Mathematics - 2018-11-13 (page 3 of 6)

Tobias Kildetoft

12:01 PM

@AsafKaragila So which one was the suspected one? That all surjections have sections?

Asaf Karagila

@TobiasKildetoft No, that is equivalent to AC.

Tobias Kildetoft

ahh, ok.

Asaf Karagila

It's easy to see why: if you have a family of pairwise disjoint non-empty sets, this is a partition of the union of the family, so there is a natural surjection, and a section of that natural surjection is a choice function.

But if you just claim "Oh there is a surjection from X onto Y, so there is an injection from Y into X", well, you've got nothing to work with. So all you can really do is prove things like well-ordered choice, which in turn is enough to prove DC, and the existence of non-measurable sets, but we know that it cannot even prove DC_{\aleph_1}. So we cannot do much more than this with current technology.

Tobias Kildetoft

@AsafKaragila Well, now you lost me in technical set theoretical stuff. But I do see what you mean in spirit (nothing to work with)

Alec

12:24 PM

A

Hello users! Can anybody explain this question?

s.harp

@Alec what is your math background?

Alec

I have completed high school last year. Right now I am doing bachelors in Physics. My mathematics background is up to the level of high school plus basics of vector calculus (gradient theorem, curl theorem,etc).

Gaurang Tandon

Hey all, what does the notation $\langle (13)\rangle$ mean in group theory? Like, I saw it in this answer. I do know that $(13)$ refers to the identity permutation, but those angle brackets confuse me. What do they refer to?

Similarly I am confused by $\langle (1,2,3)\rangle$.. :(

Tobias Kildetoft

12:40 PM

@GaurangTandon $\langle\cdot\rangle$ refers to the subgroup generated by the given element

But $(13)$ is not the identity permutation

Gaurang Tandon

@TobiasKildetoft yes, I know that bit. but how is $(1,2,3)$ generating a subgroup?

s.harp

@Alec I will start with an analogy. Consider a function on the interval $[0,1]$. It is clear that any continuous function has an anti-derivative, meaning something where if you take the derivative you get the original function. Now identify the points $0$ and $1$ to get a circle, if you want a function to be continuous on the circle you need the value of the function on $0$ and the value of the function on $1$ to be the same.

Tobias Kildetoft

@GaurangTandon The same way any element generates a subgroup

s.harp

You see that continuius functions on the circle are basically periodic continuous functions on $[0,1]$. Here the question: Does every periodic function have a perdiodic anti-derivative? Has a negative answer. Specifically the constant function $1$ does not have any anti-derivative

Semiclassical

{e,(123),(123)(123),...}

Gaurang Tandon

12:43 PM

@TobiasKildetoft In Gallian they said under permutation groups chapter that: "Of course, the identity permutation consists only of cycles with one entry, so we cannot omit all of these! In this case, one usually writes just one cycle. (The identity permutation) can be written as $\epsilon =(5)$ or $\epsilon = (1)$."

s.harp

There is a condition for a function have an anti-derivative, namely $\int_0^1 f(x)dx = 0$. By changing the space (in a small fashion) we suddenly find that we cannot always find anti-derivatives.

Your question is basically the same question, you are looking for anti-derivatives via curls. One obstruction you already know, the divergence has to be zero for an anti-derivative to exist.

But there are more.

Gaurang Tandon

@Semiclassical oh right, good point. i forgot i can compose cycle notation permutation groups

s.harp

Specifically on $\Bbb R^3$ or convex subsets of $\Bbb R^3$, the obstruction with $\nabla \vec{f} = 0$ is the only one, but for other sets you have problems.

Gaurang Tandon

but won't $(12)$ and $(13)$ be the same identity permutation group? So, both their orders should be three. So, $HK$ should also be identity permutation and thus order of $HK$ should also be 3.

Semiclassical

Another way to put the point: having gradient zero is sufficient to have a “local” antiderivative (eg valid on a convex subset of R^3)

Gaurang Tandon

12:49 PM

what's wrong in this?

Semiclassical

(13) is not the identity permutation

Nor is (12)

Tobias Kildetoft

@GaurangTandon Where did "order 3" come from here?

Gaurang Tandon

@Semiclassical then what about that statement from gallian I referenced to above? :(

Tobias Kildetoft

@GaurangTandon is the 13 in the parentheses meant to be the number thirteen?

