Mathematics - 2017-04-22 (page 2 of 3)

Astyx

12:00

$$\mathop{\LARGE\mathrm E}_{i\in S}a_i = a_1^{a_2^{\dots^{a_k^{\dots}}}}$$

Secret

Yes, much better

SBM

What's that?

Secret

although I guess one reason that is not include (besides it is very rarely used) might be because exponantials are not associative

SBM

What are we trying to figure out btw?

Secret

Me sharing a random idea that we should have an operator for exponential tower series

Simply Beautiful Art

12:02

nah

I'd honestly rather not

Secret

11 hours ago, by Benjamin

@SimplyBeautifulArt But what would $$\psi(0)^{\psi(1)^{\psi(2)^{\psi(3)^{\dots}}}}$$ be?

Astyx

Yeah you'd rather write
$$\mathop{\LARGE\mathrm E}_{i=0}^{n+1} a_i = \left(\mathop{\LARGE\mathrm E}_{i=0}^{n}a_i\right)^{a_{n+1}}$$ and $\mathop{\LARGE\mathrm E}_{i=0}^{0}a_i = a_0$

Simply Beautiful Art

@Secret Yeah, I'd rather not.

Astyx

Which you can't always define

Secret

also, I guess the nonassociativity of exponentiation makes it very hard to work with

Astyx

12:05

But then again $\sum_{i\in S} a_i$ is defined only when the familly is summable (ie the series is not half-convergent)

Benjamin

@Secret It could be helpful.

Secret

One of the major hurdle I think to formalise $\mathop{\LARGE\mathrm E}_{i\in S} a_i$, is, however how to handle the nonassociativity

For sums and products, we don't have that problem because stuff is associative

Astyx

Why $E$ though ?

You can't formulate it

For instance if $a_0=1$ it will always converge

However you can have something not converging even if there is an $a_i=1$

By the way I think I miswrote what you meant in my recursive formula above

Secret

$\sum$ is Sigma
$\prod$ is Pi
Therefore, for $\mathop{E}$, following the pattern it has to be capital epsilon

SBM

what is the discussion about?

Benjamin

12:12

$\epsilon$

SBM

epsilon ?

Astyx

Something about infinite exponentiation @SBM

Secret

@Astyx Yeah, I noticed that when starting to read about hyperoperations. It seems 1 is a fixed point for many hyperoperations.

Benjamin

@SBM Formalizing a notation for exponentiation tower series.

Astyx

Not $+$ :)

Secret

12:13

But then again, I think that might be just a subset of all exponential tower series, thus that is not very broken

SBM

infinite exponentiation sounds interesting but it's complex

Secret

just as we have different types of converging series for sums and poducts, I think having $a_0=1$ should not be too big an issue

as most of the exponential towers that converges are not necessary in that subset

Astyx

I'm sure someone crazy has already looked into it :p

Danu

12:27

@TedShifrin I wrote something, but not a complete answer.

Studentmath

Problem is I can't see how it can be rearranged that way - probably should go at another direction..

Astyx

12:44

Hi Steamy

SteamyRoot

ohi

Adam

13:06

I have a question about the game of dots-and-boxes. Anyone can help?

Harry Evans

Good morning! I have a question on how to show this.

Let $f$ be analytic on a non-empty domain $D$. If $f^2(z) = \overline {f(z)}$ for all $z \in D$ then prove that $f^3$ is constant in $D$. Deduce that $f$ is also constant in $D$.

Simply Beautiful Art

13:20

@HarryEvans

0

Q: Proving that $f^3$ is constant

Good morning! I have a question on how to show this. Let $f$ be analytic on a non-empty domain $D$. If $f^2(z) = \overline {f(z)}$ for all $z \in D$ then prove that $f^3$ is constant in $D$. Deduce that $f$ is also constant in $D$. I'm thinking of proving by contradiction (i. e., $f^3$ is non-c...

Harry Evans

13:37

Thanks, @SimplyBeautifulArt!

Simply Beautiful Art

Lol, your welcome?

Harry Evans

I forgot that simple fact! That a real-valued analytic function on a connected set (in this case, a domain) is constant!

Studentmath

13:57

Say,
$\sum_{j=-\infty}^\infty sup_{i\in N} |a_{i,i+j}|$
Is the same as
$\sum_{j=1}^\infty sup_{i\in N} |a_{i,j}|$, right?

If we agree that $a_{ij}=0, j\leq 0$

BAYMAX

14:15

collection of Borel sets is a strict inclusion in the space of Lebesgue measurable sets?

