Mathematics - 2024-10-17

leslie townes

00:47

psie: in what world would "the completion of ... on ..." not be complete? that seems like something a definition chase should answer. the borel sigma algebra is not the lebesgue sigma algebra so it would not surprise me if you do not end up with the same thing until you do "the completion of"

leslie townes

01:05

or, could you be more precise about what are "these spaces" in "i think these spaces are both not complete." i have a vague memory that somewhere in the text he explains why m on B(R) is not complete and why a product of lebesgue measures is not complete unless "the completion of ..." is rolled into what you call a product, for example.

as for how difficult that inclusion is, i would unroll your definitions of B(R^n) and mathcal L. different books decide those definitional questions differently, so this could potentially be something very specific to folland.

The Empty String Photographer

08:08

@XanderHenderson Italics are only for Italian people!

Jakobian

08:27

@psie since $\mathbb{R}^n$ is second-countable $$\mathcal{B}(\mathbb{R}^n)=\mathcal{B}(\mathbb{R})^n\subseteq\mathcal{L}_1^n$$

Jakobian

08:38

Let $C$ be the standard Cantor set. Then $\mathcal{L}_1\restriction_C$ is the power set of $C$ since $C$ is of measure zero. So there is $2^\mathfrak{c}$ sets in it. But there is only $\mathfrak{c}$ elements of $\mathcal{B}(C) = \mathcal{B}(\mathbb{R})\restriction_C$

So there exists a Lebesgue measurable set which is not Borel

The same can be said for $\{0\}^{n-1}\times C$, $\mathcal{B}(\mathbb{R}^n)$ and $\mathcal{L}_1^n$

So the inclusion above is strict

Next note that there exists $V\subseteq \mathbb{R}$ with $V\notin \mathcal{L}_1$ such as Vitali set. So $V\times\mathbb{R}^{n-2}\times C$ is in $\mathcal{L}_n$, $n\geq 2$ since it has Lebesgue measure zero

If it were in $\mathcal{L}_1^n$ then so would be $V\times\{0\}^{n-1}$, which should then imply $V\in\mathcal{L}_1$ I believe

think_meaning_builds

08:56

youtube.com/live/2xYosqVjROc?si=dPoaxgIry_-Tc8Xk

Putnam^ results

Jakobian

My last argument is a bit shaky because it relies on $\mathcal{L}_1^n\restriction_{\mathbb{R}\times\{0\}^{n-1}} \cong \mathcal{L}_1$ by the projection map

Ah yes it does work

$\pi:\mathbb{R}\times\{0\}^{n-1}\to\mathbb{R}$ is measurable, and its inverse being the inclusion is clearly measurable as a check on generators shows (in that case we need to check for the map $f:\mathbb{R}\to\mathbb{R}^n$ where $f(x) = (x, 0, ..., 0)$ where $\mathbb{R}^n$ has $\mathcal{L}_1^n$)

Ben Steffan

@Jakobian is it not enough to note that $f$ is continuous?

Jakobian

@BenSteffan no because that wouldn't mean a preimage of a Lebesgue measurable set is Lebesgue measurable necessarily

Here when I say measurable, I mean with those specific sigma algebras

Anyway we obtain $$\mathcal{B}(\mathbb{R}^n)\subseteq\mathcal{L}_1^n\subseteq\mathcal{L}_n$$ where first inclusion is strict and second inclusion is strict for $n\geq 2$. Now to see completions of both are $\mathcal{L}_n$ one just has to note that the first are - and that's the definition.

As for why those inclusions even hold, again check on generators

The argument should generalize to products $\mathcal{L}_{n_1}\otimes...\otimes\mathcal{L}_{n_k}$ but I won't dwell on it more

I also believe the argument shows $\mathcal{M}\otimes\mathcal{N}$ is not complete if $\mathcal{M}$ has non-empty sets of measure zero and $\mathcal{N}$ is not the power set

That is the product is almost never complete

psie

11:27

Ok, thank you leslie and Jakobian. I will think on this some more.

nickbros123

13:04

Greetings

Jakobian

@nickbros123 Greetings

Almanzoris

Regards

Sine of the Time

cheers

Pizza

15:42

Hi

Sine of the Time

hi

Xander Henderson

15:57

Bye.

Jakobian

@XanderHenderson see ya

Soumik Mukherjee

Hola

Gian's Pizzeria

16:14

$\Huge Hi$

Test

Xander Henderson

Stop.

Gian's Pizzeria

16:31

?

