Mathematics - 2018-10-20 (page 1 of 4)

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1:11 AM

If anyone is available can I ask if I computed this correctly? I've been so lost today with these cycle products. (1234)(12)(34)=(14)(2)(3)

?

1:31 AM

@dls its (13) (2) (4)

you do this from left to right

Some authors have the right to left convention, but even then it'd be (24)

Oh I did it from right to left and I got (134) right now.

I'll try it again

Question: If i did it from the left to right convention and I was solving for elements of a left coset. I still solve aH for a left coset right?

1:55 AM

Is anayone familiar with Operations research book by Hillier? I've checked edition 10th,9th and 7th and apparently the proposed exercises does not change in none of the editions. I am curious if this holds for the rest of the editions?

For example, the exercise 5.1-1 is the same in the 3 mentioned editions

I could not find the earlier editions (first,second, etc) online.

sorry i'm not familiar with it.

Ok

Anyone able to help on this problem: math.stackexchange.com/questions/2962656/…

Let $G$ be a group and $A,B$ be subgroups of $G$, and it is known that $AB=\{ab:a\in A,b\in B\}$ is subgroup of $G$. Then is it true that $[AB:B]=[A:A\cap B]$?

1 hour later…

3:18 AM

For the above question I have this in mind: $$a_1b_1B=a_2b_2B\iff a_1B=a_2B\iff a_2^{-1}a_1\in A\cap B\iff a_1(A\cap B)=a_2(A\cap B)$$

where, $a_i\in A, b_i\in B$

Am I correct?

4:01 AM

science.sciencemag.org/lookup/doi/10.1126/science.aar3106

"Computational efficiency independent of the size of the system"

i wonder if I can set up a forcing routine to produce strange finite sets that bijects with the naturals but also have proper subsets that can be injected into

Meanwhile, I think in ultrafinite set theory, the natural number M must necessary have supersets all of the same cardinality

4:43 AM

hello

2 hours later…

6:29 AM

0

Q: about direct sum decomposition of module

I know that M is free R module and $M_F$ free F module. I also know that how $M/M\cap N$ is free R module. I want to know why that $gs=id$ is identity map and how it yields direct sum decomposition of M?

linear-algebra commutative-algebra modules

diedidedisdwidiwesisudidiediediediedirediedideuidie

be sp prot

We are going to change some gears today and build some ultrafinite set theory

(Put on hold as a wild transition state enters the maths chat and pull my soul away)

6:59 AM

lets say we have module M_F= M tensor F over R module ;R is pid F, is field of fraction then how to show that for all x belongs to M_F there exist a in R st ax belongs to M?

A NONZERO

$$\Huge{ALL_CAPS^{ALL_CAPS^{ALL_CAPS^{ALL_CAPS^{ALL_CAPS^{ALL_CAPS^{ALL_CAPS^{AL‌L_CAPS^{ALL_CAPS^{ALL_CAPS^{ALL_CAPS^{ALL_CAPS^{ALL_CAPS^{ALL_CAPS^{ALL_CAPS^{ALL‌_CAPS^{ALL_CAPS^{ALL_CAPS^{ALL_CAPS^{ALL_CAPS^{ALL_CAPS^{ALL_CAPS^{ALL_CAPS^{ALL_‌CAPS^{ALL_CAPS^{ALL_CAPS^{ALL_CAPS^{ALL_CAPS^{ALL_CAPS^{ALL_CAPS}}}}}}}}}}}}}}}}}‌}}}}}}}}}}}}}$$

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Canvas Purged

> From: Silent: For the above question I have this in mind: $$a_1b_1B=a_2b_2B\iff a_1B=a_2B\iff a_2^{-1}a_1\in A\cap B\iff a_1(A\cap B)=a_2(A\cap B)$$
where, $a_i\in A, b_i\in B$

For this question, hmm...

but $a_2^{-1}a_1 B = B$ and $a_2^{-1}a_1 \in A \cap B$ thus:

$a_2^{-1}a_1 \in B$?

and thus $B \subset A$?

