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12:50 AM
Unfortunately I am not a professor so can barely comment. I have TA'd for several professors so far and there is not a general rule observable from my perspective so far.
 
 
2 hours later…
2:21 AM
Suppose $X$ is a set and $E_1,E_2,\dots,$ is a disjoint sequence of subsets of $X$ such that $\bigcup_{k=1}^{\infty}E_k=X$. Let $\mathcal{S}=\{\bigcup_{k\in\,K}E_k\,:\;K\subset\mathbb{Z}^{+}\}$. Prove that a function from $X$ to $\mathbb{R}$ is $\mathcal{S}$-measurable if and only if the function is constant on $E_k$ for every $k\in\mathbb{Z}^{+}$
I am stuck on the forward direction, don't see how a measurable function on disjoint set is a constant on each subset
any hints are welcome
 
2:33 AM
@Simple what have you tried?
Any examples?
 
there is an example says if $ S=\{\phi,(-\infty,0),[0,\infty)\}$ then a function
$f : R \to R$ is S-measurable if and only if f is constant on $(-\infty, 0)$ and f is
constant on $[0,\infty)$
 
The proof of that should be enlightening for this, I think.
 
I think so, but the example is unclear to me
 
Okay, well what about the example is unclear. Walk me through it from the start.
 
$S=\{\phi,(-\infty,0),[0,\infty),R\}$ is a $\sigma$ algebra with disjoint set, $f((-\infty,0))=B$ where $B$ is Borel set if $f$ is measurable
then how do we know $B$ just has one constant element
 
2:45 AM
What does it mean for $f$ to be measurable?
 
the inverse image of $f$ is a Borel set
 
So since $f$ is a function, it definitely attains some value, right? Let's say $f(x) = y\in\mathbb R$.
So take a Borel set with $y$ in it.
Call it $B$ or something.
What is $f^{-1}[B]$?
 
some $x$ in $R$
 
Since $f$ is measurable, it should be a measurable subset, no?
 
it's measurable
 
2:59 AM
Is {x} a measurable subset?
 
yes,
 
!!!
Is it really?
 
I was thinking $S=\{\{x\}\,:\;x\in\,X\}$
$S$ is measurable, $\{x\}$ is an element of $S$, $\{x\}$ should be measurable
 
$S$ is your collection of measurable subsets.
$S$ itself is not "measurable".
 
I see, the smallest sigma-algebra on X containing S is the set of all subsets E of X
such that E is countable
 
3:24 AM
Okay, so recalling that $S$ is actually that four element subset you mentioned way back there, what is the smallest sigma algebra containing it?
 
@anakhro it's $S$ itself
 
Okay, so returning to $f^{-1}[B]$, what is it?
 
You say $f(x)=y$, then $f^{-1}(B)$ is measueable subsets of $R$
 
So the conclusion should be that it is an element of $S$, right?
 
yes
 
3:35 AM
So now do it by cases. What happens if it is $\emptyset$? Can it be $\emptyset$?
 
the empty set is an element of $S$, but $f(\phi)=\phi$
 
So you conclude?
 
$f^{-1}(B)=X$
 
What is X?
 
it is R in this example
 
3:44 AM
Hey, I was wondering if this is right: $\mathbb{Z} \nsubseteq \mathbb{N} \cap \mathbb{Q}$?
 
@Abwatts What element is in Z which is not in $\mathbb N\cap \mathbb Q$?
 
negative integers?
 
Is the question mark because you are unsure?
 
Well, I am quite sure it's the case, but I just wanted to confirm that ;]
 
I think you have your answer, then. :)
It's also worth mentioning that the right side is easily reduced to $\mathbb N\cap\mathbb Q = \mathbb N$
 
3:48 AM
I see, thank you!
By the way, could you assist me with a quick proof that I tried to construct involving sets?
I am unsure if that is quite right
 
Sure!
 
Awesome! So, the question is: If$A-B = B-A$, then $A=B$. I proceeded by using proof by contradiction, like so: Assume $A \ne B$. Now, this means that there exists $x \in A$ but $x \notin B$. Furthermore, since we are given that $A-B=B-A$, we know that $x \in A-B$ and $x \in B-A$ (by definition of set equality), but this is a contradiction since $x \in A$ and $x \notin A$ at the same time, which goes against our initial assumption ($x \in A$)!
 
First little thing, you can say "without loss of generality" that there exists an $x\in A$ but $x\notin B$. It could very well be the other way around.
 