How to prove this actually?

Alessandro Codenotti

14:35

the usual proof is to show that there are $2^{\aleph_0}$ Borel sets and $2^{2^{\aleph_0}}$ Lebesgue measurable sets

Daminark

14:55

There's a very easy way to do what@Alessandro is saying if you think about it right

Well it won't be exactly that but close

Alessandro Codenotti

hey @dami

how's measure theory going?

Daminark

It's pretty well, last class we did Dominated Convergence, and then we did about half of the proof of Fubini (only in $\Bbb R^n$)

BAYMAX

hi

Daminark

Hey @Baymax!

BAYMAX

yup,hey you know how to do that!

like how do I approach that

Daminark

15:04

Show that Lebesgue measurable sets strictly include the Borel sets?

BAYMAX

yes

Daminark

So do you know the proof that there are continuum-many Borel sets?

BAYMAX

no

I think the question says collection of borel sets $\subset$ space of lebesgue measurable sets

Daminark

Well we know non-strict inclusion easily

Every open interval $(a,b)$ is Lebesgue measurable

Astyx

Hi chat

BAYMAX

15:06

yes

Daminark

Every open set is a countable union of intervals, and is thus Lebesgue measurable

But if you contain the open sets, you contain the sigma-algebra they generate (Borel sets)

Hey @Astyx!

Astyx

How are you ?

BAYMAX

sounds hard a bit

Daminark

Now, the idea behind showing that this inclusion is strict is to show that there are at most continuum-many Borel sets, but you know that any subset of a measure zero set is measurable. So, every subset of the Cantor set is measurable, but there are more than continuum-many of those

BAYMAX

borel sets are those that can be formed of open intervals

Daminark

15:09

Doing alright, how about you?

Astyx

I'm good

Nearly done with the exams of the schools that interrest me

Daminark

Ah, nice! How'd they go?

BAYMAX

hmm that seems little hard

Astyx

I don't feel they went great, but I guess I'll just wait for the admissibility results instead of relying on my impressions :p

Daminark

Oh yeah that's not easy. Not sure if that's what they're asking you to do or not. The symbol $\subset$ registers in my mind as not necessarily being strict inclusion, just inclusion

And that's a good plan @Astyx, no need for unnecessary worrying

BAYMAX

15:12

yes @Daminark that is right , strict inclusion

Astyx

Exactly my point of view

Daminark

What do you use for generic inclusion?

Studentmath

Ugh, I thought I managed it with some play with series, looked like it was going right but then it just got messy and pointless

BAYMAX

strict inclusion $\subset$

proper inclusion

Daminark

No I mean, if your inclusion is not necessarily strict/proper, equality is allowed, what do you use?

Studentmath

15:14

$\subseteq$

BAYMAX

$\subseteq$

Daminark

Oh, weird

Though I guess that makes sense

Studentmath

You use $\subset$ and $\subsetneq$, right?

Astyx

I tend to use $\varsubsetneq$ to mean $\subset$

Yeah

Daminark

Yeah

Studentmath

15:16

Yeah..

Daminark

I mean, in reality I don't often talk about strict inclusion too much. Like I rarely find myself specifying that, but that's how I do it

Studentmath

I wonder if it is country kind of notation, or university..

Daminark

I'd say it's probably closer to random distribution

Studentmath

I think I've seen both here, but then again almost everything is translated\brought in the English form
Probably

But $\subset , \subseteq$ is definitely more common here

Balarka Sen

hi chat

Astyx

15:18

Hi Balarka

Daminark

Hey @Balarka!

Studentmath

\o @Balarka

Hm, if $\sum_{j=-\infty}^\infty sup_{i\in N} |a_{i,i+j}|$ converges (and say $a_{ij}=0$ if $j\leq 0$)
Can I say:
$\sum_{i=1}^\infty \sum_{j=1}^\infty |a_{ij}|^2$
Is $\leq$ then the square of the above sum?
If I can, I can't see why..

But if I can, that would probably solve my question

Never mind, I am probably over-complicating the question

Alessandro Codenotti

Hi @Balarka

Balarka Sen

what're y'all upto

Studentmath

Trying to solve some simple linear operators question in $l^2$, failing
What's with you?

Balarka Sen

15:31

I feel ya. I was out for a while, now back and should start studying various things.

Studentmath

Anything specific?

Balarka Sen

Topology, as usual :) Studying things called foliations for now.

