Mathematics - 2012-07-15 (page 2 of 3)

nbouscal

09:00

No matter what metric you put on R, there will always be a open cover with no finite subcover.

Paul Manta

Do you guys know any general courses on undergraduate calculus that I can find on the web? I know some universities tape their courses and post them on the web.

BenjaLim

@nbouscal how come?

nbouscal

MIT OCW has online courses that would probably suit you, Paul.

BenjaLim

@JonasTeuwen Yes I know that already.

Jonas Teuwen

@BenjaLim But your statement means that both have the same continuous functions. So take a metric, and see what that means.

nbouscal

09:02

Because of the nature of the continuum. Even just looking at the open interval (0,1), you can see that there will always be an open cover with no finite subcover. I'm not sure how to better explain it for you.

BenjaLim

@nbouscal when you say open interval you are already assuming "euclidean metric"

nbouscal

No, I'm speaking of a general topological interval based on the ordering of the real line, not necessarily the euclidean metric

You can have an open interval with any simple ordering

BenjaLim

@nbouscal no need to go to general topological spaces, just metric spaces

@nbouscal And I think the order topology on $\Bbb{R}$ is the same as euclidean topology

nbouscal

The two are the same topology, yes, but that does not mean that I'm assuming the Euclidean metric. They just happen to be the same in this case, because of the nature of the real line.

If you want R to ever be compact, you basically need a way to fit it into a finite collection of boxes. There is no way to accomplish that.

Just picture intervals (n-1,n+1) as n ranges over Z. Whatever metric you happened to assign, there will still never be a finite subcover for that cover.

Jonas M.

I just wanted to remark that the trivial metric d(x,y) = 1 if x =/= y and 0 if x = y induces the trivial topology on |R^{n} (on any set, actually). Therefore, non-equivalent metrics on euclidean spaces do exist

And I think they can even be made compact, by just using a bijection from [0,1] to \R and then pushing forward the metric on [0,1] induced by the usual metric

but I really shouldn't even be here... I'm just procrastinating :(

bye

nbouscal

09:18

I'm with you on non-equivalent metrics, if I remember rightly there are quite a few.

I don't follow on compactness, though.

There isn't a bijection from [0,1] to R, because [0,1] has a least element and a greatest element, and R does not.

BenjaLim

09:46

@JonasM You are completely wrong. That is the discrete metric that induces the $discrete topology$ on $\Bbb{R}^n$. The trivial topology is not metrisable.

BenjaLim

10:15

@AsafKaragila shall we chat here?

J. M.

10:34

@BenjaLim I don't think that'll work, as he hasn't been around for quite a while. You need to link to the last message he posted here to ensure he gets pinged.

N3buchadnezzar

@JM

Did that ping you ?

J. M.

Yes?

@N3buchadnezzar No, it did not.

N3buchadnezzar

Oh, I do not think it did. You have a special username.

J. M.

I'm currently immune to those sort of pings, but it seems they'll be fixing that quirk soon...

nbouscal

@JM You can use this sort, though, right?

J. M.

10:36

@nbouscal Yes, that works.

nbouscal

This chat is actually pretty well-designed.

N3buchadnezzar

Indeed

$ \Huge \mathbf{ WABBAWABBA }$

The bad kerning, it hurts..

Martin Sleziak

10:54

@JM BTW I don't think you're the only one who gives rat's ass, as you've put it here.

Willie Wong certainly follows new tags, as can be seen from his actions, see here.

I'm pretty sure several other people check new tags regularly.

BTW I've posted a short list of removed tags here.

The list in the long thread does not contain "split tags" such as (algebra) split into (algebra-precalculus) and (abstract-algebra).

J. M.

11:17

@MartinSleziak Sigh, you're right. I'm just frustrated that it has kept popping up over the last four days...

Martin Sleziak

no problem

Speaking of tags, what would be a good tag for this: How many times clock hands coincide in a half day? Perhaps (geometry)?