Semiclassical

What about it? (1) is the identity permutation, anf so is (2)

( 13 ) would be, but not ( 1 3 )

Gaurang Tandon

12:53 PM

@Semiclassical ohhhhhh

omg

i was confusing that for so long

i was wondering where he got 12 and 13 from all the time....

Tobias Kildetoft

I still don't see where you got "order 3" from, even given this confusion

Gaurang Tandon

@TobiasKildetoft sorry, i think it was a mistake. i just saw the number 3 in $S_3$ and thought it's also the order

Semiclassical

Typically n is small enough that (13) can only be interpreted as the 1<->3 transposition

Alec

@s.harp thank you very much for the detailed information. I got your point. If you know electrodynamics basics, can you please explain this question. There are two "good" answers but they doesn't clear my confusion.

mercio

sohuldn't the identity permutation be written ()

Gaurang Tandon

12:56 PM

@Semiclassical yes...thanks

Alec

This question

Gaurang Tandon

@mercio it should imho but gallian chose to describe it by a single element. dunno why :/

Semiclassical

my own preference is just e

Tobias Kildetoft

I am with @Semiclassical here

Semiclassical

That said, I don’t think that convention originated with Gallian

s.harp

1:04 PM

@Alec here $B$ is defined on $\Bbb R^3 - \text{wire}$, the domain of the function is not all of $\Bbb R^3$. $\nabla B$ is here the function on $\Bbb R^3-\text{wire}$ that only takes on the value $0$. Because the wire is not part of your space, the condition that B has zero divergence is true. The conclusion that it has a vector-potential need not true, there is a possibility for an obstruction here because the domain is not convex.

There are ways to extend the notion of divergence to include the singular behaviour of what $B$ is doing on the wire. But this involves some more mathematics. If you did that you could formulate your demand $\nabla B=0$ in a way that would involve the things going on on that wire.

Semiclassical

Hmm, now I find myself curious about the history. The two-line notation seems to be attributed to Cauchy

But I haven’t found a source for the cycle notation

Adam

I want to know what the word used for $\mathbb G$ please.

$${\{a_{i,j}}\}\subset \mathbb G= {\{i+j,\Bigl\lfloor \frac{1}{2}\bigl\lfloor \frac{i}{j}\bigr\rfloor\Bigr\rfloor+1}\}$$

$${\{a_{i,j}}\} \not\subset \mathbb G ={\{i^{\, j/2}j,i^{\,j},j^\,i}\}$$

$$a_{i,j}\in \mathbb G \operatorname{iff} \sum_{j=1}^{N\,!}\operatorname{sgn}(\sigma_j )\prod^N_{i=1}a_{i,\sigma_{j,i} }=0$$

Where

where $\sigma_j$ is the $j^{th}$ permutation of ${\{1,2,3,...,N}\}$

$\sigma_{j,k}$ is the $k^{th}$ entry of the $j^{th}$ permutation of ${\{1,2,3,...,N}\}$

Tobias Kildetoft

@Adam That seems like mostly nonsense to me

s.harp

use words to describe what you are looking at, not a jumble of symbols that are meaningless if you dont already know the thing

Adam

1:20 PM

which symbols are you unfamiliar with the meaning of?

s.harp

your \lfoor, \rfloor , what i,j are supposed to be in the very first line etc

Alec

@s.harp: "Because the wire is not part of your space" Why? The wire lies in $\Bbb R^3$ and hence is a part of the space. Am I wrong?

Semiclassical

What’s the magnetic field at the wire itself?

Secret

Listening to my recordings God I hate my voice

Adam

ok ${\{i,j}\} \subset \mathbb N$

Semiclassical

1:22 PM

(Or, more correctly: what happens to the magnetic field of a current-carrying wire as you approach the wire?)

s.harp

@Alec because the mathematical formalism of vector calculus cannot handle the wire. You need to upgrade the mathematics to do that. If you want to stay in vector calculus you need remove the wire from your domain.

@Adam so $\Bbb G = \{ i+j \mid i,j\in\Bbb N\}\cup\{ \text{mess with floors}(i,j)\mid i,j\in\Bbb N\}$? if you do that you get $\Bbb G = \Bbb N$

Semiclassical

Isn’t the third condition for GG just the Leibniz formula for the determinant of $(a_{ij})$?