Or more generally, let $c \in A \cap B$, then what we have here based on these implications is that $cB = B$ thus $c \in B$

because if $c \in A - B$ then $cB\neq B$

5 hours ago, by Silent

Let $G$ be a group and $A,B$ be subgroups of $G$, and it is known that $AB=\{ab:a\in A,b\in B\}$ is subgroup of $G$. Then is it true that $[AB:B]=[A:A\cap B]$?

5 hours ago, by Drew Brady

Anyone able to help on this problem: https://math.stackexchange.com/questions/2962656/mapping-the-cycle-graph-into-th‌e-real-line

(important questions before purge summarised here)

7:33 AM

@LeakyNun lets say we have module M_F= M tensor F over R module ;R is pid F, is field of fraction then how to show that for all x belongs to M_F there exist a in R st ax belongs to M?
a is nonzero

@Ninjahatori why are you doing advanced things and not knowing basic things?

I am confused with small things and I know why this is true but I don't know how to write it?

then why are you learning advanced things when you are still confused with small things?

How do you prove matrix M is symmetric? Am I correct with my approach? imgur.com/a/sKQHEot

@LeakyNun can you please explain me this?

7:39 AM

@YolandaHui no. v^T v and vv^T are not the same

@Ninjahatori no. this is something you should know. otherwise you shouldn't be doing advanced things.

@LeakyNun more context here, i.stack.imgur.com/r1AEG.jpg

ok

how do I proceed if v^T v and vv^T are not the same?

so can you just check whether I am correct or not?

@YolandaHui fix the step where you turned vv^T to v^T v

7:41 AM

You are indeed the most persistent residental help vampire I have ever seen

@Ninjahatori you didn't present any proof

only gateprep is much worse

@YolandaHui There's a careful point here. (vv^T)^T is not v^T v---remember that transposition switches the order.

(vv^T)^T = vv^T

Indeed.

7:43 AM

so M^T is equal to M

therefore M is symmetric?

$(AB)^T=B^TA^T$ is such a useful rule

Seems like solid reasoning to me.

The sum of symmetric matrices are symmetric

@Fargle the bottom ll v ll ^2 is unaffected by transpose right

I wish there's a more elegant geometric way to show this. (algebraically it is easy to check using distributivity of T)

7:44 AM

Yes---why?

halp I'm studying p-adic Hodge theory, what is seven times three

@YolandaHui hint: what object is ll v ll ^2

scalar

@LeakyNun Not sure, but a quick Fermi approximation gives it as 10

and the transpose of a scalar is clear

7:46 AM

@YolandaHui Yep, transposition respects scalar multiplication. I think you've got it.

How do I say that M satisfies M^-1 = M? Just show that M^2=I?

@YolandaHui yes

user image

@LeakyNun check this please

@Ninjahatori which university do you study in?

nit

7:50 AM

aha

Is it coorect?

no it isn't

why?

spam spam spam spam spam

how do I find the minimum value of this quadratic form? with M being the previous matrix. imgur.com/a/TCa6g8S

7:52 AM

I've heard that mathematic education is very bad in india

what is fault there?

and it's precisely that country that can produce people who are doing advanced stuff without knowing basic stuff

Oct 25 '17 at 22:02, by Balarka Sen

Math as a competitive, extremely ad hoc calculation oriented thing, is prevalent in India and I suspect most of Asia

Sometimes I am wondering. How on earth people do advance things without knowing basic things,

the amount of error cancellation for a result to pop up lacking a foundation will be very huge

I won't speak for what goes on in the rest of the world, but I feel like America has an almost opposite problem. It's hard to find good resources for advanced mathematics outside of the uni system, and even then it's not at every (or, by my guess, even most) universities here

Combine that with what seems to be a fundamental failing of our mathematics pedagogy in the primary and secondary systems, and you get a mess

anyone know anything about KKT and lagrangians?

7:57 AM

Any idea on the minimum value of the quadratic form x^TMx ?

