Mathematics - 2017-08-16 (page 2 of 4)

Balarka Sen

12:01

@Daminark

Kirill

@TobiasKildetoft either I represented the task wrong, or, it is really true (that it is wrong)

Balarka Sen

I have something for you

Tobias Kildetoft

@Kirill I provided you with an example of a matrix satisfying what you wrote and with eigenvalues other than $1$

Daminark

What be?

@Balarka

Kirill

@TobiasKildetoft true

Balarka Sen

12:02

@Daminark I have made a groundbreaking observation

it will really break the floor beneath you

Daminark

Let's hear it

Balarka Sen

There is some dispute about the most fundamental definition of $\pi$ in mathematics. I propose we define $\pi$ as the total curvature of a closed surface divided by twice it's Euler characteristic.

Tobias Kildetoft

@BalarkaSen Does that make it floorbreaking rather than groundbreaking?

Kirill

@BalarkaSen the definition is my notes is more prosaic...

Daminark

What's "total" curvature?

SteamyRoot

12:05

Some crazy professor in Ghent (a Belgian university that isn't mine) wrote some giant textbook where he derives like all of calculus from the axioms.

He introduces and defines $\pi$ as $\int_1^\infty \frac{4}{1+x^2} dx$

Balarka Sen

@Daminark Integral of curvature over the surface.

Daminark

@Steamy how far back are we going when we say "axioms"? ZFC?

Kirill

@SteamyRoot the $\pi$ of Ramanujan is also great

Daminark

@Balarka integrating Gaussian curvature?

Balarka Sen

yep

Daminark

12:07

Wait hold on a second what?

SteamyRoot

@Daminark Probably not that far, nobody cares about ZFC if your focus isn't set theory :P

Balarka Sen

@TobiasKildetoft Relevant:

SteamyRoot

I haven't actually read his book(s), I have better things to do in life :P

Daminark

So you're telling me that if you integrate Gaussian curvature over any closed surface you get $2\pi$ times Euler characteristic?

Balarka Sen

SteamyRoot

12:08

@Daminark Gauss-Bonnet

Kirill

I got that: $\pi=2 \cdot \mathrm{inf} \{ x: x \ge 0, \cos x = 0\}$.

Balarka Sen

@Daminark Ah, yes, I thought you knew that.

It's a brilliant theorem.

Gauss curvature is a geometric invariant; but total curvature is a topological invariant.

SteamyRoot

Another one of those things that Gauss "knew" but he never bothered to actually publish

Daminark

@Steamy I meant more like constructing up to $\mathbb{R}$ from total scratch. Otherwise, wouldn't Spivak likely count under deriving everything axiomatically?

And huh, that sounds pretty slick

SteamyRoot

It's way worse than Spivak, that I know.

Balarka Sen

12:10

@Daminark Corollary: $T^2$ does not admit a metric which is negatively curved everywhere.

Daminark

Is the proof gonna have something to do with geodesic flow inducing vector fields?

Balarka Sen

Well, nonpositively curved.

@Daminark Brilliant idea to invoke Poincare-Hopf.

Maybe you should try to ponder on that one.

Daminark

(do note that we're still like halfway through section 2.3 in Ted's book so probably a good bit away from G-B right now)

SteamyRoot

Well, in general, You can just write $\int_M kdA + \int_{\partial M} k_g ds = 2\pi \chi(M)$ with $k_g$ the geodeisc curvature

Daminark

@Steamy yikes

Balarka Sen

12:11

Ted does it differently.

His method is more powerful because it picks up the boundary term whenever it's there (see Steamy's message). But I like P-H :)

Kirill

for those who know this: please tell me, what power of the kernel I should start with, if I create a Jordan basis

Daminark

If I had to guess how to connect those two it feels like it'd have something to do with th degree of the Gauss map

Balarka Sen

Mhm

Jasper Loy

Mhm is a palindrome.

