00:00 - 17:0017:00 - 00:00

12:04 AM
*keplers surface of revolution I mean

12:31 AM
@Danu, stranger ! I just got a Facebook reminder with the pictures from my visit to München. Hope you're doing well!

I am! How are you? We should really catch up sometime

Indeed.

Any chance of catching you at a time that is more doable for my side of the big pond? It's 2:30 AM and I really shouldn't be up by now...

Of course. I'm often around in the mornings here (so 6-ish your time). Wait for the end of the French Open, though :P

I'm psyched for Federer - Nadal, though I fear the worst!
(even if Roger gets through that though, he'll get crushed by Djokovic no doubt)

12:35 AM
Yeah, I wouldn't put much money on Fed, but I think it'll be a better match than years ago.
Yup.

unless Thiem...

Dominates

You're not picking Zverev to beat Djoker?

Anyways, Ted, here's a vague math question I wondered about today
@TedShifrin Are you kidding? :P Zverev is such a choker
my hopes for the future are on Tsitsipas

He's very exciting, actually.

12:39 AM
I kinda like Borna
Coric
beat federer like 5 months ago I think
he's 23 ish

I'm not a fan of Zverev... But oh well

I like Coric, but he's not the most scintillating player.
Re your question, @Danu, of course you're supposed to have a self-adjoint linear map from $\Lambda^2$ to $\Lambda^2$. I guess I'd have to sort that out in your situation.

@TedShifrin That's the same as saying it should lie in $S^2(\Lambda^2T^*)$ right (after lowering the two indices you've raised)?

Sounds right.

I gotta see that fed-nadal match

12:45 AM
That's not sufficient though, because one needs Bianchi. In fact, the Kulkarni-Nomizu product of two two-forms already lies in $S^2(\Lambda^2 T^*)$, but is not in general an algebraic curvature tensor
There should be some nice way to understand the possible ways of making algebraic curvature tensors out of two two-forms, purely in algebraic terms, but idk how to do that

@TedShifrin if you have time after Danu's question.. I'd like to understand fibrations a little better

@Danu: One of my math friends has run an REU (undergraduate research experience) numerous summers on these sorts of questions. Would you like his name/email to contact?

ex. Kepler's surface of revolution. I want to try to gain a deeper geometric understanding of fibrating this surface
this surface, as you might know, is a surface generated by revolveing the arc of a circle
or an arc with less radius than a circle

I don't know what surface that is, @Ultra. Noncompact surfaces of revolution with two ends will of course have a foliation by circles, but in general you need Euler characteristic zero (homeo to a torus) to get a foliation of a compact surface.

Why not? :) Not sure I'll actually get in touch with him, but maybe I will!

12:51 AM
OK, let me dig up his email, and I'll email you. Same e-address?
(His doctoral work was on such questions.)

I guess so :) I'm also easy to find through my University nowadays by just googling my name and math; there's another email there
Thank you! :)

I have your gmail. I'll just use that.

@TedShifrin I see. Thanks

Perfect

Sent. Schlaf gut.

12:54 AM
Received! Off to bed!

@TedShifrin what about non-contractible geodesics on a compact surface with two singular endpoints. Would this be more in the realm of systolic geometry

I have no idea what that is, nor about geodesics on such singular things, although I'm sure somebody knows about that.

using isoperimetric inequalties and such. I guess I should get better at inequalities
they are used so much in research mathematics

Unless you do just algebra, it's hard to avoid inequalities in math.

I wish I could see what my professors wrote on their letter of recommendation for me lol
but I figured it's not a good idea to request access to them

1:00 AM
Some things are better not known.

Well I got in so they must have been decent ones.

I always advised my students to say that they were giving up their chance to see what was written. People reading the rec will trust it more if it's confidential, for obvious reasons.

And that is very true.

Usually I told students I'd let them read it if they wanted, but almost no one asked. If I felt I couldn't write something good, I tried to get out of writing altogether.

Hey ted

1:03 AM
Nope.

think he was at U georgia maybe

LOL, no, I don't think so. When?