Related question you've found was tagged (algebra-precalculus)

And also (quantitative-aptitude) - but I am not really sure what that means.

J. M.

@MartinSleziak I'm not too fond of that quantitative thingie myself. I've tagged it algebra-precalculus.

Martin Sleziak

@JM Sounds good. Thanks!

BenjaLim

11:43

@MartinSleziak hey

can I ask you some algebraic topology?

It's probably a simple question

Martin Sleziak

I know almost nothing about algebraic topology.

BenjaLim

oh.....

Martin Sleziak

But you were probably not addressing me, but people in this chat in general.

BenjaLim

@MartinSleziak I know it's problem if we talk about the set of all sets

we get russell's paradox

Martin Sleziak

Yes.

BenjaLim

11:44

but what is not wrong when you talk about the category of sets?

Martin Sleziak

you can also talk about class of sets.

Some things are allowed when working with collection of all sets, some are not. (E.g. some things that would lead to Russell's paradox.)

BenjaLim

@MartinSleziak I have to say my set theory is not very deep

Martin Sleziak

Most of categories you know have class of objects.

BenjaLim

yes

Rajesh D

Does anyone have a copy of Feynman Lectures on Physics with them right now?

BenjaLim

11:46

Like the category $\mathscr{C}$ of topological spaces

Martin Sleziak

Small categories are categories where you only have set of objects.

BenjaLim

@MartinSleziak yes

Martin Sleziak

I would say that most books on category theory don't bother too much with classes.

I don't think you should worry too much about it.

BenjaLim

ok

so I should just take when you say a "class of objects"

like

topological spaces - morphisms - continuous functions

groups - group homomorphisms $\textbf{Grp}$

@MartinSleziak I realised I need to know it

because in my book it verifies that $\pi_1$ defines a functor from $\textbf{Top}$ to $\textbf{Grp}$

Rajesh D

Does anyone here know about interference?

BenjaLim

11:50

physics.se is probably a good place to ask @RajeshD

Martin Sleziak

@BenjaLim I don't think you need some deep knowledge of classes for that.

Rajesh D

yeah I am going there

BenjaLim

yeah @MartinSleziak

Martin Sleziak

Functions between classes work similarly as between sets.

BenjaLim

I am now confused between "class" and "set"

Martin Sleziak

11:51

Every set is class, but some classes are "too large" to be sets.

You know that Set is proper class.

BenjaLim

proper class?

Martin Sleziak

proper class = a class which is not a set

BenjaLim

ok

Martin Sleziak

Top is "larger" than Set, since you have a topological space on every set. So it is a proper class.

BenjaLim

@MartinSleziak when did you learn category theory?

Martin Sleziak

11:52

Grp is a proper class for the same reasons.

BenjaLim

@MartinSleziak Yes because given any set we can always put the trivial topology/discrete topology on it

Martin Sleziak

@BenjaLim I would say I started during my 4th year at the university. (I was 22.)

BenjaLim

oh wow

Martin Sleziak

There is plenty of things I do not know about categories, which would probably be useful for me to know.

BenjaLim

@MartinSleziak given any set how do you turn it into a group?

Martin Sleziak

11:54

Good question.

BenjaLim

So why is it a proper class?

It's not like in a topological space I always have several topologies available to me immediately

Martin Sleziak

Finite sets - you have $(\mathbb Z_n,\oplus)$.

BenjaLim

$\Bbb{Z}/n\Bbb{Z}$ under addition?

Martin Sleziak

yes

Perhaps for infinite cardinalities you could try additive groups of vector spaces over $\mathbb Q$.

BenjaLim

ok

seems like there are not many algebraic topologists on math.se...

Martin Sleziak

11:59

perhaps free groups are easier?

BenjaLim

not very familiar with free groups TBH

But I will learn it along the way in AT I think

Martin Sleziak

Cardinality of free Abelian group is the same as the cardinality of basis, I guess.