@Fargle right

I wonder which country has the best mathematics education that can cater both fundamentals and advanced level. Perhaps it might be in Europe given the number of European mathematicans

@YolandaHui x^T M x = x^T (I - 2 v v^T / |v|^2) x = |x|^2 - 2 |v.x|^2/|v|^2

draw a picture to see what this represents

hmm the 2 is strange

Is it correct to say that (group theory) a cycle are kind of a "prime" of permutations?

@IsanaYashiro I don't think an analogy can be correct/incorrect

8:11 AM

Do you agree

Do you like it

There are cycles of composite number of elements

@LeakyNun how do u go from x^T (I - 2 v v^T / |v|^2) x to |x|^2 - 2 |v.x|^2/|v|^2 ?

Would you smile when you heard that

@LeakyNun tell first what is wrong there?

@YolandaHui distribution, noting that x^T y = x.y

@Ninjahatori stop pretending like we're obliged to help you.

9

8:12 AM

It is absolutely correct that x tensor a belongs to M

@IsanaYashiro I think the analogy holds up in the sense that there is a unique decomposition of any permutation of a finite set into disjoint cycles. However, primeness is usually defined in terms of a ring structure, so I don't know that the analogy goes too far.

@LeakyNun I tried to plot |x|^2 - 2 |v.x|^2/|v|^2 and it seems like it has a maximum instead of a minimum??

sure it has both

don't try to plot it

16 mins ago, by Leaky Nun

draw a picture to see what this represents

remember that |x|=1

the minimum is zero?

I don't think so

8:27 AM

@Fargle: Thank you, I just found that the decomposition is not unique, but the parity

@IsanaYashiro Into cycles it should be unique up to ordering of the factors, and ordering within each cycle, because disjoint cycles commute. It's the transposition decomposition that fails uniqueness.

@Fargle lol, now you know I'm reading about transposition and made wrong conclusion, anyway thank you again

No problem

hmm... I think since cycles of any length n can be decomposed into transpositions, asking whether some cycle n can be decomposed into product of prime cycles is not going to be unique

It's cool that it seems like everyone like group theory and number theory, I got instant reply than other topics

8:30 AM

so there isn't really any analogy in the form of fundemental theorem of arithmetic if that's what you are getting at

It's possible that's largely coincidence. I'm just far more familiar with basic abstract algebra than basic analysis

That's not to toot my own horn--I am still very much a beginner to it all myself

If I shot an arrow to the number line, I can't hit any integer, not to mention about primes. Is this why anyone like number theory?

I can handle some questions in the following topics:
Calculus
Some results on real analysis or elementary reasoning
Some logic results and algebraic workings
Some transcendental number theory
Some set theory (ZF and ZFC) not including model theory
Some complex analysis calculations but less so on theorems
Some high school geometry
Some terminology in differential geometry
Some basic number theory results
Abstract algebra that is not too high level or involve polynomial rings
Some special function properties and functional equations

I dunno, I just think numbers are pretty rad.

I am more inclined to answer questions on:
Topology, infinite sets, abstract algebra, functional equations

But otherwise, the only subject which I really have a solid foundation on is linear algebra

and I don't know if it has gone rusty as it has been 3 years since I last took that course

and I have not focused much on linear algebra recently

8:37 AM

@Secret Cool, I remember my first question of linear algebra was answered by u

you said a matrix can be think of as a way to arrange information or something ... I forgot

@Secret do you recommend any book for abstract algebra?

@IsanaYashiro not really, I still don't have enough time to even start reading the abstract algebra books recommended by 0celo and Tobias

The only maths book I have read more than 1 chapter is Munkres

All my other learning is from disorganised learning from the math chat conversations, googlings of random questions and papers and so on

And a strange question: if i didn't do any problem set on the book, what's the probability that I can get the idea right? Is it the same as my arrow hit pi on the number line?

You can go very far on some subjects with intuition, but ultimately misconceptions will start to pile up

5

@Secret You meant abstract algebra book by Munkres? or

topology munkres

8:43 AM

@IsanaYashiro I personally am a big fan of Artin's Algebra.

@Secret Agree... I definitely felt that before...

@Fargle thanks, I will take a look

Mar 17 at 12:10, by Niing

Since $span(\emptyset)={0}$, could I say "$0$ is a linear combination of nothing"?

