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12:00 AM
@AkivaWeinberger Are you suggesting that, given $g(x) = h(\frac{2\pi}{N}x - \pi)$, I can simply replace, in the integral, $h(\frac{2\pi}{N}v - \pi)$ by $g(x)$?
I'm done! I didn't do great, but I think I passed!
I had no idea how to find the automorphism group of $A_5$. I didn't even recall that $A_5$ was generated by its 3-cycles... I didn't look at the proof of the simplicity of $A_n, n \geq 5$.
how do those two facts help you find the automorphism group of A5?
Hey everybody!
@Leaky well you know you'll have to permute the 3-cycles
12:02 AM
Well, if I had looked at the proof, I would have some familiarity with working with r-cycles and $A_5$.
Yeah, look at where the generators will go.
Since an element of order 3 has to go to another element of order 3, and the only elements of order 3 in $A_5$ is a 3-cycle
Turns out $Aut(A_5) = S_5$. What a shocker.
@LeakyNun this is incorrect
What's incorrect?
12:03 AM
Well, $g(v)$, but it's a dummy variable so it doesn't really matter
@AkivaWeinberger What does it mean exactly "it doesn't matter"?
@orbit-stabilizer I was replying to a thing from earlier
Feel like Akiva is being ganged up on by questions.
@EricSilva Oh I see.
So, let's have $E$ be the union of the $x$ and $y$ axis in the plane, and let $B = \mathbb{R}$
12:05 AM
@Daminark no since you broke the first rule of this chat room
@AkivaWeinberger But I have a further even more important question.
Rule 1) Never speak about math
Never ask to ask.
We have been changing the interval of integration
12:06 AM
is rule 1
I know ;) Just joking
Now, what? For me, this doesn't look like a proof whose steps follow logically...
We should convert back the variable of the integral
Anyway so there's the projection $p:E\to B$, and I think this should be a quasifibration but not a fibration, and I want to double check that this is the case
Is a good way to study for a real analysis exam to know all the big theorems and their proofs?
whats your definition of fibration?
Any map between contractible spaces should lift trivially.
12:10 AM
Honestly, this is the least convincing proof I've ever seen
Oh... Okay so that doesn't work
nevermind what I said is nonsense.
it is at most inconsistent w.r.t some statements, dude :)
Oh hmm
@nbro your proof is done $h(\frac{2 \pi \nu}{N} - \pi)$ is just $g(\nu)$
12:19 AM
@KevinDriscoll Yes, I understood, but this proof really doesn't seem to be consistent to me
What seems inconsistent? Its exactly right as far as I can tell
@KevinDriscoll We perform a change of variables in an integral which is actually a constant w.r.t the infinite sum (i.e. the Fourier series)
This change of variables should not have anything to do with the initial change of variables of $g$ and $h$, as far as I can tell
In the past, I didn't have to perform many changes of variables, so this is probably one reason why for me this "proof" looks inconsistent.
From the Fourier Theorem applied to $h$ you know $c_n$ in terms of $h$. Now all you're doing is rewriting that so that you learn what the $c_n$ are in terms of $g$
The way you did it obscures things slighty. Instead of doing the change of variables and then substituting in $g$ for something, you could have swapped $h$ for $g$ at the beginning. Then the change of variables is obvious so that you get $g(\nu)$ and not $g$ with some more complicated argument
You don't have to apply the same change of variables in both places. It isn't logically required that you do this rewriting.
But it makes the formula for $d_n$ simple if you do, and so we do it
@KevinDriscoll Ok, we're rewriting $c_n$ in terms of $g$. I understand we did that. But what confuses me is that, to achieve that, we have to change the interval of integration. So, in my head, a voice is saying that we need to change back to the original variable of the integral, i.e. $t$.
Let $(a_n)_{n\in\mathbb{N}}$ be a sequence of real numbers with the following property. There exists a real number $0 < r < 1$ and an integer $N_0$ such that $\forall n \geq N_0$, $|a_{n}-a_{n-1}| \leq r|a_{n-1}-a_{n-2}|$. Prove that $(a_n)$ convergs.
Now, since we're working in $\mathbb{R}$, it suffices to show that this sequence is Cauchy. Beyond that, I've tried the triangle inequality, and that hasn't helped - so I'm kinda stuck.
12:33 AM
@KevinDriscoll Why don't we have to change the current variable of the integral (i.e. $v$) back to $t$ (which means that we would also change the interval of integration, which means that $c_n$ wouldn't anymore look like $d_n$)?
