Mathematics - 2017-02-03 (page 2 of 3)

Secret

14:00

sorry typo: I mean there are no countable subsets

$(W,\preceq)$, the surreals and $\mathbb{R}+\omega_{\alpha}$ will be quite useful sets and proper classes to help me build some geometric intuition (if there exist a well defined one) on sets of these sizes, and thus a better intuition of different levels of continuua

@AkivaWeinberger If there is no well ordering on the cardinals due to lack of choice, then $\mathfrak{c}$ may not even be an ordinal, and thus the expression $\mathfrak{c}+\omega_1$ may not obey the ordering of the ordinals, I guess....

arctic tern

14:16

@Secret what is a "countable element"?

Secret

@arctictern that's a typo, I mean countable subset

arctic tern

you're saying W has no countable subsets? what?

Secret

not sure, I might have overlooked something...

Let me check again...

arctic tern

any infinite set has countable subsets

DHMO

@arctictern why?

why can we always construct a subset A from a set S where A has a cardinality less than S?

Secret

14:21

@arctictern I know that sounds intuitive in terms of partitions, but then you will expect W to suffer from a similar notion of "rationals are uncountable" contradiction? when some order preserving map embed it to one of its subsets of one level lower in cardinality. Or is it because the countable subsets of W are never dense, thus discrete and hence avoided this problem?

arctic tern

Let W be infinite. Pick w_1. Then pick w_2 distinct. Then pick w_3 distinct. Etc.

Get countable subset.

Order has nothing to do with this.

@DHMO S can be bijected with an ordinal, which has every other ordinal and hence every lesser cardinal as a subset.

Secret

ok nvm, yes if all subsets of W are discrete, then that will avoid the "rational is uncountable" contradiction

DHMO

@arctictern interesting.

arctic tern

@Secret have no idea what you're talking about.

DHMO

@Secret "discrete" is a topological term. topology is built on top of sets. we don't need topology to talk about sets.

Secret

14:26

> Observe that $\{r×R | r∈R\}$ is an uncountable collection of pairwise-disjoint non-degenerate ⪯-intervals of $R^2$. Hence, $(R^2,⪯)$ cannot be embedded into (R,≤R) in an order-preserving manner; if this were not the case, then RR would contain uncountably many pairwise-disjoint non-degenerate intervals, so by picking a rational number from each of these intervals, we would end up with uncountably many rational numbers ⎯ contradiction.

arctic tern

not sure how that's relevant to countable subsets of infinite sets

Secret

How does W avoid this problem as if there exists a order preserving map to embed it into one of its countable subsets, then we can conclude that the countable subset have uncountably many elements, hence contradiction in a similar manner to how if there is order preserving map from some larger set to the reals, will conclude a countable subset (the rationals) will have uncountable elements?

So the only way out is that the countable subset cannot be dense, otherwise uncountably pairwise disjoint intervals of W will contain countable elements hence result in the countabe subset have uncountable elements?

arctic tern

indeed, the existence of an embedding of W into a countable set would be a problem, because it mean an uncountable subset fits inside a countable one. you get a contradiction just from cardinality, no order to speak of. but I never said anything about embedding W into a countable subset.

Mahmoud

Hello there.

Secret

So is it possible for W to have the same ordering of the reals as long it does not contain the reals as a subset (because the answer said that cannot happen)?

arctic tern

14:34

@Secret Michael defined W to be well-ordered, so no it can't have the same ordering as the reals

@Mahmoud hello

Akiva Weinberger

@arctictern (Assuming the axiom of choice)

Secret

@arctictern ok, I see.

GFauxPas

good morning guys

Mahmoud

Greetings @GFauxPas.

GFauxPas

if the cross product isn't zero, is it invertible? left- or right- I mean

Mahmoud

14:42

The map of Mathematics probably requires a hyper-dimensional space to draw faithfully but heh, better than nothing.

GFauxPas

does that have context Mahmoud

also your link doesnt work

oh, found the answer. there is no identity for cross product. oops

Mahmoud

Correction : heh

@GFauxPas What do you mean ?

arctic tern

@GFauxPas Is what invertible? What do you mean by left or right?

