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12:04 AM
You use the product rule to differentiate $f\bar f$, right? But $f$ is holomorphic. So what does that tell you?
 
@Ted: just to be sure, the fact that $d(f \times g)_{(x,y)} = df_x \times dg_y$ is just going to come from the same fact for smooth maps on open sets, right?
 
@TedShifrin Okay, yes! Is zero
 
Oh, actually, @Mancala, it's not right.
Sure, @Fargle.
@Mancala: You need $f\dfrac{\partial \bar f}{\partial\bar z} = f\overline{f'(z)}$ for the second term.
 
@Ted I tried to be good about that, but I will have to look to find some places I can excise technical details.
Maybe in my prehistory section. But the audience also needs that.
 
Easier, split the log of the product into two pieces. $\log(f\bar f) = \log f + \log\bar f$, so $\dfrac{\partial}{\partial\bar z}(\log f+\log\bar f) = \dfrac{\partial}{\partial\bar z}(\log\bar f)$.
Yup, I think prehistory and history are very important in talks, @MikeM.
 
12:15 AM
I call it prehistory because the first result I cite was published before I was born... But well in the lifetimes of some attendees.
 
you mean before you were born? :)
oh ...
did I read it wrong or were you sly?
 
I fixed
 
I mean ... I only had one sip of martini so far.
 
I walked 6 miles or so in dress shoes today
 
Oh, yikes. That's deadly.
 
12:19 AM
My feet are very angry
 
I would be too.
 
Next time I do this I wear sneakers and pack the shoes
 
Yup. Mathematicians don't exactly dress up so much, anyhow, but good that you tried to.
I guess I don't know where you're staying. Certain places there wouldn't be public transportation to help you.
 
@TedShifrin Yes!
 
Cool, @Mancala. Also see what I wrote a few lines below.
@MikeM: Good place for dinner?
 
12:24 AM
Mongolian hot pot.
@Ted Public transit to and fro is not so bad, actually.
I'm at Porter square.
I just wanted to walk. Not smart.
 
Oh, yeah, not quite the red line, but not too far.
But probably a direct bus, come to think of it.
 
So, one thing we're wondering about, if you have a knot as a subspace of something, you are allowed to sorta unlink it through homotopy, right? Like if we're thinking about it as a loop in the larger space?
 
Well, you can only do that if it's the unknot?
 
If you are doing knot theory type stuffy you want up to isotopy, so the knot does not bass through itself.
 
Not homotopy, but isotopy.
<--- sniped
 
12:32 AM
But we want the knot to pass through itself.
 
Well, in this case I'm more concerned with homotopy
 
That is fine then
 
Then it's homotopic to the unknot, yeah.
 
Well there could be other kinds of knots if you are not just in euclidean space
 
Oh, like a non-simply-connected ambient manifold ... I have no idea where this question is coming from.
 
12:34 AM
Okay so we're good then I think. I guess just to be sure, the idea was that you had a knot inside of the solid torus and wanted to show it wasn't a retract. If you could just forcibly unknot it then it's nullhomotopic, so then rip the injection of fundamental groups criterion
 
oh, I think I know which one you're doing.
I'm confused. The criterion doesn't depend on how the thing is embedded. You just are violating $\pi_1(X)\to\pi_1(A)$ injecting, aren't you?
Oh, no, duh, it's $\pi_1(A)\to\pi_1(X)$.
But $\pi_1(A)$ is intrinsic, not extrinsic.
 
It is intrinsic but I'm not sure if the map is going to be intrinsic
 
Fair enough. You have to look at the inclusion map.
That's the whole point.
I see. I misinterpreted your sentence. Carry on.
hides in corner
 
Does anyone have a link for what it means for a group to have "natural structure"? I can't find a definition anywhere
 
Without context that sounds meaningless @bphi
 
12:45 AM
There needs to be some context for that phrase ...
LOL ...
<--- hands the microphone to @Paul :)
 
I got you in my cross-hairs ;)
 
I cede the floor ... :)
 
Show that S3/A3 has a natural group structure
 
Particularly now since I'm going to cook dinner. You're in charge :)
Ohhh ...
There's a "natural" or "canonical" group structure on $G/H$ whenever $H$ is normal in $G$.
Yours @Paul
 
The point is that a set can be given many group structures (by defining a multiplication table), but you can try to define a group structure on the set of cosets $G/H$ where $aH * bH=abH$, but this doesn't quite work, you need extra conditions on how $H$ sits in $G$, which as Ted says the condition is to be normal. @bphi
 
12:51 AM
Is the "factor group" another name for that?
 