@nbro I think you're getting confused because in the problem statement they wrote $c_n$ as an integral over $t$. And then they wrote $d_n$ also as an integral over $t$. These $t$ are not the same.
$t$ is just a dummy variable. It parameterizes the integral over which the function is periodic. We could have called it $Trump$ and done the $c_n$ over $t$ from $-\pi, \pi$ and then done the $d_n$ interval over $Trump$ where $Trump \in [0, N]$
@KevinDriscoll I don't think that's my confusion. Let me explain better.
Initially, we have the $c_n$ for $h$, which follows immediately from the theorem, that is we have:
Then, we changed the interval of integration of $c_n$ to $[0, N]$. To do that, we had to introduce a new variable, i.e. $v$
Im not sure what that means
We performed a change of variables. The region we are integrating over does not change.
12:41 AM
@KevinDriscoll Oh, that's probably my confusion then.
I am thinking physically here. If we imagine the region we are integrating over as existing in real space, say its a certain time interval.
Then we are integrating over that interval of time.
And our change of variables is just a re-parameterization of that interval
But the physical span of time we are integrating over remains the same
The only thing thats changing is how we describe it
$lim_{n \to \infty} \sup (-a_n) = \lim_{n \to \infty} \sup \{ -a_k \mid k \ge n \} = - \lim_{n \to \infty} \inf \{a_k \mid k \ge n \} = - \lim_{n \to \infty } \inf (a_n)$...QED?
I was not realizing that changing the variables is only a way to eventually more conveniently solve, in this case, an integral over the same interval of the original function, even though the interval of the integral after the changes of variables has changed.
Is that a correct proof of $\lim_{n \to \infty} \sup (-a_n) = - \lim_{n \to \infty} \inf (a_n)$?
Is that correct?
12:45 AM
@nbro I can understand the confusion, mathematically speaking. In the abstract, it isn't super clear. But if we think about the mathematics as being a representation of some feature of our universe, then it becomes more clear that what we're talking about can be thought of as different ways of describing the same thing
@nbro It still doesn't look like so, because, from what I know, once we calculated an integral with the new variable and we obtained a result, we need to re-express that result in terms of the original variable...
You can do it both ways. Let me think of a quick example
Ok VERY simple. $$\int_0^1 (2x+1) \ dx = x^2+x \vert_0^1 =2$$. Now consider the change of variables $u = 2x+1$ $$\int_0^1 (2x+1) \ dx = \int_1^3 u \ du$$
And we can evaluate this 2 different ways
Made a mistake
$$\int_0^1 (2x+1) \ dx = \int_1^3 \frac{1}{2}u \ du$$
Now we can evaluate 2 different ways
$$ = \frac{1}{4} (u(x))^2 \vert_{x=0}^{x=1} = \frac{1}{4} (2x+1)^2 \vert_0^1 = \frac{9}{4} - \frac{1}{4} = 2$$
OR we can evaluate in terms of $u$
$$= \frac{1}{4} u^2\vert_{u=1}^{u=3} = \frac{1}{4} (9 - 1) = 2$$
So we can re-express $u$ in terms of $x$ and then plug the approriate $x$ values into the anti-derivative in terms of $x$. Or we can figure out what $u$ values then $x$-endpoints correspond to. And then we can plug those $u$ values directly into the anti-derivative in terms of $u$. Same thing
Does that make things any more clear @nbro?
Side comment: A lot of people seem not to be taught (especially in high school?) to rewrite the new integral with new limits of integration, but do the substitution, do the integral, and then rewrite everything in terms of the original variables and then evaluate. Ugh.
I've found the same @Ted. I frequently have to explain to students how to do it the other way.
1:00 AM
Won't help them understand when they get to line integrals, surface integrals, etc.
@TedShifrin which is why they don't get to that stuff in the first place :P
It really hurts people because it shows up again and again in other places. Any kind of abstract manipulation, you have to re-write the limits. Including when doing sums, or multivariable calculations
Arguably not ofc, but plenty of people don't go far past calc
that's weird man
heya Eric
1:02 AM
yo yo yo
@KevinDriscoll So, in my particular case, you're saying that the final integral we obtain can then be re-evaluated using the original variable $t$ (and with the corresponding interval) or evaluated using $v$, correct?
Indeed. And the answer will be the same. You'll get teh same constants
I also notice that, at least in my high school's calc classes, limits are done very poorly, especially when it comes to applying series expansions or L'H, etc.
most high school calculus teachers aren't qualified to teach it.