Daminark

Okay so here's an idea

So let's say $V$ is a vector field on your surface

Kirill

12:24

@JasperLoy the best that I know is

Daminark

You have the isolated zeroes $x_1,\ldots,x_n$

Kirill

@JasperLoy SATOR AREPO TENET OPERA ROTAS

Daminark

Exclude disjoint balls around each $B_1,\ldots,B_n$

Kirill

@JasperLoy try also to read only the first letters, then the second letters...

Daminark

So now you know that the degree of this map on the boundary of the manifold (the edges of the balls) will be negative sum of indices

But you can extend this to the whole manifold so you have degree 0 in total

Kirill

12:27

translated as "Shephard Arepo сultivates the ground and sows" - something like that, @JasperLoy

Huy

how was your exam @Kirill

Kirill

@Huy perfect!

Huy

very good

I hope you were more pedantic than your teacher

Leaky Nun

@Kirill nice

Kirill

@Huy I've made some ugly mistakes, but they are not crucial

Huy

12:29

why did you decide to make ugly mistakes?

Daminark

Or not the whole manifold but the whole manifold minus the balls

Kirill

@Huy cause noone wants me to understand the Jordan basis well

Huy

are you studying with a textbook or only with lecture notes?

Daminark

Anyway, so this means the sum of vector field indices should equal the degree of this map away from the balls somehow, and there the vector field should be homotopic to the "vector field of unit normals"

@Balarka does this sound reasonable? Ish?

Kirill

@Huy I hate my lecture notes sometimes, but all the exercises and the exam are based on them. I try to learn with other means - books and sources I prefere.

Huy

12:31

@Kirill: I studied linear algebra with Fischer's (German) book years ago. you should have a look at it.

Kirill

@Huy I just think that changing the topic every 5th day is just crazy

@Huy I don't know what our education system thinks doing such plans

Huy

?

Kirill

@Huy Otto?

Huy

@Kirill: isn't it the teacher specifically planning the lecture?

Kirill

@Huy he adapts the plan of the land, and land adapts the plan of the country

Huy

12:33

ok

Kirill

so, Otto Fischer?

Huy

over here teachers and professors are given much more of a personal choice

no, Gerd Fischer

I think he is actually at TU right now

at your TU

;)

Kirill

@Huy I think so. I do not like Fischer and Rudin

Huy

I never did Rudin

but I was very satisfied with Fischer

Kirill

@Huy I could accept Beutelspachter, but I have another 800 books in my library...

@Huy reading Banach or Euler is much more interesting than Fischer, but I have nothing against him

Huy

12:35

what's wrong with Fischer?

Kirill

@Huy nothing, not my type

Huy

ok

you have many books for your age

Daminark

Hey @PVAL and @anon!

PVAL-inactive

howdy

Kirill

@Huy what year of study are you doing?

Daminark

12:39

How's everything going?

Huy

@Kirill: define "study"

Kirill

Studium @Huy

Huy

then, in my 6th.

Kirill

@Huy you well get a super well-talented guy from us the next year, but he will either do the 1. year again or continue

Huy

I hope you're talking about yourself

Jasper Loy

12:42

Hi @huy have you managed to get money?

Huy

@JasperLoy: I have enough to survive, but having a bit more would be nice.

Kirill

@Huy nope, I've just finished my first year and am not going to change to Zurich. Yet.

Huy

@Kirill: do you mean he will do the 1st year again at the ETH?

Kirill

@Huy he says he wants that, dunno why, he had 1,0 here...

Huy

ok

if he actually does the 1st year again, he might stumble across my name a lot :P

Kirill

12:45

btw he says you have some kind of selection during the study, is that true?

Huy

natural selection?

Kirill

oh no

like, you have to do some exams in order to be able to continue, more frequently than we do

Huy

well you have to take exams at the end of the semester and pass them to continue ^^

other than that, for mathematicians, no.

Kirill

strange

I've thought you really have some kind of natural selection, according to his words...