Ok, thank you sir
 
1:05 AM
so whats the consensus on Atiyah's proof?
 
Come on. Do a modicum of googling on your own.
Or read the star panel, which doesn't even require a click.
 
is that directed at me?
 
1:59 AM
@JoeShmo please discuss in this room and keep the discussion away from the main chat.
 
2:24 AM
Blarg
How's everyone doing tonight
 
 
1 hour later…
3:40 AM
@JoeShmo People are still discussing it? It seems like the biggest errors have been picked out already.
 
4:06 AM
You back @TedShifrin
 
Who?
 
@Ted Hello
 
Re Demonark
 
(I must say I rather liked the problem about showing the Mobius band doesn't retract to its boundary)
(And now that I think about it I probably totally knew that smoothly but still this was slick)
 
hatcher has superb exercises
 
4:17 AM
tired
no busses came for 1.5hrs so i went back to the university
 
don't get re sick, Faust
 
nfi whats going on
im not even entirely better yet
 
Yeah, that's part of why I switched to it from Rotman. I liked that Rotman was a bit more careful in a way, but many of his problems in chapters 0-2 were boring
 
Please take care, Faust.
 
i just want to go home lol
 
4:19 AM
So wait home is too far to walk and there are no buses? That's no fun
 
Preimage of a regular value should do the smooth case, Demonark. Thete's never a retraction to the boundary.
 
yeah missed 3 buses that didnt show up
its probally a 2hr walk but im not dressed very appropriately for the weather certainly not for that long of a wlak
 
Do they have Uber/Lyft there, Faust?
 
Anastasia is supposed to be off at 9pm
but she gets stuck late alot
stupid emergency room
im gunna insure my truck tommorrow this is the second time no buses have come in 5 days and 2 of those were the weekend
at least until Anastasias on a more normal shift
@TedShifrin no irregal by our bylams
 
Ok, maybe splurge on a cab ...
 
4:22 AM
ill wait till i hear from her
if she stuck for the night ill cab home
 
OK ...
 
library is warm
also think im getting better at understanding math
 
Let me try this again
 
5 classes and i likely dont have anthor hw assinment for almost 2 weeks
 
right, Demonark, mult by 2 has no left inverse
 
4:24 AM
and i got no hw left
spent the whole weekend doing algebraic topology C* algebras and analytic number theory
well i guess i did some stuff on manifolds
 
Okay yeah I guess you could still read that monstrosity, and that was the trick. But yeah I'm not sure how to prove in computer-verifiable detail that if you include the boundary, this induces the multiplication by 2 map, it just kinda looks like it wraps around twice
 
Draw the rectangle picture.
 
yea shes coming to get me
 
Yeah!
 
lebagel measure
studying the lebagel measure
 
4:29 AM
Inner cream cheese
Outer lox
 
lol
 
At some point I'd like to learn about C* algebras. Feels like a fun direction to pursue within functional analysis
No idea what they're good for but sounds fun nonetheless
 
im mostly lost
there good for dynamical systems and quatum mechanics
 
Lots of connections to geometry/topology too ....
 
kool
i like geometry and topology
w8 u told me u knew nothing about them
 
4:33 AM
I don't
 
but you are geometry
 
I've been to colloquium lectures for 40 years
 
Ted geometric Shifrin
 
Demonark, you see the 2 rigorously now?
 
H = \{ ( x , y ) \in \mathbb { R } \times \mathbb { R } : y \neq 0 \}
( a , b ) \star ( c , d ) = ( a d + b c , b d )
 
4:38 AM
Yeah, the rectangle helps for sure
 
Given that group defined by * on H. I found identity to be (0,1)
however, i am unsure of \text { If } ( x , y ) \in H , \text { what is } ( x , y ) ^ { - 1 }
 
Did you try to solve?
 