7 months ago

Jan 21 '17 at 6:17, by Secret

List of maths fields I have interest in:
1. Group theory in terms of orbits and actions
2. Zero term algebra and division by zero algebra
3. Integration in the language of abstract algebra and as a functional, symmetry of integrands
4. Optimising proofs given axiomatic systems
5. Category theory
6. Unnatural algebraic structures
7. Patterns in expanding multiplications of polynomials
8. Set of all counterexamples given a proposition
9. Tensor visualisation and intuitions
10. Numerical analysis methods to explore special regions or points of mathematical functions or systems of equations

Updated List of maths topic I have interest in:

9:15 AM

0. Philosophy of mathematics: Mathematics of extraterrestrials
1. Formal system: Rewriting systems
2. Foundations: Predicative mathematics, Correspondence between infinite and finite sentences
3. Logic: Higher order logic. Algorithms to explore the set of all possible proofs of a proposition and the structure of proofs. Optimising proofs given axiomatic systems, Set of all counterexamples given a proposition
4. Set theory: Infinite sets, intuitions on infinity, pathological results and possible applications outside set theory

3 hours later…

12:13 PM

Hello!!

Does the following statement hold?

" $M\cup N$ is a finite set $\implies$ $M$ and $N$ are finite sets. "

I think that it holds, but I don't know exactly how to prove that. Te union contains all the elements taht are either in M or in N or in both sets, so the union cannot be finite if one of the M, N is infinite, can it? But how we can prove that formally?

Finally, a question that starts with "Hello!!" that is interesting

The first question to ask is: Are you working in ZFC?

(otherwise finite can mean a lot of things)

Assuming you are working in ZFC. Then all notions of finite are equivalent and we can proceed as follows:

$M \cup N$ is finite $\implies$ there exists a bijection between $M \cup N$ and some natural number $k$. Suppose $M$ is infinite, meaning that it bijects with some infinite cardinal $\aleph_{\alpha}$. Then $M \subset M \cup N$ meaning that $M \cup N$ must contain at least $\aleph_{\alpha}$ elements, contradicting our assumption that it is finite

@Secret no need choice

12:30 PM

3

Q: The union of finite sets is a finite set

The union of two finite sets is a finite set. Let $X,Y$ be finite sets. That means that there are $n, m\in \omega$ such that $X \sim n$ and $Y \sim m$, i.e. there are functions $f: X \overset{\text{bijective}}{\rightarrow} n$, $g: Y \overset{\text{bijective}}{\rightarrow} m$. Then we disti...

elementary-set-theory

looks like my bijection proof skills is still very terrible

@Secret they want the other direction...

@MaryStar If $f:M\cup N\to\{0,1....,k-1\}$ bijective then $f[M]\subseteq \{0,1,...,k-1\}$ which, by the bijectivity of $f$, implies that $M$ is finite. Do the same thing to $N$ and you are done

1:04 PM

Weirdness of finite sets:

If you think really hard about it, the notion of finite set is actually quite self referential

@Secret you don't need the definition of finite sets to define $\Bbb N$

I am guessing you mean defining $\Bbb{N}$ as the intersection of all inductive sets, which exists by the axiom of infinity?

@Holo but that definition is as impredicative as it can get

(cont. on why I think finite sets are quite self referential)
$\omega$ is constructed by iterating the operation $A \cup \{A\}$ where $A$ is a set at the current step with base case $\varnothing$

So we have $\varnothing = 0, \varnothing \cup \{\varnothing\} = 1, \varnothing \cup \{\varnothing\} \cup \{\varnothing \cup \{\varnothing\}\} = 2$ and so on

but how do we justify the existence of one "unit" of $\cup$. the numeral $1$ does not exist yet before the step $\varnothing \cup \{\varnothing\}$

o wait...

$\{\varnothing,\varnothing\} = \{\varnothing\}$

hmm... we act pairing once to get from 0 to 1

so "one" is constructed from using the axiom of pairing "once"

I think I need to take a shower...

