Mathematics - 2020-07-08 (page 1 of 3)

@TedShifrin I suppose I just don't like reading sentences about math over the math itself. something like this line: "An ordinary differential equation (ODE) is an equation containing an unknown function of one real or complex variable x, its derivatives, and some given functions of x" just bores/makes me lose interest

I'd rather just read the mathematical construction

Ted Shifrin

There is no construction to read. I'm not continuing this discussion, though.

Thorgott

12:20 AM

not sure what kind of construction you were expecting

examples are abound and without looking, I'm gonna claim the wikipedia page has at least a couple

obliv

Yeah I guess diff eqs don't have a general form or whatever, and it's specific to the type

As I read through it more it seems to not just be a mathematical object like a group

Thorgott

i mean, it is a perfectly fine mathematical object, but not in any way similar to how groups are mathematical objects

user2103480

Can anybody enlighten me how exactly the factoring of the morphism at the bottom of page 34 here works? https://www.math.uni-hamburg.de/home/dyckerhoff/higher/notes.pdf

The morphism itself is defined by the universal property of the pushout of the diagram Δ[n]x{0} <- dΔ[n] x {0} -> dΔ[n] x Δ[1], where {0} is the simplicial set with {0}(n) = {the unique function [n]->[0]}. The morphism to the left is (boundary inclusion,id) and the morphism to the right is (id,vertex inclusion).

The UP is applied to the extension of the diagram to a commutative diagram with Δ[n]xΔ[1] on the right corner.

And I wouldn't know what to search for on SE

user2103480

1:01 AM

Oh no I now suspect this is just trivial, although I still can't fill in the details. Is it true that one would obtain Δ[n]xΔ[1] as the simplicial set generated by the (n+1)-simplices in any case?

user69608

5:48 AM

@satan29

satan 29

hi

user69608

hi

satan 29

write k+3 as integral (0 to 1 x^(k+2)dx)

user69608

@satan29 why

satan 29

and bring the sum inside the integral, since the sum doesnt vary with "x"

@user69608 The factor k+3 is causing a lot of problems. So its better we re-write it.

the question becomes: (integral from(0 to 1) ( sum(k-0 to n) nCk*(x^k/n^k)*x^2dx.

user69608

5:54 AM

oh yeah done thanks

satan 29

the" sum" simplifies to (1+(x/n)^n) using the binomial theorem

so the integral is (1+(x/n)^n)*x^2dx

and for large n, it becomes e^x*x^2 dx

which you can integrate by parts easily to get (e^x(x^2-2x+2)) evaluated at 0 and 1 to get e-2

so i believe the answer is 3?

user69608

yeah done :)

satan 29

oh ok :)

user69608

method is correct. dont remember answer

do u know a good source for such type of question? @satan29

satan 29

i had an entire matrix match on a similar question in the FIITJEE GMP

user69608

6:00 AM

oh

satan 29

thats why kind of recognized that we can rewrite 1/k+3 as an integral

user69608

6:15 AM

@satan29 want to know about fog

Drathora

6:28 AM

What's $[x]$ representing here?

user69608

@Drathora floor function

Drathora

And what do you want to know about f o g?

user69608

differentiability

Drathora

Well, think about this. To be differentiable we need to be continuous right?

user69608

@Drathora yes

Drathora

6:35 AM

But then think about what happens around the point $x = 1$

I.e. consider $ f(g(1+\epsilon))$ for a small $\epsilon > 0$

user69608

oh, that wont be differentiable at integers

Drathora

Well, it will be at $0$

But yeah, at the others it won't be

user69608

when $x^2$ is integers it wont be differentiable right?

Drathora

Correct, since for a slightly lower $x$ fog(x) will jump to the previous integer

user69608

but what does c option tells , shouldnt it be$\sqrt{I}$ in place of I?