Though in that case it makes little sense to say that that’s equivalent to $a_{i,j}\in \mathbb{G}$. I could see $(a_{i,j})\in \mathbb{G}$ tho

Alec

I see. But you stated above that "There is a condition for a function have an anti-derivative, namely $\int_0^1 f(x)dx = 0$. By changing the space (in a small fashion) we suddenly find that we cannot always find anti-derivatives." if we remove the wire, would'nt we be changing our space $\Bbb R^3$?

Semiclassical

(That a determinant vanishes is a claim about the matrix itself, not about whether each matrix element lies in a specific set.)

s.harp

Indeed, on $\Bbb R^3$ you will always find a vector-potential (anti-derivative) to your divergence free field. However on $\Bbb R^3$ minus the wire it is not always possible to find a vector-potential for every divergence free field. It might be possible for the magnetic field corresponding to the current, but that depends on the current.

There is a furhter distinction you will see in your physics classes that leads to gauge theory. It is that locally you will always find the potentials, but you will not be able to piece them together to a global potential.

That message is a bit cryptic if you are not sufficiently familiar with the field though.

Semiclassical

1:37 PM

it should also be noted that we're talking about the case of a line current, i.e. all the current is contained in a wire of zero thickness

if you take the wire to have finite thickness, then you need to replace the line current with a volume current $J=I/A$

in that case you genuinely can write down the magnetic field everywhere in space, including inside the wire itself

s.harp

yes, in that case everything works "as expected" wrt vector-calculus

Semiclassical

yeah

and that is a pretty typical way of making sense of pathological cases in physics: replace an ideal case with something less idealized

that doesn't always get rid of problems, but it's a good option

Alec

@Semiclassical One of my little misunderstandings: The line current has zero thickness. Even so, wouldn't our domain be $\Bbb R^3$ minus wire. And so wouldn't we have changed our space (in a small fashion)? and hence we cannot always find anti-derivatives?

Semiclassical

1:53 PM

that sounds right?

Alec

@s.harp: what is your opinion?

Adam

@s.harp you didn't read the predicate for membership $$a_{i,j}\in \mathbb G \operatorname{iff} \sum_{j=1}^{N\,!}\operatorname{sgn}(\sigma_j )\prod^N_{i=1}a_{i,\sigma_{j,i} }=0$$

Where

where $\sigma_j$ is the $j^{th}$ permutation of ${\{1,2,3,...,N}\}$

$\sigma_{j,k}$ is the $k^{th}$ entry of the $j^{th}$ permutation of ${\{1,2,3,...,N}\}$

$$\operatorname{sgn}(\sigma_j)=\cases{\,\,\,\,\,1&$\operatorname{the number of concatenations needed to map}\,\,{\{1,2,3,...,N}\}\,\, \operatorname{to}\,\,\sigma_j \,\,\operatorname{is even}$\cr -1&$\operatorname{the number of concatenations needed to map}\,\,{\{1,2,3,...,N}\}\,\, \operatorname{to}\,\,\sigma_j \,\,\operatorname{is odd}$\cr}$$

Semiclassical

1) I think you meant to address that to me

2) the sign of a permutation counts the parity of of the number of transpositions needed, not concatenations

Adam

yes equivalently you can state that any square matrix formed from a member of the group as the index of the entries in the matrix, it will have a determinant of 0

Semiclassical

Sure. But that would not be $a_{i,j}\in \mathbb{G}$.

$\{a_{i,j}\}_{1\leq i,j,\leq N}\in \mathbb{G}$, sure

Adam

2:04 PM

sorry wrong word yes you are right, it just meant "a change of one element's position and another"

I* just meant

Alec

Any comments on my post??? : "The line current has zero thickness. Even so, wouldn't our domain be $\Bbb R^3$ minus wire. And so wouldn't we have changed our space (in a small fashion)? and hence we cannot always find anti-derivatives?"

Semiclassical

regardless, a matrix with determinant zero is said to be singular

So I'd say $\mathbb{G}$ is the set of singular N-by-N matrices

(I mean, you still need to say what kinds of numbers the $a_{i,j}$ are allowed to be. Do you consider singular matrices with integer-valued entries? real-valued? complex-valued?)