@Secret The axiom of infinity says: there exists inductive set. Their intersection can be easily check to be the smallest ordinal. finite set is a set with bijective to initial segment of $(\omega,\in)$

1:16 PM

Actually, if we use that route, can we recover the natural numbers 0,1,2,3,4,5,... using initial segments without ever mentioning the successor operator $\cdot \cup \{\cdot\}$?

@Secret what route?

@LeakyNun I agree that it looks very impredicative

Route = Axiom of infinity -> arbitrary intersection of inductive sets to establish the smallest infinite ordinal $\omega$ -> take initial segment of $\omega$ to recover 0,1,2,3,4,...

@Secret well, this is exactly the same route. It is easy to show that if $x\in y$ and $\forall z(\lnot(x\in z\in y))$ is the same as saying $S(x)=y$

But the successor operator does not require the axiom of infinity and let as advance in infinity ordinals to $\omega+1$ and so on

I see

My question, however is pondering about a possibly ridiculous scenario (thinking maths as if it is a computer program constructed from the ground up) where the number of times an axiom is used need to be labelled by a natural number and that if the natural number 1 is not even constructed yet, then we cannot even use any axioms at all, and hence thinking that 1 is actually more weird than one expected

(Also note that for ordinals $\in\iff \subsetneq$, so initial segments is equivalent to the usual order)

@Secret Here you enter more into first order logic, which I don't know how to answer for sure. This semester I am doing a course in logic, so in few months I will be able to answer a good answer :P

1:26 PM

I see

I am sometimes wondering whether my weakness in logic is the reason why I am so prone to weirdness

@Secret why do you think you have weakness to logic?

Because it is a known fact (Leaky and user21820 knew that well, also see this)

My most common error, pointed out by my linear algebra x graph theory professor in my 2nd year undergrad is I tend to prove $P \implies Q$ while the question actually wants $Q \implies P$ and I don't realise it until being pointed out

My thinking tends to "go backwards" so to speak, and I have not mentioned how many times free associations lead to weird jumps in my reasoning

This is one of the reason I am learning logic in the logic chat room and also using Terrence Tao's interactive logic textbook

Oh, this is actually common problem for new students. Although I wouldn't call 2nd year new. I see what you are saying

The best way to describe my "logic" is that I am like a person walking forward with my head turned to the back, plus teleportations randomly

hmm... @Secret if I can teleport without restrains, i.e. I can teleport without "cooldown". Then I teleport without stopping, at a given point of time, where am I?

1:36 PM

you can be anywhere. If the space is of size n, then the probability of you being in any one specific place is 1/n

If the space is infinite, then well you have trouble because the uniform distribution does not exist in unbounded spaces

But I teleport the moment I appear, so how can I be at a place if I move instantaneously to other place?

hmm interesting, then for any specific place, the probability of finding you there is zero, yet the probability of you in that region is 1 (you are in "somewhere") I think. Actually, is it possible that since the probability of you being found in any one place is zero, then the probability of finding you anywhere at all is also zero?

Exactly, this question annoyed me for some time now

I think the first thing to check is what is the set that describe the probability of finding the particle in the region. If that set is measure zero, then no way you can integrate it to get one, is it?

But what I am suspecting is that set is not only measure zero, but it is zero

and thus giving the probability of finding you anywhere in that region to be zero as well

My measure theory is not very good to think clearly on the answer yet however

Maybe the set is not measurable?

Like, if you take the set of all points the particle was and check them

2:25 PM

44

Q: Zero probability and impossibility

I read a comment under this question: There are plenty of events that can occur that have zero probability. This reminds me that I have seen similar saying before elsewhere, and have never been able to make sense out of it. So I was wondering if zero probability and impossibility mean th...

probability-theory

1

Q: If the outer measure of a set is zero then each subset is lebesgue measurable

Okay so I'm asked to show (not necessarily prove but that would be helpful too!) that if the outer measure of a subset F of R is equal to zero .i.e m*(F)=0 then each subset E of F is Lebesgue Measurable. I'm thinking this is going to follow directly from the definition of Lebesgue measurable but...

real-analysis lebesgue-measure

In other words, the probability can be described with a set of outer measure zero. But I don't think I fully grasp neither measure theory nor probability theory to try to dab on what will happen

Me neither, sigh. It appear we won't get answers yet

Prashin Jeevaganth

user image

Can someone explain how the following expression is derived?

what kind of logical operator is $\cdot$

@Secret And.