Ah, Georgia State University

Very different kettle of fish.

I see lol

1:06 AM
Anybody here care to look at my Post:
0

Below is a problem that I did. I would like somebody to check it for me. Thanks, Bob Problem: Suppose that $z_1$ and $z_2$ are two independent random variables that are normally distributed with mean $2$ and standard deviation of $4$. Now if $z = |z_1 - z_2|$, what is the probability that $z$ ...

I think he's very talented tho

(b) is cauchy. Since it is uniformly convergent to f(x)=1.
(a) I am getting triangles with base length 2 and height 1
which is shifting toward right. as n increases

So that sequence converges very non-uniformly to what?

$f(x)=0,x\in \mathbb R$

Right.
Does the sequence converge in $L^1$? I assume that's what the $d_1$ metric is.

1:18 AM
$\int_{-\infty}^{\infty}|f_n(x)-0|dx=1$
No.

Right. So you're done.

okay. so it is not cauchy. Thank you :)

You're welcome.

3 hours later…
4:13 AM
9 hours ago, by Ryan Unger
I like how every question with Balarka starts as someone asking him something he knows nothing about but he just figures out the whole theory on the spot
Except set theory, which is not really important for most of maths anyway
But again, the knowledge of infinite topos means he will knew how to derive pretty much anything in maths, since everything can be expressed in terms of topos and categories

1 hour later…
5:31 AM
"You've earned the Tumbleweed badge." Yay?

I am the mother of all tumbleweed
I can kill a chat activity with The Word

1 hour later…
6:46 AM
why dV/dt is negative definite? Shouldn't be negative semi definite?!

7:19 AM
Morning all

Good morning
yesterday, by user76284
What’s the order type of the set of all integer polynomials, where $f \leq g \leftrightarrow \exists x \in \mathbb{Z} : \forall y \in \mathbb{Z} : f(y) \leq g(y)$?
@Secret Since you seem interested in infinities what are your thoughts on this?

7:39 AM
By the way, that should be $y > x \rightarrow f(y) \leq g(y)$.

8:06 AM
@user76284 This is not a well order

8:41 AM
$x$ does not appeared anywhere within f, g, thus other than it is not a well order, cannot say much?

It's fixed in the next message

But why is x important when it never appeared inside f(y)<=g(y)?

Because it says that $g$ eventually dominates $f$

Ah ok

The inequality doesn't have to hold everywhere, just for all $y$ big enough

8:45 AM
ok, so it is at least linearly ordered, probably dense. Need to figure if it is order isomorphic to some product order of the integers...

I don't think it's dense, what's between the constant polynomials with value $1$ and $2$?

Ah right, integers...
Yeah I cannot work this out on mobile. Might need to sit down with paper. I guess the only things we can say is its linear order type has something to do with $\Bbb{Z}$

I think the answer given to this question is wrong. Hall claims that we can find integers. But the user 4-ier only constructs fractions. Am I right?

9:17 AM
If you borrow 243 each month for 12 months at 8% per year, how much do you owe at the end of the year?
I meant in interest only
so I guess it is 1.08*243+1.08^(11/12)*243+1.08^(10/12)*243+... - 243*12? but is that right?

10:03 AM
I am working out something relating to fixed point division, and I have an expression that's (AM×2¹⁶ + AL) / (BM×2¹⁶ + BL). Assuming the result is between -1 and 1, is there a well known expansion of this kind of binomial division in terms of the single terms?
(Apologies if I'm using the word "binomial" incorrectly there.)

So I'm asked to find whether the following integral is either convergent, divergent or indefinite.
$$\int_1^2 \frac {\sqrt{x-1}}{\abs{\sin(1-x)}\tan(2-x)} dx$$
I've simplified it as:
$$\int_0^2 \frac {\sqrtx}{\sinx \tan(1-x)} dx$$

now the Idea I had was to find a function which is greater [or smaller] at each point than the integrand, easier to integrate and which also converges [or diverges].
But I'm having a hard time finding any such functions.
Is there something that escapes me? maybe an easier method?
the limits of the second integral are 0 and 1, not 0 and 2 as it is displayed

10:34 AM
(Long division was the answer to my thing!)

