuition :

As these sequence of sequences (all of them are convergent) approach a sequence, $\sup |a_n-x_n|$ becomes lesser and lesser. Now, this sup metric enforces that the limiting sequence must also be convergent.
 
hey @Daminark
@SubhasisBiswas OK, can you make that into a proof?
 
Wait. there's one more point.

Now, in order for this sequence of convergent sequences to be convergent, after a very large $n$, all of their individual limits will be almost same (Cauchy Criterion).
before getting into writing the proof, am I mistaken somewhere?
 
4:29 AM
So what do you guess will be the limit sequence's (so $\{x_n\}$) limit in $\Bbb R$?
 
What is $l$?
 
where $l$ is the limit of the convergent sequences
after a large $N$
 
The convergent sequences need not converge to the same limit.
 
example?
 
4:38 AM
I mean think about it for a second. I have a sequence of sequences $\{\{x_{n, m}\}_n\}_m$ such that for any fixed $m$, $\{x_{n, m}\}$ is convergent, and the whole sequence $\{\{x_{n, m}\}\}$ converges in $X$ in the sup metric. Why on earth will $\{x_{n, m}\}$ for each fixed $m$ have the same limit?
I don't understand what your thought process there is. You shouldn't require an example to understand this.
 
I will come back to it later. Don't answer it right now
cya
off to college
 
Talk later
 
Ehh ... Homology pisses me off.
Why is there so much of algebra ... It's fun to draw diagrams but you lose all the geometric fun in those exact sequences.
 
Nah with practice you can understand the exact sequences geometrically at the level of chains
 
5:10 AM
Balarka doing PhD level functional analysis
damn
 
?
u mock me bro
 
partially
not really mocking
 
ill get u at ur elliptic regularity game one day
not anytime soon
I need to learn proper analysis man.
 
we'll see
if you can figure out GMT, we'll all be lost
 
lmao
 
5:13 AM
you will have achieved mathematical zen then
someone needs to explain Almgren's computation of the homotopy groups of the current cycle groups
might as well be you
 
jesus
 
@BalarkaSen you still didn't answer my question
and I have another one
if $M^n$ (smooth) contains an $S^{n-1}$ (smooth, standard structure) which does not separate, can we factor an $S^1\times S^{n-1}$ (standard structure)
emphasis being standard structure
topologically it's clear
 
Yes.
 
why
if you lift the argument from Hempel it's not clear that you get the standard structure, is it?
you're gluing the tubular neighborhood of the curve
is there only one way to do that?
actually what I said isn't true, one probably also needs orientability
then one also needs to know how many $S^{n-1}$-bundles there are over $S^1$
are these things obvious?
 
@LeakyNun I have one question. Why should $x$ be a member of some simplex?
 
5:23 AM
Right, you need orientability. You're gluing a standard $D^{n-1} \times I$ to $S^{n-1} \times [-1, 1]$ standardly by the attatching spheres, which gets you a standard embedded handlebody $D^1 \times D^{n-1}$ in $M$.
@RyanUnger No, just need to figure out how many $D^{n-1}$-bundles there are over $S^1$.
There is only one which is orientable
 
@BalarkaSen why's that
 
@RyanUnger cover $S^1$ by upper and lower hemicircles, over which the $D^{n-1}$-trivializes. It only remains to count the number of ways to glue it at the equator, which is determined by a map $S^0 \to \text{Diffeo}(D^{n-1})$.
 
yeah sure
why does Diff(D^n) have two comps
I guess cuz D^n is contractible?
I don't buy that without more explanation
 
Every diffeomorphism of the ball is isotopic to a linear diffeomorphism, so the inclusion $GL_n \to \text{Diff}(D^n)$ is an isomorphism in $\pi_0$.
 
oh. is that in hirsch?
 
5:30 AM
It's obvious, right? Take $f : D^n \to D^n$, let's say $f(0) = 0$ (you can arrange this). Then $f$ is isotopic to $df_0$
Try $f_t(x) = f(tx)/t$
Ya works
 
sorry it's not obvious to me that $df_0$ maps into the ball
 
That's no problem, you can do 1/ten million*df0
 
ok sure that was the question
ugh this is in Hirsch and I was looking at it last week for another reason
thanks, I need to sleep
night
 
G'night
 
 
3 hours later…
8:32 AM
Mornin'
 
9:21 AM
Sup
Well, @BalarkaSen is the answer "yes" to the question?(that it is a closed set)
I have arrived at:
 
10:08 AM
for all $\epsilon>0$, $\exists k \in \mathbb{N}$ s.t $\sup_n |x_n^m -a_n| < \epsilon/2$ $\forall m\geq k$ [$(a_n)$ is the limiting sequence].