GFauxPas

I mean if $a \times b = c$ does there exist $a^{-1}$ such that $=a^{-1} \times (a \times b) = b$ or $(a^{-1} \times a)\times b = b$

but as I found online

such would need an identity element for cross product

but there is none

$a, b \in \mathbb R^3 \text{ or } \mathbb C^3$

DHMO

> A common example of this is the cross product of vectors; in this case, the absence of an identity element is related to the fact that the direction of any nonzero cross product is always orthogonal to any element multiplied – so that it is not possible to obtain a non-zero vector in the same direction as the original.

from Identity element on Wikipedia

GFauxPas

14:50

right

arctic tern

If $a\times b=c$ then there exists an $a'$ such that $a'\times(a\times b)=b$. However $a'$ will depend on both $a$ and $b$. When $b$ is nonzero there does not exist an $e$ such that $e\times b=b$, since $e\times b$ must be orthogonal to $b$.

GFauxPas

one of these days I have to go and continue Laplace transform stuff on PW DHMO but I got distracted

DHMO

ok

Mahmoud

What's PW ?

GFauxPas

proofwiki.org

DHMO I started my abstract algebra class, I can tell already I'm going to enjoy it :)

Mahmoud

14:51

Oh, okay.

GFauxPas

great site

DHMO

@GFauxPas congratulations! do you have any questions to share?

GFauxPas

is there a good notation for $\{f \in X^X: f \text{ is bijective} \}$?

the professor didn't remember a good notation

DHMO

@GFauxPas i'm not aware of a notation for bijectivity

SteamyRoot

Afaik there's no standard notation for it.

GFauxPas

14:53

other than the symmetric group in some cases

arctic tern

@GFauxPas Perm(X), Sym(X), S_X, Aut(X)

DHMO

@GFauxPas I do know that surjective functions can be written as $\{f \in X^X: \operatorname{Im}(f) = X \}$

SteamyRoot

Hmmm... The first 3 I've seen before; but I would advise against using Aut...

GFauxPas

he was using $S(X)$ or $\operatorname{Sym}(X)$ but he said he wasn't happy using it

isn't that only if $X$ is finite?

arctic tern

no

GFauxPas

14:56

I guess there's no reason to limit yourself to finish cases

what does automorphism mean here, I only know the meaning for two vector spaces

that is satisfies group axioms?

SteamyRoot

That's why I would advise against using Automorphism to define the set of bijections.

arctic tern

automorphism depends on what category you're working it. in the category of sets, it just means the group of bijections from a set to itself. (usually this isn't done though.)

SteamyRoot

It tends to mean that the map also "preserves the structure" of the object your map is on.

Secret

Is it possible to have a $> \mathfrak{c}$ set that has the same ordering as the reals, contain the reals as a subset, is not a proper class like the surreals or $\mathbb{R}+\omega_{\alpha}$, and does not contain order embedding that result in the rational uncountable contradiction? I think I have looked hard in MSE and cannot find anything similar?

arctic tern

@SteamyRoot that's not really "also," it's build into being a morphism in the category in the first place :P

GFauxPas

14:59

right but it requires an operation as part of its definition

arctic tern

@Secret what do you mean by "same ordering as the reals"?

SteamyRoot

If you're talking about bijections from $\mathbb{Z} \to \mathbb{Z}$ and you denote the set as $\operatorname{Aut}(\mathbb{Z})$, most people would assume you mean a group automorphism of $(\mathbb{Z},+)$.

arctic tern

@GFauxPas or topology

GFauxPas

I didn't take topology so that's out of my zone :P

arctic tern

or differentiable structure. or incidence structure. or...

Secret

15:00

@arctictern Not well ordered, has the same linear ordering as the reals in the sense of 1,2,3.... Basically the real line but with more elements, but don't become so large and become a proper class

arctic tern

"has the same linear ordering as the reals in the sense of 1,2,3" Huh?

SteamyRoot

@arctictern With "also", I meant to clarify what an automorphism is "more" than just a bijection from a set to itself.

Secret

how do we describe the usual ordering of the reals (I think that's the description I am looking for)

1<2<3... ?

GFauxPas

"the usual ordering on the reals"

probably

arctic tern

@Secret Use words.

SteamyRoot

15:02

You can describe it lexicographically by using the decimal notation of any number...

But everyone knows what you mean with the "standard order on the reals" like @GFauxPas mentioned.