I wasn't writing anything

Anyway sorry

psie

18:10

Consider the above theorem. Ignore the highlighted box, I just got the image from this question. I'm getting hung up on the very last sentence. We have that $V_j$ is a rectangle whose sides are finite unions of intervals. In the last sentence of the proof of the theorem, the author asserts that $\bigcup_1^N V_j$ can be expressed as a finite disjoint union of rectangles whose sides are intervals. How is that possible?

I see how we can make the union disjoint, but shouldn't the sides of $\bigcup_1^N V_j$ be finite unions of intervals instead of intervals?

leslie townes

okay, implicitly a "rectangle" is a subset of R^n that is a cartesian product of subsets of R (which are, again implicitly but probably by definition, said to be its "sides")? this is nonstandard terminology but it seems to be what you have in mind.

psie

@leslietownes yes, sorry, the "sides" are exactly what you say they are :)

leslie townes

if so, what is going on here may be no more than, if the subset A of R is a disjoint union of A_1, ..., A_n, then any cartesian product of subsets of R in which A is a 'side' is a disjoint union of n subsets of R in which, in the jth subset, A_j is a "side"

e.g. you can just 'expand out' a cartesian product of sets having these slightly more complicated sides as a disjoint union of cartesian products with simpler 'sides.' it might be helpful to write out what is going on in R^2

Xander Henderson

18:27

@leslietownes Yeah, that's the right idea. If a set is the Cartesian product of intervals (or of disjoint unions of intervals, but you can always reduce that a union of sets which are the Cartesian products of intervals), then it can be decomposed into a disjoint union of sets where each set is the Cartesian product of several intervals.

psie

Let's stick to $n=2$. We have $V=\bigcup_1^N V_j$ and suppose for simplicity $N=3$. Then, to make the union disjoint, we simply take $V_1,V_2\setminus V_1, V_3\setminus V_2$; these are disjoint sets and their union is $V$. Now, each $V_j$ is a rectangle whose sides are a finite union of intervals. So $$V_1=\left(\bigcup_{i=1}^k I_{i1}\right)\times\left(\bigcup_{i=1}^p J_{i1}\right).$$ There is nothing in the theorem that says that e.g. $\bigcup_{i=1}^k I_{i1}$ is an interval?

Xander Henderson

Your construction does not necessarily give you a disjoint union of rectangles.

psie

@XanderHenderson ok, so you are saying $V_2\setminus V_1$ is not necessarily a rectangle?

Xander Henderson

@psie Yes.

psie

I'm lost at sea then...

Oh, ok.

Xander Henderson

18:37

But you can carve up that union into disjoint rectangles.

psie

Ah ok, makes sense. That image was helpful.

Xander Henderson

Note that the way that I have done it here is not unique, and is not really the right way of doing it if you want something generalizable.

psie

Ok 👍

Xander Henderson

But maybe that gives you an idea of where to start?

psie

It does, yes, thanks!

Akhilesh G

19:21

Hello! Please help me with doubt.
Expressions are something we get a value for every different value of that variable. So graphs can be plotted.

Equations are basically demanding the value of variable for which it is valid.

But when I enter a Quadratic Equation in Desmos Graphing Calculator, why does it show an Parabola? Is it not just supposed to show only the roots? How can we plot grpah for an EQUATION??

Please correct if my understanding of the concept is wrong.

Ben Steffan

who can tell without knowing what exactly you entered into desmos

Akhilesh G

3x^{2}-5 = 0 is the equation, I entered in Desmos

Ben Steffan

well when I enter that into desmos I don't get a parabola :)

I get two straight vertical lines, which is... perhaps not optimal but checks out

Akhilesh G

Are those two parallel staright lines?

Ben Steffan

yes

see, desmos expects an equation in two variables here (I guess), $x$ and $y$

Xander Henderson

19:25

When you enter an equation into Desmos or GeoGebra, it interprets the input as a set, i.e. the set of all points $(x,y)$ in the Cartesian plane which satisfy the equation.

Ben Steffan

your equation does not constrain $y$, so all values of $y$ are valid

thus you get vertical lines, not points

Akhilesh G

Oh makes sense.

think_meaning_builds

@AkhileshG y = f(x) is an equation, right?

Akhilesh G

Yes

Xander Henderson

The equation $3x^2 - 5 = 0$ has two solutions in $x$, i.e. $\pm \sqrt{3/5}$, so the collection of points satisfying that equation is the collection $\{(\pm\sqrt{3/5}, y) : y\in\mathbb{R}\}$.

i.e. two vertical lines.

Akhilesh G

19:27

But

Is it not supposed to be just two points that has to be plotted? Why two staright line?

think_meaning_builds

Replace f(x) with a function and you get its graph.

Xander Henderson

@AkhileshG No, it is doing exactly what it is supposed to do.

It interprets equations as equations of two variables, and solutions as points in the Cartesian plane.

think_meaning_builds

y = f(x) = mx + b.

