1:22 AM
Hmm...If $\iota : X \to X \times X$ is given by $\iota (x) = (x,x)$, then I agree that $(id_X \times f) \circ \iota : X \to X \times Y$ is measurable...But how does that show that $graph(f) = \{(x,f(x)) : x \in X \}$ is a measurable subset of $X \times Y$? All I can see is that $graph(f)$ is the image of $X$ under $(id_X \times f) \circ \iota$...but the image of a measurable set under a measurable function is not necessarily measurable...hmmm...
0

$$3 + 5 = 8 \\ 3 + 7 = 10\\ 5 + 7 = 12 \\ 5 + 11 = 16 \\ 7 + 11 = 18 \\ 7 + 13 = 20 \\ 11 + 13 = 24 \\ 11 + 17 = 18 \\ 13 + 17 = 30 \\ 13 + 19 = 32\\ 17 + 19 = 36\\ 17 + 23 = 40 \\ 19 + 23 = 42 \\ ? = 44 \\ 19 + 29 = 48\\ ? = 50 \\ 23 + 29 = 52 \\ 23 + 31 = 54 \\ ? = 56 \\ 29 + 31 = 60 \\ ? = ... Ask me any questions you might have 1 hour later… 2:36 AM 3 Suppose S = \{(x,y) \in \mathbb{R}^2\mid y = x + 1\text{ and } 0 < x < 2\}. Question Describe the equivalence relation T on the real line that is the intersection of all equivalence relations on the real line that contain S. Suppose that$$A_1 = SA_2 = \{(x,y) \in \mathbb{R}^2\mid y = ...

Two questions: How were all of the $A_i$ determined? and from that how were the equivalence classes determined?
without even checking the definitions of those A_i, you might generally look up the concepts of 'symmetric closure' and 'transitive closure' of a relation. not because they are theoretically deep but (aside from adding in the diagonal if it happens to be missing) they capture the spirit of "toss in the stuff that any symmetric/transitive relation containing this stuff would have to include."
i'm guessing they are doing something like that.
computing the transitive closure of an arbitrary relation can be computationally, i don't know if difficult is the right word, but more interesting than the other kinds of closing up, in that you might think more about how to organize an algorithm to do it to make sure you get everything, than you do for symmetry and reflexivity.
This is the first chapter of Munkres's Topology. So though I could decipher what symmetric closure and transitive closure mean based on experience, I'm wondering if this is something I should have known prior
i want to be very clear, there is nothing about 'symmetric closure' or 'transitive closure' that would require a separate section of a textbook or special definitions.
they're just the words for the general thing that this kind of exercise is asking you to do.
Ok fair enough.
so for purposes of googling, that's what i might google. definitely not saying "oh if you haven't gotten there in the textbook, forget it, they must want you to do it some other way."
however you do it, you will be confronting those concepts in this example.
but maybe someone somewhere has some cute way of organizing the work, or speeding it up, when the relation is a subset of R^2, or something. or someone has written out a lot of examples.
if you haven't guessed from the form of the exercise, the specific example in the problem is pretty random and made up. you will not be confronting a lot of this later in the book :)
2:44 AM
I'll take a look at the concepts you suggested and try to work it out a but
No I agree I won't be confronting it. It is more so for me to just get more comfortable with getting dirty with elements and understand what is going on beyond the usual "abstract" idea
@TedShifrin Hmm...If $\iota : X \to X \times X$ is given by $\iota (x) = (x,x)$, then I agree that $(id_X \times f) \circ \iota : X \to X \times Y$ is measurable...But how does that show that $graph(f) = \{(x,f(x)) : x \in X \}$ is a measurable subset of $X \times Y$? All I can see is that $graph(f)$ is the image of $X$ under $(id_X \times f) \circ \iota$...but the image of a measurable set under a measurable function is not necessarily measurable...hmmm...
3:02 AM
user: what are your hypotheses on the sigma algebras on X, Y, and X x Y? they matter. e.g. if {emptyset, X} and {emptyset, Y} are the sigma algebras on X and Y then every function from X to Y will be measurable, but graphs of functions will not be measurable in X x Y with the product sigma algebra (outside of trivial cases e.g. where X and Y are empty, or Y is a singleton)
I don't think we can prove Twin Primes without proving Goldbach and vise versa - they're intricately linked
You need to prove both of them simultaneously for an induction argument to work
3:19 AM
@leslietownes what about all that blsh completion stuff blah done for Fubini?