[Intuitively, the terms of the sequence of sequences gets more and more identical to that of the terms of the limiting sequence]

Therefore, $|x^m_n-a_n| <\epsilon/2$, $\forall m\geq k$ , $\forall n\in \mathbb{N}$ [since the inequality holds for the supremum, it should hold for other values too].


Again, let $\lim_{n \to \infty} x^m_n=l_m$.

Therefore, for $\epsilon>0$, $\exists k^*>0, $ s.t $|x_n^m-l_m| < \epsilon/2$, $\forall n \geq k^*$
@BalarkaSen
 
 
2 hours later…
11:39 AM
I know if it was commutative ring with unity, then it would surely be a field. And char was prime.
 
 
2 hours later…
1:40 PM
trying to figure out what branch of math this might be classified under.
Consider the following table of numbers:
9 7 5 3 1
1 3 5 7 9
9 5 1 3 7
There's two properties here. 1) Each row contains the first five odd integers. 2) There's some way to add/subtract the numbers in each column such that the result is 1. (Doesn't need to be the same combination of add/subtract for each column)
I think there's a generic construction here.
First line: Start from 2n+1 and go down in steps of 2,
Second line: Start from 1 and go up in steps of 2,
Third line: Start from 2n+1 and go down in steps of 4, taking absolute values as necessary
 
1:58 PM
Let $X$ be a compact Hausdorff space and let $T: X\rightarrow X$ be an arbitrary continuous function, how can I prove that there exists a nonempty compact $K \subseteq X$ such that $T(K)=K$ ?
 
2:17 PM
@SubhasisBiswas You should be careful about the $k, k^*$ as $k$ is in general dependent on $n$ and $k^*$ is dependent on $m$. Does it cause a problem in your proof, or easily fixable?
@Eran Try $K = \bigcap_{n \geq 1} T^n(X)$
 
K can't be empty
 
By Cantor's intersection theorem, it isn't, no.
It's obvious that $T(K) \subseteq K$, but you need some argument to show equality.
 
What would be that argument
 
asking balarka for too much is dangerous
 
either way I'm clueless
I have nothing to lose :D
 
2:25 PM
The argument is straightforward -- you've already used continuity
try applying $T$ to both sides of $K=...$
 
@Ryan $f(A \cap B) \subseteq f(A) \cap f(B)$ though, so that only gives $T(K) \subseteq K$.
 
$T(K) = T(\bigcap_{n\geq 1} T^n (X)) \subseteq \bigcap_{n\geq 1}T( T^n (X)) $?
The last one is not subset of K though
 
It is, it's $\bigcap_{n \geq 1} T^{n+1}(X) = \bigcap_{n \geq 2} T^n(X)$.
 
wa-what?
less sets in the intersection ---> greater set
 
$\{T^n(X)\}$ is a decreasing sequence of sets in $X$
 
2:32 PM
Maybe you do need compactness again
 
So $\bigcap_{n \geq 2} T^n(X) = \bigcap_{n \geq 1} T^n(X)$
 
Why are they decreasing?
 
@RyanUnger Let's take $X$ to be a compact metric space first. Given $x \in K$, I can write $x = T(x_1) = T^2(x_2) = \cdots$. The sequence $\{T^{n-1}(x_n)\}$ has a convergent subsequence, say $\{T^{n_k - 1}(x_{n_k})\}$, which converges to some $y \in X$. So $T^{n_k - 1}(x_{n_k}) \to y$ implies $T^{n_k}(x_{n_k}) \to T(y)$. But $T^{n_k}(x_{n_k}) = x$ for all $k$, so what that means is $T(y) = x$. Just check now that $y$ is indeed from $K$.
 
Balarka I've got it
I'm thinking about disks again now
 
balls u mean
@Eran If that's not clear think about it for some time.
 
2:37 PM
What if X is not metric?
 
A compact Hausdorff space is almost a metric space ;) I haven't thought about the details without a metric.
Why don't you try that case?
 
I see...
I understood what you said when it's metric, however it's not metric in my question
 
ok so I buy that you're gluing $D^{n-1}\times I$ to $S^{n-1}\times I$ along the boundary balls
now because this is orientable and there's only one way to glue balls up to orientation, this is like gluing the boundary balls in the right way
 
Yeah, by a standard map $D^{n-1} \times \{0, 1\} \to S^{n-1} \times \{0, 1\}$ as well.
Right.
 
so why is the boundary of this glued guy a standard $S^{n-1}$
I mean it won't even be smooth if you do the gluing wrong
 
2:43 PM
So this glued guy is nothing but an $\epsilon$-neighborhood of the subcomplex $S^{n-1} \vee S^1$ in the standard $S^{n-1} \times S^1$
 
yes
 
so what's the beef? that has standard (n-1)-sphere as boundary
 
@BalarkaSen is $D^{n-1} $ is some sort of differential operator ?
 
