@PeterTamaroff Heine borel tells you that closed and bounded is equivalent to compact in $\Bbb{R}^n$ WITH THE EUCLIDEAN METRIC

Because if you put the standard derived bounded metric on $\Bbb{R}^n$

the whole of $\Bbb{R}^n$ is closed and bounded

but cannot be compact because the topologies are the same

@JonasTeuwen Do you know Neil Trudinger?

@BenjaLim Yes.

I've seen him around at ANU

Is he famous?

The amount of drama a math Q&A site can attract is quite remarkable! :)

@BenjaLim Cool! Well, "famous" :-).

@BenjaLim McIntosh is "famous".

@MarianoSuárezAlvarez I did not really want to get involved in the MK issue

@HenryTHorton I have written it all out on paper to my heart's content :D

@JonasTeuwen he likes to wear shorts around in summer

with shirt tucked in

@BenjaLim, that is probably a good idea :/

@MarianoSuárezAlvarez Trying to prove the trivial problem that you can always embed $X$ inside $CX$

hm?

@MarianoSuárezAlvarez what was a good idea?

@PeterTamaroff have you invaded mariano's offiec?

not getting involved in this week's drama de la semaine :)

@BenjaLim Perfect! Also sandals with socks?

@MarianoSuárezAlvarez I think it is extremely selfish that he crowds the front page with all his posts

@JonasTeuwen I have not observed his shoes

@BenjaLim Hmm... I will be there... to check it!

Although I know my supervisor wears black nike basketball shoes :D

@BenjaLim Well, that would only make sense if he gets something out of it, wouldn't it?

jonas when are you coming again?

@HenningMakholm It's called attention

the ultimate scarce resource

Given that the CW hammer falls relatively early in the process, it's unclear to me whether he has any other goal than to spite Asaf.

@BenjaLim Start 2013.

@MarianoSuárezAlvarez @HenningMakholm By the way you understood what I meant when I asked him to post on MO yes?

I told him the people on Math.se were not competent enough to answer his questions :D

I confess to not havng reading the whole meta thread

@BenjaLim Not even sure what you refer to, and I'm not terribly motivated to go look.

I mean I'm interested in commutative algebra but what he's doing is extremely selfish

I have no prpblem with the level of his or anyone's questions

if he wants to be a functioning member of society he better learn

@BenjaLim Please explain "selfish" again...?

@HenningMakholm I told him to post on MO. I told him the people on Math.se were not competent enough to answer his questions

@HenningMakholm That the front page is constantly clogged up with his questions

and he knows it does

bill told him

@HenningMakholm Anyway I will say no more I don't want to get dragged into this mess

Yeah, well, apparently the bumping logic is buggy that way.

whiel that is true

he has enthusiasm dammit

it is more or less orrthogonal to the fact that he seems to be using the site in a way which is quite not the intended one

I am al for people answering their own quesdtions

@MarianoSuárezAlvarez Not even orthogonal :D

Maybe there should be a bumping blacklist 8-).

but it is quite clear that if you are going to answer your own quesdtion, you can write the answer before posting it

Is it talk like a pirate day yet?

Say if you bump >n where n depends on your rep within n hours then blacklisted for a day.

bye guys!

that would result in people being able to do things "unattended" by the community

I'm off

@MarianoSuárezAlvarez No argument there. Well, at least to the extent on thinks the answer is finished when one submits it.

@HenryTHorton thankx - I'll shout you a six pack if you're here

That is the extents of my annoyance with his behaviour

I could not care less for what you can do without AC

but I understand that some people do care

@JonasTeuwen Easier yet, just make it so the same question can be bumped only once in 10 hours by the same user. Any later actions will be deferred till the 10 hours are up.

Most humans simply do not care for Hochschild cohomology, in turn...

@MarianoSuárezAlvarez I do

@HenningMakholm Hmm... I apparently like too complex solutions.

Heh

I want to study that in future :D

@MarianoSuárezAlvarez I confess to being pretty annoyed with his comment reactions to Asaf's latest meta thread too.