DHMO

@Secret $a \le b \iff a \subseteq b$

arctic tern

you might want the properties of "complete" and "totally ordered" and "least upper bound property"

Secret

Ok yes. So the polished question is: I want a $> \mathfrak{c}$ set, that has the standard ordering on the reals, has reals as one of its subset, is not a proper class like surreals or $\mathbb{R}+\omega_{\alpha}$, and does not have order embedding maps to result in the rational uncountable contradiction.

The issue in trying to construct this set, is that the reals are already complete, thus there seemed to be no way to put extra elements in?

arctic tern

do not say "standard ordering on the reals," it doesn't make sense to use that to describe sets that are not the reals

Secret

> has reals as one of its subset

arctic tern

15:08

Meaningless.

"Standard ordering on the reals" applies to the reals, it does not apply to sets that are bigger than the reals.

You need to say things that mean things.

Secret

In plain english, I want a real line of size $> \mathfrak{c}$ and preserve as many properties of the reals including all the subsets, but is not a proper class. I don't know how to describe that better

arctic tern

@Secret What is a "real line"? What does that mean? If it means the real number line, there is only one up to isomorphism, and does not have size $>\frak c$.

DHMO

@Secret $\mathcal P(\Bbb R)$

Akiva Weinberger

@arctictern He means that $(\Bbb R,<)$ can be embedded into it (where $<$ is the standard ordering)

arctic tern

@AkivaWeinberger Yeah, I got that part. That's not what I'm talking about.

Akiva Weinberger

15:13

@Secret I'm confused. $\Bbb R+\omega_\alpha$ isn't a proper class.

arctic tern

@DHMO that has (an isomorphic copy of) R as a subset, but it is not linearly ordered for example

DHMO

@arctictern how do you know it isn't linear ordered when i haven't given it an ordering?

arctic tern

@DHMO it comes with one. in the absence of defining an ordering the default is assumed.

Vrouvrou

hello, can i deduce $(f_3)$ from (f_1) where
$(f_1)$: $f\in C([0,\infty[\times\mathbb{R})$ and for some $2<p<2^*, c_0>0$ $$|f(r,u)|\leq c_0(|u|+|u|^{p-1})$$

$(f_3)$ $f(r,u)=o(|u|),|u|\rightarrow 0,~\text{uniformly on } \mathbb{R}^+$

Secret

@AkivaWeinberger yeah, I must confused because I saw ordinals, and I recall the ordinals form a proper class. Do your set has the standard ordering?

Akiva Weinberger

15:15

The $\Bbb R$ part has the standard ordering, yeah

@DHMO It's consistent with ZF that that doesn't have a total order.

DHMO

Can we prove the well-ordering of natural numbers under Peano's formulation without resorting to set theory?

@AkivaWeinberger but ZFC says that it has a well-order.

Akiva Weinberger

@Secret Maybe you want the surreals (not all of them, but maybe the ones with birthday less than $\omega_1$ or something)

That includes the reals

https://en.wikipedia.org/wiki/Surreal_number

Secret

@AkivaWeinberger Hmm, that might work

The cool thing about it is that DHMO taught me Dedekind cuts and conway construction behave similarly to dedekind cuts

Before DHMO taught me that I don't understand surreals

Wrote recipe into notebook

[More recipe requests] What if now I want another set $S$ that has the following properties:
1. $|S| > \mathfrak{c}$
2. $(\mathbb{R},<)$ can be embedded into it (where < is the standard ordering)
3. The whole set is complete, has least upper bound property, totally ordered
4. Is archimedian
5. Not a proper class
6. No order embeddings to trigger the rational uncountable contradiction

Akiva Weinberger

Complete and Archimedean?

What does #6 mean

DHMO

6 mins ago, by DHMO

Can we prove the well-ordering of natural numbers under Peano's formulation without resorting to set theory?

Secret

15:24

@AkivaWeinberger Not having any maps that exists such that one can prove the uncountable disjoint number of interval will lead to the contradiction that rationals are uncountable

Akiva Weinberger

@DHMO Isn't that essentially the axiom schema of induction?

DHMO

@AkivaWeinberger que?

Akiva Weinberger

The last axiom of PA

The axiom (schema) of induction

DHMO

@AkivaWeinberger I know what it is.