And?
 
I think the way to proceed
so basically we want ad + bc = 0
and bd = 1
 
4:41 AM
Yup.
 
and we want to find c,d in which this work? Is my approach correct
 
Yes
 
Ok just wanted to confirm this, thanks.'
 
Sure.
 
hi @TedShifrin
 
4:42 AM
Hi Leaky
 
so why do we care about the involution in C*?
 
Generalizing the importance of complex conjugation
 
Hatcher stahp
 
This is Bing's house.
Longgggg before Spanier and Hatcher.
 
I might end up skipping this bit, I've tried doing battle with it and my left shoulder still hasn't recovered
 
4:46 AM
Try the right.
 
@TedShifrin unpopular opinion: generalize to galois extensions
 
is there a way to send pictures here
 
And now the surface of genus g... dies
 
anyone have any idea how to approach this in a quick manner?
 
You're not going to make it far dead, Demonark.
 
4:54 AM
> in a quick manner
previously unsaid sentence in human history
 
friend of mine saw it on a kind of aptitude test with time limit. Was just wondering how one would go about solving given a time limit
 
Yes, @Sharath. What are the eigenvalues of $A$?
 
simply 1,2, ..., N-1 N
 
So what are tbe eigenvalues of $B$?
 
1^2, 2^2, ..., (N-1)^2, N^2
so is it just N^2 - N
 
4:59 AM
Yup.
 
what?
 
Oops
Typo?
 
why oops?
 
Actually now I found something which explains the 4-gon business nicely, I'm convinced
Thinking about it as a connected sum, so you take a torus square, rip it at a corner, and then attach
 
I have that in my geometry notes, I think. I have to ponder it every time.
 
5:04 AM
Yeah I tried just gluing the edges directly and it's just not making sense. Which is what happened the first time. Now I'm like o
 
oh i got the socratic badge @TedShifrin
 
Is this good for computing homology?
 
With Mayer-Vietoris ... or cellular.
Faust?
 
it means im really stupid and had to ask alot of questions
 
But Socrates taught ...
 
5:07 AM
I think the idea is Socratic dialogue, exploration by asking
 
Ah ...
 
i dunno its pretty hard to get
you got ask a question on 100 separate days with a positive overall score for the day
 
Oh on here? Cool!
You'd have two if you weren't in chat!
 
lmao
my biggest problem is definitions
i dont understand them very well somtimes
and half the time i misread ever i read
 
\text { Let } G \text { be a group and define } H = \left\{ x \in G : x ^ { 3 } = e \right\}
Show that \text { if } G \text { is abelian then } H \leq G . ( \text { I.e., } H \text { is a subgroup of } G )
How do I proceed with this proof
 
5:22 AM
well use some latex so its easier to read
 
$ \text { Let } G \text { be a group and define } H = \left\{ x \in G : x ^ { 3 } = e \right\} $
Show that $ \text { if } G \text { is abelian then } H \leq G . ( \text { I.e., } H \text { is a subgroup of } G ) $
so H is nonempty because identity exists
 
Let $G$ be a group and define $H = \{x \in G \mid x^3 = e\}$.
Show that if $G$ is abelian then $H \le G$, i.e. $H$ is a subgroup of $G$
(to see the latex, see room description)
 
he fixed it
 
he made every text in latex
 
i can read it
 
5:24 AM
it looks ugly
 
Last night dream, a strange object that is basically an uncountable version of $(\Bbb{Q}\cap [0,1])$ cartesian producted with $\omega_1$ was explored on graph paper, with the question being to show that there is an order preserving injection of the whole thing into the first segment of it, and not vise versa.
The animated proof then proceeded by mapping every point in that structure into the first [0,1]. To show the latter, it marked the segments in sequences of 1,2,3,1,2,3,1,2,3,... and showed that after the $10^4$ segment, gaps started to appear in some segments labelled 3 which eventuall
 
How do I show that if $a,b \in H
$
 
@SharathZotis all you really need to show is if you have two things say $x$ and $y$ that $(xy)^3 = e $
 
implies $a*b \in H$
Hmm
Why does it matter if $G$ is abelian?
 
well $(xy)^3= xyxyxy $
why does it matter that G is abelian
 
5:26 AM
I see
 
its really obvious now right?
 