I think everything in user's question are Borel measurable
X, Y top spaces, X x Y product top, all equipped with the sigma algebra of Borel sets
I think this is fine as long as Y has a countable basis, or something
definitely don't want to mix topology with measure theory unless I have some assumption like that, that's true
ending up in set-theory madness resulting from not assuming that is not something that seems productive
most of measure theory is "Stuff is true if everything is nice"
which is good enough for those of us who couldn't care less about geometric measure theory
4:00 AM
I think I just proved Goldbach -_-
This is getting ridiculous mon
Who wants to see the 2 paragraph proof of Goldbach?
Check my profile. It's my latest post
I argue that if not Goldbach, then this monoid doesn't generate the odd numberse
I mean this generating set doesn't generate the odd number monoid
It's solid as a rock, so far
It's actually more solid than that
I would say it's the center of a black hole
?
Take the odd numbers
it forms a monoid
:D
It's all finite products of odd primes and 1
In other words both $2\Bbb{N} + 1$ and $2 \Bbb{Z} + 1$ take your choice are both monoids, this is well known
Yeah sorry
@SoumikMukherjee did you check my post?
I need someone to check it, since I was kind enough to make it 2 paragraphs, surely someone could glance at it and find the flaw.
0

Let $M = 2\Bbb{N} + 1$ be the monoid of odd naturals. It contains all odd primes. If for some $2n \in \Bbb{N}$ we have that there exists no prime solution to $2n - p = q$, then that's the same thing as saying the submonoid $N$ generated by all $2n - q, n \in \Bbb{N}, q \in \Bbb{P}$ such that $q... Thx to whoever on mse upvoted that 4:34 AM @DLeftAdjointtoU yes What are your thoughts or questions When constructing the subset, you are considering all$2n-p$as generators, in the later argument you are saying that for each odd integer ( or better for each prime)$p$you will have a prime$q$such that$p+q$is even. But this does not guarantee that$p+q$will be the same$2n$as you started with. The$2n - q: n \in \Bbb{N}, q \in \Bbb{P}$are the generates of sum submonoid, right? Since you can generate one from any subset It doesn't have to be the same$n$that we started with - I'm not sure what you're asking Those are just dummy / internal variables *some @SoumikMukherjee Look at the third paragraph which expounds upon the first paragraph's argument that it glanced over Any$q \in N$must be of the form$2n - p$since you can't form a prime via generation ops You can't write$q = xy$, for nontrivial x,y, so... @DLeftAdjointtoU Yes, your post precise shows that any$q \in \Bbb{N}$will be of the form$2n-p$, so in particular any prime$q$will be of the form$2n-p$for some other prime$p$, but that does not imply that any even natural number will be of the sum of two primes. you've shown that$\forall p\in \Bbb{P} \exists q \in \Bbb{P}$such that$p+q=2n$, what you need to show is that$\forall 2n\in 2\Bbb{N} \exists p, q \in \Bbb{P}$such that$p+q=2n$5:00 AM Hi guys! I was recently going through a whole new proof outline of the Schröder-Bernstein, and I never seen a proof like this. However, I am stuch at a portion in the proof, which halts me from proceeding with the remaining lines of the reasoning in the proof. Any help regarding this will be much appreciated. I made a post. The link is: 1 I have seen a number of proofs of the much popular Schroder–Bernstein Theorem. But a particular proof outline interested me. My version of the Schroder–Bernstein Theorem is : Assume there exists a$1–1$function$f : X → Y$and another$1–1$function$g : Y → X.$Then there exists a$1–1,$onto... @SoumikMukherjee thx @SoumikMukherjee Hmm...Now, that I take a look, your explanation looks self-contained! But the attempt of D Left was exciting to see, I admit. @ThomasFinley yes But, now back to my question. Anyone out there who is interested to help me/solve my apparent issue ? 