BenjaLim

ok

Martin Sleziak

Does every set has a group structure?

It would be surprising if such question had not been asked here before.

BenjaLim

oh wow

so are you still sure that the class of all groups is a proper class?

wow my god arturo

he is pro at everything

including advanced set theory

Martin Sleziak

12:05

If you believe Arturo's answer (and other answers from that thread), it is indeed a proper class.

BenjaLim

@MartinSleziak Is it true that the cardinality of every set is either countable or uncountable?

Martin Sleziak

@BenjaLim Should I understand this question as: Is it true that for any set we have either $|A|\le\aleph_0$ or $|A|>\aleph_0$?

Yes in ZFC. (=if we assum Axiom of Choice)

BenjaLim

ok

@MartinSleziak But when you say of cardinality greater than aleph 0

there are different kinds of infinities if I understand correctly

Martin Sleziak

yes

BenjaLim

set theory is messed up man

Martin Sleziak

12:18

If we have Axiom of Choice, it is the same as saying $|A|\ge\aleph_1$.

BenjaLim

N_1?

Martin Sleziak

=the smallest uncountable cardinal.

BenjaLim

is that $\omega$?

Martin Sleziak

$\omega$ and $\aleph_0$ are the same sets (at least the usual representation in ZFC is the same)

Did you mean $\Omega$? It is sometimes uses as notation for $\omega_1$.

BenjaLim

ah ok

yeah

sorry my set theory is not very deep

I should take the foundations of maths course next year

Martin Sleziak

12:20

Naive set theory is sufficient for most things.

BenjaLim

:D

Rajesh D

Here If anyone interested to answer a question on Physics.SE

Rajesh D

12:41

Hey @FrankScience

Jeff

13:06

A quickie question if anyone is around: Is the opposite of 'there exists' and 'for all' the same thing? The opposite of 'there exists' is 'there doesn't exist' and the opposite of 'all' is 'none'. 'does not exist' is the same as 'none'. Yes?

Martin Sleziak

@Jeff Wikipedia: Universal_quantification - Negation and Existential_quantification - Negation.

Negation of: "There exists a flower which smells nice." is "All flowers smell bad". (If we consider bad as negation of nice here.)

Jeff

isn't it better to do the negation as 'no flowers smell nice'?

Martin Sleziak

Yes. that's correct.

Jeff

and isn't that also the negation of 'all flowers smell nice'? or not! I guess the negation of 'all flowers smell nice' is 'there exists a flower that doesn't smell nice'...

ok. i think i got it. thx

Martin Sleziak

I just wanted to stress the change between $\exists$ and $\forall$.

Frank Science

13:18

@RajeshD Hello, what's up?

Jeff

@martin TY. gotta go.

Rajesh D

nothin much

Frank Science

@RajeshD I have just re-computed my question and got the same answer.

Rajesh D

@FrankScience Did you edit the question giving your new computation?

Frank Science

@RajeshD Almost the same.

Rajesh D

13:28

@FrankScience I don't have much exp with asymptotics, but got to admit, it looks very intimidating! May be you should consider adding a short summary explaining what is actually happening there and what is the peculiarities/interesting things about this problem.

Frank Science

@RajeshD Intimidating? Little skill is needed, therefore the computation is so mechanic. The only interesting thing is that, we get an asymptotic value for the extremely large thing, as I've mentioned in the end.

Rajesh D

@FrankScience : "computation is so mechanic" (and lengthy). That's what I mean by intimidating.

Frank Science

@RajeshD Do you have any idea?

Rajesh D

@FrankScience No. Moreover this question doesn't seem to interest me as there are some other things bothering me right now. I am afraid I can't contribute anything here.

Frank Science

13:51

How can I contain multiline non-aligned equations in LaTeX?

Rajesh D

@FrankScience Here A short recipie, but have to look at other things (may be a book) for more deeper things. Looks like your question badly needs this.