The next scene then involves me and a guy who was boarding a bus in the street, then decided to jump off. The bus then recede into the horizon at an accelerating pace. I and the guy then tried to run after it, but we found ourselves get exhausted very rapidly and the surrounding houses does not seemed to change much. It is later revealed we are on $\omega_1$
 
If G is not abelian then H doesn't necessarily have to be a subgroup?
 
its possible that $xyxyxy \neq e$ if x and y do not commute
then u cant say that H is closed so its not group
 
5:28 AM
But you need a specific example ....
 
I see
 
?
whatcha talking about ted?
 
You can't disprove it with an argument. You need a counterexample.
 
well its easy once u know what ur looking for
 
Depends what. Groups you know ...
 
5:30 AM
pick $D_6$
 
How exactly do I see latex here
Ive been trying to configure mathjax, but not sure what I am doing wrong?
 
its a link u click on
set the start chatjax link as a book mark
then click on it with this page open
 
I don't think $D_6$ works.
 
shouldn't any symmetric group work as a counterexample?
 
oh it doesnt
 
5:32 AM
Good idea, @Sharath. Not quite any ...
 
circle
works
 
Huh?
 
circle is abelian..
 
$e^{\frac{2\pi}{3}}$ and $e^{\frac{-2\pi}{3}}$
oh yeah it is
its isomorphic to $\mathbb{Z} $
 
no it isn't
 
5:37 AM
Faust, you need sleep!
3
 
O.o
im missing an i
are those at least order 3?
whysit abelian
im stupid
nvm
 
@Faust "all you really need to show is if you have two things say $x and y that (xy)3=e"
@Faust "all you really need to show is if you have two things say $x$ and $y$ that $(xy)^3=e$"
 
thats for the abelian case
 
how does this imply if $x,y \in H $ that $x*y \in H$
further mustn't we show that if $x \in H$ that $x^{-1} \in H$
 
whats it mean for $x*y \in H $
 
5:41 AM
hmm
Im not sure
 
well to be in H it means $(x*y)^3=e $
 
where $x,y \in G$?
 
@TedShifrin looks like we passed the torch!
 
no where x,y are in H
 
ok that makes sense
but what about inverse
don't we need to argue that inverse exists in $H$ ?
 
5:44 AM
@TedShifrin (trying to come up with an English equivalent of a Chinese saying)
 
you can do it at the same time
or seperatly
you know $x*x^{-1} = e $
 
right
 
now you know $(x* x^{-1})^3 =e $
 
I see
 
so $ x^3 * (x^{-1})^3=e $ but $ x^3 = e $
 
5:47 AM
that makes so much more sense
thanks for your help
 
k im defintly going to bed now as im working with half a brain
 
> natural number subtraction is best understood in terms of the galois connection it forms with addition
 
gnight all
 
(Cont. Translating information from the dream) It may be helpful to consider the structure defined to be the following:
$$X = \omega_1 \times (\Bbb{I} \cap [0,1])$$
That is, a long line with all rational entries removed, or copies of $\Bbb{I} \cap [0,1]$ stuffed in between each successor ordinals
The dream claims there exists an injection $J$ such that:
$$J : X \to \Bbb{I} \cap [0,1]$$
and no surjection exists for the opposite direction
 
x^p^n-x is irreducible in FP[X] or not I guess it should be that is reason it is splitting field but I don't get concrete proof?
 
6:02 AM
@Ninjahatori try factoring out x
"it is splitting field" doesn't make any sense
I think you need to revise some basic concept
 
By factoring we get x((x^p^n-1)-1) right after that how to proceed
 
It might be helpful to compare the above mapping to the theorem stated in this MSE:
0
Q: How explicitly can we find the cantor diagonal set to directly prove that $\omega_1$ is uncountable?

SecretRecall that $\omega_1$ is defined as: $$\omega_1 = \{\text{Set of all countable ordinals}\}$$ and the cantor diagonal set is defined as: $$T = \{ s \in S: s \not\in f(s) \}$$ where $f : S \to \mathcal{P}(S)$ not surjective We note from this proof due to Carl Mummert that $\omega_1$ is uncoun...