I haven't thought about that, thanks for pointing it out! However, do you think the proof would be valid if I'll add it to my proof?
 
Technically, yes.
I have some minor complaints, but I think those are less important.
 
3:59 AM
Cool! I just started out with proofs, so I'm looking to get comfortable with writing them.
 
You are off to a good start.
 
Thanks! Well, go for it. Feel free to tell me your complaints! I'd be glad to hear them. I would like to learn from my mistakes ;p
 
They are just minor things, like I think it is "more obvious" that $x\in B-A$ implies $x\in B$, but you get a similar contradiction either way. It's really the same thing, just looking at it upside down.
@Simple sorry I stopped responding: I think you need to go for a jog and clear your mind or something. Come back to this problem after a bit with a fresh canvas and try to work at it step by step.
 
I think so, I got a headache now
 
Yeah, just come back to it later. Ping me if you need help and no one is giving it to you. The question is elementary enough that most people who know measure theory to some degree can probably help you walk through it.
 
4:20 AM
Trying to prove the snake lemma, and I'm getting hung up on the kernel maps
 
No, I know the movie, and I'm not watching it. That's cheating
Well "cheating"
It's actually only cheating with respect to the arbitrary rules I put upon myself, but still
 
@Rithaniel What do you mean you get hung up on the kernel maps
 
It's a rite of passage.
 
Well, I believe I have everything except the fact that $\text{ker}(\alpha_2)\subseteq\text{Im}(\alpha_1)$
 
4:24 AM
If you can't prove the snake lemma, you might as well give up your hopes and dreams of being a tenured professor at the ENS.
 
First, we approach $\alpha_1$ and $\alpha_2$. We define these functions as equivalent to $f_i$, but restricted to $\text{ker}(g_i)$. We immediately note that if $a_i\in\text{ker}(g_i)$, then $h_i(g_i(a_i))=0$, and, by commutativity, we know that $h_i(g_i)=g_{i+1}(f_i)$ and so $f_i(a_i)\in\text{ker}(g_{i+1})$.
Thus, these $\alpha_i$ map from $\text{ker}(g_i)$ to $\text{ker}(g_{i+1})$. Further, since each $f_i$ is a well-defined homomorphism, we know that each $\alpha_i$ is a well-defined homomorphism
Since $f_2(f_1(a_1))=0$ for all $a_1\in A_1$, we know that $\text{Im}(\alpha_1)\subseteq\text{ker}(\alpha_2)$. Further, . . .
I supposedly have already finished the difficult part, in finding the bridging homomorphism and showing that it is well defined and a homomorphism
 
4:49 AM
Ah, found the bit I was missing
The zeros in the snake diagram are important
 
5:17 AM
$f$ is measurable, by definition $f^{-1}(B)=\begin{cases}X\\\phi\end{cases}=\begin{cases}\cup_{k\in\,K}E_k\\\phi\end{cases}$ for every borel set of $R$, and $\cup_{k\in\,K}E_k$ is measurable which implies each $E_k$ is measurable
@anakhro
 
5:54 AM
the sigma algebra $S$ is $\{\phi, X\}$, this implies $f$ is constant from $X$ to $R$. Because $X$ is a union of disjoint $E_k$, $f$ is constant on each $E_k$
 
6:10 AM
No, this is totally incorrect
 
6:26 AM
There is some discussion of the above problem also in the calculus chatroom: chat.stackexchange.com/transcript/14150/2020/1/23 chat.stackexchange.com/transcript/14150/2020/1/24
 
 
3 hours later…
9:52 AM
Hey @AlessandroCodenotti
 
Hi
I greeted you yesterday or a couple days ago, but I think it was just my laptop being drunk and showing you as online while you weren't
 
Oh I think I know what happened
I think I have chat on one of my tabs open in chrome so whenever I get in it reloads
 
I'd be hard pushed to find this tab though
I have thousands of them open simultaneously
 
There should be an option to only reload tabs when you switch to them, it's nice for memory consumption if you often keep a lot of tabs open
 
9:55 AM
Yeah
 
So what kind of math have you been thinking about lately?
 