Secret

9

Q: Complete ordered field is an Archimedean field that cannot be extended to an Archimedean field

As a bonus problem, our professor of real analysis asked us to prove that the real numbers (a complete ordered field) cannot be extended into an Archimedean field, with no definition of what he meant by extending. I have tried using proof by contradiction to show that if we have some set, $\math...

Ok nvm, its impossible, the only space left behind to fill in new things are the infinite and infintesimals

Akiva Weinberger

15:26

@DHMO Suppose there is a proposition $P(x)$ such that $\exists x,P(x)$. Well-ordering means that there's a minimum $x$ that satisfies it.

Suppose there's no minimum. Clearly, $\lnot P(0)$ (since otherwise $0$ would be the minimum)

Also, if $1$ through $n$ don't satisfy it then neither does $n+1$

By induction, no $x$ can satisfy it. Contradiction.

DHMO

@AkivaWeinberger formulation of well-ordering principle in terms of Peano arithmetic: $\forall S[S \subseteq \Bbb N \implies \exists n[n \in \Bbb N \land \forall x[x \in S \implies \exists c[c \in \Bbb N \land n+c=x]]]]$ (have fun decoding this)

@AkivaWeinberger what is $P$?

Secret

Ok, so that means, $\mathbb{R}+\omega_{\alpha}$, $\mathbb{R}^{\mathbb{Z}}$, $\mathbb{R}^{\mathbb{R}}$, $(W,\preceq)$ and the surreals will keep me busy in helping me to develop more algebraic and geometric intuition for higher level infinite sets

Writes in notebook. Infinite sets discussion concluded for now...

Akiva Weinberger

@DHMO Peano is first-order. You can't have $\forall S,S\subseteq\Bbb N$.

First order means you can't talk about sets.

That's why I phrased it in terms of propositions, which are the next best thing.

Oskar Tegby

What are the equilibrium points of $x'=\cos x, y'=\sin y$? Is it only the solutions of $\cos x=0$ and $\sin y=0$?

Akiva Weinberger

For each proposition, you can talk about it in PA

DHMO

15:30

@AkivaWeinberger what is the last axiom, formally?

@OskarTegby what is "equilibrium point"?

Akiva Weinberger

@DHMO There's infinitely many, one for each proposition.

DHMO

@AkivaWeinberger it's an axiom schema?

Oskar Tegby

Equilibrium point

In mathematics, specifically in differential equations, an equilibrium point is a constant solution to a differential equation. == Formal definition == The point x ~ ∈ R n {\displaystyle {\tilde {\mathbf {x} }}\in \mathbb {R} ^{n}} is an equilibrium point for the differential equation ...

Akiva Weinberger

Yep

Oskar Tegby

It just feels to simple, but maybe that's how it is.

DHMO

15:32

@OskarTegby $\displaystyle \int \frac {\mathrm dx} {\cos x} = \int \ \mathrm dt$

Akiva Weinberger

wikimedia.org/api/rest_v1/media/math/render/svg/…

From Wikipedia

Oskar Tegby

Yeah. I tried to solve that but for $\sin y$. I got $-\log\left(\frac{\cos x+1}{\sin x}\right)=C-t$, and got kind of confused on how to continu.e

Akiva Weinberger

@DHMO Actually, I was wrong; there is a second-order version of PA (in which induction is just one axiom), in addition to the first-order version

You can't really prove anything in second order theories, though.

DHMO

@AkivaWeinberger in your proof, what is $P$?

Akiva Weinberger

(Gödel's theorem only applies to first order things, for example)

A proposition. Thing of it as me trying to show that $A:=\{x\mid P(x)\}$ has a minimum.

DHMO

15:35

@AkivaWeinberger why?

Oskar Tegby

I want to solve that for $x$, but the Wolfram said nope and I agreed.

Akiva Weinberger

That would make $n+1$ the minimum, but we were assuming (for contradiction) that there is no minimum

@DHMO I need to go

DHMO

oh, ok

@AkivaWeinberger hasta luego

Akiva Weinberger

Hasta la vista

Oskar Tegby

Take care, @Akiva!

DHMO

15:38

Oskar Tegby

I'm not sure that I follow exactly.