We did $\epsilon - \delta$ definition and BOY DID I NOT UNDERSTAND THAT
1:05 AM
but almost everyone does limits with L'Hôpital and almost no one shows people how to use Taylor polynomials to do it (even in BC calculus).
But, clearly, the original integral for $c_n$ is different from the integral we obtained for $d_n$!
But I took calc 1 at a community college that was relatively rigorous about actually teach math and science for engineers because they send lots of people to engineering schools after 2 years of community college
Eric, I told Demonark that I heard from another one of your UC compatriots in email today. Another victim of my videos :P
@KevinDriscoll x'D I can imagine
Limits are all hand-wavy at first
@TedShifrin THIS ANNOYS ME OT NO END! Students come to me and they're like "How do I show this limit?" And it takes me freakin 10 minutes to figure out how to do it using their tools because I would do it in 30 seconds with Taylor series
1:06 AM
@Ted I didn't realize until today that IVT can be phrased as a theorem of degree theory.
@KevinDriscoll x'D
Oh yeah, it's the dimension 1 case of the diff top version of the argument principle, @MikeM. I always mentioned that in the G&P course.
Oh we learned how to do it using taylor polynomials I think... It involved big o
@Kevin: We've had a few people show up here with crazy limits with nothing allowed ... and it required some super tricky trig identities to do 'em.
For me, I've simply solved a lot of limit problems on mse as well as looked at many answers, so I'm flexible with how I solve limit problems most of the time
1:08 AM
@Ted I mentioned that today, as well as how to generalize the usual proof to certain polynomial in H and O.
H and O ?
(Akiva helped clarify some of my poorly-worded statements)
the other two real normed division algebras
DogAteMy is good at words.
OH. H and O.
H for Hamilton, I imagine
1:13 AM
@Ted Do you know/see an argument that surfaces in S^2 x S^2 have trivial self-intersection?
Anyway, thanks a lot guys for all your valuable and precious help! I've not taken a serious nap since the last WW. I probably could have solved this alone (by searching around and revising a few concepts) if I wasn't so tired.........
@nbro I invite you to do a simple example, like the square wave periodic on $[0,N]$ and teh corresponding re-scaled and re-centered version of the square wave periodic on $[-\pi, \pi]$ to see that in fact the sequence of constants that you get for one are exactly the same as the sequence of constants for the other. The thing that changes is which harmonic (AKA which exponential) they go with. But the sequence of constants is identical.
I wanted to prove in a self-contained way that there's no degree 1 map $S^2 \times S^2 \to \Bbb{CP}^2$.
So cohomology isn't self-contained?
1:14 AM
but i guess it's crucial that every surface is homologous to a sum of the fibers
not for the differential topology class
Ah, gotcha. I need to go cook and go play bridge, but I like that question. I think it should be doable.
Oh, some other sucker is doing 225B next quarter (forms etc); I'm doing "foundations of geometry". I think that's usually Euclid.
@TedShifrin bid 3NT and alert the other pair that opening 3NT means you're a badass
You got tired of this course after 2+ years?
The guy twachinf it says he wants to do transformation groups a la Klein.
1:15 AM
it's usually axiomatics.
oh, the education majors will love that :P
Kevin: My partner is a many-times-over life master.
OK, I need to go cook. Mike, I'll let you know if I figure out an argument.
@KevinDriscoll Yeah, I guess it would be a helpful exercise. But the coefficients are also defined in terms of the exponentials...
Can someone help me understand this: If we have the tensor algebra of module M over K, why is 1(tensor)m = m, where 1 is the unit of K.
@KevinDriscoll That is, those coefficients do not really seem to be constants, i.e. they depend surely on $n$. I am not sure what's the accepted definition of coefficients, but ...
@TedShifrin Oh, very cool! I'm just a casual player myself, but Patty Tucker very nicely donated her time to teach people here, so I know how insanely good people with that level of experience can be
@nbro Ya $c_n$ is a constant. The $c_n$ as a group are a sequence of constants indexed by the natural numbers.
Yeah :)
1:28 AM
@nbro But anyway the coefficients are defined as
'going with' a certain exponential
and that correspondence will shift from one to the rother
but if you just ignore the exponentials and write down the sequence of coefficients, they will be hte same
@brot Yes, my mistake. I assumed the sequence was monotone because of the $m<n$ bit. What you can also say is. Suppose $a_n$ is bounded by $M$. Consider $[-M,M]\in\mathbb{R}$. Note that the number of elements of $\{a_n\}$ is limited if we are to preserve the condition of "spacing" between any two elements of the sequence.