10:50 AM
Hi
From the book "Prime Numbers A Computational Perspective", I am having trouble understanding the following definition
"A negative integer D is a fundamental discriminant if the
odd part of D is squarefree, and |D| ≡ 3, 4, 7, 8, 11, 15 (mod 16)."
What is meant by "odd part"? Are the comas between the numbers meant to be interpreted as "or"?

@northerner divide by 2 until it becomes odd

aka square of an odd number doesn't appear as a divisor of D
that's a great book by the way

Ok great. So |D| ≡ 3, 4, 7, 8, 11, 15 (mod 16) means the remainder must be 3 OR 4 OR...15 correct? Idk how to interpret the ,

11:06 AM
yes it's "or"

11:18 AM
Let $V$ be a finite dimensional vector space over $\mathbb R$ and let $f$ and $g$ be
two non-zero linear functionals on V such that whenever $f(x) ≥ 0$, we also
have that $g(x) ≥ 0.$ Which of the following statements are true?
a. $ker(f) ⊂ ker(g).$
b. $ker(f) = ker(g).$
c.$f = αg$ for some $α > 0.$
I know (b) and (c) are true.

woops nonzero functionals

while choosing the 100% accurate answer. why should I choose (a)?
@AlessandroCodenotti yes

@BalarkaSen is this definition equivalent to the one on Wikipedia or are they different? en.wikipedia.org/wiki/Fundamental_discriminant

11:47 AM
\o @Secret

o/

wazzup pal

As usual

coolio

This is what Artin in his book Algebra says about Gaussian integers:
I was wondering if the comment 'as many as four choices for $\gamma$' is for $\alpha=2+i$, or any nonzero Gaussian integer $\alpha$? Also, can we see that 'as many as four choices for $\gamma$' from this kind of geometric argument?

11:53 AM
I guess there are four units in $\Bbb Z[i]$
1, -1, i, -i
@Silent (i.e. units don't change the divisibility properties of a ring element)

You have a complex number $x=\frac\beta\alpha$ and you want a Gaussian integer $z$ such that $|z-x|<1$.
This is the same as looking at a square in the square grid and requiring that $x$ is in the distance less than one from one of the corners.
If you are near the middle of the square, you can choose any of the four corners.

12:10 PM
What's your opinion about Artin's Algebra book? @MartinSleziak

This is approximately what I meant:
@skullpatrol I do not know that book well enough to be able to give any reasonable judgement on it.

@MartinSleziak, Wow! Thank you so much! Those comments of yours made me think hard and allowed me to see what simple algebraic manipulations can do geometrically! (Sorry for my late appreciation, I am a slow learner.)
So, there can be one, two, three or four choices for $\gamma$, but there has to be at least one.

I am not sure whether it is possible to have three choices.
I'd guess one, two or four are possible.

Yeah I also doubted that after I wrote :)
@MartinSleziak So, your figure is about one choice only, right? because we are dealing with $|z-x|<1$ and not with $|z-x|\le1$

@Silent I used $|z-x|<1$ since you want $|r|<|\alpha|$. (Strict inequality for the norm of the remainder.)

12:26 PM
@Silent well two ring elements $a, b$ are called "associates" if $a \mid b$ and $b \mid a$
they are related by multiplication by units

Given a sequence of groups $G_i$ is there always a space $X$ with $\pi_i(X) = G_i$?
@BalarkaSen

@Silent so for instance $1 + i$, $-(1 +i)$, $-1 + i$ and $1 - i$ are all associates

I see. So, how units don't change the divisibility properties of a ring element?

@LeakyNun Sure, take a product of $K(G_i, i)$'s

great thanks

12:37 PM
There's a construction called the Postnikov tower which recovers the homotopy type of $X$ as a twisted product of $K(\pi_n(X), n)$'s

@MartinSleziak I am sorry but I think that three choices is also a possibility: See this.