Huy

not anymore, since 2012 already, I think

Kirill

12:47

doesn't matter

Huy

we used to have "graded homework" where you could theoretically be blocked from taking exams if you didn't do your homework

but that doesn't exist anymore

Jasper Loy

Oh @Huy I think you haven't heard me sing before: youtube.com/watch?v=4n5vi6ktZnQ

Huy

@JasperLoy: actually, I have, I was still active when you started the channel. good work. :)

Jasper Loy

@Huy I remember you played the guitar right?

Huy

yes, I do play the guitar.

Jasper Loy

12:50

Do you have some video of you playing something?

@Kirill OK, and what language is that exactly?

Huy

yes, but nothing from the recent years.

Kirill

@JasperLoy it is something from the old, maybe latin, maybe old greek, do not want to disinform you. I got that from my teacher in the school when I was 13, I was astouned that day.

Leaky Nun

@JasperLoy it is Latin

latin square, if I remember correctly

Jasper Loy

@LeakyNun Oh I thought a Latin square was just a magic square where all the numbers added up to the same in each row and column.

Huy

@JasperLoy: if you want, I can send you something I recorded with a girl just a year ago. it's not very good, but it's the most recent and we actually practiced it a bit.

(email)

Leaky Nun

12:56

Sator Square

The Sator Square (or Rotas Square) is a word square containing a five-word Latin palindrome: S A T O R A R E P O T E N E T O P E R A R O T A S In particular, this is a square 2D palindrome, which is when a square text admits four symmetries: identity, two diagonal reflections, and 180 degree rotation. As can be seen, the text may be read top-to-bottom, bottom-to-top, left-to-right, or right-to-left; and it may be rotated 180 degrees and still be read in all those ways. The Sator Square is the earliest dateable 2D palindrome. It was found in the ruins of Pompeii, at Herculaneum, a city buried...

@JasperLoy I remembered wrongly

Jasper Loy

@Huy Oh OK, if you want, we can say more private things over emails in future too. My email is in my profile.

Huy

@JasperLoy: I also have a recording of myself playing and singing from 2012, which is not bad. but I'm not good at singing.

Kirill

@Huy I am good at piano playing)

Huy

I play the piano too, but I'm not good at it.

Kirill

@Huy I've made the music study first, now math

Huy

12:58

I see.

my music teacher wanted me to study music, but I studied maths instead.

Jasper Loy

@LeakyNun I had a book on magic squares long ago. I learned how to make magic squares of any odd order. It's very simple actually, the stepladder method.

Kirill

"argue by contradiction"

Leaky Nun

@JasperLoy I also read about it a long time ago lol

Jasper Loy

@LeakyNun Oh I didn't know you use lol too, lol. I thought you were a serious guy, lol.

Leaky Nun

@JasperLoy lol

Kirill

13:01

and what do serious guys say?

Jasper Loy

They say serious things and never use lol.

Huy

@Kirill: here you can witness my insane piano skills: soundcloud.com/ihuy/mein-song-5/s-CDyQB

Kirill

sos - "standing over here seriously"? not "laughing out loud"?

Jasper Loy

WTF, LOL

Kirill

@LeakyNun I've bookmarked it

@LeakyNun not now

@LeakyNun, and, you've named it German!

Jasper Loy

13:05

@Huy I will be waiting for your email then.

Leaky Nun

@Kirill wrong pings?

Kirill

@LeakyNun I've heard that! What do you mean?

Leaky Nun

why are you pinging me?

Kirill

@LeakyNun what is that? I do not know the word.

Leaky Nun

it is when you write @ followed by a user name

Tobias Kildetoft

13:07

@Kirill To ping someone is to use the @theirname to make their chat make a sound

Kirill

@LeakyNun so , maybe because I was talking to you I think

@TobiasKildetoft thank you

I have pinged that, heard, and said you have named the audio in German

ping, @LeakyNun

Huy

@JasperLoy: I sent you the thing where I sing and play the guitar. unfortunately, my singing isn't as good as my guitar playing.