Or at least I assume so. Seems like $\oplus$ is xor and $+$ is or.

$\oplus$ is xor is standard. I was not sure about $\cdot$ until now

$(a \land \neg b) \lor (\neg a \land b) =$

if $\text{value}(a)=\text{value}(b)$ then $F \lor F = F, T \lor T = T$

else if $\text{value}(a)=\neg\text{value}(b)$ then $T \lor F = F \lor T = F$

wait that does not sound right, let me check...

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I really have to think how to do this without going through the whole truth table as it gets unwieldy if there are more than 3 variables

Still, that does not really answer Prashin's question on how exactly we get this expression in the first place

other than perhaps the functional completeness of the set $\{\land,\neg\}$ might provide us clues

Anyway:

$\neg(b \oplus c) \lor (b \oplus d) = \neg((b \land \neg c) \lor (\neg b \land c)) \lor (b \land \neg d) \lor (\neg b \land d) = (\neg(b \land \neg c) \land \neg(\neg b \land c)) \lor (b \land \neg d) \lor (\neg b \land d) = ((\neg b \lor \neg\neg c) \land (\neg\neg b \lor \neg c)) \lor (b \land \neg d) \lor (\neg b \land d) =$

$((\neg b \lor c) \land ( b \lor \neg c)) \lor (b \land \neg d) \lor (\neg b \land d) =$

2:57 PM

@Secret you \\ sometimes... the line is too long

edit timeout

Lol

will keep that in mind next time

$\neg(b \oplus c) \lor (b \oplus d) =\\ \neg((b \land \neg c) \lor (\neg b \land c)) \lor (b \land \neg d) \lor (\neg b \land d) = (\neg(b \land \neg c) \land \neg(\neg b \land c)) \lor (b \land \neg d) \lor (\neg b \land d) =\\ ((\neg b \lor \neg\neg c) \land (\neg\neg b \lor \neg c)) \lor (b \land \neg d) \lor (\neg b \land d) =((\neg b \lor c) \land ( b \lor \neg c)) \lor (b \land \neg d) \lor (\neg b \land d) =$

also I got stuck, I am not that ready to deal with lattices yet

btw, regarding that teleport question:

4

Q: How can the probability of each point be zero in continuous random variable?

I know this is duplicated but I think the question is a bit different and needs different answer. How can CDF be continuous and have derivative at each point that is not equal to zero but the probability at each point is zero? Why not say for example if you want to choose a real number between...

random-variable continuous-data

any CDF will do, because probability of finding the particle any specific point is zero yet the whole thing integrates to one

This is because each singleton is measure zero yet a line segment made of continuumly many singletons in that order is nonzero

3:06 PM

I see

So that means, if the teleportation is not biased anywhere in a finite region, then the CDF is the uniform PDF

it is only when infinite that we have some biasing because the uniform distribution in an unbounded space does not exist

11

Q: Why isn't there a uniform probability distribution over the positive real numbers?

Apparently, the solution to the Card Doubling Paradox is that a uniform probability distribution over the positive real numbers doesn't exist. Can anyone explain why this is the case and what probability distributions can exist over the positive real numbers (it seems that this would be quite lim...

probability-theory

hmm... the inequality reminds of those in Bernstein and Vitali sets, so why we cannot assign the uniform pdf to a nonmeasuable set...

Probability is not my strong side sadly

currently checking. I suspect the outer measure of any proported "uniform infinite distribution" will not fit the definition of a non measurable set hence even those are out

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3:21 PM

Oh nvm that idea! It is impossible for my question to be a non measurable set

There exists a model such that every subset of reals is measurable(in ZF)

Axiom of choice

In mathematics, the axiom of choice, or AC, is an axiom of set theory equivalent to the statement that the Cartesian product of a collection of non-empty sets is non-empty. Informally put, the axiom of choice says that given any collection of bins, each containing at least one object, it is possible to make a selection of exactly one object from each bin, even if the collection is infinite. Formally, it states that for every indexed family ( S i ) i ∈ ...