So it seems that I miscounted and the possibilities are as above. (With 1 for position in the vertex of the square.)

wow!! Thank you very much.

12:57 PM
@Silent you can think of it in terms of (rational) integers; in the integers clearly $a$ and $-a$ have the same divisibility properties (for example they have the same prime factorisation, though this description only makes sense when your ring of integers is a UFD)
in fact two ring elements $a$ and $b$ are associates iff there exists a unit $u$ for which $a = ub$

Let $K$ be a compact Hausdorff space, and let $C(K)$ be the space of continuous complex valued functions equipped with the supremum norm. If one speaks of strong convergence in this context, does one mean convergence with respect to the supremum norm?

@user193319 My guess would be that if we're working with some normed space, strong convergence means convergence w.r.t. the given norm. (As opposed to weak* convergence, for example)
Of course, it's possible that this term is used also in some meanings I am not familiar with.

That's what I figured.Thanks for confirming!

1:34 PM
In Rudin's Functional Analysis, he states that the continuous dual of $C(K)$ equipped with the sup. norm can be identified with the space of all regular complex Borel measures on $K$. Does anyone have reference for the proof of this?

@ÍgjøgnumMeg, Thank you very much. Now I get it.

@user193319 This is equivalent to stating the pairing $C(K) \times M(K) \to \Bbb R$ taking a pair $(f, \mu)$ where $f \in C(K)$ and $\mu \in M(K)$ is a regular Borel measure, and spitting out $\int_K f d\mu$, is a nondegenerate pairing. This is the Riesz-Markov-Kakutani theorem

Oh, wow! What a lovely theorem!

1:51 PM
I believe the idea to show that $M(K) \to C(K)^*$ is surjective is, given any continuous functional $\phi : C(K) \to \Bbb R$, define $\mu(U)$ to be the supremum of $\phi(f)$ where $f$ varies over continuous functions with values in $[0, 1]$ which vanish outside of $U$ (so these approximate $\chi_U$, roughly speaking)
Then you extend $\mu$ over all Borel subsets to get the corresponding measure to $\phi$

Ah, I see. Very intuitive.

2:03 PM
Hmm... very curious:
It seems the information of the coefficient of the polynomial can be partially recovered somewhat from the resulting curve:
For example, the asymptotic behaviour of the polynomial is basically the same as when you truncate all the lower order terms. This can be easily explained as $x \to \infty$ then $cx^3+ax^2+bx \to cx^3$
which means, the coefficient of the highest order term can be recovered by inspecting the graph
Likewise, near the neighbourhood of the origin, the behaviour of the polynomial matches that of its corresponding linear term
meaning that the slope of the tangent of the polynomial at the origin will give you the coefficient for $x$
This means, if the polynomial has no linear term, then it will be flat locally at the origin. If in addition the polynomial has no constant term, then the origin will be a root
Now if the same logic holds for all other $x^n$, we might just have a way to predict where the roots of an arbitrary polynomial will be
The behaviour of the linear term can be proved as for all $n > 1$, pick any $x \in (-1,1)$ and the higher order terms will quickly go to zero, and only the linear and constant term survives
This desmos also showed how polynomials form a vector space. Since we knew the behaviour of the linear and highest degree term, note how that behaviour remains untouched when we add other terms of $x^n$
If we have some algorithm that can pin down the condition for $x$ such that $x^n$ is invariant, then one can in theory deconvolute any polynomial by inspection

2:25 PM
If an integrand is unbounded on a given interval, can we say that its corresponding improper integral diverges? are there exceptions? (like odd functions on a [-a,a] interval?)
when can I be confident enough to say that it diverges?