(n-1)-sphere yes. But why standard
 
Because what's happening at the boundary is you're gluing a standard $I \times S^{n-1}$ to a standard $S^{n-1} \times \{1, -1\}$, which is the same as taking connected sum of two $S^{n-1}$'s by a standard tube.
The standard sphere is the identity element in the group of exotic spheres :3
 
2:50 PM
@BalarkaSen, is my proof alright?
 
I wrote some comments above, which you should look into.
 
ok viewing this as a connected sum is probably smarter. So the boundary of the nbhd of the S^{n-1} is two spheres but then adding the S^1 makes a connected sum
 
ok thanks
 
@BalarkaSen Re: that description you suggested for that circle diff thing
One review article contains the following remark plus a footnote:
“In that paper [of Thurston], earthquake theory is devel- oped in the setting of the universal Teichmüller space, that is, the space parametrizing the set of complete hyperbolic metrics on the unit disk up to orientation-preserving homeomorphisms that extend continuously as the identity map on the boundary of the disk.“
 
2:56 PM
Aha
These hyperbolic geometers keep talking about the earthquake map which I have no clue what it is but it's a great name lmao
 
Footnote (will edit): “A common trend is to call DiffC.S1/=PSL.2;R/ the physicists universal Teichmüller space and QS.S1/=PSL.2;R/theBersuniversalTeichmüllerspace. Here,DiffC.S1/denotesthegroupoforientation- preserving homeomorphisms of the circle and QS.S1/ its group of quasi-symmetric homeomorphisms, of which we talk later in this text.”
 
@BalarkaSen, I don't see any problem with it (maybe I am missing something obvious). The final form becomes $|a_n-l_m|< \epsilon$ and $m,n ≥ g(m,n)$. Where $g(m,n)= \max\{m,n\}$.
 
But yeah this seems to agree with my thought that Diff+S^1/PSL2R is the space of orientation preserving diffeomorphisms of the circle which extend conformally to the hyperbolic disk
 
This should not be a problem since "the place of choice of $m,n$ is a common region where the inequality holds for both of the independent $m$ and $n$
 
Blah, I can’t edit fast on my phone
 
3:01 PM
There's probably some description of this thing as the space of tuples (complex structure on upper hemisphere, smooth structure on equator, complex structure on lower hemisphere) modulo when they glue appropriately (when they glue to a complex structure on S^2?) but both me and Mike were unsure how to formulate it.
 
And, the keyword is "for all $m$ and $n$". We don't need to look back and watch if we are missing out any of those $m$ and $n$ s, after a certain $g(m,n)$
 
@BalarkaSen *modulo when they extend conformally...
 
What perplexes me is that I’ve seen multiple assertions about the “physicists” UTS as being the set of (orientation preserving) diffeos, normalized to have three fixed points on the unit circle
But don’t Mobius transformation generically have just two fixed points?
 
@BalarkaSen , I don't understand 99.99% of the discussion here, but is $PSL2R$ the projective special linear group over the reals?
 
3:07 PM
@SubhasisBiswas You mean $g(m, n) = \max\{k, k^*\}$. But why is this sufficient to conclude $a_n$ and $l_m$ converge (to the same limit)?
 
Hmm, maybe it’s fine. Can I write a Mobius transformation that sends three arbitrary z to 0,1,infinity? If I can do then the claim makes sense
 
It's not a good idea to have a constant depending on your variables hanging around in your computations.
 
Oh, I guess I can invert that.
 
Yeah you can do that
There is a unique Mobius guy which sends a given trio of points to another given trio of points
 
Ahah. That actually explains my other confusion as well: the only Mobius transformation with three fixed points is the identity
Ok, not confused about that any more.
 
3:16 PM
@BalarkaSen How can I show that K \subseteq T(K) ?
I understood why they are decreasing it was the most stupid question I have ever asked in my life @BalarkaSen
 
@SubhasisBiswas This is sort of subtle from a symbolic point of view; here's what: $k^*$ might depend on $m$, but $k$ doesn't (that's what convergence in sup metric means). That's the key idea.
 
@BalarkaSen since $\sup_{m,n} |a_n -l_m|≥ \sup |a_n -l_n|$
So, $(a_n) \to (l_n)$
I get the idea behind it, but I can't communicate this properly.
Is this good enough?
 
@Eran OK, I think the sequential proof can be slightly abstracted out. Fix some $x \in K$ and look at $A_n = T^n(X) \cap T^{-1}(x)$. $\{A_n\}$ is a decreasing sequence of compact sets again in $X$, so contains some $y \in \bigcap A_n \subset K$. That is your $y \in K$ such that $T(y) = x$ :D
 
thanks
 
@SubhasisBiswas That's not good enough, $\sup_{m, n} |a_n - l_m|$ might be large. You want $\sup_{m > k, n > k^*}$ but then again you run into dependence issues. Use what I pointed out: $k$ doesn't depend on $n$, it's a universal constant.
 
