Mathematics - 2012-07-15 (page 1 of 3)

user19161

1:51 AM

@JonasTeuwen I just noticed that your name contains TW just like TWK. That makes you the next great harmonic analyst!

2:12 AM

@JasperLoy If there any theoretical importance of the "characteristic" metric $$d(x,y)=\begin{cases} 1\; ; \;\text{ if } x\neq y\cr 0\; ; \;\text{ if } x = y \end{cases}$$

Henry T. Horton

Hey Piotr

That metric is good for counterexamples

Peter Tamaroff

@HenryTHorton Oh. Well, now I have to prove three things

If $a\in X$, then
$(1)$. $\{a\}$ is a nbhd of $a$.
$(2)$. $\{a\}$ constitutes a basis for the system of nbhds at $a$.
$(3)$. Every subset of $X$ is a nbhd of its points (i.e. it is open)

@HenryTHorton Maybe those are counterexamples to something???

Henry T. Horton

No, they are just properties of the discrete metric

(I call your "characteristic" metric the discrete metric -- it generates the discrete topology)

Peter Tamaroff

@HenryTHorton OK. I'll let you know how it goes.

@HenryTHorton It just reminded me of $\chi_A$

@HenryTHorton Yes.

@HenryTHorton To prove $(1)$ I can smply choose $0<\delta <1$ and $B(a;\delta)=\{a\}\subset \{a\}$

Henry T. Horton

Yes

Peter Tamaroff

2:24 AM

@HenryTHorton How are you related to math?

Henry T. Horton

@PeterTamaroff Math is my wife.

Peter Tamaroff

@HenryTHorton For a long time now?

Henry T. Horton

Yes, we were married when I was 11

I just finished my first year in a Ph.D program now

Peter Tamaroff

@HenryTHorton Oh, with what orientation?

Henry T. Horton

What do you mean by that?

Peter Tamaroff

2:27 AM

@HenryTHorton I mean.... what are you planning your thesis to be about?

Henry T. Horton

Gauge theory of some sort... I'm just a kid though, I won't even do my oral exams for another year or two

Peter Tamaroff

@HenryTHorton A kid?

Henry T. Horton

I mean I am just in the beginning stages of my Ph.D, so I haven't begun any real work yet

Peter Tamaroff

@HenryTHorton Oh, OK.

In this metric space given $B(a;\delta)$, if $d\geq 1$, then $B(a;\delta)=X$, isn't this right?

Henry T. Horton

Yes, if $\delta \geq 1$

Peter Tamaroff

2:33 AM

Kind of a bipolar metric here.....

Rajesh D

Hi

@HenryTHorton Are you a physicist?

Henry T. Horton

Nope

Rajesh D

Gauge theory?

Henry T. Horton

Mathematical gauge theory -- the study of principal bundles, connections on them, curvature of connections, PDEs relating such things, etc

The inspiration comes from physics

Rajesh D

ok i see

Henry T. Horton

2:42 AM

But the mathematical techniques are actually quite useful in studying low-dimensional topology

Rajesh D

ok

@HenryTHorton What is a good book to get introduced to Differential Geometry? starting from the very very basics :-)

Henry T. Horton

That is a hard question... most books are either too elementary or too advanced

What kind of differential geometry are you interested in learning about? Do you know?

user19161

@RajeshD Differential geometry can mean many things? What exactly are you talking about?

user19161

@RajeshD If you know advanced calculus already and want to study the differential geometry of curves and surfaces and then generalize to riemannian geometry in dimensions n, take a look at Kuhnel's Differential Geometry: Curves, Surfaces and Manifolds.

user19161

@RajeshD The first half covers curves and surfaces; the second half covers manifolds. It is an excellent book published by the AMS.

Henry T. Horton

2:56 AM

DON'T TELL ME WHAT TO DO JLO

user19161

@HenryTHorton Sorry, I meant to ping Rajesh instead of Prof Horton.

Henry T. Horton

@JasperLoy That's Dr. Professor Horton to you

Peter Tamaroff

@HenryTHorton You're not a PhD yet! =D

user19161

@RajeshD Do Carmo has Differential Geometry of Curves and Surfaces as well as Riemannian Geometry. They are ancient and slow-moving IMHO. If you want a very complete treatment with lots of exposition, use Spivak's five volumes Comprehensive Introduction to Differential Geometry.