Leaky Nun

have you confused me with Huy?

Kirill

@LeakyNun yes, sorry

Jasper Loy

@LeakyNun It's very easy to reply to the wrong line, especially when you have a moving target, when the lines move when someone says something new, lol.

Leaky Nun

13:18

@JasperLoy got it

Secret

13:35

How does the reals gain a total order, as it is constructed by dedekind cuts or cauchy sequences hence using elements from $2^{\Bbb{Q}}$ and $\Bbb{Q}$ is not well ordered

given we knew that to construct a total order on a powerset, the set itself has to be well ordered?

Tobias Kildetoft

@Secret The reals do not have any well-order that we can describe

Nor does a set being well-ordered suffice to get a well-ordering on its powerset in any direct way.

Secret

I know the reals have no well order, but it does has a linear order, so how is the linear order constructed?

Tobias Kildetoft

@Secret The reals do have a well-order, assuming choice

And there are uncountable well-ordered sets that can be constructed without choice

Daminark

@TobiasKildetoft How does that go?

Tobias Kildetoft

I don't see the issue with defining the usual order on the reals

@Daminark I don't recall the details. It is in Munkres. It is something like taking the set of well-orders on the naturals up to some equivalence and giving these an ordering

Daminark

13:38

And @Secret when you construct $\mathbb{R}$ via cuts, ordering is just via subset

Tobias Kildetoft

Note that assuming choice, this constructed set will have the same cardinality as the reals, but the bijection to the reals will take some amount of choice to obtain

Daminark

Ah, I see

Secret

is the usual order on the reals a lexicographical order or something different, because in our exercise yesterday on trying to define a lexicographical order on $\{0,1\}^{\Bbb{R}^+}$ that fails and Leaky said it is because the reals don't have a well order hence the powerset cannot have a linear order?

Martin Sleziak

@Daminark I think that you do not need AC to get $\omega_1$; the smallest uncountable ordinal. (And, as Tobias wrote, this is supremum of all countable ordinals.)

Secret

which is why I am a bit suprised that the reals have a linear order (the usual order, which is lexicographical if it is viewed as a binary expansion) since the rationals hve indefintely decreasing sequences hence no well ordering

Tobias Kildetoft

13:41

@Secret The binary expansion has the naturals as "base", rather than the rationals

Leaky Nun

@Secret but the rationals are still linearly ordered

@Secret but the natural numbers do have a well-order, so the powerset can have a linear order

the usual order can be somehow regarded as a lexicographical order, just be reminded that 0.999... = 1

Secret

Do the set of all dedekind cuts form a well order without choice. I can see each cut it only produces two subset, but neither necessary have a minimum unless the cut occured right at a rational?

but well ordering requires all subsets to have a minima

Leaky Nun

@Secret the set of all dedekind cuts is basically the set of all real numbers, right?

Secret

It is, but I am trying to understand how it avoids the infinitely decreasing sequence issue as the rationals contains many of these no matter which dedekind cut it is contained in

Leaky Nun

@Secret they have an infimum

the real numbers are not well-ordered

are you confusing between dedekind cuts and the set of all dedekind cuts?

a dedekind cut can be thought of as a real number

Secret

13:49

The set of all dedekind cuts should form the real number line.

I think my confusion came from the following:

Daminark

@MartinSleziak That's nifty

Secret

20 hours ago, by Leaky Nun

@TobiasKildetoft when we are finding the smallest element that differ

20 hours ago, by Tobias Kildetoft

@LeakyNun Ahh, in order for this to give a total ordering, right. I think you can define it without this, but it will not necessarily be total.

Leaky Nun

@Secret you were talking about the reals as $10^\Bbb N$

Secret

so, I then possibly mixed up necessary and sufficient conditions for $2^S$ to have a linear order, $S$ needs a well order

Leaky Nun

@Secret so what is your question now?