Sorry I meant number of elements of $\{a_n\}\in[-M,M]$
1:49 AM
Question: If a person is providing an answer that is assuming something which is not necessarily being assumed in the initial question, like "whole numbers are being used," for example, is that a potential issue?
You can split it up into cases
case 1) Assume x is a natural number. case 2) assume x is not a natural number
Well, yeah. The question being referred to regards Euler's identity, and the answer was stating that it always simplifies to either 1 or -1, which is not necessarily true.
I don't have enough reputation to comment and point that out, though.
Can you link it?
A: Simplyfying Algebra with $e$ function

user8663905Well, we know that $e^{ni} = \cos(n) + i\sin(n)$. We also know that $\sin(-n) = -\sin(n)$ and $\cos(-n) = \cos(n)$, so we can figure that $e^{-ni} = \cos(n) - i\sin(n)$, and thus $e^{ni} - e^{-ni} = 2i\sin(n)$. This is a classic extended definition of $sin$, where $sin(x) = \frac{ie^{-ix} - ie^...

but n is indeed an integer
2:00 AM
Why do we assume that? n might be being used as a nondescript variable. It isn't specified.
because when we use "n" we usually intend it to be an integer
"Usually" being the main reason we shouldn't assume, particularly if the person asking the question might not be familiar with these conventions.
2:15 AM
So... I was tidying up old files and then I saw this note I wrote for myself back in 2013:
> What is worse than an essential singularity?
A mad, increasingly or decressingly oscillating function
That good old problem
Did you know I once made the mistake of using $x$ for complex variables?
I now use $z$ and whatever greek letters fit the context.
In any case, the usage of $n$ being integer or natural is pretty common.
In linear algebra, is the rule for the determinant and switching rows in a matrix usually that it is switching ANY two rows, or any two adjacent?
I remember being taught adjacent but it is the same either way, I guess.
switching any two rows swaps the sign
2:19 AM
Ah, yeah, I'm still getting comfortable with symbol conventions, myself. Like, I will often try to use "x+yi" because I relate the complex plane to parametric equations.
meanwhile I once again forgot how to prove a set is compact. Need to revise
@user8663905 That's perfectly fine AFAIK
switch row i<j results in 2(j-i)-1 adjacent swaps, so I was just checking if it was normal to hear it as any two rows.
Assuming x and y are real, which are the usual conventions
2:20 AM
@anakhronizein aka any transposition is an odd permutation
Q: The growth rate of $\epsilon\text{-LCM}$ of two real numbers

PyRulezWe define $\epsilon\text{-LCM}(a,b)$ for $a,b \in \mathbb R$ as the least non-negative real number such that there exists non-zero $n,m \in \mathbb Z$ with $|\epsilon\text{-LCM}(a,b) - na| \le \epsilon$ and $|\epsilon\text{-LCM}(a,b) - mb| \le \epsilon$. This is always defined if $\epsilon \gt 0$...

O.o interesting looking
@Secret ur always revising, just like me and my big number programs lol
That's a good way of looking at it.
Saves me from the mental calculus.
Seems you forgot to change the bounds after the substitution. — Simply Beautiful Art 21 secs ago
> What do the Real Numbers Really Look Like?
What does the real line look like if you look at it really close up? We think
of it as a continuum of points extending indefinitely in two directions, but what if you could look at it under a microscope and were able to turn up the magnification higher and higher. What would you begin to see? You might be disappointed since you will never get to a stage where you would see, “one rational number, three irrational numbers, one rational number, ... “ The real numbers are self-similar, they “look alike” no matter what the scale. So how do we “visuali
Well... until you decided to use an uncountable magnification, then suddenly everything turned to discrete points (there are infinitesimal gaps between any two reals in the hyperreals)
If countable infinity is the notion of the point you should get to if you go arbitrarily close to it in steps, then uncountable infinities are the notions you should get to if the aforementioned approaching becomes a one step process
That's one way to visualise $\omega_1$ despite its inherent lack of describable structure:
> You zoomed pass points so fast that any points that take countably many steps to reach takes only one step, and you use this newfound speed to try to get as close as you can to the next possible limit point
So, any uncomputable functions with growth rate comparable to $\omega_1$ should exceed any countable uncomputable and computable function right at the first step
(To be proved...)
and that's the hallmark of the notion of uncountability: It is something so big that no countable process can reach it
2:42 AM
@Secret wait wat do you mean by that
I mean, let $g,h$ be uncomputable functions where $g$ is comparable to the fast growing hierarchy $f_{\omega_1}$ while $h$ is comparable to the fast growing hierarchy $f_{\alpha}$ where $\alpha < \omega_1$. Then I suspect $g(1) > h(1)$ now matter what $h$ is
@MikeMiller, seems totally wrong. Plus $\Bbb CP^2$ and $\Bbb P^1\times\Bbb P^1$ are birational. I’m at bridge now.
