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12:08 AM
Congratulations to former chatter @Xander on being elected moderator. It was a tight fight, I see.
 
 
1 hour later…
1:26 AM
math.stackexchange.com/questions/3778134/… – I wonder if I should post an answer which simply points out that all non-discrete finite topological spaces are also non-Hausdorff.
Technically that doesn't answer the question at all; in order to have a complete answer to the question, you'd also need to have a "naturally occurring" example of a finite topological space which is not discrete.
 
@TerranSwett There's actually plenty depending upon what is "natural". Take any finite simplicial complex, collapse the open faces to points.
That's a finite topological space which models the simplicial complex, in the sense that they have isomorphic homotopy groups.
 
 
1 hour later…
2:37 AM
@Lelouch It might be interesting to note what happens to quasi-isometries "in the limit". Suppose $(X, d_X)$ and $(Y, d_Y)$ are metric spaces and $f : (X, d_X) \to (Y, d_Y)$ is a $(K, \varepsilon)$-quasi-isometry, so $K^{-1} d_X(a, b) - \varepsilon \leq d_Y(f(a), f(b)) \leq K d_X(a, b) + \varepsilon$.
Scale the spaces $X$ and $Y$ by some large constant $\lambda >> 1$, so $d'_X = \lambda d_X$ and $d'_Y = \lambda d_Y$ are the new metrics. Then $f$ becomes a $(K, \varepsilon/\lambda)$-quasi-isometry with respect to this scaled metrics
If $\lambda \to \infty$, then supposing $(X, d_X)$ and $(Y, d_Y)$ converge in appropriate Gromov-Hausdorff sense to some limit $(X_\infty, d_X^\infty)$, $(Y, d_Y^\infty)$, $f$ induces a $K$-biLipschitz map between them
Eg scaled limit of the Cayley graph of $\Bbb Z^n$ is $\Bbb R^n$, so any quasi-isometry of $\Bbb Z^n$ gives a bilipschitz homeomorphism of $\Bbb R^n$ in the limit.
Heh I guess this gives a short proof that $\Bbb R^n$ is not quasi-isometric to $\Bbb R^m$ if $m \neq n$; both the spaces are stable under scaling
This is not a good perspective for hyperbolic space, because if you scale a $\delta$-hyperbolic space by $\lambda$ you get like a $\delta/\lambda$-hyperbolic space, so as $\lambda \to \infty$ you get a $0$-hyperbolic space... what is known as an $\Bbb R$-tree
Horrible objects in general
But anyway in the limit a quasigeodesic will be a bounded Lipschitz distortion of a tree path in this $\Bbb R$-tree, or something like this
And that is why Morse lemma is true I think
 
3:35 AM
@robjohn yes sir
 
 
3 hours later…
6:34 AM
@BalarkaSen or you can compute their asymptotic dimension
 
 
1 hour later…
7:43 AM
True @Alessandro
 
8:15 AM
What is an example of a sequence which has a limit but does not converge?
 
8:39 AM
@Infinity could you define the two terms?
 
For example you have a_n = (1/n) which has the limit of 0 as n approaches infinity. Ist ist enough to claim that a_n is convergent?
In order to prove that a sequence is convergent one should prove that it is 1. bounded and 2. monotonic. Does 'having a limit' contains both of these conditions in itself?
 
8:55 AM
those are not definitions
@Infinity do you know the precise definitions for "has a limit" and "converges"?
 
9:06 AM
@Yuvraj was that a question on an exam?
 
@Leaky Nun Convergence means: A function approaches a final and constant term as the input approaches infinity and limit is that constant value the the function approaches
Now I see that having a limit automatically means that the sequence converges
But then comes up another question: Why in math textbooks it is usually said, in order to prove convergence we should prove that it is 1. bounded and 2. monotonic? Why not simply say: If a_n has a limit then it converge?
 
9:34 AM
@robjohn yes
 
user434058
10:17 AM
@Yuvraj $f(x)$ is a constant function. Any continuous and differentiable function satisfying $f(x)=f(kx)$ where $k\in \mathbb R^+$ and $k\neq 1$ needs to be a constant function.
 