Akhilesh G

And coefficient of y = 0?

Can we say that?

Xander Henderson

I don't understand your question.

Are you asking what happens if the coefficient of $y$ is zero?

Akhilesh G

19:34

Yes

Xander Henderson

What are the solutions to the equation $0 = mx + b$?

Akhilesh G

-b/m = x

Xander Henderson

@AkhileshG No, be more explicit.

The equation is an equation in two variables. Solutions are ordered pairs of points.

Akhilesh G

@XanderHenderson Oh

Xander Henderson

@think_meaning_builds I don't think that is helpful.

Akhilesh G

19:36

@XanderHenderson So solution to that equation is (-b/m, 0). Is it correct?

Xander Henderson

No.

Unless you are asking a slightly different question than I think you are asking.

The equation $0 = mx+b$ has infinitely many solutions. These solutions are of the form $(x,y) = (-b/m, y)$, where $y$ can be any real number.

This is a vertical line in the plane.

There is another, slightly different question, i.e. what are the solutions to the system $$\begin{cases} y=0 \\ 0 = mx + b\end{cases}$$

think_meaning_builds

I don't think he's asking that question.

Xander Henderson

This has only one solution, i.e. $(x,y) = (-b/m,0)$.

But these are two different things.

Akhilesh G

Oh

think_meaning_builds

18 mins ago, by Akhilesh G

Hello! Please help me with doubt.
Expressions are something we get a value for every different value of that variable. So graphs can be plotted.

Equations are basically demanding the value of variable for which it is valid.

But when I enter a Quadratic Equation in Desmos Graphing Calculator, why does it show an Parabola? Is it not just supposed to show only the roots? How can we plot grpah for an EQUATION??

Please correct if my understanding of the concept is wrong.

Xander Henderson

19:40

@think_meaning_builds Yes, I read that, and responded to it, above.

Akhilesh G

So this is (0)y = mx + b. Solutions for are infinite = (-b/m, y) where y belong R.
Where as the asnwer which I was giving to is Solution for (0)y = mx + b at y = 0.
Am I correct?

Xander Henderson

@AkhileshG I don't understand your question.

Akhilesh G

I am trying to understand the difference

Xander Henderson

If you think of the equation $0=mx+b$ as an equation in a single variable ($x$), then there is exactly one solution, and that solution is $x = -b/m$. But you can think of that equation as an equation in more variables, say in two variables, $x$ and $y$.

If you think of the equation as an equation in two variables, then the value of $y$ is free---you can choose $y$ to be anything you like. So the set of solutions is $\{(-b/m,y) : y\in\mathbb{R}\}$.

This set of solutions, thought of as a subset of the Cartesian plane, is a vertical line.

think_meaning_builds

the general equation for a line is ax + by = c @AkhileshG

Xander Henderson

19:44

When $y=0$, $x=-b/m$. When $y=3$, $x=-b/m$. When $y=\pi$, $x=-b/m$.

Akhilesh G

@XanderHenderson If I think of it as single variable, it would just be a point? For two variable it would be a staright line since it is valid for all 'y'?

Xander Henderson

@AkhileshG If you think of it as an equation in one variable, then the solution is a point on the real number line (not in the two-dimensional Cartesian plane).

Akhilesh G

Oh right

Understood

And I have two more doubts

What is the difference between roots and solutions?

Xander Henderson

@AkhileshG A solution is a set of values for a collection of variables which, when substituted into an equation, makes that equation a true statement. For example, $(1,2)$ is a solution to the equation $x+y=3$.

Sine of the Time

@AkhileshG you can't find solutions under a tree

Ben Steffan

19:48

@SineoftheTime unless it's chrismas and you're a chemist

Xander Henderson

A root (of a function $f$) is an ordered tuple $x$ such that $f(x) = 0$. For example, if $f(x) = x-3$, then $x=3$ is a root of $f$.

Equations have solutions, functions have roots.

(generally speaking---in many contexts, this distinction doesn't matter, or won't be imposed)

5

Q: The difference between 'solution' and 'root'

I am wondering about the difference between the following demands: Prove that P(x) has at least one root. Prove that P(x) has at least one solution. Are they the same? The background to my question: Let $c \in \Bbb R$. Prove that the equation: $$\frac{1}{\ln x} - \frac{1}{x-1...

functions

Akhilesh G

@XanderHenderson Oh good itt

think_meaning_builds

Not all problems have solutions.

Akhilesh G

@XanderHenderson But I feel understanding the meaning would make me understand the subject better, to the core.

And another question is
What is the difference between
3x + 5 = 2 and 3x + 3 = 0. Is there any difference in any way?