Astyx

Taking exams for my part

Secret

Doing my PhD chemistry stuff, plan to have jobs running before I zip back quickly to ordinal collapsing functions

Ayoub Rossi

Solving differential equations...

Studentmath

15:37

Well, that all sounds very successful :P

Daminark

Trying to find whether the typical continuous function sends a first category set to
1) A first category set
2) A measure zero set

Alessandro Codenotti

my guess is neither

Daminark

I say that first category is probably more likely

I'm mostly sure that 2 is false

Like, you can have a residual set of measure zero

Take a sequence fat Cantor sets of measure tending to 1, and then take their union. It will be first category set of full measure in $[0,1]$

So like, measure and category don't seem to correspond all too well

So I don't see a reason to believe that 2 is true, even the identity doesn't satisfy that, you know?

Part 1 is more likely

(Actually I'm completely wrong, turns out it's true)

Alessandro Codenotti

wait what are you talking about now? 2 is false

Daminark

That a typical continuous function sends a first category set to a measure zero set

Apparently that's actually true

Alessandro Codenotti

15:46

what does "typical" mean?

Daminark

Residual in the space of continuous functions

Studentmath

Ugh, think I'm gonna have to give up and try again next weekend..

Studentmath

16:01

Well, sadly that's it - g'night and good week all!

Simply Beautiful Art

@Studentmath Good night!

@Secret lol

Secret

Yeah I just read a very lengthy article on how to do NBO calculations (basically, calculating the maximally occupied orbitals in a molecule to work out the charge distribution within the molecule)

I am currently setting up the calculations

BAYMAX

16:19

Let us discuss Reisz Fischer Theorem

?

may be @Daminark @AlessandroCodenotti

Alessandro Codenotti

Riesz-Fischer*

I'm supposed to know it, but i don't really do, what's bugging you?

BAYMAX

I mean let us discuss from scratch

step by step

Daminark

Is that the $L^2$ thing?

BAYMAX

Let $(f_{n})$ , $n \geq 1$ be a cauchy sequence in $L_{p}(X)$

yup in a general setting its $p$ here@Daminark

Daminark

Note that we haven't reached $L^p$ spaces yet

BAYMAX

16:28

ok

Daminark

So you'd be walking me through it :P

BAYMAX

:) may be @AlessandroCodenotti can help me and you too.

now we have to prove that every cauchy sequence converges in $L_{p}(X)$

Daminark

Oh so it's showing that $L^p$ is complete

BAYMAX

Yes

that is the theorem

so if we can show that there is a subsequence which converges then the sequence too converges and we will be able to prove the statement

so let us construct a subsequence

Alessandro Codenotti

@Daminark Take $(X,\mathcal{A},\mu)$ a measure space and let $f:X\to\Bbb R$ be a measurable function. We define a norm $||f||_p=(\int_X |f|^pd\mu)^{\frac1p}$ and then put all of the functions for which this "norm" is finite in a space of functions

Do you note anything wrong with this "norm"?

Daminark

16:33

In the $\ell^p$ case you need that $1 \le p$, otherwise it breaks, which probably is true here as well

BAYMAX

yes

what can be wrong when $p<1$?

I heard triangle rule

need not hold but how ?

Daminark

I think that's what dies

Intuitively it makes sense for just $\ell^p$

BAYMAX

yes but how can we show that?

Alessandro Codenotti

yeah, $p$ is bigger than $1$ (can also be $\infty$, but that's slightly different)

Daminark

The unit ball in a normed space has to be convex

So if you drag $p < 1$ you lose that

Maks

16:34

Hey !

BAYMAX

any light explanation,that is little tough though

Daminark

I mean that's the proof by picture :P

BAYMAX

like suppose it holds that is let $1<p$

then can we arrive at some contradiction of trangle rule

oh

Alessandro Codenotti

anyway that norm doesn't work because if you pick a function which is $0$ $\mu$-almost everywhere it's $p$ norm will be $0$, so it's not an actual norm

BAYMAX

$0 \mu$

means measure zero functions right@AlessandroCodenotti

?

Alessandro Codenotti

16:37

We solve this problem by setting $L^p=X/\sim$ with $X=\{f:X\to\Bbb R|\quad ||f||_p<\infty\}$ and $\sim$ is the equivalence relation $f\sim g$ iff $f=g$ $\mu$-almost everywhere

Maks

I have a question
I'm supposed to graph $f(x) = 4sin(x) + 1 - x$
$f(x) = 4sin(x) + 1$ was easy, but how am I supposed to take $-x$ ??
It changes all the graph, is there a rule that says what subtracting the variable from the function does to the graph?