10:30 AM
@FakeMod he former condition implies that it's a constant function, meaning $f(x) = 3$. The integral evaluates to $6$ which is the last option
how to prove that ?? I saw that will be constant due to even no s are coming to constant but I didn't answered because I can't prove it for odd
16 mins ago, by FakeMod
@Yuvraj $f(x)$ is a constant function. Any continuous and differentiable function satisfying $f(x)=f(kx)$ where $k\in \mathbb R^+$ and $k\neq 1$ needs to be a constant function.
@robjohn hi sir
 
@Infinity using the theorem "bounded + monotonic => converges" can avoid actually finding the limit
@Yuvraj $\lim_n f(x/2^n) = \lim_n f(x) = f(x)$, $\lim_n f(x/2^n) = f(\lim_n x/2^n) = f(0)$
 
@Yuvraj Note that this condition also implies that $f(x) = f(x/k)$ for all $x$ as well. (Why?) Therefore you can assume without loss of generality that $k < 1$. From there, you can consider $f(k^nx)$ for higher and higher $n$, and use the continuity of $f$, and I leave the rest to you.
 
so $f(x) = f(0)$
it's easier to see if you try to picture it
 
10:45 AM
where limit n tends to infinity?
 
Indeed.
 
got it
@Fargle what about this?
$f(x)=\sin(\log_a(x))$,
can we prove same fo this
 
No, it doesn't satisfy that property
And it's obviously not constant: $f(a^{\pi/2}) = 1 \neq 0 = f(a^\pi)$
 
 
1 hour later…
12:17 PM
Is there a good way of seeing that $\mathbb{R}[X,Y]/(X^2+Y^2+1,aX+bY+c)\cong\mathbb{C}$ (where $a,b$ aren't both zero) without doing the ugly computations
 
1:08 PM
@BalarkaSen Nice, but at the risk of being wrong, doesn't this work to show that $\mathbb{R}^m$ and $\mathbb{R}^n$ are not quasi-isometric ? It suffices to show that $\mathbb{Z}^m$ and $\mathbb{Z}^n$ are not quasi-isometric. So basically, assume $m < n$. consider the ball of radious-$r$, in $\mathbb{Z}^n$. By the ifrst property of quasi-isometry, the image of $\mathbb{Z}^m$ inside this ball contains $\Theta(r^m)$ lattice points.
On the other hand, by the second properrty of quasi-isometry, this contains $\Theta(r^n)$ lattice points. Which is possible iff $m = n$
it's probably equivalent to what you said, I can't say becaues I don't understand how are you defining convergence of metric spaces
 
@Thorgott WLOG $a \ne 0$, then $X = b' Y + c'$, so $X^2 + Y^2 + 1$ becomes $(b'Y + c')^2 + Y^2 + 1$
which obviously has no real solution, so it is an imaginary quadratic
so the quotient is C
@Lelouch the risk of being wrong is always a risk worth taking
 
@Lelouch Yes, growth rate of a group is a qi-invariant
That is another way to do it, what I said is different
 
how is the metric space convergence being defined in this case
 
pointed Gromov-Hausdorff convergence
 
thanks, i'll look it up
 
1:16 PM
You say two compact metric spaces $X$ and $Y$ are $\varepsilon$-close if there is a map $f : X \to Y$ which is a $\varepsilon$-isometry, i.e., $d_Y(f(x), f(y))$ is bounded above/below by $(1 \pm \varepsilon) d_X(x, y) \pm \varepsilon$ and is $\varepsilon$-surjective, meaning every point of $Y$ is in $\varepsilon$ distance from $f(X)$
A sequence of compact metric spaces converge Gromov-Hausdorff to some metric space X if for any 1/n the tail of the sequence is 1/n close to X
 
Basically $f$ is $(1 + \epsilon, \epsilon)$ quasi-isometry ?
 
yes
For complete metric spaces $(X, p)$ and $(Y, q)$ with a choice of a basepoint you compact balls around $p$ and $q$ in the respect metric spaces
Thats pointed-GH
So for example $(1/n \,\Bbb Z, n |\cdot|)$ converges pointed Gromov-Hausdorff to $(\Bbb R, |\cdot|)$, centered at the origin
Easily checked from above definition
 
yeah, that's a nice way of putting it without making the computation explicit
guess that's as straightforward as it gets
 
$(1/n \,\Bbb Z, 1/n \, |\cdot|)$ I mean
Also above I should have written $d_X = \lambda d_X'$ or something
 
1:41 PM
If you downvote something, it shows:
But if you upvote something:
It doesn't show: "Please consider adding a comment if you think this post can be worsened".
8
Alas, the hypocrisy.
 
truly, 'tis a society we reside within
 
Lmao
 
You should have run for mod elections
We need big new ideas around here
 
1:57 PM
Nechayev intensifies
 
2:44 PM
Hello, is it generally true that a principal ideal of a ring (left, right, or two sided) are the same as cosets of the multiplicative group of the ring in itself?
 