@Secret That still doesn't make too much sense. At the most, I'm comfortable with some meaning behind $f_{\omega_1^{\rm CK}}(n)$, but not $f_{\omega_1}(n)$...
Do we expect the fast growing hierarchy to be able to go beyond $\omega_1$ because of issue of $\omega_1$ is it lacks a fundamental sequence, so there will be trouble in expanding $f_{\omega_1}$?
Because from what I learnt, fast growing hierarchy $f_{\alpha} > f_{\beta}$ if $\alpha > \beta$
Since $\omega_1 > \alpha$ where $\alpha$ is countable, I should expect the fast growing hierarchy should follow similarly, but I might be wrong because it might be possible it breaks down after $\omega_1^{CK}$
@Secret Very much so.
I imagine $f_{\omega_1^{\rm CK}}$ is well-defined if we choose to use an uncomputable notation for ordinals, but I've yet to seen anything like $f_{\omega_1}$
Anyways, good night.
2:53 AM
(To others who may care) Anyway, the above is a minor digress as currently I am reading about the nested interval theorem to remind myself on how to prove the compactness of closed intervals in the reals
Hmm... so the outline of the proof combining these sources is as follows:
1. The reals are linearly ordered. Pick a closed interval $[a,b]$.
2. Generate a sequence of intervals by picking some $c \in [a,b]$ and $d \in [a,b]$ such that $c \leq d$
3. All bounded sequences will converge to its infimum or supremum. Therefore these are contained in any interval constructed
4. In the case of creating nested intervals by halving, the infimum of the sequence with 1st term a and supremum of the sequence with 1st term b coincides, to some $y$. Since [y,y] is be definition a singleton, we have the required result
Bolzano-Weierstrass Theorem
1. Pick a [a,b]
2. Generate nested intervals by halving a and b
3. By nested interval, the arbitrary intersection exists and is some number y
4. Let U be a neighbourhood of y defined by $(y-r,s-y)$. Pick one of the closed intervals I form the nested intervals such that $I\subset U$. Regardless of $U$ there are uncountably many points in these intervals, hence y is an accumulation point. Since [a,b] is arbitrary, it followed every real number is an accumulation point
Proving [a,b] is compact
... and I don't want to use contradiction...
1. Let $C$ be a cover of $[a,b]$
2. Any such $C$ must obey the property that $\{a,b\} \subset C \cap [a,b]$
3. Since the arbitrary union of any open set is open, it follows there exists no sequence of unions of open intervals with infimum a or supremum b that can contain a or b.
4. Now consider some open interval $(c,d)$ such that $c < a, d > b$. $(c,d)$ can be covered by some infinite cover $C$.
5. By Bolzano-Weierstrass Theorem and that arbitrary union of open intervals are open, all points on the real line are accumulation points, therefore $C$ cannot consists of a subsequence of open intervals with some supremum or infimum $x \in [a,b]$ that includes $x$. (In more intuitive terms, the open cover cannot contains an infinite "tail" converging to some point x in [a,b] without leaving x uncovered)
6. Therefore there are no countable sequence of open intervals within $[a,b]$, and thus any infinite sequence in the infinite cover $C$ must be located outside $[a,b]$
7. Therefore the union of open intervals within $[a,b]$ is finite. Meanwhile for the region $(c,e)$ and $(f,d)$ where $e < a,b > f$, the infinite bounded sequence of intervals can form a union to give finite open intervals as a sub cover. for said region.
8. Therefore $[a,b]$ always have a finite sub cover and is compact
Q: Prove that some set is compact directly from definition

aaaaaaLet $A$ be a subset of $R$ which consist of $0$ and the numbers $\frac{1}{n}$, for $n=1,2,3,\dots$. I want to prove that $K$ is compact directly from the definition of compact. So, given any open cover of $A$, I should be able to find a finite subcover. Proving a set is compact is much difficult...