Secret

13:52

First I wan to clarify: Is it necessary and sufficient that for $2^S$ to have a constructible linear order, $S$ must be well ordered?

If not, then what else do I need?

Leaky Nun

@Secret I think it's sufficient but I'm not sure if it's necessary

Secret

(an example of a construtible linear order will be something akin to the usual order of the reals)

So, viewing reals as $10^{\Bbb{N}}$ I can see how it can be linearly ordered, but what I am not sure is that how can the set of dedekind cuts $2^{\Bbb{Q}}$ avoids an infinitely decreasing sequence

Leaky Nun

@Secret I don't know enough to tell you anything about constructibility

@Secret no, the dedekind cuts cannot be thought of as $2^\Bbb Q$

you can't just randomly place element in a set

i.e. if a rational number is in the larger set, then all rational numbers larger than it must also be in the larger set

Secret

Ah I see, it's more restrictive than $2^{\Bbb{Q}}$

Leaky Nun

@Secret but their cardinalities are the same :)

Secret

13:57

Interesting, so $2^{\Bbb{Q}}$ cannot have a linear order (because you can find an infintie decreasing sequence), but $\Bbb{R}$ can?

Leaky Nun

@Secret $2^\Bbb Q$ can have a linear order of course, because it bijects with $\Bbb R$

finding an infinite decreasing sequence makes it not well-ordered

linear order = total order, not well-order

Martin Sleziak

@Secret If you have a well-ordering on $\mathcal P(X)$, cannot you simply get a well-ordering on $X$ by putting $x\le y$ $\iff$ $\{x\}\le \{y\}$?

In the other words, you have bijection between $X$ and $\{\{x\};x\in X\}$ and the latter is a subset of a well-ordered set.

Leaky Nun

@MartinSleziak linear order, not well-ordering. "$2^S$ to have a constructible linear order"

Martin Sleziak

I think that getting a well-order on $\mathcal P(X)$ from a well-order on $X$ is a more difficult of the two implications. (Some kind of lexicographical order is taken there, right?)

Secret

Let me try to clarify the situation with a table:

Martin Sleziak

14:03

@LeakyNun Sorry, I misread.

Martin Sleziak

14:13

It seems that AC (and so also well-ordering principle) is strictly stronger than: "Every set can be linearly ordered."

The latter is also called ordering principle. MO: Are all sets totally ordered ?

So you can't prove in ZF: "$\mathcal P(X)$ is linearly orderable" $\implies$ "$X$ is well-orderable".

Secret

The fact that $2^{\Bbb{Q}}$ is linearly ordered suggest the proposition is sufficient but not necessary, but then what is the ordering on $2^{\Bbb{Q}}$ explicitly?

Leaky Nun

@Secret just to make sure: what is the dedekind cut for $\sqrt2$?

@Secret lexicographical.

Secret

$\{x^2 > 2\} \vee \{x^2 < 2\}, x \in \Bbb{Q}$ if I recall

Leaky Nun

@Secret $\Bbb N$ and $\Bbb Z$ are not linearly ordered by lexicographical ordering

Daminark

Not quite @Secret

Faust7

14:19

ZFC

Leaky Nun

@Secret As Daminark said, this isn't right.

Faust7

Mornign leaky

Leaky Nun

@Faust7 hi

Faust7

glad someones on the ball to give me small brain a headache about something this morning

Leaky Nun

@Faust7 what was it?

@Secret no, this still isn't right

Faust7

14:20

don't you remember teaching me ZFC? thought u were going to hit me with a stick lol

Lexigraphical is means theres a total order on the poe set no?

Leaky Nun

@Faust7 did I?