Adam

ok but I want to know what you refer to $m_{i,j}$ as

an arithmetic function that when defining the (i,j) th entry of a matrix , will generate a matrix in that group i mean

Semiclassical

What?

the notation I typically see is $M=(m_{i,j})$

So $M$ is the matrix, and $m_{i,j}$ is the $(i,j)$th matrix element

I also see $M_{ij}$, but this is probably a bit abusive

I guess your point is that you can interpret $m$ as a function with domain $\{1,2,\ldots,N\}^2$

But ehhh. That's not really different than just $M$ itself

s.harp

@Alec I was away for a while. Yes our domain is $\Bbb R^3$ minus wire. This removes from us the guarantee of finding a vector-potential for $B$.

However, I believe you will still be able to find a vector potential for any line current, this potential needs to be defined on $\Bbb R^3$ minus the wire also.

Semiclassical

2:14 PM

yeah. it's the difference between "any arbitrary magnetic field" and "the magnetic field generated by a line current"

now would be a good time to give an example, of course

but I'm coming up dry

s.harp

@Semiclassical replacing hte wire with a wire of thickness $\epsilon$ give sthe same magnetic field outside the wire

Semiclassical

point

s.harp

excuse me, I think that may not be entirely correct

Semiclassical

no, I think it's right

s.harp

well, you get a potential otuside for arbitrary thickness

Semiclassical

2:20 PM

outside the thick wire, you're still able to apply ampere's law

Secret

how about parellel wires with current flowing in same direction as an example?

Semiclassical

I mean, the potential will certainly fail to exist at the wire itself

Alec

"However, I believe you will still be able to find a vector potential for any line current, this potential needs to be defined on $\Bbb R^3$ minus the wire also." Can we do that with just vector calculus?

Semiclassical

yeah, and the construction being given should suffice

replace the infinitely-thin wire with one of thickness $\epsilon$ and appropriate volume current. the integral formula for the vector potential should then work

then take the limit $\epsilon\to 0$ to get a vector potential which is valid away from the wire

(There's probably some subtlety here as far as how arbitrary of curves are allowed, though)

Mike Miller

@LeakyNun He'll be back, just with a different name.

Leaky Nun

2:24 PM

@MikeMiller so where's his good question 1 hour later

Semiclassical

@s.harp I guess the obvious question here is: on the domain R^3 with the z-axis deleted, what's an example of a solenoidal field which has no global vector potential?

Secret

sorry my comment has lagged due to slow internet

Leaky Nun

@Semiclassical is this related to de Rham cohomology

s.harp

@semiclassical $(x,y,0)/(x^2+y^2)$

Mike Miller

Oh @LeakyNun I forgot about that

s.harp

2:33 PM

@LeakyNun yes

Mike Miller

The underlying question was: Banach algebras are ubiquitous, but where are the Hilbert algebras?

Alec

By introducing infinitely-thin wire, how can we escape from the problem of $\nabla \cdot \vec{B}$ inside the wire?

Mike Miller

I anticipated the answer was "There essentially aren't any"

Semiclassical

@s.harp seems legit

Mike Miller

I was wrong

Though observe that this fella is not complete under the inner product...

It just has certain Hilberty behavior

Semiclassical

2:39 PM

Pseudo-Hilbert, I guess

Mike Miller

I strongly prefer Hilberty

Semiclassical

Lol

So what would a Hilberty space be by comparison to a Hardy space?

(The difference is that only one of them is a setup for a bad pun)

Secret

@s.harp Is that the one where the vector potential should be at z=0 except that it is excluded from the space thus you cannot find it there?

Hardy space

In complex analysis, the Hardy spaces (or Hardy classes) Hp are certain spaces of holomorphic functions on the unit disk or upper half plane. They were introduced by Frigyes Riesz (Riesz 1923), who named them after G. H. Hardy, because of the paper (Hardy 1915). In real analysis Hardy spaces are certain spaces of distributions on the real line, which are (in the sense of distributions) boundary values of the holomorphic functions of the complex Hardy spaces, and are related to the Lp spaces of functional analysis. For 1 ≤ p ≤ ∞ these real Hardy spaces Hp are certain subsets of Lp, while for p ...

wut?

Will Hunting

I prefer Hilbertian to Hilberty.

Mike Miller

@Secret The joke is that it's Hard with a y on the end

Will Hunting

2:53 PM

Geezis.