Rajesh D

Thanks @Jasper

user19161

3:01 AM

@RajeshD Finally, if you want to study many of the things associated to what we might call differential geometry, go for Lee's three volumes Introduction to Topological Manifolds, Introduction to Smooth Manifolds, Riemannian Manifolds. However the real bible of differential geometry which is way too advacned and expensive is Nomizu and Kobayashi's two volumes of Differential Geometry.

Peter Tamaroff

@JasperLoy Jasper, can I ask you for some validation?

user19161

@PeterTamaroff Sure, but remember I am only a lunatic who knows almost zero math and zero LaTeX.

Rajesh D

@HenryTHorton I am looking for an introduction to basic concepts in DG. Like Riemeannian geometry, but before that I want to develop intuitive insight into things like contravariant derivatives, metrics, Tensors, and tangent spaces and bsic things like that

Peter Tamaroff

Let $f:\Bbb R \longrightarrow \Bbb R$, so that $f(x)=1 $ when $x>a$, $=0$ if $x \leq a$. Then $f$ is continuous at any $x \neq a$. PROOF

Rajesh D

I want to get to the Einstein equations first quickly

Peter Tamaroff

3:04 AM

Let $x' < a $, $\epsilon >0$ be given.

user19161

@RajeshD Kuhnel treats Einstein equations.

Rajesh D

ok

Peter Tamaroff

Choose $\delta \leq d(x',a)$ . Then whenever $|x-x'|<\delta$ we have $|f(x)-f(x')|=0<\epsilon$.

Similarily, let $x'>a$. Choose $\delta \leq d(x',a)$. Then whenever $|x-x'|<\delta$ we have $|f(x)-f(x')|=0<\epsilon$.

user19161

@PeterTamaroff That is fine, but why not choose a specific delta, say the distance between x' and a itself?

Henry T. Horton

Half the distance would be better

Peter Tamaroff

3:08 AM

@JasperLoy Yeah, it is the same.

Any $\delta$ equal or less to works.

Henry T. Horton

At least one of them would need to be strictly less than the distance from $x'$ and $a$

The $x' > a$ case

user19161

@PeterTamaroff But when you write a proof, it is best to be to the point and not bring in unnecessary stuff. Just take note.

user19161

@HenryTHorton The inequality is already strict in the less than delta part so not necessary.

Peter Tamaroff

Finally, $|f(x)-f(a)|=1$ if $x<a$, so no delta will work for any $\epsilon < 1$.

Henry T. Horton

Oh $\displaystyle \iint \mathbf{F} \cdot \mathbf{dS}$

user19161

3:11 AM

@HenryTHorton WTF?

Henry T. Horton

"Oh flux"

user19161

@HenryTHorton OIC, instead of O fuq

Rajesh D

@HenryTHorton You haven't answered my question

user19161

@peter Are you clear now?

user19161

@RajeshD That sounds intimidating! He will answer if he is free and has something to say.

Rajesh D

3:19 AM

@JasperLoy lol Sorry I didn't mean that way @Jasper

@JasperLoy : I have started Do carmo but I really don't like it that much

Peter Tamaroff

@JasperLoy Guess so. Moving on.

Rajesh D

He assumes many things for example he think we know exterior product or thinks his appendix is a good introduction but I don't think so

user19161

@RajeshD See, I told you. :-)

Rajesh D

I get a feeling that tthis book is a bit messy

Peter Tamaroff

@RajeshD Is it Massey's?

Rajesh D

3:22 AM

no "messy", "..."

DoCramo

I need an alternative solution

user19161

@RajeshD I suggest Kuhnel for you.

Rajesh D

k downloading

Peter Tamaroff

@JasperLoy Would you say this is a hard excersise? "Let $f:X\to Y$ , $X$,$Y$ metric spaces. Let $a\in X$ and $\mathcal B_{f(a)}$ be a basis for the system of nbhds at $f(a)$. Prove $f$ is continuous $\iff$ for each $N\in \mathcal B_{f(a)}$, $f^{-1}(N)$ is a nbhd of $a$."

I have several theorems characterizing continuity in terms of nbhds.

Rajesh D

@PeterTamaroff What is nbhd ?

Peter Tamaroff

Particularily, $f$ is continuous at $a \in X$ $\iff$ for each nbhd $M$ of $f(a)$, $f^{-1}(M)$ is a nbhd of $a$.

@RajeshD A neighborhood.