Secret

wolframalpha.com/input/?i=-ln((cos(x)%2B1)%2Fsin(x))

The log part has some interesting looking graph

Oskar Tegby

True.

DHMO

@OskarTegby $\log(\tan(x/2)) = -\log\left(\frac{\cos x+1}{\sin x}\right)=C-t$

Oskar Tegby

It says something about solutions there, so I'm happy with that. I just wonder how the complete computations looks like.

Yeah @DHMO

How come the sign changes?

Secret

15:42

$x'=\cos x, y'=\sin y$, that's 2 separable ODEs in t, right?

DHMO

@OskarTegby because i took reciprocal inside

Oskar Tegby

What's that in this case?

I just don't see how that gives a change of sign.

I feels that I'm missing something obvious

DHMO

@OskarTegby -lnx = ln(1/x)

Oskar Tegby

Of course. Yes. Sorry!

Now I follow.

To get the answer we just solve $\log(\tan(x/2))$ for $x$. Right?

DHMO

yes

Oskar Tegby

15:47

Okay. Great! Thanks! I'm sorry for my blunder there. Insert excuse about Friday here.

Secret

$\ln (\tan x+\sec x)=t$
$-\ln(\cot y+\csc y)=t$

$\implies ln (1+ \frac{1}{sin(x)cos(y)}+\frac{1}{cos(y)}+\frac{1}{sin(x)})$

Oskar Tegby

I feel you.

Secret

wolframalpha.com/input/…

Get some semi nice plots, not sure if it is because computation time exceed or they are realy that nice

Ok that's some really jagged function...

DHMO

@Secret wait, you're chinese?

Secret

I am born in HK, but live in aust

DHMO

16:00

:o

so you speak cantonese?

Secret

indeed

DHMO

i've never known...

you know i'm from hk right

Secret

yup you told me before. I don't know otherwise

DHMO

this is so scary

Secret

this is the internet after all

Mahmoud

16:03

Hi !

DHMO

@Mahmoud i just met my compatriot here

for so long yet never know

Mahmoud

@DHMO What a small world !

I hope to discuss something :

I recently came across these spooky words : ''The discipline of Mathematics can govern the deeds of men''.

How can this possibly make sense ?

DHMO

I have no idea.

Mahmoud

(Assuming it's true)

Vrouvrou

Can someone help me please

Mahmoud

16:10

The word ''men'' is meant to include all people, not just Mathematicians, but most people actually hate Maths ..

Vrouvrou

$N$ is Homeomorphic to the unit sphere, how to explain that $N$ separate the space into two components the interior containnig the origin and the exterior ?

Akiva Weinberger

@Vrouvrou $N$ is homeomorphic to the unit sphere, not necessarily equal?

Isn't this the Jordan curve theorem for $3$ dimensions?

Vrouvrou

no just homeomorphic

i don't know this theorem

Akiva Weinberger

Well, actually, if all we know is that it's homeomorphic to the sphere, then we don't know that the origin is in the interior @Vrouvrou

Balarka Sen

Jordan-Schoenflies

Akiva Weinberger

16:14

@BalarkaSen Isn't Schoenflies the one that fails in 3D

Balarka Sen

I don't remember. I think in 2D Scoenflies says both sides are disk

Akiva Weinberger

You want the Jordan–Brouwer separation theorem

Mike Miller

That's true. Schoenflies only fails if you don't assume locally flat.

Akiva Weinberger

@Vrouvrou en.wikipedia.org/wiki/Jordan_curve_theorem

Balarka Sen

@AkivaWeinberger Ah right

I never remember the names

Semiclassical

16:17

hi chat

Alessandro Codenotti

Hi @Semi @Balarka @Akiva and everyone

Balarka Sen

Hey

Mahmoud

Hey @Semiclassical @AlessandroCodenotti.

Vrouvrou

@AkivaWeinberger i asked a professor he told as it is homeomorphic to the sphere one compenent is inside and the auther is outside

SteamyRoot

Well, assuming the origin isn't on $N$ itself, you can just define the component containing the origin as the "interior", and the other as the exterior.

Mike Miller

16:24

theintercept.com/2017/02/02/…

wdika

Hope i am in the correct room. Can Hamiltonian path be a simple cycle too?

Astyx

Hi chat

DHMO

@Astyx bonjour

Astyx

Comment vas-tu ?