Mike Miller

You can think of the Hardy space of the circle as being either: the restriction to the boundary of functions which are holomorphic on the interior; or the set of L^2 functions whose Fourier series coefficients are nonzero only on the non-negative powers of $e^{ix}$.

Semiclassical

I encountered Hardy spaces while trying (and failing) to understand the Szego-Widom formula

Mike Miller

One needs to be more careful of course

Semiclassical

c.f. encyclopediaofmath.org/index.php/Szeg%C3%B6_limit_theorems

s.harp

I found this question interesting today: math.stackexchange.com/questions/2996580/…

Semiclassical

3:02 PM

@MikeMiller the linkage to Toeplitz operators is where I came across them

Mike Miller

@Semiclassical Right, this is the simplest case of an index theorem

Semiclassical

yeah

that's definitely related to what I was looking at

Will Hunting

3:38 PM

@Semiclassical cf. is short for confer in Latin which actually means compare with, but I think most people really want to use v. which is short for vide in Latin and means see.

en.oxforddictionaries.com/definition/cf.

en.oxforddictionaries.com/definition/vide

en.oxforddictionaries.com/definition/v

A history of how I came across cf. I browsed through Lang's math books and he uses cf. everywhere. That made me look it up, and I realised he didn't really mean compare with. He really meant see instead.

Haha, I can't put a link here that ends in a full stop!

That's why even though the first link I copied was correct, it goes to the wrong page.

Please add a full stop to the first link to see the correct link!

kthxbai

Anonymous

@WillHunting Using "cf." to mean "see" is pretty common these days. It should enter the Oxford dictionaries soon ;)

Anonymous

3:53 PM

C.f. you already have: collinsdictionary.com/dictionary/english/cf :P

Semiclassical

lol

cf = c-f = see for instance :P

(i know that's not how it actually works, but I half-believe that's what people think it means)

neraj

Problem - How to find a Homomorphic map in this case - Let S1 and S2 be two sets. Suppose that there exists a one-to-one
mapping J of S1 into S2 . Show that there exists an isomorphism of
A(S1) into A(S2), where A(S) means the set of all one-to-one mappings
of S onto itself.

Mike Miller

The Latin is not correct, true, but the point is to have clarity and conciseness. "cf" for see is a well known "abbreviation", and including something else would cause more confusing.

neraj

I thought of few but because mapping J is not onto i am facing difficulty

Semiclassical

my fall-back for questions about linguistics is always this SMBC: smbc-comics.com/comic/know-your-linguistic-philosophies

Rithaniel

4:04 PM

Perhaps it's a person saying "consider" with a lisp: "confider"

Semiclassical

works for me

Will Hunting

4:21 PM

@MikeMiller I have always wondered why they don't just use "see". It's only three letters and everyone understands it.

Mike Miller

Tradition, mostly.

No good argument for it (unlike against your v. ;) )

Will Hunting

I have never used any Latin abbreviations in my own writing. I don't like using eg and ie.

I prefer just writing for example and that is.

Anyway, lots of people confuse the two.

And I keep seeing these grammarly.com ads everywhere.

Semiclassical

@WillHunting same, ugh

though I can appreciate them more than certain other ads I frequently see

Will Hunting

Sure it's free to install and use, but I don't like it. In fact, the grammarly.com advertisement itself shows me some things that I already don't like about it.

lush

hi

Esha Manideep

4:26 PM

Can somebody help me with this. Thanks !

artofproblemsolving.com/community/c281671h1254644p6481902

Semiclassical

no, I don't want to watch PolicyEd 'educational videos' and I don't give a flying f*** about league of legends

Mike Miller

@lush mystical greetings

Semiclassical

The Abel conference in honor of Langlands starts here tomorrow. Looking forward to sitting back and letting the math-i-don’t-understand wash over me

lush

@MikeMiller thx lol

Mike Miller

absolutely strange traveler

@Semiclassical i just dont go to things i wont have a chance of understanding

Semiclassical

4:36 PM

Lol

There’s a nonzero probability I’ll get bored and not stay for all of them

lush

me?

Alec

@Semiclassical Please can we continue our discussion?

neraj

friends any idea in - How to find a Homomorphic map in this case - Let S1 and S2 be two sets. Suppose that there exists a one-to-one
mapping J of S1 into S2 . Show that there exists an isomorphism of
A(S1) into A(S2), where A(S) means the set of all one-to-one mappings
of S onto itself.