I might not be thinking of the same thing you're thinking of, but the answer looks like "no": $\Bbb Z^\times = \{1,-1\}$, but principal ideals in $\Bbb Z$ are $n\Bbb Z$, which aren't cosets of $\{1,-1\}$.
 
@MikeMiller It would be unfair with the competitors: How would they compete with such ideas from the future? (2:25)
 
coset in which sense?
the units aren't an additive subgroup
orbits under the natural multiplication action, I guess
 
3:03 PM
I suppose we would have to assume the ring has multiplicative inverses
 
So it is a skew field?
Sorry, I'm not really following
 
3:21 PM
I don't think I know enough about this stuff to ask meaningful questions haha, thanks for your help. Got to learn more before I ask this again
 
 
2 hours later…
4:58 PM
https://math.stackexchange.com/questions/3780293/calculate-int-0-infty-frac-log-x-dxxaxb-using-contour-integra
I have a question regarding taking the limit with residue
in this example at z=0, the residue is 0, so no problem with taking the limit when epsilon->0
on the selected answer @ (1) the author just has written that the integral vanishes as epsilon tends to 0.
my question, how to handle a case when the circle is not 0. I mean the limit won't be 0.
 
5:22 PM
@hashman: No, $0$ is a branch point of the function, so residue doesn't even make sense. I don't understand your question. The whole reason the residue theorem works is that if you have a meromorphic function $f(z)$ with a pole at $a$, then the integral of $\int f(z)\,dz$ around a little circle centered at $a$ is precisely $2\pi i$ times the residue of $f$ at $a$.
 
5:44 PM
so what the residue theorem won't work at 0?
@TedShifrin? I can't understand why? for each circle there is an environment with a pole
around 0
 
No. The log function is not a well-defined function in any neighborhood of 0 (deleting 0). You need to understand branch points.
That's why they use the keyhole contour.
 
6:16 PM
Hi Ted
 
Hi @Alessandro
and @MikeMiller
 
Hey @Balarka
 
Howdy, nerds.
 
Hey @Fargle!
 
Do you want to think about some topological dynamics with me? I don't know if I'm missing something or the book is missing an hypothesis (I assume it's the first)
 
6:19 PM
How goes it?
 
I don't know anything but tell me
 
So the setting is a continuous $G$ action on $X$, $G$ is a topological group and $X$ an Hausdorff space. The action is called (topologically) transitive if any open set has dense orbit, and point transitive if there is a point with dense orbit
It's obvious that point transitive implies transitive
 
Ya ok
 
And the book says that if $X$ is second countable then the converse also holds
I see that with some extra assumptions ($X$ compact, or more generally Baire)
But the book so far has been very careful in stating explicitly when $X$ is assumed compact and it's not an underlying assumption
The point is that if $Y\subseteq X$ is the set of transitive points (points with dense orbit), then $Y=\Bigcap UG$ where $U$ ranges over a basis for the topology of $X$, so if $X$ is second countable $Y$ is actually a comeager $G_\delta$ set, but I don't see why must it be nonempty without extra assumptions
 
Hm yeah this is not clear to me I'd have to think
 
6:27 PM
But I don't have a counterexample either so it's probably true because of reasons I'm missing
 
Hi, demonic @Alessandro, a @Balarka, @Fargle
 
Heya @Ted, how's your morning going?
 
Helloes
 
Just fine, thanks. Just back from a walk.
Heya @Tobias
 
isn't it non-empty, because you assumed there is a transitive point to begin with?
 
6:36 PM
No I'm assuming the action is transitive and I want to show it is point transitive
Hi Tobias
 
clears throat
 
I greeted you earlier!
 