I guess you need to avoid equations like $A = B = C =..$ in a single line.

Frank Science

@RajeshD No. $A=B, C=D, E=F$.

Frank Science

14:16

@RajeshD Thanks.

Rajesh D

14:27

@FrankScience You cannot have $A=B=C=D$, instead it should be $$\begin{align} A &= B \\ &=C \\ &=D\\ \end{align}$$

especially in simplifications

Frank Science

@RajeshD I knew it.

@RajeshD \begin{equation}\begin{split}A&=B\\&=C\\&=D\end{split}\end{equation} is more appropriate.

Rajesh D

Why not implement it in your question. I am sure it will garner you more attention from users. Lot of people avoid questions simply because they are not in a convinient form to read.

Frank Science

@RajeshD If you used that, the tag would be added automatically in the real-world $\LaTeX$.

Rajesh D

@FrankScience Which tag?

Frank Science

@RajeshD Just like \begin{align}A&=B\tag1\\&=C\tag2\\&=D\tag3\end{align}

Rajesh D

14:32

Hmm...there are ways to avoid it, I don't recall them. May be someone could help

Frank Science

@RajeshD I know \nonumber helps, but the more appropriate one is split environment for equation.

@RajeshD Because it's exactly one equation, not many equations.

Rajesh D

@FrankScience What ever you need to use the above said ^^. Your question badly needs it I guess.

Frank Science

@RajeshD Now I'm editing the postscript offline. I'll edit the posted question. When it's justified, I'll add some explanations for each equation, but only when it's justified.

Rajesh D

ok

Lets see how it comes.

Frank Science

@RajeshD You know editing online will slow down the system. I'm using an obsolete computer.

Rajesh D

14:37

Oh, ok. I can understand.

Frank Science

14:48

@RajeshD Awful. In $\LaTeX$, I use fleqn, but here it's not.

alan2here

numbers, vectors, matracies, thease are all "groups"?

whats the generic turm, I hear they are all part of a larger thing.

and that cirtain operations, like addition by definition can occur between any of them

although there may be a differnt implmentation, for example the process for adding vectors is diffent to that of numbers.

Rajesh D

15:05

@FrankScience Why use gather?

Frank Science

@RajeshD non-aligned equations. align does not seem reasonable.

user19161

@jonas How was your birthday?

Frank Science

@JasperLoy I re-computed and got the same answer.

Jonas Teuwen

@JasperLoy Grading.

Rajesh D

@FrankScience IMO & AFAIK only align should be used. It looks good that way. And one should avoid chained equations $A = B = C$, whatever the situation may be. Also putting things in double dollars $$..$$ is good practice. When ever it looks clumsy use line breaks. This should do the trick. Whatever, include all the explanation inside the answer, and hope someone will take care of the type setting.

user19161

15:13

@FrankScience Congrats!

alan2here

maybe I should start a question, but I think I may need to improve my terminology first, and it's bound to be duplicate :(

user19161

@JonasTeuwen Sad. Go have a beer later!

Rajesh D

@alan2here May try asking here informally first.

user19161

@alan2here Not clear what your question is, but maybe you are just thinking of the abstract group?

Frank Science

@RajeshD I think $a=b=c$ is needed when it's a very immediate one. and \begin{align*}a&=b=c\\&=d\end{align*} seems acceptable when $b=c$ is more easier than $c=d$.

Rajesh D

15:16

@FrankScience I am talking about the appearence Frank, not the logic. "Leave the logic and take care of the appearance"..This is the mantra I guess

Frank Science

@RajeshD And double dollars should be avoided in the real-world $\LaTeX$, which is introduced in plain $\TeX$. Instead, use \[...\].

unNaturhal

Good morning people!

Jonas Teuwen

@JasperLoy Na, tired.