@Simone There are exceptions.

mmh.
because I'm dealing with that situation, and I'm having a hard time proving either convergence or divergence

For instance, you can integrate $1/r$ over the unit ball (minus the origin) in $\Bbb R^3$

The Idea I had was to find an integrand that is greater at all points and then prove convergence that way.
or a smaller one to find divergence
@Danu that example is beyond my abilities

anyway, given P,Q polynomials, if deg(P) > deg(Q), then P eventually dominates Q
7 hours ago, by user76284
@Secret Since you seem interested in infinities what are your thoughts on this?
so it is definitely not a well order, but what kind of linear order is to be investigated. My suspicion might be the product order $\prod_{i=1}^n \text{ord}(\Bbb{Z})$

2:36 PM
@Simone $$\int_{B^3\setminus\{0\}}\frac{1}{r}d V=\int \frac{1}{r}r^2\sin\theta d r d\theta d\phi=4\pi \int_0^1 r dr$$

thanks @danu

After getting these datasets in, I will try to figure out how given a polynomial P, find the set $X$ such that $P(X) = \mathscr{o}(x^2)$

@Simone also consider $\int_0^\infty\frac{1}{(x+1)\sqrt{x}}\,dx$
It's unbounded on the left.

@anakhro but $\int_0^a \frac{1}{\sqrt x}$ converges, so I'd imagine that the integral you wrote converges aswell
right I had an exception right there
$\int_0^a \frac{1}{\sqrt x}dx$ converges even if unbounded

3:00 PM
@Simone I was responding to your question "If an integrand is unbounded on a given interval, can we say that its corresponding improper integral diverges?"

I know

The integrand is unbounded on the interval (0,a], but the integral converges.

correct

So the answer to your question is a succinct "no".

That's right, thanks

3:02 PM
Just as a single-variable answer to your question.

Would you say that a good strategy to find the convergence [divergence] of a given, maybe tough, integral is to find an integrand which is greater [smaller], and easier, at all points than the original one to then evaluate its convergence [divergence]?

Comparison tests are useful, yes.
But not always the easiest.

how many subgroups of order $3$ does a nonabelian group of order $39$ have?

Attempt: $3 \times 13=39$. Let the number of such subgroups be $s$. So, $s \mid 13$ and $s=3k+1$ for some $k \in \mathbb{Z}$. $13$ is a prime and the only possible value that $s$ can take is $1$.

So, $1$ should be the answer?

@SubhasisBiswas I feel like that leads to a contradiction.
Because then it is the unique Sylow 3-subgroup.
And then it's normal.

so it is normal
can a non abelian group of order $39$ exist?

3:12 PM
Yes.
The subgroup of order 13 is also normal.
So...

it is the direct product?
of $Z_3 \times Z_{13}$?

Which is bad, no?

@anakhro in which sense?
@Thorgott, hello sir

The non-Abelian group is a product of Abelian groups...

@anakhro what do we conclude from this?

3:15 PM
Well is that not a contradiction?

@anakhro are you saying that a subgroup of order $3$ in this case cannot exist?

@SubhasisBiswas I am saying that if you assumed that there is a unique Sylow 3-subgroup, then it leads to the original group being Abelian.

@anakhro that is not possible
$13$?

We must have $n_3 = 1$ or $n_3 = 13$
the former gives the abelian group $C_{39} = C_3 \times C_{13}$

@LeakyNun $n \neq 1$

3:25 PM
the latter is the semidirect product $C_{13} \rtimes C_3$, which is non-abelian

So, 13 must be the answer

right
18 mins ago, by Subhasis Biswas
how many subgroups of order $3$ does a nonabelian group of order $39$ have?

Attempt: $3 \times 13=39$. Let the number of such subgroups be $s$. So, $s \mid 13$ and $s=3k+1$ for some $k \in \mathbb{Z}$. $13$ is a prime and the only possible value that $s$ can take is $1$.

So, $1$ should be the answer?
so clearly you missed $s=13$

@LeakyNun yes. I missed the possibility
"1" is not the only possibility.

Hey @SubhasisBiswas ! How's it going?

@Thorgott not well. constantly getting demotivated by the knowledge that how little I know

3:38 PM
It might not help much, but all I can say is that nearly everyone feels that way and it only gets worse the more you go down the rabbit hole. Try thinking about the progress you've already made rather than the progress you still have to make.