Drathora

6:46 AM

What's "R ~ I" representing here?

user69608

real number,integer

@Drathora exclude integers from real numbers

Drathora

ah

Well then I'd say no, since there are discontinuities at the square roots of integers also right?

user69608

@Drathora yes

NoseBleed

8:26 AM

Today I got question in numerical analysis

I will ask it later

It is about something like decimals

too abstract to digest

related to machine number

NoseBleed

8:56 AM

It's binary machine number and decimal machine number

NoseBleed

9:25 AM

May be I will ask later. I really don't have energy to grasp any idea right now.

Alex

9:41 AM

@loch Maybe I can ask you about the content some time, depending on how much you got through

To understand the second section, I've gone off to read Paul Balmer's paper on Witt groups

loch

@Alex the answer is not much esp. with the technical stuff :p

Alex

The first section is definitely rather technical :')

I think it's sections 2,4,10, and 11 that I mainly want from this paper, but I'll ask my adviser tomorrow about it

loch

i just know that to see the stuff 'in action', you might want to look at the stuff that wickelgren did on counting things over non-algebraically closed fields

Alex

This maybe? swc.math.arizona.edu/aws/2019/2019WickelgrenNotes.pdf

Do you have a specific paper/notes in mind?

loch

yes

no there arent a lot of notes available

if any

(as far as im aware) - but i think the papers are pretty readable (definitely more readable than the levine paper lol)

Alex

9:49 AM

Well that goes without saying :')

WeavingBird1917

Made a funny visualization of factors: desmos.com/calculator/0ssikt8mo7

UmbQbify -Key20-

@WeavingBird1917, taking too much time to load

Oh wait, that's cool, you can make an elevator gif xD

geocalc33

0

Q: Evaluate $ \int_0^1 e^{\frac{1}{\log(\theta)}} ~d\theta $ in polar coordinates

The objective is to: Evaluate $$ \int_0^1 e^{\frac{1}{\log(\theta)}} ~d\theta $$ in polar coordinates. Using cartesian coordinates: The integral, where $K$ is the modified Bessel function of the second kind, $$ \int_{0}^{1} e^{{\frac{1}{\log(x)}}} \, dx =2K_1(2), $$ Can be evaluated using the ...

WeavingBird1917

@UmbQbify-Key20- Lol, I think it looks cooler backwards though.

UmbQbify -Key20-

10:04 AM

I don't actually understand what's happening, but it goes in reverse sometimes

WeavingBird1917

The number being divided (a) is being varied across a range, when it reaches the end of the range it reverses. The full size versions are f3 and f4. The horizontal lines are actually meant to be points on the right side of them, which represent the remainder as a divides by x.

Manolis Lyviakis

if a function is even

is it true that the integral from -00 to +00 is 2 times the limit from 0 to +00
?

UmbQbify -Key20-

10:20 AM

Try it for cosine

Manolis Lyviakis

$ \int_{-00}^{+00} cos(x) dx $

it is indeed 2 times the limiti to +00

right?

Martin Sleziak

10:37 AM

@ManolisLyviakis It works - if the integral from 0 to +\infty exists. (So $\cos x$ is not an ideal example.)

You can also check the answers here: How come $\int_{-\infty}^\infty \! e^{-|t|} \, \mathrm{d} t = 2\int_{0}^\infty \! e^{-|t|} \, \mathrm{d} t$?

Manolis Lyviakis

ye

thats how i did it

Martin Sleziak

A related question about odd functions: Can we use symmetry rules in improper integrals?

geocalc33

11:43 AM

how many definite integrals yield the same result when swapping the variable from cartesian to polar? (not changing the coordinates)

math.stackexchange.com/a/3749735/460999 (relates to this answer...also did the answerer even use polar coordinates?)

feynhat

What is the quotient of $S^\infty \times \Bbb R^2$ under diagonal action of $\Bbb Z/2$?

$\Bbb Z/2 = \{1, a\}$ acts on $\Bbb R^2$ by negation: $a\cdot (x, y) = (-x, -y)$

...and on $S^\infty$ by the usual anti-podal action.

Can I compute the fundamental group? I mean $S^\infty\times\Bbb R^2$ is simply-connected. I think the diagonal action is covering space action. So, $\pi_1((S^\infty\times\Bbb R^2)/\Bbb Z/2) = \Bbb Z/2$.