@Eugene Hi.

Eugene

3:39 AM

hi

Peter Tamaroff

You know about topology?

Eugene

erm. point-set yes

Peter Tamaroff

What would "point-set" be?

I mean, in contrast with "General"

Eugene

dunno

same i guess

Peter Tamaroff

OK, nvm. See above, please.

Eugene

3:43 AM

ugh

i think this is a fairly standard exercise

Peter Tamaroff

It's OK if you don't feel like doing it, though,

Eugene

but i'm too lazy to think about it now i guess

i think you're supposed to use the if U is open then f inverse U is open thing

that implies continuity

Peter Tamaroff

Oh, I haven't gotten to open sets yet.

I have the following characterizations of continuity:

Eugene

?

weird

Peter Tamaroff

$$\eqalign{
& f\left( {B\left( {a;\delta } \right)} \right) \subset B\left( {f\left( a \right);\delta } \right) \cr
& B\left( {a;\delta } \right) \subset {f^{ - 1}}\left( {B\left( {f\left( a \right);\delta } \right)} \right) \cr
& f\left( N \right) \subset M \cr} $$

Wher $M$ is a nbhd of $f(a)$ and $N$ is a nbhd of $a$.

The last one would be "for every $M$ of $f(a)$ there is a corresponding $N$ of $a$ such that...."

or equivalently $$N \subset {f^{ - 1}}\left( M \right)$$

Eugene

3:49 AM

yah that's about the same

Peter Tamaroff

Yeah. THe chapter works through them to get to a neater and neater theorem.

Eugene

read this en.wikipedia.org/wiki/…

Peter Tamaroff

The definite one is "$f$ is continuous at $a \in X$ $\iff$ for each nbhd $M$ of $f(a)$, $f^{-1}(M)$ is a nbhd of $a$."

Eugene

yah

that's about the same

so you just use the second property (i think) of basis

that there is a ball contained in the intersection

Peter Tamaroff

@Eugene Sorry, I have at hand only the definition of the basis. What do you mean?

Eugene

4:03 AM

what does your definition say?

Peter Tamaroff

@Eugene "A collection $\mathcal B_a$ of nbhds of $a$ is called a basis for the nbhd system at $a$ if every nbhd $N$ of $a$ contains some element $B$ of $\mathcal B_a$."

Eugene

oh

@PeterTamaroff en.wikipedia.org/wiki/Base_(topology)

Peter Tamaroff

@Eugene That means any nbhd $N$ of $a$ is the union of certain $B$ of $\mathcal B_a$?

@Eugene I'm still working in Metric Spaces though, not Topological Spaces.

Eugene

@PeterTamaroff i'm not sure of this.

Peter Tamaroff

@Eugene OK. I'll try and write something and get back to you.

Eugene

4:11 AM

@PeterTamaroff sorry i'm not so helpful.

Peter Tamaroff

@Eugene Well, you're a number theorist, not a topologist!

Eugene

@PeterTamaroff that's true. i'm also doing homework right now. maybe @JasperLoy can be of more help

user19161

@PeterTamaroff Well, this is trivial stuff you know!

user19161

@Eugene Good evening sir! Did you go out for booze?

Eugene

@JasperLoy nope

Peter Tamaroff

4:18 AM

@JasperLoy I have proved the $\Rightarrow$ part. Now I have to prove the $\Leftarrow$ part.

I have to prove that given an arbitrary nbhd $M$ of $f(a)$, then $f^{-1}(M)$ is a nbhd of $a$.

I have that is the case for any nbhd of $f(a)$ in the basis.

@JasperLoy I guess my problem is I don't know much properties about bases.

The only thing I know is that if $M$ is an arbitrary nbhd of $f(a)$, then $B\subset M$ where $B$ is some element of the basis.

user19161

@PeterTamaroff I think you mean here that f is continuous at a right?

Peter Tamaroff

@JasperLoy Yes.

user19161

@PeterTamaroff You mean now you are assuming that f is continuous at a and then proving that?

Peter Tamaroff

@JasperLoy No, no. I already assumed $f$ continuous at $a$.

This means that if $M$ is a nbhd of $f(a)$ then $f^{-1}(M)$ is a nbhd of $a$.

So that if $N\in \mathcal B_{f(a)}$ then $N$ is a nbhd of $f(a)$, so $f^{-1}(N)$ is a nbhd of $a$.