@Thorgott a very good advice. Been reading 10 pages of Artin daily. :D
thinking about skipping the "Group theory" part a little bit to go to the linear transformations and then getting back to it.
@Thorgott learning new things almost everyday as I go.

I know that feeling. I feel like I've learned a lot in the last few semesters, but, even still, there is clearly much more that I have ahead of me.

Currently running 20 calculations that makes my laptop very hot

here's a question. A continuous function maps a compact metric spaces to another compact metric space. So, is the preimage of a compact metric space under a continuous function always compact?

My answer is "No". Take $f: \mathbb{R} \to [-1,1]$, where $f(x)=\sin(x)$.

Luckily, I placed an extra fan so it does not melt down

3:47 PM
@Secret wolfram mathematica?

Not sure if that's a good idea. You can get a lot of fundamentals about linear transformations essentially for free from corresponding facts in group theory. Not sure how Artin does it though.

nah, some python script that crawls data from 1000 files
it's part of some statistical analysis

@SubhasisBiswas A constant function should also work as counter-example

@Thorgott damn.
so simple
it also acts as a counter example to : Does a continuous function always map open sets to open sets?

However, the preimage of a compact set under a continuous mapping is compact when the domain is compact (in the context of metric spaces). Though, of course, this is not particularly revelatory.

3:55 PM
@Thorgott sounded like this.

lol

I forget. Is specifying metric spaces for that statement strictly necessary? I want to say it's true in general.

it also holds for compact subsets of the codomain is what I meant, though that's still a triviality
@Rithaniel It fails when the codomain is not Hausdorff

non Hausedoff is weird

Ah, my topology course assumed Hausdorff-ness fairly often, so that explains that.
Maybe I should request a course in non-Hausdorff topologies.

4:00 PM
@Rithaniel can you explain Hausdorfness in a bit of elementary terms?

@Rithaniel wot

slereah has friends
:P
For me, I work with general topology, nets and so, thus whether it is hausedoff or not usually does not get assumed

Well, the simplest way of describing the Hausdorff quality is to say that "points are closed"

unless it needs some property to be hold in "nice spaces"

Hausdorff is a very convenient assumption in most cases.

4:01 PM
Let's see if I can recall the formal definition of it, though:

@Rithaniel aha..almost like the notion of "disconnected space"

Being Hausdorff is equivalent to the uniqueness of limits, which is what makes it a very nice property.

"Given any two points there exists an open set containing the first but not the second and an open set containing the second but not the first."
Now, gonna run to lunch.

@Thorgott since each point can be closed off by some open set and disjoint from the other?

hausedoff: points can be separated by open sets

4:04 PM
@SubhasisBiswas from the other open set
you need two open sets at the same time

@RyanUnger yes yes... got it :D

Indeed

it's very important, there are spaces where you can do this with one open set, but not two at the same time
these have some stupid name like T0 or T1

It sounds like, many rooms are nearby, with many people in each room. But each room has its door closed.

only nerds remember all of those names
what

4:05 PM
no interaction between people from one room with the another.

no, that would be a totally disconnected space
or a discrete space

I once made an infographic of the Tx spaces, I forgot where I put it

stupid question. Is a disconnected space hausdorff?

apparenrtly not
it's T1 but not T2 according to google
2

Recall a space is totally disconnected if the only connected subsets are singletons (one-point subsets). Is a totally disconnected space, Hausdorff? I think it is true since if $a$ and $b$ are two distinct points, they can be separated two disjoint open sets, since the main space is total...

Totally disconnected is different from disconnected, no?