5:18 AM @ThomasFinley I don't think that you should edit so much of your question, you can keep your attempts and write down the progresses in your edit. @SoumikMukherjee oh! I was unaware of that. But I thought it would make it cluttered and made the post fresh as new. But what do you think about the proof given out there. hi, does anyone know why this is true: if$ f : \mathbb{C} \rightarrow \mathbb{C}$is lipschitz continuous with lipschitz constant$M $, and$f = 0$outside$\{ |z| \leq \delta \}$, then$h(r, \theta) = f(r \exp(i \theta))$is lipschitz continuous with lipschitz constant$M(1 + \delta)$? I can see that$h$is lipschitz continuous, but I dont know how to exactly get this constant. there should be a simple geometric argument here also$h : \mathbb{R}^2 \equiv \mathbb{C} \rightarrow \mathbb{C}$thanks for any suggestions in advance 5:34 AM @ThomasFinley Are you asking what the chain$C_y$is? @SoumikMukherjee To be precise, I am asking for a proof-explanation for the part d. rep. theory is hard as hell! Can't understand the lecture @SoumikMukherjee I have edited the post. I hope this makes things clearer. 5:50 AM$C_y$can be assumed as$C_{g(y)}$, where$g(y)$is in$X$So you are comparing two chains$C_x$and$C_{g(y)}$where both$x$and$g(y)$are in$X$@SoumikMukherjee And as$y\in C$and$y\in C_{g(x)}$, so,$C_{g(x)} =C$? 6:17 AM @SoumikMukherjee Oh, I made a typo there. I was talking about$C_{g(y)}$. But thanks, I got it. 2 hours later… 7:51 AM :) 1 hour later… 8:55 AM Is$\infty \in \mathbb{N}$? I tested positive for COVID-19 this morning. So that sucks. Maybe some of the regulars here who know me might want to know, hence this post. The self isolation frees up a lot of my time. I'll probably spend it doing Maths :) 9:10 AM @Shaun Not nice , but nothing compared with my situation since middle of February. My 86 year old mother cannot leave her bed since then and moreover we lose much much money because the insurances barely pay anything. The chance that she can at least ever walk again is nearly zero. Oh no! I'm sorry, @Peter. That sounds awful! I hope she recovers soon. 9:37 AM @PlaceReporter99 no @Shaun I hope it's over soon. @Peter Sorry to hear that. 9:59 AM I have given the torus$\Bbb{R}^2/\Bbb{Z}^2$and a matrix which has eigenvector$(1,-2)$. I want to identify the torus with the unit square. How can I draw the eigenspace there with my eigenvector? @robjohn can you maybe help me? I am not sure what you mean by "draw the eigenspace". It seems that the torus is already identified with the unit square by its definition. Perhaps I am just not getting what you're asking. @robjohn In the lecture we have drawn a line resp two lines in the unit square and our assistant told us that this is the eigenspace given by$(1,-2)$and I don't get how he got to that lines using the eigenvector We have drawn this here: Oh, that is one line on the torus and he told us that this green lines are the eigenspace. and I don't get how he could draw this. Could you maybe help me understanding this? the line is all real multiples of the eigenvector 10:09 AM But where is (0,0) in this square? any of the corners all of the corners are identified since you're modding by$\mathbb{Z}^2$But I mean if I take the left corner below to be (0,0) then I need to go 1 in x direction and -2 in y direction? but that takes you out of the square @robjohn Ah I see because the corners are either(1,0), (0,1), (1,1) and this is all equal to (0,0) mod Z^2 yes 10:11 AM @robjohn right thats my confusion ahaa so whait if i start as he did in the left upper corner, then I draw it as I said, but I can only do this up to the boundary of the square so i land at the point (-1,1/2) because then I do mod 1 and start again at the point (0,1/2) and I go to the right lower corner? Is this correct? @robjohn you go from the upper left corner$(0,1)$go$\frac12$in$x$to$\left(\frac12,0\right)$that equals$\left(\frac12,1\right)$and continue from there to$(1,0)$and you're back where you started I think this is the same as I wrote but I wrote (0,0) instead of (0,1) for the upper left corner yeah, I am considering$[0,1]\times[0,1]$as the unit square okey thanks a lot!! 2 hours later… 12:01 PM Can someone take a quick look at wolframalpha.com/… ? So I've finally figured out a way to solve my problem and got an answer equal to the answer here. What I'm wondering is, is this form okay? There are some terms that I can't put inside the sum like$\ln16$and$(x-2)[1/2 + \ln4]$. Would it be okay if I just left those as is and stated the answer as it is shown on the page? 12:27 PM They don’t Can someone with Mathematica please help me? What answers do you get for these two one-liners? Limit[Sum[Log[2*k], {k, 1, n}] - Sum[Log[2*k - 1], {k, 1, n + 1/2}], n -> Infinity] Limit[Sum[Log[2*k], {k, 1, n}] - Sum[Log[2*k - 1], {k, 1, n}], n -> Infinity] (**) The reason is that others and I suspect that the answer to the first one-liner is a bug, at least in the current version of Wolfram Alpha and Mathematica 8.0.1. Is this true: $$\lim_{n\to \infty } \left(\sum _{k=1}^n \log (2 k)-\sum _{k=1}^{n+\frac{1}{2}} \log (2 k-1)\right)=\frac{1}{2} \log \left(\frac{\pi }{2}\right)$$ ? @MatsGranvik I'm getting$\frac{1}{2}\log(\frac{\pi}{2})$for the first one and$\infty$for the second one. @ephe Great! What Mathematica version are you using? That is what I got too in Mathematica 8.0.1 and Wolfram Alpha. 12:44 PM @MatsGranvik Sorry for the delay, my PC froze for a minute there.$version gives me 13.2.0 for Linux x86 (64-bit) (December 12, 2022)
@ephe What do you think, is it a bug answer?
Or a feature.
@MatsGranvik I think that's a bug
1:00 PM
@MatsGranvik is it $\log_{10}$ or $\log_e$?
X- connected, A -connected subset of X. Then the complement of any component of X-A is connected.
(the complement is to be taken in X)
@MatsGranvik desmos seems to suggest that the answer to that is $-\infty$
Let C be a component of X-A. Let f: X-C--> {0,1} be a continuous map. Since A is a connected subset of X-C, $f|_A$ is constant.
How do I prove $f$ to be constant?
@PlaceReporter99 $\log_e$
@Shaun indeed, spend it on math. I will want to know your progress during this time. Wish you to get well.
1:15 PM
the knots get pretty wild
@Koro Continuous function maps connected sets to connected sets. You already wrote that the complement of $C$ is connected.
It is to be proven that X-C is connected.
@Koro it's not true I think
I see, I didn't write that there.
@Jakobian NOT true if complement is taken in X-A but true if the complementation is done in X.
ah
1:22 PM
any ideas on how to show f to be constant. Showing that will prove connectedness of X-C.
But I want to prove it using cont. function based definition of connectedness.
anyone know how to solve this math.stackexchange.com/questions/4712349/… ?
@Koro it seems to me like you want to extend $f$ to a continuous function on $X$, and to prove it's continuous, use gluing lemma?
applied to $f\restriction_{X\setminus A}$ and $f\restriction_{X\setminus\text{cl}(C)}$
oh sorry those don't cover X
How would you extend $f$ to all of X?
1:44 PM
If $\sum_{n=1}^\infty a_n$ converges and $a_n<0$ then $\sum_{n=1}^\infty -a_n$ also converges hence $\sum_{n=1}^\infty |a_n|$ converges. Then since $|(-1)^n a_n|=|a_n|$ we have that $\sum_{n=1}^\infty (-1)^n a_n$ converges absolutely.

Is there an obvious fault in my reasoning here?
@Koro the length of this makes me think the proof with continuous functions is going to be hopeless
at least if we don't want to fill up a whole page
I feel like I made a mistake here because I didn't even use the fact that $a_n<0$
1:58 PM
@ephe If $a_n\lt0$ (for all $n$) and $\sum\limits_{n=1}^\infty a_n$ converges then $\sum\limits_{n=1}^\infty(-1)^na_n$ does converge absolutely.
Hello, I have a question about Fourier inversion theorem. In this question, two approaches are mentioned about existence of inverse Fourier transform for $f\in L^2$. It appears to me that using tempered distributions is not only simpler, but also covering a broader situation.