That is the first half of the proof.

user19161

@PeterTamaroff I mean you are assuming that f is continuous at a and then trying to prove that the inverse of a neighbourhood is a neighbourhood right?

Peter Tamaroff

4:29 AM

@JasperLoy No, I have that as a theorem.

1 hour ago, by Peter Tamaroff

@JasperLoy Would you say this is a hard excersise? "Let $f:X\to Y$ , $X$,$Y$ metric spaces. Let $a\in X$ and $\mathcal B_{f(a)}$ be a basis for the system of nbhds at $f(a)$. Prove $f$ is continuous $\iff$ for each $N\in \mathcal B_{f(a)}$, $f^{-1}(N)$ is a nbhd of $a$."

user19161

@PeterTamaroff OK, your message confused me.

Peter Tamaroff

@JasperLoy Sorry.

user19161

@PeterTamaroff See, this is where the confusion comes from.

Peter Tamaroff

@JasperLoy Oh. OK. Well, I have to prove that if for each $N \in \mathcal B_{f(a)}$, $f^{-1}(N)$ is a nbhd of $a$, then $f$ is continuous at $a$.

This translates into "if for each $N \in \mathcal B_{f(a)}$, $f^{-1}(N)$ is a nbhd of $a$, then given an arbitrary mbhd $M$ of $f(a)$, $f^{-1}(M)$ is a nbhd of $a$."

user19161

Oh sorry, I am confused because there are neighbourhoods and there are bases.

Peter Tamaroff

4:33 AM

@JasperLoy What I can start with is that $N \subset M$ for some $N$ in the basis and some nbhd $M$ of $f(a)$.

so $f^{-1}(N)\subset f^{-1}(M)$, so if $f^{-1}(N)$ is a nbhd of $a$, so is $f^{-1}(M)$. Done.

;)

user19161

@PeterTamaroff OK, let's do it!

user19161

@PeterTamaroff What is your definition of basis of system of nbhds that you are using?

Peter Tamaroff

31 mins ago, by Peter Tamaroff

@Eugene "A collection $\mathcal B_a$ of nbhds of $a$ is called a basis for the nbhd system at $a$ if every nbhd $N$ of $a$ contains some element $B$ of $\mathcal B_a$."

@JasperLoy

That is, a collection of nbhds of $a$, $B_a$ is a basis for the nbhd system at $a$ if for any $N$ a nbhd of $a$, $B \subset N$ for some $B \in B_a$

For example, a basis can be $B_a=\{B(a;\epsilon):\epsilon \in \Bbb Q\}$

user19161

@PeterTamaroff Is your definition of nbhd of x a set containing an open set containing x?

Peter Tamaroff

@JasperLoy A set $N$ is called a nbhd of $a$ if it contains an open ball centered at $a$.

I'm dealing with metric spaces nwo!

This is section $4$, Continuity. Next section is Limits, and the next one is Open sets and Closed sets.

user19161

4:42 AM

@PeterTamaroff Ah yes, that's why Mendelson is not exactly my cup of tea. He could have brought these in later in a more general setting!

Peter Tamaroff

@JasperLoy Bases?

@JasperLoy He redifined nbhds in topological spaces, though.

user19161

@PeterTamaroff Your proof is fine, except that you should not say "for some nbhd M" because M is already fixed!

user19161

@PeterTamaroff That is what I meant, yes.

Peter Tamaroff

@JasperLoy Hehhe OK!

user19161

@PeterTamaroff Yeah. Writing mathematics is just like writing English. You should write what you really want to mean instead of something else!

Peter Tamaroff

4:47 AM

@JasperLoy I watched Numb3rs today.

They said something like "People that are constantly tied to abstract thinking (i.e. mathematicians) are more prone to mental disease".

user19161

@PeterTamaroff Oh? What is that?

Peter Tamaroff

Or something of the sort.

@JasperLoy A series.

About a mathematician that is recluted by his brother, an FBI agent, to help him solve crimes.

Quite the rad thing to do.

►

Guess who is the mathematician

user19161

@PeterTamaroff Wait, did you mean recruited?

Peter Tamaroff

@JasperLoy Yeah. Does "reclute" mean anything?

user19161

@PeterTamaroff Not in my vocabulary.

user19161

4:52 AM

@PeterTamaroff I just watched a bit. I guess the guy showing the flowers. He seems pretentious!