Which thread? 8-).

@JonasTeuwen You are really out of the loop....

I mean, up to and including "you should start a meta thread instead of starting this meta thread"?

hehe

yeah, that was fun

sort of designed to make Asaf crazy :)

And several cases of asserting his right to answer his own questions -- I'm pretty convinced that by now he must be deliberately ignoring explanations of what it is about his editing patterns people don't like.

I don't mind his behaviour on the meta thread, because it is pretty public

It is, in a way, self defeating

It wouldn't be unless people like me let it annoy us, now would it?

Just doing a public service here.

;)

It's not a shining example of informative question titles, at least...

What is going on? Asaf posting and answering his own questions again?

@N3buchadnezzar We're talking about the guy who is a compulsive editor.

no

I have to prove that the limits of sequences in $(\Bbb R^k, d)$,$(\Bbb R^k, d')$ and $(\Bbb R^k, d'')$ are all the same, where $d$ is the $\max$ metric, $d'$ is the Euclidan metric and $d''$ is the taxicab (I think) metric.

Are we talking about chirality or politics here?

The last one is $$\sum_{i=1}^k |x_i-y_i|$$

@PeterTamaroff Show that the topologies are the same..

@JasperLoy What is chirality?

@HenningMakholm XD I'm still in metric spaces. I haven't gotten to open sets yet.

@PeterTamaroff Handedness. I think the term is more popular in chemistry, though.

Is it not possible to ban Makato for being a weirdo and nuisance? I mean..

All my definitions are based on open balls, not open sets.

@PeterTamaroff Well, you could show that an open ball of either kind contains an open ball of each of the other two kinds.

@HenningMakholm That is clever. I have already proved that in fact.

The connection to limits should just be a matter of symbol juggling then.

I'll go and eat.

BBL

oh crap makoto posted an answer to this? math.stackexchange.com/questions/171090/…

@PeterTamaroff hey

you have 10k goggles yes?

Can I see what his answer was?

@BenjaLim No he didn't. He posted two.

@HenningMakholm Can you printscreen and post the image here?

Now I really really really really really think he is using MATH.SE as his own theatre

In indonesian we call this id.wikipedia.org/wiki/Sandiwara

@HenningMakholm yes?

Just ban hi, simply enough

@BenjaLim Still thinking. Not quite sure what the general ethics of that works out to.

I just wanna see what his reply was.

Stack exchange is a private owned site, it can select who they want as their users and whom they do not want. No reason is needed.

@BenjaLim WHich answer?

@PeterTamaroff that one

@N3buchadnezzar Omg dictator

you have 10k goggles can you post his answer here ?

@JonasTeuwen Douches should be removed, dont need no democracy for that.

Admitt it, you feel the same way.

ahahahahahahahahhahahahahahahahahahaa

this is really his own macbeth here

So now I get why he posts all these questions

@JasperLoy Do you exist, bro?

@JonasTeuwen I think I explained that here.

@JasperLoy What do you mean by "his own Theatre"?

@JasperLoy Yea bro... Scary.

@HenningMakholm Bloody markdown, I had the same problem some hours ago 8-).

I think he exist but sometimes he doesn't really want to compute itself. Don't you @JasperLoy?

@JasperLoy a bit yeah

Hell yeah, I .htaccessed my website with some deny all, I like deny all.

Like iptables's DROP.

Little Bobby Tables.

@JasperLoy I flunked them ALL!

Except maybe four?

Or included whiskey coupons

@JasperLoy It annoys all the grumpy old people on the site

@JasperLoy I think it also has to do with his bad manners

WTF! 4 classes I am interested in this fall are at the same time... who made the schedule for the grad classes!?

@JasperLoy That is how it is in every social group. They all have their norms(un written rules), and certain ways to interact with each other. When someone breaks the common style, shit hits the fan.

We can only hope

There's about 20 graduate math classes next semester, but these 4 could conceivably be taken at the same time in any combination

@BenjaLim What do you mean by his own Macbeth ?????