> If $\eta, \theta$ are ordinals and there is an injective, order-preserving, non-surjective map from $\eta$ to $\theta$, then there is no order-preserving injection from $\theta$ to $\eta$. In particular, $\eta \neq \theta$.
 
@Ninjahatori that's why you need to revise some basic concepts
 
Perhaps, there is an analogous theorem between dense linear orderings as well
 
so we get two polynomials g(x)h(x) for our original f(x)
 
6:07 AM
Clearly an order preserving injection $I: \Bbb{I} \cap [0,1] \to X$ exists in the form of the identity map.
 
Neither of them is unit also in FP[X]
 
But does there exists a surjection in the same direction:
We can see that this is impossible in ZF. This is because it is well known there exists no surjection between $\Bbb{R}$ and $\omega_1$, therefore that must hold for all subsets of $\Bbb{R}$
A surjection however exists in ZFC+CH, and it is also order preserving (because the reals and its subsets can be well ordered under the axiom of choice, and CH demands $\mathfrak{c} = \aleph_1$)
 
@Secret If a surjection exists in ZFC then ZF cannot prove that no surjection exists.
 
right, I should have said "it is consistent that in ZF no surjection exists"
Now as for the other direction, in ZF, it is consistent that there is no injection between $\omega_1$ and $\Bbb{R}$. As a result it is consistent that there is no $J$ that is injective. Meanwhile for ZFC+CH, such $J$ exists via the axiom of choice.
Therefore, it is consistent in ZF that $X$ is not smaller than $\Bbb{I}\cap [0,1]$ (because an injection exists from $\Bbb{I}\cap [0,1]$ to $X$ in the form of the idenitty map) while in ZFC+CH, the two are order isomorphic by the axiom of choice
wait a minute... that does not sound right, let me quickly check the long line entry...
2
Q: Show the $\omega_1$-line (long line) is not homeomorphic and not order isomorphic to $[0,1)$

AndrewThis is based on Ex. 6.4.6 in Stillwell's "Real Numbers." Using previous exercises, it was established that one can construct for any countable ordinal $\gamma$ disjoint half-open intervals $[a_{\alpha}, a_{\alpha+1})$ for all $\alpha\lt\gamma$, with the properties $a_{\alpha}\lt a_{\beta}$ iff ...

But what we have here instead is an uncountable dense subset of (0,1) mapping to an uncountable number of such dense subsets of (0,1), so how will this fail...?
In ZFC+CH, the cardinalities of the domain and images are identical because $\aleph_1\aleph_1=\aleph_1$
The relative complement of $X$ wrt the long line, however is not order isomorphic to $\Bbb{Q}\cap [0,1]$ because we would ran out of rationals after mapping countably many of them
I wonder if they are related...
 
6:57 AM
Hmm...
Let $A \cap [0,1] = A_{[0,1]}$
$$X = \omega_1 \times \Bbb{I}_{[0,1]} = \omega_1 \times (\Bbb{R}_{[0,1]} - \Bbb{Q}_{[0,1]}) = \omega_1 \times \Bbb{R}_{[0,1]} - \omega_1 \times \Bbb{Q}_{[0,1]}$$
And our range is $\Bbb{I}_{[0,1]}= \Bbb{R}_{[0,1]} - \Bbb{Q}_{[0,1]}$
So clearly we have the long line in the expression and we already knew it is not order isomorphic because we would have ran out of rationals long before
Likewise, no order ismoprhism exists between $\omega_1 \times \Bbb{Q}_{[0,1]}$ because of the same reason
But then that will mean we are specifically subtracting away the discontinuous points of said monotone function, so the result should be order isomorphic?
Might check with the set theory guys when they got on again...
13
A: Linearly ordered sets "somewhat similar" to $\mathbb{Q}$

JDHThis is a great question! There are a large uncountable number of these orders. Consider as a background order the long rational line, the order $\mathcal{Q}=([0,1)\cap\mathbb{Q})\cdot\omega_1$. For each $A\subset\omega_1$, let $\mathcal{Q}_A$ be the suborder obtained by keeping the first point...