Daminark

Oh! Lol I didn't see the even bigger problem lol. OK this makes sense

BAYMAX

yes@AlessandroCodenotti

how can we construct a subsequence that converges then?

step1

Choose $n_{1} < n_{2} < n_{3} <...<n_{k}<...$ such that $\forall k$,$||f_{n_{k+1}}-f_{n_{k}}||_{p} < 2^{-k}$

Gunelle

hi

BAYMAX

hi@Gunelle

Kirill

16:44

Hello! What does it mean to say that the product $a_kb_k$ is small, when $a_k^2$ and $b_k^2$ are small?

small in what sense?

pilko

in the sense of converging to zero I suppose

you have $0\le ab\le (a^2+b^2)/2$ too.

sorry in absolute value.

Kirill

does it imply the existence of some $C$?

pilko

uh. what is the context ?

Gunelle

Anyone working on graph theory?

pilko

ask your question, if someone is a graph theorist he will answer, maybe even when you are gone looking at the chat history.

Gunelle

16:52

Alright, thanks.

Kirill

@pilko the context is Cauchy-Schwarz. In your example $C$ seems to be $\frac{1}{2}$. It is from the book: being given that $\sum_{k=1}^{\infty}a_k^2 < \infty$ and $\sum_{k=1}^{\infty}b_k^2<\infty$ show that $\sum_{k=1}^{\infty}\mid a_kb_k \mid < \infty$.

I was just confused with the words that the product is small.

but in sense of converging to zero (if we build some sequence) it makes sense

Gunelle

It's known that circular interval graphs (proper circular arc graphs) aren't commonly bipartite; but in the case of even cycles or disjoint unions of paths they are. A paper says that a bipartitioned circular interval graph could be obtained from any circular interval graph by partitioning its vertex set and deleting every edge with both endvertices in the same class.
Does anyone have a clue on how is the partitioning done?

Kirill

@Gunelle no, but I propose the existence of some other book that explains the way it should be done

pilko

@Kirill yes, $(b_k-a_k)^2\ge 0$ allows to show $a_kb_k\to 0$ directly but you really need cauchy-schwarz for the serie.

Kirill

@pilko ok, thank you. I was naivly following the text without the idea in backmind to think about the sequence :) Or serie, depends.

Gunelle

17:12

@Kirill I've been looking for a while actually. But thanks anyway!

Ted Shifrin

@Maks: How do you use calculus to help with graphing?

hi @Kirill ! Hope all's well.

BAYMAX

Hi @TedShifrin

Actually I was thinking like how triangle inequality is violated in the case of $L^{p}$ spaces when $p<1$ ?

like it violates a property of norm

daminark suggsted to consider a convex ball

Danu

17:29

Hi @Ted

You made me waste a bunch of hours with that question, and I didn't even completely answer it in the end :P

Kirill

@TedShifrin Hello Mr. Shifrin! In some sense it is. I have failed the exam, but it does not matter.

BAYMAX

@Daminark you know some reference where the case $p<1$ is dealt with?

Maks

@TedShifrin ... I'm ashamed to answer I dont know

Daminark

I don't know any particular references, probably Googling it will work?

@Baymax

Also the canonical textbooks may very well have it, like Folland, Stein-Shakarchi, Rudin, etc

BAYMAX

ok

Daminark

17:45

Anyway, I gotta head out and do my pset for analysis, and then study for a midterm, so see you around!

Kirill

by the way, what does it mean to say that an infinite sum ist smaller than infinity?

Secret

it means the infinite sum is bounded

Kirill

@Secret $\sum_{k=1}^{\infty}a_k^2<\infty$. If this one is bounded, then there is an upper and lower limit. Ok, we have the zero donwstairs, but we have no limit on the top. Why is it bounded then?

Secret

Is the positive terms of $a_k^2$ something of the form $\frac{1}{k^p}$ where $p>1$? If yes, then by p series test the series will converge

arctic tern

18:00

What you just said is "This sum is bounded. So there is an upper and lower limit. We have zero downstairs, but no limit on top. Why is it bounded then?" It's bounded because we assumed it's bounded.

Ted Shifrin

@Maks: Think about where $f'=0$, $f'>0$, $f'<0$, sign of $f''$. Didn't you do this kind of thing in class?

SBM

Oh, increasing functions, really?

Ted Shifrin

@Kirill: Sorry to hear that. If there are specific things I can help with, let me know.

Kirill

@arctictern you mean, this notation $<\infty$ stands for the existence of some limit?