A more elegant proof here
typo: more elegant proof strategy
3:44 AM
@Ted Ah, I make this mistake every time in forgetting about the diagonal. Even* self-intersection.
That’s the cohomological fact that’s actually true.
and... I still have no idea how to prove the lebesgue outer measure of a closed interval is its length without looking circular
for my course evaluation i was prompted to write what were my main books I used to study the material, I said math.se and WP
@GFauxPas up for a sequence question?
3:59 AM
We want to prove $b-a$, but then $b-a+2\epsilon$ appeared in the first line
sure, though i dont knwo how helpful ill be
4 hours ago, by orbit-stabilizer
Let $(a_n)_{n\in\mathbb{N}}$ be a sequence of real numbers with the following property. There exists a real number $0 < r < 1$ and an integer $N_0$ such that $\forall n \geq N_0$, $|a_{n}-a_{n-1}| \leq r|a_{n-1}-a_{n-2}|$. Prove that $(a_n)$ convergs.
4 hours ago, by orbit-stabilizer
Now, since we're working in $\mathbb{R}$, it suffices to show that this sequence is Cauchy. Beyond that, I've tried the triangle inequality, and that hasn't helped - so I'm kinda stuck.
4:22 AM
not sure
4:42 AM
@MikeMiller They don't. Take a look at the diagonal copy of S^2
That has self-intersection number 2
4:55 AM
Oh nevermind didn't see your message correcting that
Strange vectors and matrices inspired from the notion of datasets and data structures:
Let some dataset be the space $D=\Bbb{R}^3 \otimes \Bbb{R}^2 \otimes \Bbb{R}^2$. For example, the 3D vector space is a vector field of wind velocity, and the 2D vector space are precipitation with time and temperature with time
We can consider some vector $v \in D$ the represent a datapoint, given by the 3 tuple $(a,b,c)$ where $a \in \Bbb{R}^3, b,c \in \Bbb{R}^2$
@Balarka Surely there's some nice proof.
I can't come up with one off the top of my head, strange
Now, suppose we are interested in the wind velocity in the x direction, and the average temperature, precipitation between Monday and Thursday. This becomes an operator $\phi : D \to \Bbb{R}^3 \otimes \Bbb{R} \otimes \Bbb{R}$
Therefore $\phi (v)$ is an element wise inner product within each vector space, given by:
$$\phi (v)= \left( proj_{\hat{x}}(\vec{v}_1),\sum_{i=1}^4 (v_2), \sum_{i=1}^4 (v_3)\right)$$
I have to run to school now.
5:10 AM
More generally, we can think of linear operators where one of the dimensions are irregular
thus we can have something like a stack of matrices in the space $\bigotimes_{i \in I}\Bbb{R}^i$ where $I=\{1,2\times 2,3 \times 3,4 \times 4,3 \times 3,2 \times 2,1\}$
We can also consider the continuum case of a function of operators given by the space $\prod_{i \in J}\phi_i$ where $J=[a,b]$
In such scenario, the input are functions and the output are also functions and each $\phi_i$ can have a different domain and range
For example, let $f$ be a real function. Then $\phi (f)$ is also a function. Thus $\phi$ acts as a in general, nonlinear and not necessary continuous map on the space of all functions
Also unlike usual nonlinear functionals, $\phi$ can have a very complicated shaped domain due to the domains and ranges of all $\phi_i$ are not necessary the same
As my real and functional analysis get better, I will try to investigate $\phi$ in more details. But for now, I am still trying to figure out how to prove the Lebesgue outer measure of a closed interval is indeed closed
I have the Bolzano-Weierstrass Theorem for real closed intervals proved, and I am trying to prove the Lebesgue measure of closed intervals without having the expression b-a appearing anywhere except the conclusion of the proof
I also have the compactness of [a,b] proved.
But using the definition of Lebesgue outer measure, I am not sure how I can get the notion on how the measure of an open interval is derived, and that is needed to do the proof of the closed interval case
It might seemed like a stupid question or that I overlooked something, but I am trying to justify why the lebesgue outer measure of open intervals (a,b) is b-a
Is that really a definition, or is a theorem?
5:52 AM
Is there a name for integral equations of the form (I+CD)u = f. Here u and f are functions. D is a differential operator, C is a convolution operator.
It's almost a fredholm equation, but there's the differential operator
$U$ is unknown and $f$ is given?
then this is a kind of integro-differential equation
But since $I$ is a convolution, you should be able to do a fourier transofrm
and that will lead to some kind of ordinary differential equation in the fourier transform of $f$
Oh Im sorry I completely misread things
I guess $I$ is identity then
That's right
So then the $CD$ term is going to look something like $\int_{-\infty}^{\infty} c(x) D(f(x-y)) \ dy$?