 
7:13 AM
hello
what do i do if my question is not answered
 
6
Q: Is there at least one irrational number with the property that it cannot be defined by a finite string of information?

John-Mark AllenOk, so maybe that wasn't the best way of phrasing the question, but I think it's specific enough. Let me explain myself a bit more below in case I am wrong. So I'm assuming (although I've never checked) that the irrational numbers are defined as simply all reals that are not rational. I'm askin...

5
Q: Characterizing $\omega_1$-like dense linear orderings

Ioannis SouldatosI recently came upon the following theorem which was attributed to J. Conway: For each $A\subset \omega_1$, let $\Phi(A)$ be a linear ordering of type $\sum_{\alpha<\omega_1} \tau_\alpha$, where $\tau_\alpha$ is $\eta$ (the order-type of the rationals) whenever $\alpha\notin A$, and $1+\eta$ whe...

 
7:29 AM
Hi guys. Is anyone able to confirm something for me. Given a function $f(x(t),t)$. Why are the total and partial derivatives with respect to $t$ the same?
 
$$\frac{df}{dt}=\frac{\partial f}{\partial x} \frac{dx}{dt} + \frac{\partial f}{\partial t}$$
 
Thanks @Secret I know the chain rule. However, I am reading a book that calculate the partial derivative exactly like the total and I am wondering under what conditions they are equivalent
 
@Rumplestillskin In this case you have a function of a single variable, so aren't they literally the same by definition?
 
@TobiasKildetoft I know that a function of a single variable will have the same partial and total derivative. However, they are suggesting that the partial and total derivative with respect to time of $f(x(t),t)$ are the same?
 
@Rumplestillskin well, when seen as a function of $t$ that is a function of a single variable
 
7:36 AM
@TobiasKildetoft Sure! I agree
@TobiasKildetoft however it isnt. In fact, even mathematica calculates the total and partial derivative of $f(x(t),t) $ to be the same and I am wondering what I am missing?
 
it isn't what?
 
@TobiasKildetoft it isn't a single variable function.
 
yes it is, when seen as a function of $t$
 
@TobiasKildetoft the function is $f(x(t),t)$
 
you need to also be able to vary $x$ if you want it to not be, and then sure, the total derivative will change
 
8:23 AM
I'm unsure how closedness works in sequence spaces, and what the fact that every second element is zero means in this context. @TobiasKildetoft: Aren't you good with these things? :)

"Let $E=\ell^1$. Consider $$X=\{x=(x_n)_{n\geq1}\in E:x_{2n}=0\ \forall n\geq1\}$$ and $$Y=\{y=(y_n)_{n\geq1}\in E:y_{2n}=\frac{1}{2^n}y_{2n-1}\ \forall n\geq1\}.$$ Check that $X$ and $Y$ are closed linear spaces."
 
8:37 AM
I forgot what the open sets are in $\ell^p$
 
8:51 AM
But observe $x+x' = (x_n +x'_n)$ and $x_{2n}+x'_{2n} = 0$ Hence $x+x' \in X$ for all $x,x' \in E$
A similar thing happens for $y+y'$
Since both $X,Y \subset E$ it follows the sum and multiple of any pairs of sequences in $X$ and $Y$ respectively always converge to some limit $L_x, L_y$ for each $x,y \in E$
Therefore, there always exists a closed ball of radius $L_x+L_{x'}$ and $L_y+L_{y'}$ ... <broken argument>
 
@OskarTegby No, I am not really good with those things. As I said some time ago, my idea would be that probably projection to the even coordinates is a continuous map (but I have no idea if this is actually true).
 
@TobiasKildetoft: Okay, but what would that mean?
Inverses of closed sets are closed under continuous maps. Right?
 
I cannot find a finite union of closed balls that is equal to $X$or $Y$. This is because let the limit of $x$ be $L_x < \infty$ an $x'$ be $L_{x'} < \infty$. Then the limit of $x+x'$ is $L_x+L_{x'} < \infty$ but can be arbitrarily large
 
9:31 AM
Okay. So if I can show that the map is continuous, then we know that it's a closed set if the original set was closed. Which it is because...?
 
@OskarTegby Presumably the topology on these sets is Hausdorff?
 