Ted Shifrin

@Danu: Time is never wasted when you're learning something :P

Maks

18:02

@TedShifrin Uh, that's a good idea

Ted Shifrin

glares at Maks :D

SBM

@TedShifrin true

Kirill

@TedShifrin oh, thank you!

Maks

@TedShifrin <.<

@TedShifrin from 1 to 10, how familiar are you with numerical analysis ?

From 1 to Ted*

Sergey Dylda

Hello, people. Is there anyone famillial with exterior algebras and differential forms?

Maks

18:07

From Maks* to Ted*

Balarka Sen

Hi @Ted

Ted Shifrin

Not familiar much, @Maks. I've never studied it or taught much (other than an occasional tidbit).

Yes, @Sergey.

heya @Balarka

Sergey Dylda

@TedShifrin Could you please look at this question of mine. It either I don't understand something or there is some trick with it.
https://math.stackexchange.com/questions/2246837/exterior-algebra-and-differential-forms-as-equivalence-classes

Ted Shifrin

The comment that is now there is the right way to think about it, @Sergey. The algebraists' way of thinking of the quotient space isn't convenient. Indeed, we want to think of the standard way of embedding $\Lambda^k V^*\hookrightarrow \otimes^k V^*$.

Alessandro Codenotti

Hi @Ted

Ted Shifrin

18:14

Hi @Alessandro

Vrouvrou

Hello, i have for $\Omega$ open we define $$f(x)=d(x,\partial\Omega)$$ and $$d'(x,y)=\sup\left(d(x,y),\left|\dfrac{1}{f(x)}-\dfrac{1}{f(y)}\right|\right)$$ where $d$ is a distance , how to prove that if $d'(x,y)=0\Rightarrow x=y$

Sergey Dylda

@TedShifrin What do you mean by saying "right way"? I prefer to think of it as formal as I can. What step of my construction is wrong?

Balarka Sen

@Ted Seems I have to learn single variable calculus for high school.

Ted Shifrin

@Sergey: Formal isn't always best. But working with equivalence classes is not good for working with differential forms. You need (locally) to choose a good smooth representative, and the way to do that is by choosing a canonical skew-symmetric tensor, as the comment suggests. What is your problem with that?

Poor baby, @Balarka :D

Vrouvrou

someone have an idea please?

Ted Shifrin

18:18

To be honest, @Sergey, I'm not going to read everything you wrote.

Bottom line is that to work with differential forms on an open set of a manifold or open set of $\Bbb R^n$ you need a globally "smooth" way of choosing representatives of those equivalence classes.

Sergey Dylda

@TedShifrin But even if we use such an injective map with choosing a representative, the result will be in T(V) and thus will be an infinite sequence. How is that a tensor?

Vrouvrou

i have for $\Omega$ open we define $$f(x)=d(x,\partial\Omega)$$ and $$d'(x,y)=\sup\left(d(x,y),\left|\dfrac{1}{f(x)}-\dfrac{1}{f(y)}\right|\right)$$ where $d$ is a distance , how to prove that if $d'(x,y)=0\Rightarrow x=y$

If i suppose that $d'(x,y)=0$ then if $d'(x,y)=d(x,y)$ then as $d$ is a distance we deduce that $x=y$
but if $d'(x,y)=\left|\dfrac{1}{f(x)}-\dfrac{1}{f(y)}\right|=0$ then how to deduce that $x=y$ please ?

Ted Shifrin

I don't have any idea where you're getting infinite sequences. Once you pick a particular $k$-tensor, it can be built only out of $\ell$-tensors for $0\le \ell\le k$. And, in fact, we prefer to write the basis $k$-alternating tensors as built out of wedge products of $1$-tensors.

I think you need to sit down and play very concretely with tensors and alternating tensors, @Sergey, and get some computational skills with them.

Sergey Dylda

@TedShifrin T(V) is construced as infinite direct sum, and by definition elements of direct sum is the elements of cartesian product, thus elements of T(V) are inifinite tuples.

Ted Shifrin

You're doing the entire tensor algebra. Stop and look at stuff of degree $1$, $2$, $3$, individually.

Actually sit down and figure out what's going on for a $2$-dimensional real vector space $V$, concretely. And understand why, for example, $\Lambda^3 V = 0$ and $\Lambda^2 V \cong\Bbb R$.

Secret

18:26

Hey Ted, I have a topology question: Is there a "Short Line"?