Peter Tamaroff

@JasperLoy He tries to solve P=NP

►

user19161

@PeterTamaroff Oh I like the girl hehe!

Peter Tamaroff

@JasperLoy She's nice. Not a super hot woman like those fake CSI programs.

Hehehe

user19161

@PeterTamaroff I prefer the girl next door look!

Peter Tamaroff

@JasperLoy Yeah, me too.

user19161

4:57 AM

@PeterTamaroff Do you have a gf now?

Peter Tamaroff

@JasperLoy Nay.

Do you?

user19161

@PeterTamaroff OK. Never mind. Just do the math and then all the girls will be impressed, hehe.

Peter Tamaroff

@JasperLoy I guess I have a small fanbase already =P

user19161

@PeterTamaroff Good for you!

Peter Tamaroff

@JasperLoy Fuck. I just wrote "a positive rational integer".

LOL

user19161

5:01 AM

@PeterTamaroff You probably wanted positive integer or positive rational number instead.

Peter Tamaroff

@JasperLoy yes, the latter.

user19161

@PeterTamaroff Note that natural numbers may or may not include 0, depending on your choice of definition.

Peter Tamaroff

Halmos includes it. I guess I'll side with him. Whatever.

user19161

@PeterTamaroff I prefer to include it, following Bourbaki as well.

Peter Tamaroff

@JasperLoy Did you read Buorbaki?

user19161

5:05 AM

@PeterTamaroff I browsed through some of their books. I like their General Topology.

Peter Tamaroff

@JasperLoy What is the difference between "General" and "Point Set" Topology?

I know there is one called "Pointless" (LOL) Topology.

user19161

@PeterTamaroff No difference to me. Though set topology or set-theoretic topology seems to be something else.

user19161

@PeterTamaroff Oh yes, Ilya mentioned it to me that day.

user19161

@PeterTamaroff Similarly, modern algebra pretty much equates to abstract algebra.

BenjaLim

5:46 AM

hey

Peter Tamaroff

@BenjaLim Bennnnnnnnn

I'm doing some Topology here,

BenjaLim

ok

yes?

Need help?

Peter Tamaroff

@BenjaLim How are you?

BenjaLim

not bad

Peter Tamaroff

@BenjaLim I guess no, for now.

BenjaLim

5:48 AM

what are you doing in topology?

Peter Tamaroff

@BenjaLim Proving no finite collection of subsets of $\mathbb R$ can be a basis for the system of nbhds of $a\in \Bbb R$.

BenjaLim

what do you mean by system of neighbourhoods? @PeterTamaroff

Peter Tamaroff

@BenjaLim That's a name in the definition.

BenjaLim

huhuhuhuh?

I don't understand......

@PeterTamaroff ah ok

Peter Tamaroff

A collection $B_a$ of neighborhoods of $a$ is called a basis for the system of neighborhoods at $a$ if for any $N$ a nbhd of $a$, there is a $B$ in $B_a$ such that $B\subset N$.,

BenjaLim

5:51 AM

@PeterTamaroff what is the topology on $\Bbb{R}$?

Peter Tamaroff

@BenjaLim I only have metrics for the moment, and open balls.

BenjaLim

ah ok :D

actually

your problem is not true in the trivial topology :D

user19161

@PeterTamaroff Never had one in my entire life.

Peter Tamaroff

@JasperLoy But sure you fancied someone sometime.

user19161

@BenjaLim He's just doing metric spaces at the moment.

Peter Tamaroff

5:53 AM

@BenjaLim In $d(x,y)=1$ for $x\neq y$, $d(x,y)=0$ for $x=y$?

user19161

@PeterTamaroff Yes, many many hehe.

Peter Tamaroff

@BenjaLim I proved it by contradiction.

I actually drew the "proof" in my board and then I put it in words.

BenjaLim

I think you can use compactness

user19161

@PeterTamaroff You use a board? I prefer pen and paper.

Peter Tamaroff

@BenjaLim $\{a \}$ is a basis for the system of nbhds, yes.

user19161

5:56 AM

Unless it's a really big chalk board.

BenjaLim

Your neighborhoods are open balls yes?

Peter Tamaroff

@JasperLoy The good thing is that I can write stuff with no compromise in the board. I just erase it and move on. In paper I have to use correction fluid and it messes things up.

@BenjaLim Well, open balls are nbhds, but a nbhd of $a$ is a set containing an open ball about $a$.