6:07 AM
Kinda. u(x) and f(x) are real functions of two variables. f(x) is smooth with compact support. The equation is

u + ln|x| * (b \cdot \nabla u) = f

b is a constant vector
So, sorry where the convolution operator?
Oh sorry. That's what I meant by *
Ok, cool, ya then what I wrote is correct with $c(x) = \log{x}$ and $D = \vec{b} \cdot \nabla $
So in that case one can integrate by parts I think and return this to an ordinary kind of integral equation
$\vec{b} \cdot \nabla(u) = \nabla \cdot \vec{b} u$
so then $\log{x} \nabla \cdot \vec{b} u(x) = \nabla \cdot (\vec{b} \log{x} \ u(x)) - \vec{b} u(x) \cdot \nabla \log{x}$
and the first term will give some boundary contribution
and the second looks like it'll lead to some kind of singular kernel in the integral
@PhilipHoskins Does that sound right?
6:21 AM
The integral is over the whole space. So you would need u to decay. Something like O(1/|x|^2) should be sufficient, right?
I mean it doesn
t have to decay, there can be a boundayr contribution
but if you want there not to be, then ya
Well, anyway, my goal is to solve the equation numerically using GMRES. I've done some numerical experiments that suggest it works and I know there are some papers on this. I just realized I've been referring to this equation as fredholm for the past couple weeks and that's not quite right
After you carry out this integration by parts, it seems like it becomes a singular fredhold equation
with kernel -b \cdot \nabla ln|x|, right?
Yes, although I'm not sure if people use the term Fredholm when talking about multi-dimensional integrals
6:37 AM
It sounds reasonable enough
It's also a weakly singular kernel, so it should be a compact operator
6:55 AM
That Im totally ignorant about. I can do the practicalities of some integral equations. The more general stuff I have no clue
3 hours later…
1 hour later…
10:46 AM
Did you work out your problem @trynalearn?
Equation 1: Log37 = X
Equation 2: Log10 = Y
I want to calculate $\int_{\sigma}\left (-y^3dx+x^3dy-^3dz\right )$ using the fomula of Stokes, when $\sigma$ is the curve that is defined by the relations $x^2+y^2=1$ and $x+y+z=1$.

We have that $$\int_{\sigma}\left (-y^3dx+x^3dy-^3dz\right )=\int_{\sigma}\left (-y^3, x^3, -z^3\right )\cdot \left (dx, dy, dz\right )=\int_{\sigma}f\cdot d\sigma$$ with $f(x,y,z)=\left (-y^3, x^3, -z^3\right )$, right?

From the formula of Stokes we have that $$\int_{\sigma} f\cdot d\sigma=\iint_{\Sigma}\left (\nabla \times f\right )\cdot N\ dA$$
what is the relationship between x and y ?
how do I go about solving for them ?
Do you have an idea for my question @LeakyNun ?
10:50 AM
anybody ?
1. There is a maximum and a minimum
2. We start at the minimum and the maximum can be reached
I.e. the maximum is a successor of some number
3. We also require every number is a successor of some number, except the minimum
4. There are only successors and the minimum, and between any two successor there are no numbers
5. Some extra condition to rule out something like:
Meanwhile I am still have trouble figuring out $\lambda^*(a,b)$
How to prove the Lebesgue outer measure of an open interval (a,b) is b-a?
I need that to prove $\lambda^*[a,b]=b-a$
Does someone of you have an idea about my question: math.stackexchange.com/questions/2553638/…
11:06 AM
This is why this chat room deserved to be flooded. IT is SOOO BORING with the downtime increasing per day
Even though all the floods by me are accidental and has been managed by the new rooms
Currently no one is in the chat room except machines
11:19 AM
What about myself? I am neither machine nor human. I literally does not exist
How do you define the Lebesgue measure @secret
The infimum of the open cover of the given set that is a union of open intervals. Let me try write it symbolically:
$\lambda^* (E) = \text{inf}\{C=\bigcup_{i \in I} J_i, E \subset C, J_i = (a_i,b_i)\}$ where $I$ countable
How is that a number? You're taking the inf of a family of sets, what does that mean?
11:34 AM
actually, there's one thing that seems basic but I am not terribly sure. Since measure theory aim to generalise the notion of length and size, is the notion of length of an interval defined before we have a measure?