4:11 PM
found it
open sets in blue, closed sets in red

@Thorgott totally different

all subsets with more than one element are to be disconnected
when it is totally disconnected
@MatheinBoulomenos pun intended ?

it's a good exercise to check that totally disconnected does not imply discrete
you have to chase through the defintions

4:16 PM
$[0,1]$ is not homeomorphic to $\mathbb{R}$, right? Since no continuous bijection can exist...
one is compact and the other is not

@RyanUnger Wouldn't the Cantor space work as counter-example?

the easiest example is just the rational numbers
@SubhasisBiswas yes

Non T0: Anything can happen
T0: At least one point can be separated by open sets
T1: Points can be separated by open sets
T2: There exists two disjoint open sets which points can be separated
T2.5: There exists two disjoint compact set which points can be separated
Complete T2: No open sets containing points overlap with each other
T3: A closed set and a point can be separated by open sets
T3.5: A closed set and a point can be separated by disjoint open sets
T4: Closed sets can be separated by open sets

@Thorgott but totally disconnected means that every pair $\{x,y\}$ is disconnected
since connected component are open, this means $\{x\}$ is open
so why isn't $X$ discrete

I don't think connected components are necessarily open.

4:26 PM
informal question

Are there spaces worse than non T0, but not indiscrete?

in how many ways the complex plane can be interpreted?

$\aleph_1$
Next question

In how many ways maths chat can be interpreted:

@Secret I don't understand your definition of T5 and T6. what do you mean by "open sets which contain no other points"?

4:32 PM
@RyanUnger If the Cantor space were discrete, it would be homeomorphic to $\mathbb{R}$ with the discrete topology (since any two discrete spaces with the same cardinality are homeomorphic). However, the Cantor space is compact whereas $\mathbb{R}$ with the discrete topology is not (consider the open cover of singletons), so this can't be.

@Secret memory overflow. Kindly restart the program.

T5: completely normal and T1. Actually the diagrams does a very bad job at capturing T5, T6
T6: perfectly normal and T1
So basically, that lone dot in the diagram of T5 is trying to capture the T1 bit of the axioms

heyo

what's occurring

4:37 PM
which, kinda showing the limitation of the diagram as I tried to recover the Tx axioms by looking at the diagrams without reading the actual definitions, and get the wrong interpretations, it is suffice to say that my diagrams on T5, T6 are misleading

Even shorter, if the Cantor space were discrete, it couldn't be compact due to the open cover of singletons.

hi @BalarkaSen @ÍgjøgnumMeg
“Anything that happens, happens.
Anything that, in happening, causes something else to happen, causes something else to happen.
Anything that, in happening, causes itself to happen again, happens again.
It doesn’t necessarily do it in chronological order, though.”

Hi @Mathein

One thing that puzzles me about these topology classifications is how are the numbers were decided, e.g. 2.5 and 3.5

@Mathein y0

4:44 PM
and yet there are no 0.5, 1.5, 4.5

@Mathein weißt du ob man in Heidelberg um ca. 500€ eine Wohnung kriegt? :D

Pr(X)=1: Anything that happens, happens
Causality is transitive: Anything that, in happening, causes something else to happen, causes something else to happen.
Self reinforcing feedback loop: Anything that, in happening, causes itself to happen again, happens again.

@ÍgjøgnumMeg wenn du damit leben kannst, nicht ganz zentral zu wohnen, ja, denke schon

Statistical events: Anything that happens, has some chance of happening
Quantum events: There are things that happens that follow the rules of linear algebra with interesting correlations
Cauchy demons: There are things that happens without a source

@Mathein cool :) ich bekomme so 850€ im Monat vom DAAD und wenn ich mehr als 30% davon für Miete ausgeben muss zahlen sie so 250€ dazu, also such ich mir a schöne wohnung... hahaha

4:49 PM
and I can go on and on and on on that...

@Mathein ich werde ökonomischer Migrant! Hoffentlich hassen mich alle Rechtsradikaler

:) und wie läufts bei dir so? Du wurdest vermisst!

ja, mir ging es ja nicht so gut, hatte ich schon erwähnt, glaube ich. Mache ein Urlaubssemester dieses Semester

Also burnout oder wie? :/

4:58 PM
ja quasi

Schon kacke, gut dass a kleine Pause machst aber :P

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