If $G$ acts on $X$ and $Y$, and one of them is a covering space action, then is the diagonal action on $X \times Y$ also a covering space action?

Thorgott

it only acts on the factors separately, so shouldn't the quotient be $\mathbb{R}P^{\infty}\times\mathbb{H}^2$

feynhat

12:05 PM

What do you mean by 'its acts on the factor separately'?

What is the quotient of $S^1 \times \Bbb R$ under diagonal action of $\Bbb Z/2$?

I thinks its a disc not $S^1 \times \Bbb R_+$.

Thorgott

12:17 PM

oh, I see where I'm wrong

the actions on the factors aren't independent

user69608

@CalvinKhor @satan29

feynhat

@feynhat Its not going to be a disc though. I was actually thinking of $S^1 \subset \Bbb R^2$, and applying the reflection action (not the anti-podal action).

Calvin Khor

what is this and why is the typesetting so bad?

also, hi?

feynhat

ugh... no.

Calvin Khor

12:33 PM

$$\operatorname{\ell im}_{n\to\infty}\limits $$ lol

3

Thorgott

lol

feynhat

> वास्तविक संख्याऒं

Does that mean real numbers?

Calvin Khor

how on earth lol

feynhat

सीमा = limit... lol. अभिसारी = convergent... lmfao. What are these words?

ABCD

1:13 PM

This is hindi language.

anakhro

I think feynhat of all people could have guessed that. :P

NoseBleed

I have learn basic hindi and I have no idea what you are writing

shima=limit?

abhishari=convergent?

It's time to go to bed. No questions for today

Yuvraj

@robjohn hi sir

how are you?

feynhat

1:31 PM

@feynhat This is false btw.

Yuvraj

yes

anakhro

@feynhat what are you up to?

feynhat

Oh hi. Just trying to figure out the quotient $(S^\infty \times \Bbb R^2) / (\Bbb Z /2)$

anakhro

As a Lie group?

feynhat

No, just the topological space. (I don't even know how you put a smooth structure on $S^\infty$).

anakhro

1:41 PM

Me either. Is it the antipodal action on S^\infty, and then what action on R^2?

feynhat

$(x, y) \mapsto (-x, -y)$.

anakhro

Oh so technically the same as S^\infty.

robjohn

:54882438 $\sin^2(x)+2\cos(x)+1=3-(1-\cos(x))^2$ which has a range of $[-1,3]$

feynhat

yea

@ABCD Yes, I know. I am a native speaker. Its just that I've never read math in that language. Those words seem funny being used in math.

@anakhro Btw, do you know more than half of people in the country do not speak Hindi? I am pretty sure Balarka doesn't (or at least, is not a native speaker, and speaks with an atrocious accent).

anakhro

Yes, but I think significantly more than that can identify hindi.

feynhat

1:55 PM

Oh yeah. That might be true, since Hindi appears in a lot official documents, bills etc.

@BalarkaSen Help me see this. Consider the finite cylinder $S^1 \times [-1,1]$. There is a $\Bbb Z/2$ diagonal action on this cylinder (it acts on both $S^1$ and $[-1,1]$ anti-podally). What is the quotient space?

Calvin Khor

identify but not understand? Even I can lol there are signs in hindi in my country

feynhat

I think the fundamental domain will be $S^1 \times [0, 1]$.

What happens at $S^1 \times \{0\}$ ?

anakhro

It's always ugly when you aren't gluing only on boundaries. :P

Are you able to think about doing it on the boundary to begin with, and then kind of zipping it up along the seams?

It already sucks imagining the boundary version so that's probably a bad suggestion, actually.

Mike Miller

2:12 PM

@feynhat In what sense are you trying to identify the quotient?

It is a certain 2-plane bundle over $\Bbb{RP}^\infty$.

anakhro

How do you conclude it is a plane bundle so quickly?

feynhat

It will be great if we can figure out what exactly the space is, it would be great.