Oh, you greeted me when I wasn't here :D
 
clears throat
 
Just my last message was there.
Better get that checked, @Edward.
Quit the cigs.
 
6:40 PM
I don't smoke cigarettes!
haha
also yo everyone
 
clears throat back at Edward
 
hope you're wearing a mask
 
user434058
Yo!
 
Hi @Fakemd
mod
@Fakemod
 
user434058
Hi @Edw
 
6:42 PM
Yo, @FakeMod
 
user434058
That will ping you. (Only three letter are needed, so you can ping me by calling me @Fak)
 
@Fak off
weeey
Why must macadamia nuts be so expensive
 
user434058
@TedShifrin Hi @TedShifrin, how's your day goin'?
 
user434058
@EdwardEvans Which accent is that? :P
 
Oh, double ping. Doing great, thanks, and you? What are we here to discuss today?
 
6:44 PM
@Fakemod I guess a cockney or essex accent
 
user434058
@TedShifrin Well,... I just came here to "Yo" back to Edward.
 
O, yo.
 
ok, I see why Baire would suffice
but there must be something missing otherwise, no?
 
7:16 PM
That's what I'm wondering
I might ask on MSE
 
7:31 PM
for $U$ open with dense orbit, $UG$ intersects any other $U^{\prime}G$, of course, but it doesn't seem clear why they all intersect simultaneously
 
8:21 PM
I'm having trouble understanding which shape you get when identifying points on one torus to another torus
 
Please be less specific
 
gluiing the outer surface of a torus to the outer surface of another torus bijectively
 
8:50 PM
@Alessandro Yeah I think this is just false. Take $f : S^1 \to S^1$, $f(z) = z^2$, and restrict the system to the dyadic rational points on $S^1$; then it's still a chaotic system but every orbit is periodic I believe
As in, it's topologically transitive
 
But that's not invertible
There is no implication in either direction for semigroups actions
 
Oh you want invertible
I dunno
 
Yeah semigroups are too ugly
For semigroups you don't even have that point transitive implies topologically transitive, look at $\{0\}\cup\{1\n\mid n\in\Bbb N\}$ and $f(1/n)=1/(n+1)$
 
Sure yeah
 
9:06 PM
I'll think about it some more and just ask on MSE if I can't figure it out
I found it in another book too but with super strong assumptions (like compact Polish $X$)
 
Uh, there isn't even a topological-dynamics tag on MSE
There is one on MO, with a grand total of 26 questions lol
I'm having troubles finding any example of an action by a group which is topologically transitive but not point transitive
 
There's a wild homeomorphism $f : S^1 \to S^1$ which leaves a minimal Cantor set invariant, that might be relevant
I don't know if it is
Everything outside the Cantor set has dense orbit I think
 
Uh where can I read about that?
 
Don't; it is probably not going to be helpful
 
9:18 PM
Maybe but it sounds interesting regardless
 
It's called Denjoy's example or Denjoy homeomorphism
 
the Denjoy is short for "Don't enjoy"
 
It's don't enjoy but with a scottish accent
 
hilarious
 
9:21 PM
Which part of the UK are you from? @Edward
 
@Alessandro The South West from a city called Plymouth, which is on the border with Cornwall (more people seem to know what Cornwall is)
 
@BalarkaSen thanks, I'll add it to the list of "weird topological nonsense that I want to learn about"
@EdwardEvans what if I don't know where either is? Let me look that up on google maps
 
hahaha
 
I was in the UK only once, and I stayed in Aberdeen the whole time
 
Cornwall is historically its own country, but it's now just a county in England
Also fair, Aberdeen is like "nice Scotland"
along with Edinburgh
basically everywhere other than Glasgow is nice Scotland
controversial
I've signed up for 3 lectures and 2 seminars for the Wintersemester
but one of the lectures is just modular forms again, and I should pass it this time lol
 
9:24 PM
Ah I see, it's on the southwest tip basically
 
Right
 
@Alessandro Can I say some crazy nonsense
 
Oh this is unrelated, but have you seen Peaky Blinders? @Edward
@BalarkaSen sure! Well maybe, what kind of nonsense?
 