And open balls are nbhds of all of its points.

BenjaLim

Ok

user19161

@PeterTamaroff You are not submitting it or keeping it for yourself. Why bother?

Peter Tamaroff

@JasperLoy Because I waste paper.

I'm actually running out of paper now. I have not much left.

user19161

5:58 AM

@PeterTamaroff Better than wasting correction fluid and breathing in smelly things.

Peter Tamaroff

So before writing out anything I prefer to use pen or use the board.

BenjaLim

Ok

Peter Tamaroff

$(a-\epsilon,a]$ is not a nbhd of $a$, right?

BenjaLim

no

@PeterTamaroff ok

Peter Tamaroff

@BenjaLim So what are you doing? What time is it there?

BenjaLim

6:03 AM

it's 4pm

actually

YOu just took the intersection yes?

now

Peter Tamaroff

@BenjaLim The intersection of what?

user19161

@PeterTamaroff What is your time there?

BenjaLim

all the finitely many basis elements

Peter Tamaroff

@JasperLoy 3:05 am

BenjaLim

@PeterTamaroff

Peter Tamaroff

6:05 AM

@BenjaLim Not really. I just constructed a nbhd of a for which no element of the basis is contained in.

user19161

@PeterTamaroff 2 pm.

BenjaLim

ok

@PeterTamaroff how did you construct that neighbourhood?

Peter Tamaroff

@BenjaLim First I stated that any element of the "basis" is an interval (closed, open, mixed, whatever).

BenjaLim

ok

no wait

you have problem

user19161

Anyway there is a result that says that an open set in R is a countable disjoint union of open intervals. Obviously this fails in higher dimensions. I thought I would mention it. It usually goes by the name of Lindelof.

BenjaLim

6:09 AM

you want the intersection of two basis elements to be another baiss element yes?

Peter Tamaroff

@BenjaLim ?

@BenjaLim Not really. That hasn't brought up

Let me finish writing the proof.

BenjaLim

@PeterTamaroff ok

Peter Tamaroff

Then I argued: since there are finitely many, there is a $m$ such that for any interval $(a,b)$ (or closed or whatever), we have $m\leq b$ regardless what $b$ is. Similarily, there is an $n$ such that $n \leq a$ regardless what $a$ we chose. Then I choose $\epsilon$ and $\delta$ such that $a+\epsilon <m$ and $a-\delta <n$. The interval $(a-\delta,a+\epsilon)$ is a nbhd of $a$ but none of the intervals is a subset of it.

BenjaLim

yes

but there's a slicker way to say it

by the way in a basis, you usually want the elements of the basis to be elements of the topology as well, i.e. open

The way I would put it is

Suppose there are only finitely many basis elements

Peter Tamaroff

@BenjaLim I don't even know what a topology is. (I mean I know it is a collection of sets but I have no idea what it is)

BenjaLim

6:13 AM

take their intersection

which is non-empty because they all contain the point $a$

and which is open

you can choose $\epsilon$ so that $B_\epsilon(a)$ is completely contained in the intersection

and clearly there is no basis element that is in that ball. Why?

Peter Tamaroff

@BenjaLim Oh, sure.

BenjaLim

slick enough?

Peter Tamaroff

@BenjaLim I was actually going to take the lub and glb, but then decided for two glb. If I had chosen the former I'd had ended with your proof, but more wordy.

BenjaLim

@PeterTamaroff when you work with topology

you don't have a metric

so things like sup, inf don't come in a lot

then you have to think about taking intersections

Peter Tamaroff

@BenjaLim Right.

BenjaLim

6:17 AM

@PeterTamaroff actually

this problem can be generalised as follows

suppose you have a metric on $\Bbb{R}$

let us take it now to be the euclidean metric

@PeterTamaroff Actually you have got me thinking

is there a metric on $\Bbb{R}$ that makes it compact?

Peter Tamaroff

@BenjaLim Hahaha well that is good.

BenjaLim

I know there is a topology on $\Bbb{R}$ that makes it compact.

Peter Tamaroff

Now I need to prove that given a point $a$ in a metric space $X$, there is a countably infinite collection of neighborhoods of $a$ such that they are a basis.

Or in other words, there is a family $\{B_n\}_{n\in \Bbb N}$ which is a basis.

BenjaLim

Just take the collection of all open intervals about $a$ with rational endpoints.