So why does one need the other method, using Plancherel's Theorem and Carleson's Theorem? Am I missing something?
@ephe yes, you did. You said that $\sum\limits_{n=1}^\infty(-a_n)=\sum\limits_{n=1}^\infty|a_n|$ converges, which is absolute convergence.
@robjohn Oh! That made me figure out the gap in my knowledge. I thought $\sum_{n=1}^\infty a_n$ and $\sum_{n=1}^\infty -a_n$ both converging implied that $\sum_{n=1}^\infty |a_n|$ converges (for all sequences that is) which is false. Thank you!!
@ephe those definitely do NOT imply absolute convergence. They are equivalent and the second tells you nothing more than the first.
@robjohn Yes those implying each other was what I was thinking when writing the proof which made me question my whole absolute convergence knowledge. Thanks for the verification!
2:15 PM
@leslietownes I guess we assume that $X$ and $Y$ are Borel spaces.
2:27 PM
@MatsGranvik I don't know how you evaluate the $n+\frac12$ in the upper limit any differently than $n$, but the difference of the sums is the log of $2^nn!\frac{2^nn!}{(2n)!}=\frac{4^n}{\binom{2n}{n}}\sim\sqrt{\pi n}$ as shown here. Thus, I don't think the limit is correct.
gotta take the dog to the park. BBL
Hi, bye, @Ted
2:42 PM
@user193319 what's a Borel space?
what's the idea behind the proof that there's $2^{2^{|X|}}$ ultrafilters on infinite set $X$
apparently this is Pospíšil theorem
which I'll write as Pospisil since no such fancy letters on my keyboard
did someone just experiment or is there something deeper going on here
hmm I guess it's not as unintuitive to start with $\mathcal{P}^{<\omega}(\mathcal{P}^{<\omega}(X)) \subseteq \mathcal{P}(\mathcal{P}(X))$
and being a filter does sort of rely on finite subsets
math.stackexchange.com/a/83540/476484 I think this argument by Brian Scott is a more elaborate version of the proof I'm thinking about
3:14 PM
I like how everyone online calls Brian Scott a horrible teacher but his answers were actually amazing many times to me
He's amazing.
I believe he's excellent as a teacher too.
3:49 PM
I don't know, but there's definitely many factors to it, he may be better as a teacher for mathematicians than engineers etc.
4:00 PM
You are right. Another user told me to look at the plot:
(*start*)
ListLinePlot[
Table[Sum[Log[2*k], {k, 1, n}] -
Sum[Log[2*k - 1], {k, 1, n + 1/2}], {n, 1, 120}]]
(*end*)
which diverges, therefore the limit does not exist.
4:10 PM
Suppose that A is NOT a path connected subspace of B. Then B is also not path connected.
Therefore, I don't see the point of proof of corollary 3.8 here kconrad.math.uconn.edu/blurbs/topology/connnotpathconn.pdf
nvm
0

This is a problem from Pugh's Real Mathematical Analysis [Chapter 5]. If $C$ us a smooth simple closed curve in the plane, show that it is the union of finitely many arcs $C_l$, each of which is the graph of a smooth function $y = h(x)$ or $x = h(y)$, and the arcs $C_l$ meet only at common endpoi...

4:44 PM
@Jakobian Probably a standard borel space
@Koro you probably wanted to say $A$ is dense in $B$ as well. But yeah, this isn't true
even if we assume that A is connected
Yeah, I guess they mean standard Borel space. I don't like the bare bones "Borel space" when referring to them though
5:28 PM
hello, how we solve this equation $\sin(x+\frac{\pi}{4})=\sqrt{2}/4$ without using calculator
@Koro If B is path connected then every subspace of B is path connected!
@robjohn I like the asymptotics you gave.
@SouravGhosh that's be nice
d*
5:54 PM
interesting that by showing that $\{0, 1\}^{\mathcal{P}(X)}$ has a dense subspace of size $|X|$ for $X$ infinite, we can show that there's $2^{2^{|X|}}$ ultrafilters on $X$
it reminds me a little of that argument which shows that discrete space of size continuum is realcompact by considering ultrafilters on $[0, 1]$
I like the argument of saying something about ultrafilters on a set $X$ by considering some topological space
so cool
6:15 PM
Hello. Can someone help me with this question?