Second from last in "Statements consistent with the negation of AC"

Ok, such "uniform infinite distribution" object where each singleton has probability p < 1 cannot be nonmeasurable since any interval (a,b) is a subset of it and that has measure (b-a)p and the complement of this has full measure (which is 1 for probability measure). The intersection of the first one is (a,b) while the second one is empty thus we have: (b-a)p = (b-a)p + 0 which holds

Thus ruling out that it cannot be a nonmeasurable set, the rest of the argument follows in that answer and hence it is ruled out, or branded as an improper prior

o wait

no I only tried one example

1

A: Why isn't there a uniform probability distribution over the positive real numbers?

The problem stems from measure theory. Let $V^1, V^2, V^3, ...$ be a sequence of Vitali sets such that the collection $\{V^n : n \in \mathbb{N}\}$ partitions $[ \,0, 1) \,$. It is then possible to create a mapping $k$ from each set $V^n$ onto $n$ such that $( \,\forall x \in V^n)\,(\,k(x) = n) ...

This answer gives the full context: The uniform infinite distribution is expressed as a union of nonmeasurable sets that partition the real line

the simplest such is the vitali set

@Holo That will probably mean in that model, the solution to your problem will mean if you are teleporting in infinite space, then it must be more likely to find you in a certain nonzero region than elsewhere (even though finding you in any spot is zero), as you will need nonmeasurable sets to make the uniform pdf of infinite support to exist

@Secret I see

3:36 PM

Lol I like Vitali sets more now

Lol, I don't know if to like or hate them more :)

3:53 PM

is atiyahs proof of RH correct?

no

but he is one of the greatest mathematicians of our times

surely he knows better

then you

is he senile?

is that what your saying

76

Q: Is there a way to discuss the correctness of the proof of the RH by Atiyah in MO?

I just made a question in MO to discuss the correctness of the proof provided by Prof. Atiyah for the Riemann hypothesis (link here: Is there an error in the pre print published by Atiyah with his proof of the Riemann hypothesis? ). I understand that's usually outside the scope of MO, but this ti...

discussion specific-question on-topic

@Loffen If your conviction is that nobody in here could know better than him, why did you ask here?

in The h Bar, Jun 14 '16 at 13:10, by John Rennie

Well I guess that's all the proof we needed that whoever is using the Loffen account is not posting in good faith.

Loffen is a known troll, best to press the ignore button

I hate ignore in general, but with one important exception: If the person is a known troll, go ahead and ignore and ghost them however you want

You know I am dead serious when you realise that I and my collaborators literally spend 10 years of preparation in political, art, science, history, philosophy, cultural spheres to tune everything in the world just right so that both ghosting and its antithesis trolling will be completely forgotten by the human civillisation by hopefully 2020

Nothing in this world, even paedophillia, murder, rape, <this is NSFW>, will beat our determination to erase ghosting and trolling

Anyway, enough iluminati nonsense. Back to math mode

4:15 PM

@Loffen *than

4:42 PM

user image

Hmmm...

pretty sure this only make sense inside an integral

It will of course be integrated later

But as an intermediary step, it's a bit sloppy I guess

well it is still justified via the chain rule, so it should be fine

and differential forms are well defined and common in electromagnetism

Hmm but the magnetic field is discontinuous at a surface current (tangentially at least)

Not that it makes any difference in this case

Akiva Weinberger

4:57 PM

Hey, question

If you have a hyperbola, what's the name of the other hyperbola that's on the other side of the asymptotes?

Hi, DogAteMy. It has no name of which I'm aware.

Akiva Weinberger

Like, if your first hyperbola is $(\frac xa)^2-(\frac yb)^2=1$, I'm talking about $(\frac xa)^2-(\frac yb)^2=-1$

Hi

Right, I was about to say you're changing the sign of the "level" ... note that in three dimensions, you change from hyperboloid of one sheet to hyperboloid of two sheets and vice versa, so really very different families.

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