Rajesh D

@FrankScience yes but double dollars are most welcome on this site

@FrankScience Anyway if your computer is slow don't struggle with these things, just add all the explanation, someone will take care of the alignment and such stuff.

Frank Science

@RajeshD I use it in this site when I edit by hand, but my answer is cloned from my $\TeX$ file.

Rajesh D

15:19

^^

Frank Science

@RajeshD And this, will cost me much time to do. I'll rebuild the logic of the proof bottom-up.

user19161

@FrankScience If you have problems with TeX maybe check it out on the TeX site.

alan2here

@JasperLoy Thanks, this could be the case

@JasperLoy I've just asked math.stackexchange.com/questions/171152/…

@JasperLoy and seen your reply. I've tried sources like Wikipedia but don't find them greatly clear.

user19161

@JonasTeuwen Have fun bro. I'm off. I think I will use xypic for my arrow diagrams...

robjohn

17:15

Arrgh! This be a quiet Sunday.

unNaturhal

:p

@robjohn can I ask you a question?

(noob question)

robjohn

@unNaturhal certainly

unNaturhal

@robjohn :)
If I'm studing a function (real-valued function of real variables), assuming my Domain is $\forall x \in \mathbb{R}\backslash \{1\}$, and the Domain of first derivative is the same, I can say that there aren't points where the function is not differentiable?

robjohn

@unNaturhal If it is differentiable at every point of its domain and you include the domain when describing the function, then yes.

@unNaturhal For example $\frac{1}{1-x}$ whose derivative is $\frac{1}{(1-x)^2}$

unNaturhal

@robjohn Ehm... sorry but I haven't understood..

robjohn

17:28

@unNaturhal The function $\frac{1}{1-x}$ is defined and differentiable at all points of $\mathbb{R}\setminus\{1\}$

@unNaturhal So there are no points of $\mathbb{R}\setminus\{1\}$ at which $\frac{1}{1-x}$ is not differentiable

@unNaturhal However, it would be more natural to say that it is differentiable at all points of $\mathbb{R}\setminus\{1\}$

unNaturhal

@robjohn I have understood *_*

robjohn

@unNaturhal :-)

unNaturhal

@robjohn In practice, if the $D$ and $D\,'$ coincide, we can say that there aren't point of the DOMAIN where the function is not differentiable

robjohn

@unNaturhal I would say that the function is differentiable at all points of its domain (avoiding the double negative).

unNaturhal

@robjohn Yeah, it's more correct :) Thanks :D

Peter Tamaroff

17:41

►

robjohn

@PeterTamaroff I'll have to watch what I say here.

Peter Tamaroff

@robjohn Watch it. It is kinda funny.

robjohn

@PeterTamaroff I watched it, that is why I commented about watching what I say.

Peter Tamaroff

@robjohn Oh, hehe OK.

Peter Tamaroff

18:13

This is silly, but. Let $X$ be a metric space. Let $a,b$ be different points in $X$. Prove there are nbhds $N_a$ and $N_b$ such that $N_a \cap N_b =\varnothing$

PROOF

Let $\gamma = d(a,b)$

Choose $\epsilon=\delta < \gamma/2$

Then $B(a;\epsilon)$ and $B(b;\delta)$ are nbhds of $a$ and $b$ and they are disjoint.

anon

yes

wait, no

don't you mean $B(b,\epsilon)$?

nevermind

Peter Tamaroff

@anon Well, $\epsilon=\delta$ XD

Just a stupid choice of letters form my part.

Jonas Teuwen

18:36

@PeterTamaroff You mean you need to prove that it is Hausdorff? Prove that it is normal!

Peter Tamaroff

@JonasTeuwen Whaaaaaat? One day your kids will have to prove spaces are Tamaroff, and you'll see!

Jonas Teuwen

@PeterTamaroff "Prove that the only topology on the space is the trivial topology (i.e. Tamaroff)".

anon

I was just about to say that.