@Faust7 lexicographical ordering is a type of ordering

Faust7

Don't remember jokinly

Leaky Nun

and "total order on poset" doesn't make much sense because "poset" is partially ordered set

Faust7

no you did a good job considering im an undergrad

Leaky Nun

@Faust7 I can't find the message history of me teaching you ZFC

Martin Sleziak

14:23

@Secret Isn't actually this what you were looking for. If more clarification is needed, we can discuss in set theory room - so that we do not have several discussion here going on at once.

Leaky Nun

edit: I can find it now

Faust7

sorry when i say a totally ordered poset i thought it mean that everything in the poe was related to something else

Leaky Nun

@Faust7 what is poe?

Faust7

a binary operation that is reflexive antisymmetric and transitive

?

on a set

*sorry been 4 months since i did much set theory but that still sounds right O.o

Leaky Nun

alright

Faust7

14:26

but i thought there was a special case of a partially ordered set that had a total order

like everything

every*

Leaky Nun

@Faust7 in some sense you're right. never mind.

Faust7

xRy or yRx

im going to go look up a totally order set

Secret

$\{x^2 < 2, x \in \Bbb{Q}\}$?

Leaky Nun

@Secret close, but still not quite right

Faust7

i see lexicographical order is not the same as a total ordering on a poset

Leaky Nun

14:29

@Faust7 lexicograhical ordering is an ordering defined for cartesian products

Faust7

can they be infinit products or finite only?

Leaky Nun

@Faust7 usually finite only, but you can generalize it to well-ordered infinite sets

Akiva Weinberger

@Secret What's the question?

Faust7

Thats intresting

Leaky Nun

@AkivaWeinberger dedekind cut of $\sqrt2$

Akiva Weinberger

14:30

Ah

So you want to write the set of all rationals less than $\sqrt2$ without explicitly mentioning $\sqrt2$ itself, or something?

Well Secret's last thing doesn't include, say, $-100$

Faust7

@LeakyNun is this sort of like the very subtle difference between $\mathbb{Z}$ being cyclic and $ \mathbb{Q}$ not being cyclic despite the fact that they are isomorphic?

Leaky Nun

@Faust7 is what?

Secret

but $(-100)^2 > 2$?

Leaky Nun

@Secret this is why your answer is wrong

$-100 < \sqrt2$

Faust7

@Faust7 usually finite only, but you can generalize it to well-ordered infinite sets

Leaky Nun

14:33

@AkivaWeinberger precisely. Less than or equal to (doesn't make a difference anyway).

@Faust7 no, they aren't isomorphic.

Secret

but $\{x < \sqrt{2}, x \in \Bbb{Q}\}$ is circular since we want to define $\sqrt{2}$, ugh, I need to recall...

Daminark

So @Secret, think about this

Why doesn't $\{x\in\mathbb{Q}: x^2 < 2\}$ work?

Secret

all negatives will shoot pass 2, and thus be excluded

Leaky Nun

@Secret bingo

Daminark

Well, most negatives. Now, what's a way to fix that? No need to be clever, just ad hoc

Faust7

14:35

@LeakyNun im a doorknob clearly they must have the same number of generators if they are isomorphic my statement in itself was a complete contradiction.

Leaky Nun

@Faust7 no, you aren't a doorknob, but yes, your statement in itself was a complete contradiction.

Faust7

^^ thanks

Daminark

Keep in mind that the left-hand cut is a set

In particular, overlap doesn't hurt

Secret

$\{x^2 < 2 \vee x \leq 0, x \in \Bbb{Q}\}$

Daminark

Bingo

Secret

14:38

(so much easier to draw the diagram of (the envelope of) that set!)