DHMO

bien merci

Mahmoud

16:26

Are homeomorphisms exclusive to mapping between topological spaces ? Can we call a real valued function that ?

Astyx

Changed your profile pic ? @Mahmoud

@DHMO Any news of these integrals ?

Semiclassical

@wdika Any Hamiltonian path which is in fact a cycle is a simple cycle.

Mahmoud

Yes, @Astyx You're the first person to notice :D

Vrouvrou

@AkivaWeinberger what is the relation with my question ?

Semiclassical

(If it's a Hamiltonian path that's not a cycle, then of course it can't be a simple cycle in particular.)

Astyx

16:27

I doubt I am :p

I might be the first to point it out

Mahmoud

Yes, precisely, buy we can't tell since I can't read people's minds, at least not through a virtual chat room anyway :P

wdika

@Semiclassical I am trying to solve that exercise. " Given an undirected graph G(V,E) and a subset of the edges F, is there a simple cycle on G that goes through all edges of F? Show that this problem is NP-Complete"

DHMO

@Secret 你識唔識打中文？

wdika

And i was thinking that this graph contains an undirected hamiltonian path. So that means its Np-Complete. Do you think i am on the right way?

Mahmoud

@DHMO The translation doesn't make sense.

Secret

16:32

@DHMO 識，不過打中文好煩

Semiclassical

Could not tell you. I know nothing useful about complexity of graph theory problems.

Secret

this is why I use english online

DHMO

@Mahmoud few translators can translate cantonese

Mahmoud

@DHMO Okay ._.

DHMO

@Secret 你用乜嘢輸入法？

Secret

16:34

@DHMO Currently 冇裝輸入法，using Google translate mainly to type word by word

DHMO

@Secret :o 我用倉頡可以打得好快

不如你學倉頡

wdika

@Semiclassical Ohh ok thanks anyway!

Secret

I have tried, my 倉頡 is still shit

later...

DHMO

@Secret practice makes perfect

@Secret 再唔係你可以用廣東話拼音輸入法

Ramanujan

@Mahmoud is it iqra in you profile in Arabic?

Mahmoud

16:50

@Ramanujan Yes indeed :)

DHMO

@Mahmoud what for?

Mahmoud

@DHMO It means ''read'', and it is the first spoken word of the Quran, to the greatest prophet, it highlights the importance of knowledge.

DHMO

@Mahmoud I see

@Secret do you listen to cantopop?

Akiva Weinberger

@Vrouvrou That link is the 2D version. If you go to the "generalizations" section you find the version for all dimensions.

> Let $X$ be a topological sphere in the $(n+1)$-dimensional Euclidean space $\Bbb R^{n+1}$ $(n > 0)$, i.e. the image of an injective continuous mapping of the $n$-sphere $S^n$ into $\Bbb R^{n+1}$. Then the complement $Y$ of $X$ in $R^{n+1}$ consists of exactly two connected components. One of these components is bounded (the interior) and the other is unbounded (the exterior). The set $X$ is their common boundary.

SteamyRoot

17:14

It's an interesting question on what surfaces (or $n$-manifolds) the ($n$-dimension analogue) of the Jordan Curve Theorem still holds :O

Balarka Sen

@SteamyRoot For all smooth embedded surfaces in R^3.

Generalizing, all smooth hypersurfaces in R^n

Alessandro Codenotti

What does (un)bounded mean on a generic surface?

SteamyRoot

All smooth embedded surfaces?

Alessandro Codenotti

Don't you need genus 0?

SteamyRoot

A general curve on a torus need not separate the torus in 2 connected components though

Balarka Sen

17:24

In R^n

So you were asking for curves in surfaces, I misunderstood.

SteamyRoot

Ah, yeah... otherwise the question becomes a lot less interesting, judging by your answer :P

Balarka Sen

You probably need at least 2g embedded essential curves in a surface to disconnect it.

Nah, garbage. 3 curves disconnect a genus 2 surface

Vrouvrou

i undersatand thank you @AkivaWeinberger

Alessandro Codenotti

Can't you define the genus of an orientable surface as how many circles you can cut before disconnecting it?

SteamyRoot

You'd definitely need to work with essential curves

Balarka Sen

17:34

In fact just 1 does... I am silly

SteamyRoot

Otherwise, taking any point and a really tiny circle around it; disconnects any surface.