I mean some expression like $\mathbb{RP}^\infty \times \text{(something)}$.

Otherwise, I would like to compute $\pi_1$.

Mike Miller

It is not a product, it is a fiber bundle. The most explicit you will get is identifying that fiber bundle. And honestly what you've described is more or less the most explicit description. If $\tau$ is the tautological line bundle (whose unit sphere bundle is $S^\infty$), then your plane bundle here is $\tau \oplus \tau$.

@anakhro Because $\Bbb Z/2$ acts freely on the first factor, one has an open cover of this space by things that look like $((U \times \Bbb Z/2) \times \Bbb R^2) / (\Bbb Z/2)$. That quotient is just $U \times \Bbb R^2$.

If $G$ is finite and acts freely on $X$ and in some arbitrary way on $Y$ then $(X \times Y)/G$ is a $Y$-bundle over $X/G$.

I will leave it to you to calculate $\pi_1$ from the fact that it's a fiber bundle over $\Bbb{RP}^\infty$ with fiber $\Bbb R^2$.

@feynhat The same fact tells you that this gives you a $[-1,1]$-bundle over $\Bbb{RP}^1$. Can you identify which one? You know a couple.

feynhat

2:28 PM

Mobius bundle?

Mike Miller

Maybe!

anakhro

Nice, thanks for that elaboration.

@MikeMiller have you ever heard of something like the "discriminant" of a manifold?

Mike Miller

Nope

Context?

anakhro

There is this definition I read of "discriminant points" of a contact manifold $(M,\xi)$. A discriminant point of a contactomorphism $\phi\colon M\to M$ is a point $p\in M$ such that $\phi(p) = p$ and $(\phi^*\alpha)_p = \alpha_p$ for one (hence all) contact forms $\alpha$.

The discriminant of $M$ is then the set of all $\phi$ with discriminant points.

But I don't quite understand the intuition behind using the word "discriminant". Unless it's completely separate from every other use of discriminant in math. :P

Mike Miller

What's the point/what's it used for?

anakhro

2:37 PM

Well it's used with a certain type of Maslov index for a contact isotopy which is the intersection of a subspace of the discriminant and the contact isotopy.

<-- doesn't really understand the point but is trying to go through a paper.

This Maslov index has a bunch of applications, e.g. a special case of Arnold's conjecture for contact manifolds.

Also, contact non-squeezing re-proof.

Mike Miller

Yeah, I don't know. Sorry about that.

I'm sure there's a good reason for calling it that but I'd have to be more immersed in the literature I'd guess.

anakhro

It's fine, no worries. As long as I am not missing a painfully elementary comparison to a "discriminant" that I should of already.

feynhat

@MikeMiller I only know how to do that when we have a covering space. But in that ^ case, the fiber is not discrete.

anakhro

It might just be that they are more distinguished than regular fixed points.

Mike Miller

@feynhat You don't know of the long exact sequence in homotopy groups of a fiber bundle? It's probably worth knowing.

@anakhro I would suspect it's only etymologically linked to "discriminant" as in $b^2-4ac$

anakhro

2:43 PM

Yeah.

Mike Miller

In the same way that the kernel of a linear map (== solution set to some linear equations) and the kernel of a differential operator (== something which generates the space of solutions by convolution) are both linked to the English word kernel (== the core of something).

anakhro

What have you been reading about lately, @MikeMiller?

Mike Miller

20th century French colonialism. A friend also suckered me into a reading group for Hegel's phenomenology but it wouldn't be fair to say that I understand anything or that I'm spending a lot of time on it.

feynhat

Is there a group structure on $\pi_0$?

Mike Miller

$\pi_0$ of what?

feynhat

2:50 PM

Fiber space, say (as in our case) $\Bbb R^2$.

Mike Miller

Any finite set can be a fiber of something

You're basically asking (afaict) whether a random given set has a group structure

Which, yes, but not naturally so

If you want $\pi_0 G$ to be a group you probably want $G$ to be a group, in which case $\pi_0 G = G/G^e$, where $G^e$ is the path-component of the identity

feynhat

Yeah. I meant, what does $\cdots \to \pi_1(E) \to \pi_1(B) \to \pi_0(F) \to \pi_0(E) \to 0$ being exact means.