Xander Henderson

@AkhileshG Sure, there is a difference. They are different equations.

What kind of difference are you looking for?

Akhilesh G

19:52

But they have same solutions

Xander Henderson

Yes, they have the same solutions. That doesn't mean that they are the same.

The equation $x=-1$ also has the same solution.

So does the equation $(x+1)^2 = 0$.

Akhilesh G

Is the only difference in them is the way the are expressed?

think_meaning_builds

3x + 5 = 2 and 3x + 3 = 0 are equivalent equations because they have the same solution set @AkhileshG

Xander Henderson

@AkhileshG In mathematics, what is the "same" and what is "different" is a matter of what features you care about. The two equations are not the same equation---look at them! They are clearly different. But if the only thing that you care about is the set of solutions, then the two equations are equivalent.

Akhilesh G

Is there any special characterstic that we would use in mathematics with these equivalent equations?

Xander Henderson

19:59

I don't understand your question.

think_meaning_builds

{-1}

^that is the common characteristic because it satisfies both.

Akhilesh G

@XanderHenderson This clarifies my doubt

Thank you everyone for helping!

Jakobian

20:36

@AkhileshG characteristic?

Well... equations isn't something one usually formally defines, I would guess

So it can depend on your definition

Its more of an intuitive notion

@BenSteffan that's such a good joke

Ben Steffan

ty :)

ModularMindset

21:08

Can anyone recommend a book/reference that covers the interactions between the mapping class group of a surface and the Ihara zeta function of an embedded graph on the surface?

ephe

22:40

Consider the set $\Omega$ of all finite sequences with values in $\{1, 2, \ldots, N\}$ for some $N\in\mathbb{Z}^{+}$. We wish to define a probability space on $\Omega$ using the Borel sets on $[0,1]$. Namely define $f\colon \Omega\to[0,1]$ to be the function that sends every sequence to a real number by associating it with its $N$-ary expansion, i.e., $f(w)=\sum\limits_{n=1}^{\infty} \frac{w_n}{N^{n}}$. Then define the sigma-algebra on $\Omega$ as $\mathcal{F}=\{f^{-1}(B)\mid B\in\mathcal{B}\}$ where $\mathcal{B}$ is the Borel sigma algebra on $[0,1]$ and define the probability measure $P(A…

Jakobian

@ephe why not?

It is 1/2

ephe

@Jakobian I can't really see how :/

leslie townes

22:57

lots of ways to see it. for example in your general model the event w_k = 1 is the disjoint union of N^{k-1} atomic events where you specify the first k-1 entries arbitrarily (in N^{k-1} ways) and then also specify w_k = 1. each of those atoms has probability 1/N^n by the formula you just wrote (in "we note that any cylinder...")

i guess generally try to convince yourself that you get a probability of 1/N^n for any 'cylinder' event that specifies any n fixed entries of the sequence, not necessarily just the first n entries of the sequence. and maybe start there with n = 1

if this measure is coming out of a reference (it seems like something one wouldn't just stumble upon by tinkering around by hand), maybe just reading further in the reference may also clarify

you might also just think about how the graph of the function f_k(x) = [the kth 'digit' in the N-ary expansion of x] looks as, a graph of a real valued function of the real variable x

ephe

@leslietownes Oh okay I see. I thought all I got from a fixed coordinate was some connected interval. I didn't notice the jumps. Thank you!

psie

23:22

Reading about denseness of simple functions in $L^1(m)$, where $m$ is $n$-dimensional Lebesgue measure on $\mathbb R^n$. It is also stated that if $f\in L^1(m)$, there is a continuous function $g$ that vanishes outside a bounded set such that $\int |f-g|<\epsilon$. The author doesn't really prove anything here, so I'm reading a proof of this here.

In that answer, it says that we can always approximate an indicator function by continuous functions. How come?

In particular, $\chi_R$ where $R$ is a rectangle whose sides are intervals, then there's a continuous function $\varphi$ such that $$\left | \chi_{R}- \varphi \right |< \epsilon.$$

I know how to construct such a continuous function if we are in $\mathbb R$.

I guess we can just apply this to $\chi_R=\chi_{I_1}\chi_{I_2}\ldots$, where each $\chi_{I_j}$ is a function depending only on one variable.

Here $R=I_1\times I_2\times\cdots$.

psie

23:42

And products of continuous functions are continuous.

leslie townes

23:55

yes, without checking the details that is exactly the style of argument you often see. one or more stages of successive approximation to go from indicators of intervals to indicators of more general measurable sets to simple functions to some larger collection of functions

your 2.40(c) way up above is a version of this, relating general finite measure subsets of R^n to 'nice' subsets made out of products of intervals

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