I have another question too. The problem formulation goes as follows.

"Let $E=\Bbb{R}^n$, $P=\{\vec{x}\in\Bbb{R}^n:x_i\geq0\ \forall i\in\{1,\dots,n\}\}$, and $M$ be a linear subspace of $E$ such that $M\cap P=\{0\}$. Prove that there's some hyperplane $H$ in $E$ such that $M\subset H$ and $H\cap P=\{0\}$."

Here, $M^\perp=\{f\in E^\ast:\langle f,x\rangle=0\ \forall x\in M\}$. I don't really see how the set $M^\perp\cap\text{Int }P$ makes sense. $M^\perp$ consists of functions and $\text{Int } P$ of points. How can we take the intersection?
Okay. Cool! I'll look into that. I don't remember how to prove that something is Hausdorff.
 
 
2 hours later…
11:17 AM
hey chat
 
Hey lattice
 
I need some reference: Let $f:[a,b]\to\mathbb{R}$ have a bounded positive slope (in my case it varies between $1/2$ and $2$). I'm approximating $f$ by some function $g$. The approximation order is given by $\lVert f-g\rVert_\infty = \mathcal{O}(h)$, where $h$ is the step width of some regular grid used in the construction of $g$.
Now it is somehow clear to me that due to the bounded slope, $g^{-1}$ is approximating $f^{-1}$ with the same approximation order, i.e. $\lVert f^{-1}-g^{-1}\rVert_\infty = O(h)$.
But I was asked to do this more exactly
I was wondering if there is some well-known result that I could use in here
I thought there would be some theorem in a more general situation, saying something like: "If the determinant of the Jacobian is bounded and strictly positive, then the inverse..."
 
N..
12:00 PM
Hi. With all the news about the Riemann Hypothesis and what it implies, I would like to ask a big-list question If you prove $X$, then you have also proved $Y$ that links theorems to other results they imply. But I'm not sure if that is acceptable on Math.SE. Any ideas? Has it been done?
 
a list of big lists!
 
N..
@LeakyNun yeah, I was curious which open question is the most "impactful"
 
maybe you can make a list of lists that don’t list themselves! :p
 
N..
that would only drive Russell mad
 
@N.. sounds like a cool idea, sounds hard to implement though
good community wiki
 
12:17 PM
There are multiple questions like that on math.se and MO for specific theorems (like RH), so should search for them before asking
I don't think collecting a list of $X$ consequences $Y$ works for math.se
 
N..
@PaulPlummer well I’d need to know what theorems to search for, hence the list ...
 
Since the Renaissance, every century has seen the solution of more mathematical problems than the century before, and yet many mathematical problems, both major and minor, still remain unsolved. Prizes are often awarded for the solution to a long-standing problem, and lists of unsolved problems (such as the list of Millennium Prize Problems) receive considerable attention. Unsolved problems remain in multiple domains, including physics, computer science, algebra, additive and algebraic number theories, analysis, combinatorics, algebraic, discrete and Euclidean geometries, graph, group, model, number...
probably a place to start, who would have thought that you could use the internet to search for information
And now you can travel down some wiki holes
 
N..
Nice, thanks! Now if only there was a site where you could ask questions to experts who might already have some answers
4
But I take the point that it might not fit well on math.se
 
12:38 PM
How to Show that in a valuation ring every radical ideal is prime.
 
@LeakyNun Are you here
 
Let, $G=GL_n(\mathbb{R})$ , the group of all invertible $2\times 2$ matrices and $H=\{\begin{bmatrix} 1&0\\0&1 \end{bmatrix}, \begin{bmatrix} -1&0\\0&-1 \end{bmatrix}\}$. How can I show $G/H$(qutiont group) is abelian or not?
 
@KenOno Find two matrices $A$ and $B$ such that $ABA^{-1}B^{-1}$ is not in that subgroup.
 
@TobiasKildetoft Thank you but why that helps?
 
What would it mean for that element to be in the subgroup (when we see this in the quotient)?
 
12:51 PM
Yeah not abelian
 
How come none of the Latex in chat shows up for me? Is there something you guys have installed on your computer?
 
@CaptainAmerica16 See the room description
 

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