What about $S^1 \times [0, 1]/(z, 0) \sim(z^2, 1)$, and then acting by $\Bbb R$ on this space, where $s \cdot (z, t) = (z, s + t)$?
 
I haven't, but it's fairly popular lol
 
9:28 PM
I wanted to know if people in Birmingham really talk like that lol
 
Look at only the periodic orbits
 
Loool the Birmingham accent is like.. the most hated accent in the UK
 
isn't that action topologically transitive as well
 
Hmm ok let's see
 
@Alessandro yeah they actually speak like that
just watched a video lol
 
9:30 PM
That's amazing thanks
 
Also, it's referred to as the "Brummie" accent
 
@BalarkaSen Ok so this is like a twisted torus and the action moves point along the big circles?
 
yeah this is some nonsense like this
 
Ah ok and then you restrict to the periodic orbits because the full action is point transitive
 
right
I have no idea if this is actually topologically transitive though
 
9:36 PM
I'm not sure either
 
I'd hazard a guess that it may be? flowing a little interval on the meridian makes it larger and larger as it returns to the meridian
because $z^2$ stretches arcs
 
The problem is that it sounds plausible, but for compact Hausdorff spaces topologically transitive does imply point transitive
(because they are Baire)
 
but the union of the periodic orbits is not compact
 
Oh right, sorry
Makes sense
I had already forgotten that we restricted the action
 
yeah; you can alternatively start with the 2^blah roots of unities in $S^1$ than $S^1$ itself
z^2 is topologically transitive on that
if it's true it should be easy to prove that if $T : X \to X$ is top transitive then the $\Bbb R$-action on the mapping torus of $T$ is top transitive
if it's false it would be false lol
 
9:39 PM
Right lol
Seems like a very good candidate to me, but I'll need to think about it
 
same
 
I'm afraid I need to sleep first, my brain doesn't work anymore
 
I am going to bed as well lol
cya
sorry for terrible candidate counterexample
 
It's an ugly question, I'm expecting an ugly answer
Good night
 
10:16 PM
0
Q: Group law table logic: what we can learn about a finite group given (dihedral) symmetries of its law table?

AbstractAlgebraLearner This image you will recognize as a re-indexing of the group law for $\Bbb{Z}_6$, together with a coloring that indicates the equal subgrids of the whole grid. The group of symmetries (geometric; of the table itself) include reflection across the two diagonals and (if the table were presented squ...

Got a good one for you
 
0
Q: Are there other mathematical frameworks of artificial general intelligence apart from AIXI?

nbroAIXI is a mathematical framework for artificial general intelligence developed by Marcus Hutter since the year 2000. It's based on many concepts, such as reinforcement learning, Bayesian statistics, Occam's razor, or Solomonoff induction. The blog post What is AIXI? — An Introduction to General R...

 
@nbro there's one called Orange
allows you to layout workflows graphically and do anything from random forests to NNets e.g. there is a NN widget that you plug into
I could help you wire something up
Python-based
 
@AbstractAlgebraLearner AGI != AI
 
Oh I c, it doesn't do that
 
AGI is artificial general intelligence, i.e. strong AI
 
10:21 PM
There's TensorFlow and all that graphics card-based stuff by NVIDIA
 
i.e. something that hasn't yet been implemented really
no, that's not AGI
 
What are you applying it to?
 
not applying it to anything
I want to know if someone is working on that
 
researching?
Do you need a coding partner?
 
no
why do you ask this?
I need someone to review my thesis though
 
10:24 PM
Because I'm an expert in Python, less so C++ / Qt Framework. I can throw together a nice GUI'd app in a day (for desktop)
 
are you familiar with variational inference?
 
Nope
What is variational inference?
 
an approach to approximate Bayesian inference
now maybe you will ask "what is BI"? :D
 
Oh, is that hard to compute just like HMM probabilities?
 
It's just the Bayes' rule
BI can be hard yes, that's why we use approximate solutions, like VI
I should probably say that in most realistic scenarios BI is too difficult
 
10:28 PM
I would model the system using an HMM :D
 
I wish there was fast algorithm to perform exact Bayesian inference in almost all cases
I think the world would be fundamentally improved
 
There is, a random number generator that just happens to be correct
:>
I lol'd
 
lolz
 
link to your thesis?
 
can't share it with u if u don't know about it
 

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