Ehh dunno. Mostly Galois theory, also reading Grauert-Remmert's Stein spaces. Thinking a little bit about statistics
Do you know anything about measured foliations by chance
 
I don't know anything about foliations of any kind
 
Ah ok. Fair
I think the idea is they are foliations with a measure equipped along the transverse direction
 
I know a little about measures though :P
I'm thinking about doing a gmt course next term
 
10:01 AM
Cool idk any GMT
 
I'll go to the first lectures and see if I have the background for it
The professor is a PDEs person I think, but I asked him about the syllabus for gmt and it seems interesting
Unfortunately he does not plan on using Federer. What a missed opportunity
 
Lmfao
 
try checkmating with a bishop and a knight
 
I learned how to do that last year
That does not imply I remember how to do it now
 
the interactive lesson on lichess is very instructive
 
10:09 AM
That might be the only piece of chess knowledge I actually learned OTB
 
10:45 AM
How do you express density in $R^2,R^3,...$? does it even make sense?
With a lexicographical ordering?
 
10:57 AM
With a topology
 
meaning "neighborhoods"?
Are you saying it takes balls to define density in $R^n$? sorry I couldn't resist.
 
 
2 hours later…
1:00 PM
Hi guys from Math
Suppose I have some summation, and I wish to treat this sum as a sum of residues
The residues will be those of a function with poles on the real axis, the rational function which I wish to relate it to, will also have poles on the real axis
I'd like to choose the correct integration path here
I'm aware of the Matsubara frequency summations
 
1:13 PM
0.75=(0.9-x)/(0.7-x)..How to solve this
Please help
 
@user586228 whats the problem with it?
 
1:26 PM
Hmmm, Meg is now Edward. Perhaps I should drop the name Rithaniel
 
 
1 hour later…
2:39 PM
HI all! TIL that being sequentially compact doesn't necessarily imply being compact, what about sequential continuity?
 
It does not imply continuity
Sequential continuity and continuity are equivalent in first countable spaces though
 
@AlessandroCodenotti ok just like with sequential compactness and compactness
sounds good
 
3:24 PM
@StupidQuestionsInc yes
 
@Shootforthemoon I don't want to clutter up the comments on the post, so I am going to respond here. You said "Btw I mean it fits the definitions of course, but they come even after a simple proof such as x=0.999... - - >10x-x=9x=9 hence x=1."
This is not a proof. This is a heuristic, which is often used to justify the statement in an elementary setting. However, a lot of work needs to be done before this is actually a proof.
In particular, the notation $0.\overline{9}$ must be defined before you can start working with it.
 
Also sequentially compact implies countably compact, but this is not a very useful property
 
As I said in my comment to your question, the notation $0.\overline{9}$ is usually defined as a series: $$\sum_{n=1}^{\infty} 9\cdot 10^{-n}.$$
If this is how the notation is defined, then $0.\overline{9} = 1$ because this series is convergent, and can be computed explicitly via an $\varepsilon$-$\delta$ argument.
If you have some other definition in mind, then you need to start by making that definition implicit. Once you have done that, then we can discuss whether or not $0.\overline{9} = 1$.
 
in defense of that simple proof, I'll note that it's basically the same as the proof of the geometric series using stability and regularity of the summation method here
so it's an argument which works for any summation method with those as properties, and in that regard is neutral as to what method you use
(in that vein, I wonder if one could define $0.\overline{ab\ldots c}$ as the decimal number such that multiplying by 10 gives $a+0.\overline{b\ldots c}$ and adding 1 gives $1.\overline{ab\ldots c}$? but this gets back to the point of definition being important.)
 
3:58 PM
@Semiclassical Like I said, the argument can be made rigorous, but one has to do the work, which starts with an appropriate definition.
 
sure. my point is that you don't actually need a lot in order to get 0.999... = 1, to the extent that a definition which didn't yield that would have to be in some sense unreasonable
(to say it differently, my intuition would be: one can choose to define 0.999... in such a way that it's 1, or you can define it in such a way that it's not a real number. but you can't define it in such a way that it's a real number other than 1.)
 
4:33 PM
@Semiclassical Sure. I have no objection to this statement.
I think any objection I have to your statements is simply a matter of degree: I actually think that it takes a bit of work to define the real numbers in the first place. If we accept that the real numbers are there (or if we believe that they don't require much work to build), then one does not need "a lot" to get $0.\overline{9} = 1$.
 
5:11 PM
As many interesting constructions of the reals as there are, I'm not sure if there's any that doesn't require a bit of hands-on work
 
5:27 PM
@XanderHenderson sure. maybe the way I'd say it is this: I'd anticipate that it's not too hard to argue that, if 0.999... is to be a rational number, it's got to be 1. where things get dicey is replacing "rational number" with "real number"
 
 
1 hour later…
6:42 PM
@XanderHenderson Eventually I was asking what is the sense behind the definitions we gave and therefore the physical interpretation of the result we obtained. We got an equality between symbols. We should be able to see if that final result has some concrete meaning or is just abstract.
Who tells us that a limit is really accurate in defining for example real numbers? When the distance (theoretical or physical) between two numbers is zero, with the limit we conclude that the two objects are the same.
 