Mike Miller

If you're worried about applying the fiber sequence near the tail end here, when the fiber is connected you should just read it as ending $\pi_1 E \to \pi_1 B \to 1$, the last thing the trivial group (so that $\pi_1 E \to \pi_1 B$ is surjective).

Jaakko Seppälä

Let $X$ be a topological space. Is it always true that if $A\subset X$ then $\operatorname{int}\bar A=\operatorname{int}A$?

Mike Miller

When $F$ is disconnected and your fiber sequence isn't of the form $G \to X \to X/G$ then it's going to be more irritating to parse through the end of the sequence. But you're not in that situation, so I wouldn't worry.

Alessandro Codenotti

2:54 PM

@JaakkoSeppälä no

Jaakko Seppälä

What would be a counterexample?

Alessandro Codenotti

$\Bbb Q\subseteq\Bbb R$

Thorgott

Consider a field $k$ and the $k$-algebra $k[x,xy,xy^2,...]\subset k[x,y]$. If this were finitely generated (as $k$-algebra, now and in the forthcoming) by elements $f_1,...,f_n$, then it would also be finitely generated by the monomial terms appearing in $f_1,...,f_n$, because any polynomial expression in $f_1,...,f_n$ is a polynomial expression in their monomial terms. So assume it is generated by some $xy^{a_1},...,xy^{a_k}$ and pick $n>\max a_i$. Looking at any polynomial expression in $xy^{a_1},...,xy^{a_k}$, we see that any monomial term containing the power $y^n$ necessarily contains …

Alessandro Codenotti

There's even counterexamples with $A$ open, that's why the concept of regular open set is interesting

feynhat

Oh, okay. So, $1 \to \pi_1 E \to \pi_1 B \to 1$ is exact, where $E = (S^\infty \times \Bbb R^2)/(\Bbb Z/2)$ and $B = \mathbb{RP}^\infty$. So, $\pi_1 E = \Bbb Z /2$.

anakhro

3:03 PM

@MikeMiller ask yourself: would a true friend sucker you into a reading group based on anything Hegel wrote?

Mike Miller

Yes, just not the phenomenology.

The trick though is to read it without pretending you understand.

@feynhat Right. Another way to see your conclusion is that your space is a vector bundle over $\Bbb{RP}^\infty$, and every vector bundle def retracts onto the zero section.

feynhat

oh duh. a vector bundle deformation retracts to the base space by by linear homotopy.

satan 29

@user69608 the thing inside the bracket approches 1, while the power approches $\infty$. reminiscent of e.

Mike Miller

@feynhat The really fancy observation is that if $Y$ is $G$-equivariantly homotopy equivalent to $Y'$ (and $G$ acts freely on $X$), then $(X \times Y)/G$ is homotopy equivalent to $(X \times Y')/G$. And if $G$ is any group of linear maps, the straight-line homotopy is an equivariant homotopy equivalence from your vector space to a point.

That's the fancy way of observing the vector bundle thing here. But it's unnecessarily fancy of course.

anakhro

@MikeMiller Heh.

user69608

3:11 PM

@satan29 yeah how to do after that?

satan 29

@user69608 if the thing inside the bracket is a, and the exponent is b, then your expression is : $(1+(a-1))^{\dfrac{1}{a-1}*{b*(a-1)}}$

feynhat

@MikeMiller Bookmarked.

Mike Miller

Well, you do you.

satan 29

@user69608 so the limit boils down to e^ (limit of (b(a-1))

Thorgott

@Thorgott I'm starting to believe it just doesn't fail

satan 29

3:20 PM

@user69608 now let x=1/t. so b(a-1) becomes (after writing 1 as n/n) :$\dfrac{n[(a_1^t-1)+(a_{2}^t-1).....]}{nt}$

which is a standard limit

$ln(a_{1})+ln(a_{2))...... =ln(a_{1}a_{2}.......$.

so e raised to this power is simply $a_{1}a_{2}..$

i.e D

Balarka Sen

@feynhat You can take the fundamental domain to be (the upper hemisphere of $S^1$) x [-1, 1].