Nikola Tesla here !

Alec your profile image is cool

lush

4:55 PM

@neraj: what have you tried so far/ where are you stuck?

Gaurang Tandon

Hey all, in Gallian it's given that:

> For each positive integer $n$, the set $n\mathbb{Z} = \{0, \pm n, \pm 2n, \pm 3n, . . .\}$ is a subring of the integers $\mathbb{Z}$.

However, my question is that in Z we had the unity element 1, however, we don't have any unity element in nZ. How can then it be a subring?

Even on Wikipedia under subring test it is mentioned that "for any ring R, a subset of R is a subring if it is closed under multiplication and addition, and _contains the multiplicative and additive identity of R_ ."

Mike Miller

@lush just saying weird things

lush

@GaurangTandon There are different definitions of rings actively used. The most common one is "ring = commutative ring with 1". Other common definitions leave out the assumption that a ring is commutative and/or that it has a 1.

If you don't demand a ring to have a 1, nZ is a subring (in fact: every ideal defines a subring), if you want rings to have a 1 nZ will be a ring only for n = 1 (again: in this case an ideal will be a subring exactly if its the whole ring, e.g. the ideal generated by 1)

neraj

5:10 PM

@lush i am not able to find the homomorphic map .

I am not able to find the homomorphic map because J is not necessarily onto.If J was onto we have define a map in which each symbol in an element x belonging to A(S1) could be replaced by corresponding in S2 by using the map J.

Gaurang Tandon

@lush i guess i should stick to my textbook definitions then...thanks though :)

neraj

@lush can you give me some hint

lush

@neraj: you said J is 1-to-1, so bijective. Why isn't it onto then?

Mike Miller

@GaurangTandon When following that textbook, at least. But it is more common to say that a subring necessarily includes the same unit as the larger ring.

neraj

No one to one does not means bijective

one one and onto means bijective

Gaurang Tandon

5:13 PM

@MikeMiller ok

lush

ok so you just want an injection from S1 into S2

I'll go eat sth for the moment

neraj

@lush Homomorphic map

Do you know about it ?

lush

5:26 PM

@neraj: So you are really looking for an isomorphism (of groups) A(S1) -> A(S2)?

neraj

@lush You got it

How i proceed ?

lush

@neraj: are you sure 1-to-1 is only meant to mean injective? If it's injective only the claim is wrong

neraj

@lush Google it

Yes i am sure

s.harp

lol

neraj

@lush Claim is not wrong

It is question from Herstein

Anderson Felipe Viveiros

5:29 PM

Hello, guys. Do we know what a sequentially compact space lacks in order to be metrizable?

lush

@neraj: it's wrong though: Take S1 = emptyset. Then A(S1) = 0, the trivial group ('cause there is only one map, emptyset -> emptyset). The emptyset injects into any other set, so your statement would imply that A(S) is the zero group for any set S

Tobias Kildetoft

@neraj If there is just an injective map from one set to the other, you just get an injective homomorphism between the groups

s.harp

@neraj one to one in this case does refer to bijectivity, otherwise the statement is just wrong

Anderson Felipe Viveiros

We know that in metric spaces, sequentially compact is equivalent to compact, but this is not the question...

Mike Miller

This is the same guy who was using sockpuppets to support each other yesterday after mass-pinging people.

s.harp

5:30 PM

@Anderson there are a lot of things missing, for example HAusdorff, paracompact etc

Anderson Felipe Viveiros

I always end up finding this on Google, when I try the keywords sequentially compact, metric spaces...

Tobias Kildetoft

@MikeMiller Who?

Anderson Felipe Viveiros

Yes, exactly. @s.harp What would be these things that are missing?

Mike Miller

@TobiasKildetoft the one spamming questions they don't understand? :)

Tobias Kildetoft

@MikeMiller Could you narrow it down to less than 10 people?

Mike Miller

5:31 PM

the one doing that right now

I was just trying to say that perhaps commenting here isn't the most valuable use of time, but of course it is up to you

Tobias Kildetoft

@MikeMiller Thanks, I do appreciate the warning.