More precisely, is length $(a,b) = b-a$ a theorem or a definition?
We just say that the "natural" measure of $(a,b)$ is $b-a$ and to define the Lebesgue measure we take the inf of the natural measures of the coverings
Ah I see...
That's just how the Lebesgue measure is defined though, there's no reason to ask $\mu((a,b))=b-a$ for a generic measure $\mu$
(question left for later) whether for a generic measure $\mu$ there's always some set choose measure is defined instead of calculated and we use these as building blocks to compute the measure of other sets in the $\sigma$-algebra.

Thanks, I think I should have all the ingredient I need to prove $\lambda^*[a,b]=b-a$ now. Having done the proofs on the compactness of closed intervals in reals, the nested interval theorem and the Bolzano-Weierstrass Theorem for closed intervals, I should have everything to do the above proof using these and topological arguments. Will write out the proof afte
You can also prove that the singletons have measure zero first and use $[a,b]=\{a\}\cup(a,b)\cup\{b\}$
12:00 PM
hmm... let's see:
$\{a\} = [a,a]$ is compact since it is a closed interval in the reals. Any open cover of $\{a\}$ will contain the point $a$. For every open cover $C$, there's always an open interval $(b,c)$ where $b<a<c$ that is contained in $C$. The infimum of $(b,c)$ tends to zero as $b,c$ get arbitirarily close to $a$ thus the sequence of nested open intervals converges to $a$. Therefore the natural measure of $\{a\}$ is $\lim_{c\to a}c - \lim_{b\to a} b = a-a=0$. Thus any singleton has zero Lebesgue measure
Therefore the Lebesgue outer measure of $[a,b] = \{a\} \cup (a,b) \cup \{b\}$ follows from subadditivity of the measure. Hence $\lambda^*[a,b] \leq 0+b-a+0=b-a$ and equality is ensured because the union is disjoint
Now for rationals: $\Bbb{Q} \cap [a,b]$ :
Proof: The rationals are countable. Let $f$ be an enumeration of rationals given by the sequence under the Cantor pairing function $\{\frac{a+b}{2},\frac{a+b}{2}+1,\frac{a+b}{2}-1,\frac{a+b}{2}+\frac{1}{2}, \frac{a+b}{2} -\frac{1}{2},\frac{a+b}{2}+2,\frac{a+b}{2}-2,...\}$
if $\frac{a+b}{2}$ is rational, this sequence give all the rationals in $[a,b]$. Thus we can then take the union of each rational here and then use countable subadditivity to establish $\lambda^*(\Bbb{Q}\cap [a,b])=\sum_{i=1}^{\aleph_0}0 = 0$
if $\frac{a+b}{2}$ is irrational, then (thinking...)
If $\frac{a+b}{2}$ is irrational, pick any rational $p > \frac{a+b}{2}$ and then use $f$ to generate a sequence. Once again, all rationals in $[a,b]$ will be enumerated and thus the proof proceed analogously
thus $\lambda^*(\Bbb{Q}\cap [a,b]) = \sum_{i=1}^{\aleph_0}0=0$ as required
12:27 PM
Q: Examples of $A$ halts If $B$ does not halt?

mickIm looking for examples of $A,B,C,D$ such that : $A$ halts If $B$ does not halt. $C$ does not halt If $D$ halts.

To compute $\lambda^*(\Bbb{Q})$ we let $\frac{a+b}{2}=0$ and $a,b$ be unbounded. Therefore all rationals and its subsets re of measure zero
@MaryStar you have cylinder intesected by a plane, find the bounds and do the line integral
Now to figure out how to do $\Bbb{I} \cap [a,b]$...
We knew that uncountable subadditivity does not hold thus $\sum_{i=1}^{\beth_1}0\neq 0$, so we need to approach differently
The question is then, how direct we can get in the proof before the proof of using $\Bbb{R}=\Bbb{Q} \cup \Bbb{I}$
Schematically, I am looking for a proof:
12:46 PM
that is, I want to think of the computation entirely in terms of covers
I guess the first step is to work out whether $\Bbb{I}\cap [a,b]$ is compact, or even countably compact
A cover $C$ can be countable by having countably many open intervals converging to some rational. Since there are countably many rationals, there are at least countably many such sequences of open intervals that converges to them (1 or 2 more if the endpoints a or b are rationals). Hence such $C$ form a countable cover
Now to check whether $\Bbb{I} \cap [a,b]$ is countably compact:

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