Jonas Teuwen

@anon Idiots think alike. Also geniuses. Which one do you prefer?

I will add that as a definition to my topology syllabus. "The Tamaroff topology".

Peter Tamaroff

Now I need to prove given $n$ metric spaces $(X_1,d_1),\dots,(X_n,d_n)$, then by converting $X=\prod_{i..n}X_i$ to a metric space $(X,d)$ by setting $d(x,y)=\max_i \{d_i(x_i,y_i)\}$, the ball about $x=(x_1,\dots,x_n)$ is the product of balls $B_1,\dots,B_n$ about $x_1,\dots,x_n$

Jeff

18:39

Hey, anyone here heard of Predicate Logic? If so, why is my answer being marked wrong:

Sonny was a better quarterback than any of the others

Let Sxy be "x was a better qb than y"
My answer: (^x)Qsx (for all x, Sonny was a better quarter back than x)

Several other answers were marked wrong, too. No explanation provided.

anon

I prefer geniuses on my team and idiots on the other teams, naturally.

Peter Tamaroff

@anon Maybe you can convert some idiots if they are in your team.

Jonas Teuwen

@PeterTamaroff What is there to prove...?

Jeff

@PeterTamaroff that's quite the task! :D

@PeterTamaroff no matter how hard people try, I'm still an idiot. :D

Peter Tamaroff

@JonasTeuwen A ball on $X$ is the product of balls on $X_i$, $i=1..n$

@Jeff Hehehe, probably.

►

anon

18:45

Prove inclusion in one direction, and then the other direction.

Peter Tamaroff

@anon AoE

robjohn

@Jeff who's marking them wrong?

Jeff

computer based training

anon

AlmOst Everywhere? Age Of Empires?

Jeff

i don't have time to find the page atm

robjohn

18:46

@Jeff What's the name of the program?

Peter Tamaroff

@anon Axiom of Extension hehehe

Jeff

@robjohn it's somewhere on the site thelogiccafe.net

somewhere in chapter 6 of thelogiccafe.net/PLI

robjohn

@Jeff Does it give you what "Sxy" is, or are you providing the abbreviations?

Jeff

they gives it

robjohn

@Jeff Is that the only abbreviation they give, or is there a predicate for "x is a quarterback"

Jeff

18:51

they give abbrevs: Qxy, 's is better quarterback y' and and s, 'Brandon'.

supposed to use letters after 'v' for variables

@robjohn it sounds like you think there's a program flaw? the answer is right, no?

robjohn

@Jeff There doesn't seem to be enough predicates to symbolize that statement.

Jeff

@robjohn Really? OK, let me go find the page again and see that I've given it to you right. Personally, I think they are wrong and I was just looking for confirmation from someone who already knows the subject. Give me minute to find it.

robjohn

@Jeff I have written a program similar to that, and so I was trying to find out if you were using that. :-)

Jeff

did you write it in HTML for thelogiccafe.net? :D

@robjohn Click the following link, click 'Start', do problem 8.

thelogiccafe.net/PLI/Ech6T1g2.htm#start

robjohn

@Jeff No, mine is in Java for UCLA.

@Jeff I have pop-ups blocked, so I didn't get the whole problem.

Jeff

18:59

that page didn't use popups <confused>

robjohn

@Jeff I don't know the syntax they expect, but with the predicates they give, $\forall xQsx$ should seem to be correct.

@Jeff It told me that my browser blocked a pop-up

Peter Tamaroff

@anon Darn, just realized Halmos uses $$\mathop{\Huge\bf{ \times }}\limits_{i=1}^n$$ for $$\prod_{i=1}^n$$

Jeff

@robjohn that's a good enough answer for me. I'm just using that site for myself (not taking the course), so I don't really care if my syntax doesn't match theirs.

...also, I have the right syntax. :D thx

Peter Tamaroff

19:29

@JonasTeuwen Are you around?

Matt N.