9 hours ago, by Secret

I am not sure, recall that a long line is $[0,1]\times \omega$, thus it seemed longer than the real line due to the topology effectively ballooning up each point in the interval into a countable set.

Therefore, a short line will be something like $[0,1]\times ?$ where ? will be some number or set such that the resulting line is still an uncountable set, but "shorter" because e.g. infinite many points in the interval [0,1] get identified into one single point still within the interval

(NB $\omega$ should be $\omega_1$)

Ted Shifrin

Actually, @Secret, your definition of the long line isn't right.

But I have no idea what you mean by short. What's shorter than $[0,1)$?

Other than a point.

And, actually, you're not thinking of the long line correctly, IMHO. You get an interval $[0,1)$ for each element of $\omega_1$; think of laying them down horizontally, and consecutively, thereby forming a (long) line.

Back later.

Sergey Dylda

@TedShifrin Sorry, I prefert to think of it as $T(V)$ -> $\Lambda(V)$ -> $\Lambda^k(V)$ and not the other way around.

Kirill

What does it mean, to normalize a sequence? Any sources where I can read about it?

Secret

sorry you are right, I misremebered by swapping the roles of [0,1) and $\omega_1$ in your setence in my original statement

I suspect a short line will be like laying down suitable number of intervals and identifying the points within the interval somehow so that it takes shorter time to move from [0,1) compared to the orignal [0,1) interval. Not sure how that can be done though

If this is a countable set and not an interval, then the idea might be like instead of each step takes you through 0,1,2,3,... you instead go 0,2,4,6,8,... or larger leaps

But since an interval is uncountable, I am not sure if there is a good way to phrase it other than it takes less time to walk from 0 to 1 for this interval after being treated by some equivalence class operation

Balarka Sen

@SergeyDylda I am not entirely sure what your question is. The comment below writes down an injection $\Lambda^k V \to T^k V$.

That gives you a way to represent differential forms as tensors, yeah?

Sergey Dylda

18:40

@BalarkaSen I already figured it out. I was confused about the injective map $\Lambda(V) \to T(V)$ on entire tensor algebra. However, if that so, we still define forms as equivalence classes or as images of classes under injective map?

Balarka Sen

We can define them as both equivalently.

It's more conventional to define them as the usual quotient of tensors (on the tangent bundle), however.

Sergey Dylda

@BalarkaSen Is there any proof that this injective map is unique? I could possibly pick up any representative from each class in many ways. Is this restricter by construction itself?

restricted

Balarka Sen

Also, I don't see what's to be confused about the injective map on the entire tensor algebra. What the comment wrote is an injection for any grade $k$, and makes up an injective map on the entire algebra.

Astyx

Hi again chat

Balarka Sen

@SergeyDylda Unique is not the right word; you mean well-defined. You can check that by hand.

Sergey Dylda

18:46

@BalarkaSen, Because image of injective map will be infinite tuple, and to extract tensor structure of k-orded you would need to project that tuple by $proj_k$.

Balarka Sen

Image of injective map is a subalgebra, not an "infinite tuple".

Sergey Dylda

@BalarkaSen I mean subalgebra of elements of T(V) , so subalgebra of sequences

Mike Miller

You think too formally about what a direct sum is.

Better, I think, to think T(V) as finite sums of tensors

Balarka Sen

Right, you shouldn't think of direct sum of vector spaces as a space of sequences/tuples, but as a space of linear combinations.

But I see what you mean now.

I don't see why that's a problem, though.

Danu

@TedShifrin ...riiiight :P

Sergey Dylda

18:50

@MikeMiller Thank you, but no. What do you mean by "adding" tensors form different vector spaces?

Balarka Sen

Formal linear combination.

Sergey Dylda

@BalarkaSen Then we would need to introduce polynomial ring R[X] of formal combinaions as free-module, that would complicate things.

Balarka Sen

No you don't. Saying formal linear combinations requires exactly the same effort as saying tuples, because they are one and the same thing.

Sergey Dylda

@BalarkaSen No, tuple is just a special case of set, constructed from ordered pairs. And formal linear combination is either an element of free-module or expression of symbols in language of model/type-theory

Balarka Sen

Ok. It seems answering your question is beyond the scope of me or any other person in this chat. here is a better place to look

Sergey Dylda

18:58

@BalarkaSen Thank you, I'll look at it.

Kirill

@Gunelle I have some from Ruohonen, Liu, Busacker, Berg, Wilson, Distel, Tatt, Ore, Harary. Maybe your point can be mentioned in some of them.

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$Mathematics$ Mathematics