Then simply observe the sides are getting glued like a Mobius strip :P

feynhat

(balarka ninja)

Balarka Sen

Sniped. Also, my Hindi accent is fine.

satan 29

nice :)

feynhat

ok Bolorko

Balarka Sen

3:25 PM

Lmfao

@feynhat There's a cyclotron in Saha Institute in Calcutta. It doesn't work, do you know why?

feynhat

umm no. Why?

Balarka Sen

Because V x B = 0 according to Bengalis

feynhat

loooooooooooool

Vinícius Machado Vogt

0

Q: Wikipedia: "Classification of discontinuities"

I've added an information to the Wikipedia article "Classification of discontinuities", but now I am in doubt if it's correct: "Intuitively, a function is continuous if (and only if) it has no breaks, jumps, vertical asymptotes, or wild oscillations." Is it correct? (ignoring the fact I give no p...

calculus

0

Q: Cryptography puzzle

The two strings below describe the same English word: 7415963852 1475369963 What word is it? Just for fun! Credits: Hindemburg Melão Jr.

cryptography

Anyone?

Anonymous

I'm thinking of $\mathbb R/\mathbb Z$ as the half-open interval $[0, 1)$. Is there any sense in which this $[0, 1)$ can be considered as a "circle" by joining its "endpoints"? I mean if the interval were $[0, 1]$ I suppose there is some terminology for gluing the endpoints but I'm not sure about $[0, 1)$.

Mike Miller

3:40 PM

Can you make more precise what you're asking

$[0,1)$ only has one endpoint so it's hard to pin down what that would mean

Anonymous

@MikeMiller Umm, well first of all does it make sense to represent $\mathbb R/\mathbb Z$ as $[0, 1)$?

Mike Miller

What does "to represent" mean

There's a natural set bijection between them

Anonymous

@MikeMiller That is representing the cosets of $\mathbb Z$ in $\mathbb R$ by taking a single representative element from each coset.

Anonymous

@MikeMiller I guess that's what I mean

Mike Miller

The composite $[0,1) \hookrightarrow \Bbb R \to \Bbb R/\Bbb Z$ is a bijection

Because every number is (uniquely) of the form $n + r$ where $0 \leq r < 1$ and $n \in \Bbb Z$

Anonymous

3:43 PM

However, in the construction of the Vitali set they're apparently representing $\mathbb R/\mathbb Z$ with the interval $[0, 1]$ so I'm a bit confused

Anonymous

From what I can understand, there is no reasonable bijection between $\mathbb R/\mathbb Z$ and $[0, 1]$. Is that right?

Mike Miller

I don't see anywhere that they use such a bijection. It also doesn't seem important to me, because this is about measurabiity, and chucking out a single point doesn't change whether or not a set is measurable

Balarka Sen

Isn't the Vitali set about picking a coset representative of each coset in $\Bbb R/\Bbb Q$?

user69608

@satan29 acha mistake milgaya ,thanks

Jaakko Seppälä

Let us equip the space $X=\{1,2,3,4\}$ be the topology
\[\{\emptyset,X,\{1\},\{2\},\{1,2\},\{3,4\},\{1,3,4\},\{2,3,4\}\}.\]
Consider the partition $X=a\cup b\cup c$ where $a=\{1\}$, $b=\{2,3\}$, $c=\{4\}$. How can I compute the quotient space corresponding to this partition?

Anonymous

3:47 PM

@BalarkaSen Yes, or rather the cosets of $\mathbb Q/\mathbb Z$ in $\mathbb R/\mathbb Z$

Thorgott

same thing

Balarka Sen

Just phrase it like, picking a coset representative of each coset in $\Bbb R/\Bbb Q$ lying in $[0, 1]$

Anonymous

Oh, Wikipedia writes this as a note: The notation R/Z is somewhat ambiguous. If Z is understood to be a group acting on R via addition, then the quotient is the circle. However, if Z is thought of as a subspace of R, then the quotient is a countably infinite bouquet of circles joined at a single point.