3:37 PM
Okay. I am trying to write down the piecewise function corresponding to the following graph...Let $r \in [0,1]$. From $0$ to $r$, I want to the function to be $y=x$; then a vertical line connecting $(r,r)$ to $(r,0)$; and then $0$ from [r,1]$.
That middle portion is giving me trouble.
I don't know why it is giving me so much trouble...
 
@SubhasisBiswas Also, yeah, that's it.
 
Also what?
 
PSL2R is indeed a projective special linear group
 
As I go through my proof again, I realize that I entirely missed the point you mentioned. Very subtle indeed the fact that $k^*$ does depend on $m$.
How do you even do this magic. Were you born with this or is it just a result of devotion and dedication to the subject
 
Can anyone help me solve this? I am stuck at finding/solving variance
 
3:47 PM
i mean these are standard subtleties in analysis; i learnt by making mistakes
it's all in Rudin if you do convergence of sequences and functions
 
:50360971
How do you know the intersection T^n(X) \cap T^-1(x) is not empty?
 
@Eran 'Cuz $x \in K \subset T^n(X)$
 
right
 
I am stuck at finding the variance term inorder to find the error term
 
Now, at some point (in the proof) I mentioned that the inequality holds when $m≥k$ and for all $n$. Doesn't this remove the dependency of $k$ from $n$?
 
3:59 PM
@Subhasis Yes, the point is, $k$ never depended on $n$ in the first place. For all $m \geq k$, $\text{sup}_n |a_{m, n} - a_n| < \epsilon$.
That quantity in the sup doesn't depend on $n$. So when you extract out $|a_{m,n} - a_n| < \epsilon$ for all $m \geq k$ and for all $n$ from there, it's the same $k$. No dependence on $n$.
That's why convergence in sup metric is called "uniform convergence" by the way, contrary to "pointwise convergence" of sequences, where there is a dependence.
(Pointwise convergence just means $\lim_{m \to \infty} a_{m, n} = a_n$ for all $n$)
 
Okay. I am trying to fix my proof. What we have so far: $k$ depends on $m$ but not on $n$. $k^*$ depends on $n$ and $m$. Right?
We need to find such a constant $u$, that whenever $m,n≥ u$ implies the inequality.
???
 
4:19 PM
Let's write it all out precisely. Say $\{a^1_n\}, \{a^2_n\}, \cdots$ is a sequence converging to $\{a_n\}$ in sup. $\{a^m_n\}$ has limit $l_m$ for any fixed $m$. Then your technique is: $|a_n - l_m| < |a_n - a^m_n| + |a^m_n - l_m| < \sup |a_n - a^m_n| + |a^m_n - l_m|$. There is a universal constant $k$ such that $\sup |a_n - a^m_n| < \epsilon/2$ whenever $m > k$. There is a constant $k^* = k^*(m)$ depending on $m$ such that $|a^m_n - l_m| < \epsilon/2$ whenever $n > k^*(m)$.
Combining we have that for all $m > k, n > k^*(m)$, $|a_n - l_m| < \epsilon$.
How do you argue from here that $\{a_n\}$ converges, now? Remember that that was the goal.
 
what are you proving?
 
My stupidity :p
 
I asked him to prove convergent sequences are closed in $\ell^\infty$.
 
Wait @BalarkaSen, I am trying. Don't tell the answer.
 
Sure take your time
@RyanUnger Some drama ensued in JEE chemistry room lol
 
4:27 PM
how do I access that
do I need an indian ip
 
possibly my dude
 
to show that z/2 lies in the middle point of z
i've calculated length of z and of z/2 and showed the angle is the same
is this approach correct?
 
Consider a ring called Stack Exchange. Then its maximal ideal is JEE
Everything just get sucked into This blackhole whenever it touches JEE, the abomination of all abomination of exams in this world
 
harsh
 
Okay, we fix one $m_0>k$. Now, we can argue that $|a_n -l_{m_0}|< \epsilon$, for all $n≥k(m_0)$
 
4:33 PM
That is true.
 
(yeah, I do felt an ideal basically behave like a blackhole, because you can multiply something outside the ideal with something inside it, and you fell into the ideal, never to see the light again lol)
 
Multiplicatively absorbing?
 
well, zero is a trivial ideal
(absorbers in general, yes)
 
4:59 PM
@BalarkaSen so what happened
seems everything got deleted
 
some guy told some other guy he doesn't know any math
 
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