Mike Miller

There are relevant things on the starboard to the right though not much discussion of the sockpuppeting, but it becomes pretty clear if you look at the questions they're asking and way they talk when at least 2 are around

s.harp

@anderson i cannot reply right now but sequentially compact is not actually necessary for metric

many metric spaces are not sequentially compact

Mike Miller

@s.harp many compact metric spaces? I thought the two notions of compactness were the same in that case

Anderson Felipe Viveiros

Yes, because they are many metric spaces that are not compact, they are the same in that case

^^'

s.harp

5:35 PM

@mike in metric sequentially compact \iff compact

Mike Miller

right

I guess you're trying to understand sufficient conditions for sequential compactness ~ open cover compactness

s.harp

well i understood his question as : if a space is sequentially compact, waht else do i need for metric?

Mike Miller

i see, I think there is essentially nothing one can say about sufficient conditions there that don't just ultimately ignore the sequential compactness

If you just want sufficient conditions for sequential = open cover, then "first countable Hausdorff" suffices, iirc

but i don't remember why i think that and in partic don't have a proof offhand

Anderson Felipe Viveiros

Actually, I'm trying to show that a certain topological space is metrizable. This is space is sequentially compact (namely, it is the space of varifolds with uniformly bounded mass, on a compact riemannian manifold... if anybody knows something about)

s.harp

here is an old question about the compact iff seq. compact from me :D math.stackexchange.com/questions/2319734/…

(slightly related)

Mike Miller

5:39 PM

@AndersonFelipeViveiros When I want to show that something is metrizable, in practice I either use Urysohn or try to write down the metric.

There is the Bing metrization theorem but I have never found it useful, only interesting.

Anderson Felipe Viveiros

Varifolds are measures on the grassmannian bundle of the base manifold $M$... The convergence is the weak convergence and this yields a topology...

Mike Miller

@s.harp From the deleted answer I see my claim first countable hausdorff => "sequential cpt ~ open cover cpt" is false

@AndersonFelipeViveiros can you prove Hausdorff? how about stronger regularity properties, like "can separate any closed set from a point by open sets" or "can separate any two closed sets by open sets"?

neraj

Do @otherpeople also think same as @lush @TobiasKildetoft @s.harp That question means bijective .Read following - services.artofproblemsolving.com/…

How to find a Homomorphic map in this case - Let S1 and S2 be two sets. Suppose that there exists a one-to-one
mapping J of S1 into S2 . Show that there exists an isomorphism of
A(S1) into A(S2), where A(S) means the set of all one-to-one mappings
of S onto itself.

Mike Miller

I don't know much about varifolds, so I am unlikely to be able to actually produce a proof for you

Anderson Felipe Viveiros

No problem...

I just wanted to know if there is a kind of known result about sequentially compact + (what?) implies metrizable

Mike Miller

5:44 PM

I really doubt sequential compactness helps here, is all

Anderson Felipe Viveiros

Sufficient conditions, so that I have a direction to follow in my proof ^^'

Hausdorff is one I need, for sure

Mike Miller

What I'm pointing you towards are classical metrization theorems that have little to do with compactness in any sense

Anderson Felipe Viveiros

Ok @MikeMiller, @s.harp thank you guys

Mike Miller

Good luck!

Anderson Felipe Viveiros

o/

Mike Miller

5:45 PM

@AndersonFelipeViveiros Wait, maybe I have something

What is weak convergence precisely?

For every measurable $X \subset \text{Gr}(M)$ we have $\mu_n(X) \to \mu(X)$?

If that's so, then what is the $\sigma$-algebra? Just Borel, aka the one "generated" by open sets?

Here is a proof strategy: found a countable collection of $X_i$ so that weak convergence on arbitrary $X$ is guaranteed by the same for all of the $X_i$. Define $d_i(\mu, \mu') = |\mu(X_i) - \mu'(X_i)|$. Argue that if $d_i(\mu_n, \mu) \to 0$ for all $i$, then $\mu_n \to \mu$ weakly.

Now assemble these into a single metric a la the argument for Frechet spaces.

Maybe finding countably many $X_i$ is hard but it doesn't feel like it should be.

s.harp

ok, now I can read things, sorry was halfway between two things

ninja hatori

0

Q: about flatness of module

I need help in (iii) implies (iv) and (v) implies (1). In (v) implies (i) I understand the proof until the use of artin-rees lemma but after that I don't understand it ? Can someone explain it?