@Lukas What are you using for blog?
 
7:36 PM
@Alessandro ciao! uso wordpress
 
What is a CW complex of finite type?
 
@user193319 a CW complex with only finitely many cells
 
Ah, okay. So that entails that it is finite dimensional?
 
yes
 
Is there a finite dimensional CW complex that is not of finite type?
Gluing an infinite number of discs together?
Or taking the disjoint union of an infinite number of discs?
 
7:43 PM
yes, both of those examples work
 
Cool. Thanks!
 
@LukasHeger Capito, supporta LaTeX nativamente?
 
@AlessandroCodenotti Sì! Però non supporta tanti "package" quanti MathJax
 
tanti* quanti* (usiamo package anche in Italiano in questo contesto, non credo che ci sia una traduzione, ma lo usiamo come se fosse un plurale)
 
grazie
 
7:56 PM
But that's very interesting, thanks! I have some nice maths that I wanted to write down in LaTeX for myself, and I was thinking I might as well put it somewhere where other people can read it too
 
e si deve scrivere "$latex $" ogni volta, è un po' dispensioso
 
deve*
Hm capito, un po' scomodo ma non eccessivamente dai
 
if you use wordpress with LaTeX make sure to choose a theme with the property that the wordpress LaTeX guesses your background colour correctly
 
Otherwise you get ugly gray boxes around the LaTeX?
 
there's some algorithm that guesses the background colour of your theme, but if that doesn't work, all the LaTeX is rendered into an image with a different background colour which looks bad
yes
I recently fixed that on my blog
 
8:02 PM
I see
I'll look into it after the exams, right now those keep me busy enough :P
 
in bocca al lupo per gli esami!
 
Grazie!
Ne devi sostenere anche tu a breve?
 
Sì, ho due esami orali
 
Per che corsi? Immagino cose di algebra o teoria algebrica dei numeri?
 
una dei spazi adici (adic spaces, I'm only guessing the Italian words) e una di funzioni L
 
8:05 PM
Capito
(I don't know the Italian name either, that's stuff I've never heard of before!)
 
è "funzioni L" oppure "L-funzioni"? È "L-functions" in inglese e "L-Funktionen" in tedesco, ma "funzioni L" suona più corretto
 
corretto*
Purtroppo non ne ho idea
Funzioni L a quanto pare. Il trucco che uso di solito è aprire la pagina di wikipedia in Inglese (L-functions) e cambiare la lingua. Non funziona sempre ma stavolta mi è andata bene
 
Purtroppo non esiste una pagina di wikipedia italiana dei spazi adici
Però esiste "Numero p-adico", perciò pensa che "spazi adici" è corretto
oppure "spazi adichi"?
no, il plurale è numeri p-adici
 
@LukasHeger adici
 
 
2 hours later…
10:00 PM
Salut @Ted
 
Salut @Lukas
 
Weird question, but do you like poetry?
 
Although I studied it in college (mostly in French), it's not something I have spent much of my adult life reading, I confess.
 
I like it very much, I even wrote a sonnet
Hey @Edward
 
Hey @Lukas et al
 
10:03 PM
Oh, you changed your name to Edward, and now I'm al?
 
Right
Al Shifrin
Al 'Mighty' Shifrin
 
Sounds inspiring.
 
Truly so
@Lukas finding Hecke theory quite difficult atm lol
 
it's not the easiest part of modular forms, that's for sure
it'll all make sense when you work out how a Hecke operator acts on the Fourier expansion of a modular form
 
Yeah I think so too, I'm just stuck in the middle of all of the double coset junk atm
 
10:08 PM
Just be thankful there are only two sides ... or you'd have triple cosets :P
 
lol ouch
 
Hi @Ted @Lukas @Edward
 
maybe if you have some tertiary group-like operation
 
Hey @Alessandro
 
Ciao @Alessandro
 
10:10 PM
hi Demonic @Alessandro
 
@Lukas also I was just in the Luisenpark in Mannheim and they have some cool light exhibitions
Winterlichter halt
 
yeah I saw that
not the Winterlichter themselves, but the fact that they are there
 
Ah right
Yeah they had a cool projection through water
 

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