Balarka Sen

That's irrelevant yeah forget about that

Anonymous

I'm basically trying to think of $\mathbb R/\mathbb Z$ as a circle

Mike Miller

3:49 PM

Ignore that note completely, they should not have written that

Balarka Sen

It is a circle.

Mike Miller

They're trying to clarify potential confusion but they've only created it lol

Anonymous

@BalarkaSen But how is $[0, 1)$ (which is in bijection with $\mathbb R/\mathbb Z$) a "circle" if one of its endpoints is missing?

Balarka Sen

$[0, 1)$ is also in bijection with the circle, $t \mapsto e^{2\pi i t}$.

Thorgott

a circle doesn't need endpoints

Anonymous

3:51 PM

5

Q: How to construct a bijection from $(0, 1)$ to $[0, 1]$?

Possible Duplicate: Bijection between an open and a closed interval How do I define a bijection between $(0,1)$ and $(0,1]$? I wonder if I can cut the interval $(0,1)$ into three pieces: $(0, \frac{1}{3})\cup(\frac{1}{3},\frac{2}{3})\cup(\frac{2}{3},1)$, in which I'm able to map poin...

real-analysis

Anonymous

Oh, I think it makes sense now

Anonymous

We can always construct a bijection between $[0, 1)$ and $[0, 1]$

Balarka Sen

You don't want to be putting random bijections between different objects when doing math.

Then everything is a circle.

Clearly not true

Thorgott

this doesn't matter

this how it be

Anonymous

@Thorgott Okay. Let's do this step by step. How do you say define or construct a circle given an interval like $(0, 1)$?

Thorgott

3:53 PM

I wouldn't

Mike Miller

By focusing so much on this you are obfuscating the point that the Vitali set is trying to make

Anonymous

@Thorgott Can this be made rigorous?

Thorgott

3 mins ago, by Balarka Sen

$[0, 1)$ is also in bijection with the circle, $t \mapsto e^{2\pi i t}$.

Mike Miller

A measurable set less one point is still measurable. A non-measurable set less one point is still non-measurable.

Thorgott

that's rigorous

Mike Miller

3:54 PM

You can write out the same construction as on that Wikipedia page with $[0,1)$ instead and you produce a non-measurable set

Or with $(0,1)$

Thorgott

missing a non- at the end, Mike

Anonymous

@MikeMiller The Vitali set isn't really my main point of concern here. I do understand that countable sets are measure zero :P

Mike Miller

Well, it's pretty unclear what your main point of concern is.

Thorgott

that said, it's suboptimal to think of $\mathbb{R}/\mathbb{Z}$ as $[0,1)$, because it obscures the topology of the space

if you don't want "too many representatives" for each equivalence class, think of it as $[0,1]/\sim$ where $\sim$ is the equivalence relation only identifying $0$ and $1$

Anonymous

@Thorgott Right, so even something like $(0, 1)$ is in bijection with $[0, 1]$. Then by gluing together $0$ and $1$ we can make a circle, yes?

Thorgott

3:58 PM

$0$ and $1$ are not elements of $(0,1)$

Balarka Sen

Why do you want to do this? $\Bbb R^2$ is in bijection with $S^1$. What is to be gained from knowing that?

Anonymous

@Thorgott I guess that is one of my confusions. What exactly do you mean by it obscures the "topology of the space" ?

Balarka Sen

$\Bbb R/\Bbb Z$ is homeomorphic, as a topological space, to $S^1$.

$[0, 1)$ is not homeomorphic to $S^1$

Thorgott

when you write $[0,1)$, you, I and everyone else thinks of this as subspace of $\mathbb{R}$ with the corresponding subspace topology

now refer to Balarka's comment

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$Mathematics$ Mathematics