Daminark

I mean yeah, I just think of cuts via having the picture of the full real line already there

And just saying okay, $a = "(-\infty,a)\cap \mathbb{Q}"$

Secret

Indeed, that's quite intuitive

Daminark

It's just that doing it explicitly requires being sneaky sometimes

Algebraic numbers aren't too bad since you basically just do what you did right now

But I think for transcendental numbers the you mostly can't do things explicitly

user8469759

Hi guys, I've been reading about the open mapping theorem in functional analysis

and I was wondering

Daminark

Or at least have no good reason to, the point is to know that in principle it can be done

user8469759

14:42

if I can "relax" the hypothesis to topological vector spaces instead of requiring these to be banach spaces

Daminark

And that the dedekind construction gives a field with the right ordering. Then just take it from there

user8469759

namely... this is the statment I know

Secret

26 mins ago, by Martin Sleziak

So you can't prove in ZF: "$\mathcal P(X)$ is linearly orderable" $\implies$ "$X$ is well-orderable".

user8469759

Let $E,F$ two banach spaces and $\varphi$ a continuous linear mapping surjective then open sets are mapped into open sets

can't I just say instead

$E,F$ two topological vector spaces and $\varphi$ linear mapping then open sets are mapped into open sets?

Daminark

@user8469759 so I remember looking this up

And you can't do that directly

The best generalization I know of is this

You define something called an $F$-space

Which is a (real/complex) vector space with a metric satisfying a few properties

Secret

14:45

Hmm... if we cannot proof $\mathcal{P}(X)$ is linearly orderable $\implies$ $X$ is well orderable, then by contrapositive, we cannot prove $X$ is not well orderable \implies $\mathcal{P}(X)$ is not linearly orderable...

Daminark

First, scalar multiplication is continuous with respect to this metric and the standard one on ($\mathbb{R}$/$\mathbb{C}$)

Second, addition is wrt the metric

Third, translation invariance

And finally, the F-spaces should be complete

Secret

@LeakyNun @Daminark (NB discussion will continue in set theory room)

Daminark

A Banach space is automatically an F-space, for reference

So it turns out you have this theorem

So let's say $T:X\to Y$ is a continuous linear operator, where $X$ is an F-space and $Y$ is some topological vector space

Then either $T(X)$ is first category, or it's all of $Y$

If it's all of $Y$, then $T$ is an open mapping and $Y$ is an $F$-space

user8469759

@Daminark is your argument taken by Rudin's functional analysis?

Daminark

I think it should be there

I had gotten this from Wikipedia though

user8469759

14:49

but

I was wondering

if there's a counter example

for my case

because I assume if I take $E,F$ finite dimensionals

and linear mapping are described by matrices

my argument would work

Alessandro Codenotti

@Secret the second thing, $X$ not well orderable implies $\mathcal{P}(X)$ is not linearly orderable is false, "every set can be linearly ordered" is stricly weaker than "every set can be well ordered" (which is AC) over ZF

user8469759

not sure about the infinite case though

Daminark

@user8469759 I think it breaks in finite dimensions as wll

Alessandro Codenotti

So you just need to find a non well orderable set in a model of ZF+every set can be totally ordered

Daminark

Let's take $E = \mathbb{R}$ with the discrete topology

That's a topological vector space

Addition and scalar multiplication are immediately continuous because everything is continuous

user8469759

14:51

ok

Daminark

Now, let $F = \mathbb{R}$ with the moral topology

Moral meaning the standard one

And let $T:E\to F$ via $T(x) = x$

This is a continuous linear operator

It's surjective

But it's not an open mapping

A single point is open in $E$, and its image is a single point in $F$, which isn't open

user8469759

ah ok

Alessandro Codenotti

Is every $\sigma$-algebra on a set $X$ the Borel $\sigma$-algebra of a topology on $X$?

user8469759

I was thinking both $E,F$ with the standard topology

I didn't think of changing the topology

Daminark

Yeah, that makes them Banach spaces so you do have the open mapping theorem. But by screwing up the topology you can break it

user8469759

14:55

I see, so the version I know is actually a specific version of what you mentioned

Daminark

@Alessandro in the case of the Lebesgue $\sigma$-algebra on $\mathbb{R}^n$, would the density/denjoy topology do it?

Alessandro Codenotti

I'm not sure which topology you're talking about

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$Mathematics$ Mathematics