19:40

Sometimes I get really dumb comments.

Peter Tamaroff

@MattN I don't even....

@MattN Do you know some topology?

anon

just ask your question brah

Peter Tamaroff

@anon I provided a proof but I think it is wrong.

anon

k

Peter Tamaroff

$(1)$ Let $B_i=B(a_i;\delta_i)$ be open balls about $a_i \in X_i$, with $i=1,\dots,n$. Then

$$B = \prod\limits_{i = 1}^n {{B_i}} = \left\{ {\left( {{x_1}, \ldots ,{x_n}} \right):\begin{cases} d_1(x_1,a_1)&<&\delta_1 \cr &\vdots &\cr d_n(x_n,a_n)&<&\delta_n\end{cases}} \right\}$$

Conversely, $$B\left( {a,\delta } \right) = \left\{ {\left( {{x_1}, \ldots ,{x_n}} \right):\mathop {\max }\limits_{1 \leqslant i \leqslant n} \left\{ {{d_i}\left( {{x_i},{a_i}} \right)} \right\} < \delta } \right\}$$

That is the proof of

> Let $(X_i,d_i)\;,\;i=1,\dots,n$ be metric spaces. Convert $$X=\prod_{i=1}^n X_i$$ into a metric space $(X,d)$ by setting $d(a,x)=\max\limits_{1\leq i \leq n}\{d_i(a_i,x_i)\}$.

>$(1)$ Prove that an open ball in $(X,d)$ is the product of open balls from $X_1,\dots,X_n$ respectively.

anon

19:47

that's correct

Peter Tamaroff

@anon OK.

Now I'm proving $(2)$ Let $a_i\in X_i$, and let $\mathfrak B_{a_i}$ be a basis for the neighborhood system at $a_i$ [viz. every neighborhood $N$ of $a_i$ contains some neighborhood $B$ of the basis]. Let $\mathfrak B_a$ be the collection of all sets of the form $B_1\times \cdots \times B_n$ with $B_i \in \mathfrak B_{a_i}$. Prove that $\mathfrak B_{a}$ is a basis for the neighborhood system at $a=(A_1,\dots,a_n)\in X$.

It seems to follow from $(1)$.

anon

yeah

If N is a neighborhood of a, then N contains an open set (ball) B containing a, and B will decompose as a product of "indexed balls", each of which is a neighborhood of the $a_i$ component in the $i$th space and hence contains some element from the basis. now put things together.

Matt N.

Ah, anon is helping. Good : )

anon

well, N contains an open set, which will contain a ball, anyway

Peter Tamaroff

@anon Actually my definition of nbhd of $a$ is a set $N$ containing an open ball about $a$. Open sets are 2 sections away still.

anon

19:59

well then it works out just a tad quicker

if you really wanna learn general topology, topology without tears is the way to go. I didn't get terribly far but it was exceptionally straightforward.

Peter Tamaroff

@anon I'm studying this to read Rudin's easier.

Peter Tamaroff

20:23

What on Earth is a Class XI student?

Old John

Hi folks

anon

class $\xi$, no idea

Peter Tamaroff

@OldJohn Hey.

Old John

@anon Maybe class $Xi$ ? :P

Peter Tamaroff

@anon Just google it.

Bunch of Suicides, Rapes and other crazy news about them.

Old John

20:26

Darn - MathJax doesn't seem to like upper case Greek :(

anon

no, you simply forgot the slash. $\Xi$

Old John

@PeterTamaroff Ah - darn again :(

@PeterTamaroff Hi Peter - how are things?

Peter Tamaroff

@OldJohn Good. An awfull Sunday though. The sky is grey and it is cold.

Old John

Cold and grey here too - but we still had a barbecue

anon

I like it when the sky is gray and stormy and windy but the air is warm.

Old John

20:29

@PeterTamaroff Working on topology today?

Peter Tamaroff

@OldJohn Nice. Though it will never top an Argentinean barbecue..