If $C$ is a smooth simple closed curve in the plane, show that it is the union of finitely many arcs $C_l$, each of which is the graph of a smooth function $y = h(x)$ or $x = h(y)$, and the arcs $C_l$ meet only at common endpoints.
any hints regarding where to start solving this?
My Attempt : Intuitively, you can "cut" the curve at the points where you have a vertical or horizontal tangent, but how to know that the number of these points are finite?
@Jakobian you, lover of topology, can you help?
It's not the kind of topology that I like
6:30 PM
@Jakobian ok
7:00 PM
@FarhadRouhbakhsh unless you assume somerthing like regular that statement is very likely false
its a smooth curve...
OK, I understand your contention. "Smooth curve" should mean a smooth submanifold, not just a curve with a smooth parametrization.
Also draw a picture to illustrate that your approach does not work. It is possible to have infinitely many points with a vertical tangent.
7:15 PM
@fantasie I don't entirely understand your question. The approach using the space of tempered distribution proves Plancherel's theorem, which the first answer quotes, not proves. Carleson's theorem is referred to explain how the Fourier inversion does not hold pointwise, but only pointwise almost everywhere.
Pointwise almost everywhere is stronger than $L^2$, and cannot be proved by means of the theory of distributions.
@BalarkaSen Ok, I also thought about implicit function theorem. But I encountered that my curve is a function from [0,1] to R^2. In the statement of implicit function theorem you need to have a function with domain in R^n * R^m. (That's what is written in Pugh's book Real Mathematical Analysis). So I need to find a function with two variables. On the other hand I know that I am discussing about x & y, so perhaps my function should be related to them. The problem is how to define the function.
So, the answer to wether inverse Fourier transformation exists relies on how one decide functions are equal, in cases of a.e. equal or $L^2$-equivalent, different answers arrises. Right?
@s.harp Its an exercise from Pugh [Chapter 5]. Naturally, it shouldn't be wrong. But I am not familiar with manifolds, regular curves, etc. so I can't comment.
https://math.stackexchange.com/questions/4712503/is-a-smooth-simple-closed-curve-the-union-of-finitely-many-arcs
What is the definition of a smooth curve in Pugh?
@fantasie "Equality" does not exactly make sense. You mean a.e. convergence, L^2 convergence, etc.
Implicit Function Theorem is not directly applicable, as we're not given the curve as a zero-set of a smooth function.
7:29 PM
@BalarkaSen Alright, I get it. Thank you!
Yeah, I wasn't sure what Farhad meant by a smooth curve, hence the question.
@BalarkaSen Smooth means being of class C^r, r times differentiable continuously. Of course r can also be infinity.
@Farhad Here is the right approach: Show that at each point, the tangent line projects isomorphically onto (at least) one of the coordinate axes. Use this and the Inverse Function Theorem to show that nearby it is a graph of a smooth function on an interval on that axis.
I assume Pugh means a $1$-dimensional submanifold.
I am too lazy to go looking.
7:31 PM
I think you mean $\gamma : [0, 1] \to \Bbb R^2$ is smooth if $\gamma$ is $C^r$, $r \geq 1$ and $\gamma' \neq 0$.
This approach shows up on literally dozens of MSE posts because people do not know properly how to show that $y^2=x^3$ is not a submanifold.
@Farhad A smooth curve need not be parametrized.
@TedShifrin Lets say its isomorphic, then how does it help?
Exactly where in Pugh is this exercise?
I want to partition the curve into a finite number of arcs. How does talking about 1 tangent line (horizontal or vertical) help me do that? Or talking about its isomorphisms
Ted is giving a different approach.
The partitioning approach does not work directly, as I mentioned. Figure out why, and then you can worry about fixing it.
7:36 PM
I want to see the precise exercise. Which problem?
Chapter 5. Exercise 50
part d
Yeah, I see. So what is his definition of a smooth curve in the plane?
Anyhow, if the tangent line at $p$ projects isomorphically onto the $x$-axis, then you get an open interval around $p$ which is a graph over the $x$-axis. You can write the curve as a (finite) union of such curves. They overlap. So you make closed arcs which have their endpoints in common.