Balarka Sen

@AlessandroCodenotti You can disconnect the genus 2 surface by a single essential curve

SteamyRoot

Ooooh...

Alessandro Codenotti

What's an essential curve?

Balarka Sen

Not nullhomotopic. Does not bound a disk on either side.

OK, yes, the minimal number should be g - 1.

Any surface is connected sum of g torii. g - 1 curves (boundary of the disks you're connected summing along) disconnect it

SteamyRoot

17:37

Hmmm...

Balarka Sen

For g > 1

SteamyRoot

Right, but, doesn't that disconnect the original surface into $g$ punctured tori?

Balarka Sen

Yes. (some once punctured, some twice)

Why's that a problem?

SteamyRoot

Ah, I was only really thinking about disconnecting in $2$ components

Hmmm, actually, the number of punctures could be anywhere up to $g-1$ if you connect all other handles to a central one.

A single curve, yes, but a single essential curve? :O

For any surface of genus $\geq2$, it does.

Balarka Sen

Sure, any of those g - 1 I mentioned are essential.

g > 1

Semiclassical

17:42

Ugh, I can't get my brain to work right now.

Not sure why.

SteamyRoot

Poke it

Balarka Sen

I think we're going around in circles. The interesting question to ask here is what is the maximum number of disjoint curves you can cut out from the surface without disconnecting it.

That should be g.

Secret

@DHMO nope, there aren't much of these in aust

DHMO

@Secret but you can use youtube and spotify

Secret

I did listen to some in the past, but really not much

DHMO

17:44

I see.

Alessandro Codenotti

@Semiclassical have you tried switching it off and on again?

2

Semiclassical

@AlessandroCodenotti I suspect what I need to do in this particular problem is switch off my math brain and switch on my physics brain :P

SteamyRoot

Would be nice if you could switch like that without rebooting your brain

Semiclassical

(which in this case really means "think of this in terms of linear algebra rather than algebraic geometry")

user189740

Would a tree have any vertices in its block-cutpoint graph?

SteamyRoot

17:54

W-when was it decided that linear algebra was physics rather than math?

4

Semiclassical

Quite arbitrarily. I'm not being terribly serious.

SteamyRoot

Actually... if that's the case, maybe I should try to get a double PhD...

Semiclassical

What I have in mind when I say that is that the problem emerges from the eigenvalue problem for a certain Hermitean operator. That's bread-and-butter in quantum mechanics.

So I should think of this particular question in terms of the original formulation in terms of linear algebra, which is directly physical, rather than the algebraic-geometry reformulation I made of it.

aaanyways, back later

Topologicalife

Hi.

How do I prove $\overline{\sigma_p} \subseteq \sigma_p \cup \sigma_c$ where $\sigma$ is the continuous spectrum of a bounded operator in a hilbert space.

Err, where $\sigma_c$ is the continuous spectrum, and $\sigma_p$ is the point spectrum

Vrouvrou

18:10

How to prove that a set has a small ball around the origin ?

SteamyRoot

I think we'll need more information on the set...

Balarka Sen

By hand? If it's open it has that

Topologicalife

Yeah.

Akiva Weinberger

That's the definition of a neighborhood of the origin

so you'll have to give more info on the set, yeah

Balarka Sen

If it's metric you can equivalently look for balls

Vrouvrou

18:17

@SteamyRoot i have this set $N=\{u\in X\setminus\{0\}, I'(u)u=0\}$ it is the Nehari manifold , in the book he says that $N$ separate $X$ into two compenents, the compenent containing the origin contain a small ball around the origin

so what we must to prove using the functional $I$

to prove that there is a ball around the origin

?

Akiva Weinberger

Is $N$ is closed?

Vrouvrou

Yes it is the inverse image of \{0\} by a continuous function !

Akiva Weinberger

Then its complement is open, right?

Vrouvrou

yes

Akiva Weinberger

What does it mean for a set to be open?

Vrouvrou

18:30

it is a neighbohood of all it's point

so it contains a ball

containing the point

@AkivaWeinberger

But why the compenent containig the origin must be in the complement of N ?

someone here ?

Daminark

This midterm in analysis has proven that I can't ODE

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$Mathematics$ Mathematics