Peter Tamaroff

@BenjaLim This is not $\Bbb R$ anymore. It is just a metric space $X$.

BenjaLim

6:21 AM

@PeterTamaroff Actually if all metrics on $\Bbb{R}$ are equivalent then what I said is not true.

@PeterTamaroff all metric spaces are first countable, so you just take all balls of radius $1/n$ about $a$

Peter Tamaroff

I think balls centered at $a$ with radius $1/n$

will do

@BenjaLim "first" countable?

BenjaLim

First-countable space

In topology, a branch of mathematics, a first-countable space is a topological space satisfying the "first axiom of countability". Specifically, a space X is said to be first-countable if each point has a countable neighbourhood basis (local base). That is, for each point x in X there exists a sequence U1, U2, … of open neighbourhoods of x such that for any open neighbourhood V of x there exists an integer i with Ui contained in V. Examples and counterexamples The majority of 'everyday' spaces in mathematics are first-countable. In particular, every metric space is first-countable. To see ...

Peter Tamaroff

I don't know about topological spaces dude!

BenjaLim

hahahahhahahaha

Peter Tamaroff

@BenjaLim And this is asking me to prove precisely that!

BenjaLim

6:24 AM

there is nothing to prove

really you just stated the "proof"

Peter Tamaroff

@BenjaLim It is an axiom?

BenjaLim

I would rather say it's an observation

@PeterTamaroff actually

Peter Tamaroff

@BenjaLim Why? Mendelson is asking me to prove it....

BenjaLim

I don't believe all metrics on $\Bbb{R}$ are equivalent @JonasTeuwen what do you think

Peter Tamaroff

@BenjaLim You don't belive things in math.....

BenjaLim

6:26 AM

@PeterTamaroff well..

Peter Tamaroff

@BenjaLim Are you asking if $(\mathbb R^n,d)$,$(\mathbb R^n,d')$ and $(\mathbb R^n,d'')$ are metrically equivalent?

@BenjaLim I'm just being pedantic =D

BenjaLim

you say two metrics are equivalent, if they generate the same topology :D

Peter Tamaroff

@BenjaLim I say two metrics are equivalent if there is a bijection $f$ such that $d(f(x),f(y))=d'(x,y)$

BenjaLim

I have never seen that definition before.

@PeterTamaroff en.wikipedia.org/wiki/Equivalence_of_metrics

@PeterTamaroff I should go

Peter Tamaroff

@BenjaLim I'll go and sleep....

BenjaLim

6:36 AM

yes bye man!!

Peter Tamaroff

@BenjaLim Only 130 pages for "Locally Compact Hausdorff"!

Hehehehe

BenjaLim

that's not such a difficult definition

@PeterTamaroff As I always say

there is no rush to do things quickly.

For example,

I would never do algebraic number theory without galois theory

without computing tons of Galois groups and stuff

Peter Tamaroff

@BenjaLim Hm. OK. Bye byes.

BenjaLim

bye

Jonas Teuwen

8:35 AM

8-).

@BenjaLim $\mathbf R$ is finite dimensional... So...

Do you have a topological reason for the equivalence? We just have $d \sim d'$ for all metrics.

@PeterTamaroff That point metric is quite handy for counterexamples! 8-).

Old John

@JonasTeuwen Happy "completed orbit" day!

Jonas Teuwen

@OldJohn Thanks :-).

@BenjaLim Check out "all norms on finite-dimensional spaces are equivalent", so I wonder what your statement is precisely :-).

BenjaLim

8:54 AM

@JonasTeuwen not every metric comes from a norm

Jonas Teuwen

@BenjaLim Thanks. Yes.

@BenjaLim So...? You want to have the same topology.

BenjaLim

yes

Jonas Teuwen

So I wonder: what do you mean with same topology...?

BenjaLim

they generate the same open sets

Jonas Teuwen

They have the same continuous functions...?

BenjaLim

8:55 AM

@JonasTeuwen I was just wondering if there is a metric on $\Bbb{R}$ that makes it compact.

so if all metrics are equivalent

Jonas Teuwen

@BenjaLim In the topological sense?

BenjaLim

then there is none

yes @JonasTeuwen

So

are all metrics on R equivalent in the topological sense?

Jonas Teuwen

@BenjaLim A metric space is compact if it is complete and totally bounded.

Transcript for

$Mathematics$ Mathematics