@TedShifrin let me look up again for the definition
7:40 PM
So the main point is to apply the Inverse Function Theorem to $\pi\circ\phi$ (where $\phi$ is your local parametrization around $p$). This is a map from an open interval in $\Bbb R$ to $\Bbb R$.
I looked again. The definition of smooth is being of the class C infinity. Previous definition was wrong, pardon.
The definition in Pugh's book
Yeah, but what is a smooth curve? :D
I suspect Pugh may have been sloppy here.
Let's say it's a topological subset homeomorphic to the circle that is covered by smooth parametrizations.
That's all you need to do what I suggested.
@Jakobian A Borel space is a type of measurable space built from a topological space. More precisely, given a topological space $X$, you can form the so-called Borel $\sigma$-algebra which is the smallest $\sigma$-algebra containing all the open sets. Then $X$ together with its Borel $\sigma$-algebra is called a Borel space.
@TedShifrin hahaha a curve in the plane in this question is a function from [0,1] to R^2. Being a smooth curve means that this function is of class C infinity. Clear?
@TedShifrin I don't think so. Graph of $y = |x|$ admits a $C^1$ parametrization, $\gamma(t) = (-t^2, t^2)$ for $t \leq 0$ and $(t^2, t^2)$ if $t > 0$. Can be made $C^\infty$ as well.
7:46 PM
Not correct, though.
Right, we need the parametrizations to be regular — nowhere-zero derivative. And it's not clear that every smooth closed curve is the image of a single parametrization. Maybe piecewise-smooth.
@FarhadRouhbakhsh The point is the result is not true if that is the definition of a smooth curve.
Yes, Balarka is correct.
The parametrization needs to be a diffeomorphism.
Ufff, I am confused
And this is needed to proceed with my proof.
You need a tangent line at every point, in particular. With your definition, you might not have that.
7:49 PM
Well, I gave a bit more detail in further messages.
I know Pugh very well. I took courses from him in graduate school. He is not like Rudin. He gives excellent conceptual understanding and very interesting exercises, but he does make little errors. :)
I remember seeing an interview with Pugh and Smale on the panel, and they were showing Pugh's wire frame models of the sphere eversion. One point he said Smale's proof was nonconstructive and did not give any clue as to how one could realize the eversion. Smale shuffled around in his chair uncomfortably and said, "Um, not really"
Thought that was pretty funny
@TedShifrin Why if you project $p$ isomorphically onto the $x$-axis, and after doing this for all of the points in the curve, then you can write the curve as a finite union of such curves? Why finite?
Hint: Compactness.
Yeah I guessed that's it. You get an open cover, and this has finite subcovers
@BalarkaSen I got lost again. These covers are diffeomorphic to intervals in the x-axis. So are you talking about compactness of the curve itself (I mean its image in R^2) or compactness of the domain [0,1] ?
Compactness of the curve.
Ted's hint gives a cover by tiny arcs around each point of the image of the curve in $\Bbb R^2$, which are either graph over the $x$-axis or over the $y$-axis. Choose a finite subcover by compactness of the image of the curve in $\Bbb R^2$.
8:05 PM
@user193319 I see. Pretty reasonable name, but controversial
@user193319 leslie's measureable spaces are Borel spaces according to your definition
equipped with indiscrete topology
@BalarkaSen ok, I need to review how implicit function theorem works to make sure I understand his approach fully.
someone gave some other definition involving a metric, or at least metrizable topologies. i believe the result with that definition.
Standard Borel space you mean?
i didn't mean to insert myself too deeply into that discussion, my only point was that some version of the question had the flavor of "why doesn't this come right out of the definition of measurability, isn't XXX measurable" and my answer was "it doesn't, whether or not that function is measurable"
We can even assume standard Borel if necessary.
I'm working through these notes: mathweb.ucsd.edu/~aioana/orbit%20equivalence.pdf
8:10 PM
Standard Borel spaces have really simple structure, they should be just $\omega^\omega$
Specifically, working through exercise 1.9.
well not $\omega^\omega$, also $\omega$ and stuff like that
when X = Y = R at least there's some version of an argument you can do which is, the graph of f from R to R is the inverse image of {0} under the function (x,y) mapsto y - f(x).
it seems that if the sigma algebras come from nice enough metric topologies you can do something similar, where you intersect sets where d(y,f(x)) is in some basic open set, or something.