Old John

@PeterTamaroff Best barbecues I ever had were done by Persians

Peter Tamaroff

@OldJohn Yep. Still on it. I did some Naive Set THeory yesterday but it saturated me, and I somehow understood a little about arbitrary products and projections (which is why I consulted Halmos' book) so I went back to Topology.

But I will also move on with Halmos' text, since I haven't gotten to the Axio of Choice yet, which is section 15. I'm in section 9.

(Families)

Old John

@PeterTamaroff Did a bit of topology myself today - in association with $p$-adics

@PeterTamaroff Tpology in ultrametric spaces is failrly crazy :)

Peter Tamaroff

@OldJohn Ultrametric?

anon

20:32

non-archimedean

where the strong triangle inequality holds

Old John

@anon Yep - if you change the standard metric property from the triangle law to ... what anon said

Managed to prove today that all the open balls are also closed, and all the closed balls are also open :)

so I think the topology is totally disconnected, but not the discrete topology (I think)

anon

Singletons will not be open sets, so not discrete.

Old John

@anon Thats what I thought

But I'm having problems visualising what goes on with the boundary of an open ball - seems to be a bit weird

Peter Tamaroff

@anon Oh, John mentioned it before.

Was it $d(a,b)\leq \min\{d(a,c),d(c,d)\}$¿

Old John

@PeterTamaroff $\max$ rather than $\min$, I believe

(and that last $d$ should be a $b$)

from the strong triangle law it turns out that every point in an open ball has an equal right to be called "the centre"

Peter Tamaroff

20:43

@OldJohn Why?

Old John

@PeterTamaroff It follows from considering the inequality defining the ball in combination with the strong triangle law - not sure I can give any intuitive reason why it should be so - it just follows from the inequalities

Jonas Teuwen

Grading... grading! I hope I finish before the end of my year orbit! 8-)))).

What intuitive reason?

Old John

@JonasTeuwen For why any point in an open ball in an ultrametric space can be regarded as the centre

Hope you finish the grading soon - and as painlessly as possible

Jonas Teuwen

Yes, that is because of how it allows triangles right? :-).

The intuitive reason... to me...

Old John

@JonasTeuwen Yes, it all follows from the inequalities for the ball and the strong triangle law

Jonas Teuwen

20:49

Hmm, it feels very duh Im Kopf.

Old John

@JonasTeuwen ??

Jonas Teuwen

@OldJohn That it feels very natural to me :-).

But maybe my head parses triangle inequalities intuitively 8-))).

anon

@PeterTamaroff Suppose $x\in B(a,\epsilon)$. Then $$y\in B(x,\epsilon)\iff |y-x|<\epsilon \\ \iff |y-x|=|(y-a)+(a-x)|=\max\{|y-a|,|a-x|\}<\epsilon \\ \implies |y-a|<\epsilon\iff y\in B(a,\epsilon)$$ and hence $B(x,\epsilon)\subseteq B(a,\epsilon)$. By symmetry we have the other inclusion and therefore equality of balls, so everything inside the ball is a "center." This works for both open/closed balls.

Old John

@JonasTeuwen Ah OK - still feels a bit strange to me - I am struggling to visualise what goes on with the boundary of open balls

Jonas Teuwen

@OldJohn Hmm... Keep thinking! :-)))).

Old John

20:52

It seems to be that if you take a ball of smaller radius with centre on the boundary, then the whole of the smaller ball is contained in the boundary of the bigger ball (I might be wrong - I often am!)

Jonas Teuwen

@OldJohn What topic did you do your PhD in?

Old John

@JonasTeuwen Boundary behaviour of functions - mainly fine continuous or continuous - in euclidean space. Asymptotic paths, mainly

fine continuous in the sense of the coarsest topology making all the subharmonic functions continuous

as here: en.wikipedia.org/wiki/Fine_topology_(potential_theory)

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