@TedShifrin Ok, so you are defining a function from an open interval in R to R, and use that for the implicit function theorem. But as I said earlier, Pugh says the function used for implicit function theorem needs to have a domain R^n*R^m, so for example name the first n coordinates x, and the last m coordinates y. Then the theorem talks about writing for example x as a function of y near a locus point f(x_0,y_0) = z_0. But your function is one variable only. I dont understand...
$(\omega^\omega, \mathcal{B}(\omega^\omega)$ and $(\mathbb{R}, \mathcal{B}(\mathbb{R}))$ are isomorphic in fact
measure theory gets really simple when you only care about nice spaces I guess :)
@user193319 you probably want to use theorem 1.3
split between the countable and uncountable case
8:18 PM
i zapped all of that stuff about ugly measure spaces out of my mind. my advisor made me learn a bunch of it and i took my revenge by forgetting it.
@TedShifrin You can see in this image. f needs to have a multivariable domain, at least 2 variables, to be permitted to get applied in the theorem. But your function is one variable. Or if is it a thing which is clear but I don't get it, you can help me. Thanks
"
So the main point is to apply the Inverse Function Theorem to $\pi\circ\phi$ (where $\phi$ is your local parametrization around $p$). This is a map from an open interval in $\Bbb R$ to $\Bbb R$.

"

@leslietownes was it anyhow useful?
@BalarkaSen Do you have any comments regarding my objection?
jakob: i dunno that it was affirmatively useful but it kept me from making a fool of myself
8:39 PM
So the reason of my confusion was that @BalarkaSen talked about implicit function theorem, so I looked that up to understand what @TedShifrin said. But Ted himself said Inverse Function Theorem. Nonetheless, If you want to use Inverse function theorem you need to have that the derivative of that function is non-zero everywhere. Is that what Pugh missed in writing the question @Ted?
@FarhadRouhbakhsh If you read more carefully, you'll see I said INVERSE function theorem both times.
Yes, ok, I see you saw that finally :)
Yes, as Balarka and I both said, you need to know that your parametrization has nowhere-vanishing derivative.
@TedShifrin I saw that before. Barkla said IMPLICIT XDD
The 1-dimensional inverse function theorem is not usually quoted, because supposedly it's just calculus.
But that's what you need here.
I think I already said so, but with regard to your finiteness question, that's just by compactness of the curve.
Ok, thanks for your guidance and clarifications, sir. If you have time, you can also write it as an answer in the link
https://math.stackexchange.com/questions/4712503/is-a-smooth-simple-closed-curve-the-union-of-finitely-many-arcs

But if you don't have time, perhaps like Pugh writing his book with haste and errors, its also ok. My pleasure to talk with you, professor.
Perhaps you should start by editing your post to include what we have established is the correct version of the definition you intend to be using for the smooth curve.
9:22 PM
9:36 PM
+1 👍
9:47 PM
everyone makes mistakes
more like charles pyeeeeewwwww.
Did you know that $ab/\gcd(a,b)^2$ is a group law on nonzero integers?
So integers are strange.
But there are perfect ones among them.

1 hour later…
11:14 PM
new question
0

Consider $$f(1) = g(1) = 1$$ $$f(2) = A,g(2) = B$$ $$f(3) = 1 + A,g(3) = 1+B$$ And for $n>3$ : $$f(n) = f(n-1) + f(n-2) + f(n/2)$$ $$g(n) = g(n-1) + g(n-2)$$ where we take the integer part of the fraction $n/2$ so $5/2 = 2,7/2=3,...$ Let us define this limit that always converges: t(A,B) = \lim...

11:29 PM
and a soft question
0

What if we generalize and modify the Weierstrass elliptic function for quaterions ? So our function could have $2,3$ or $4$ periods. How would that theory be like ? Is matrix representation the key here ? Noticed I picked quaterions because they have no nonzero zero-divisors.