@HenningMakholm COuld I bother you for a sec?

Good night guys!

@JonasTeuwen Cya!

@PeterTamaroff There is something wrong with your most recent post, you have a stray double dollar sign

@PeterTamaroff In the 3rd sentence of your proof

@HenryTHorton Fixed. Thank you.

@HenryTHorton Do you have a word on it?

seems fine to me

@anon IS there a "slicker" way to prove it?

I don't imagine so. I think it's slick enough.

It's the "textbook" way.

@PeterTamaroff what?

if I can ask

I <3 groupprops

@anon What are groupprops?

groupprops.subwiki.org

@anon Hurkyl mentioned something about the topologies being the same.,

@anon Cool!

Is there one for topology?

Yes they're the same topology, but establishing that is practically the point of the exercise.

@leo See my latest question.

@anon Right. But if I exhibit a continuous one-one mapping from $(R^k,d)$ onto $(R^k,d')$ then that means the metrics are equivalent right?

BTW, $x^3-2x^2+1$ is a fairly good approximation of $\sin(\pi x)/\pi x $ in $[0,1]$.

+1 there. Seems good

Error is less than .013

I think you mean bijective rather than one-to-one, but I'm not sure offhand on your question.

@anon I said oneone and then onto.

It is the same.

Nevermind though.

Oh, I didn't see onto.

@PeterTamaroff topospaces.subwiki.org/wiki/Main_Page

If you're asking about a groupprops for Topology

@HenryTHorton Cool!

It's not as in depth as groupprops, since the guy who does the site studies group theory

@anon if the map is inyective, $d'(f(x),f(y))\gt 0$ whenever $x\neq y$ and the proof of can follow that lines

Anyone take logic? How do I prove the theorem (that is, first, what do I assume): $[~D \vee (A \cdot B)] \rightarrow [(J \rightarrow ~A) \rightarrow (D \rightarrow ~J)]$?

(...and why won't my stinkin' mathJax render?)

It should say [~D v (A * B)] --> [(J --> ~A) --> (D --> ~J)]

You need a bookmark from here: math.ucla.edu/~robjohn/math/mathjax.html

\neg = $\neg$, \sim = $\sim$

@HenryTHorton it finally rendered. it just took about 5 minutes to do so.

$[~D \vee (A \cdot B)] \rightarrow [(J \rightarrow \sim A) \rightarrow (D \rightarrow \sim J)]$

@JasperLoy no point in refreshing now. it's rendered. i'm more interested in the proof.

why is B hangin there by itself? and what does * (\cdot) stand for?

wait it's wrong!

sigh

$[\sim D \vee (A \cdot B)] \rightarrow [(J \rightarrow \sim A) \rightarrow (D \rightarrow \sim J)]$

I don't see a $B$ hanging by itself. \cdot means 'and' (or &).

NOW it's right (whew)

$\vee$ for or but no $\wedge$ for and? i am disappoint. I only see one B in the whole thing, which worried me.

well, i was just following the syntax of the book. I can write it that way, too:

$[\sim D \vee (A \wedge B)] \rightarrow [(J \rightarrow ~A) \rightarrow (D \rightarrow ~J)]$

@anon "Son, I am disappoint."

@anon People do that. Get over it.

ain't even mad

@anon yeah, there's only the one $B$. I just double checked it.

yeah, it works out. I can explain why it's true in words but I wouldn't know how to tackle symbolic proof.

i was helpin' someone when she gave up. she took the textbook with her! now i have no idea what to do. :D i only know i have to make assumptions.

@anon: let's hear it in words. maybe that will trigger something in me

darnit, the damn \sim keeps going away:

$[\sim D \vee (A \cdot B)] \rightarrow [(J \rightarrow \sim A) \rightarrow (D \rightarrow \sim J)]$

last time (dammit)!

$[\sim D \vee (A \wedge B)] \rightarrow [(J \rightarrow \sim A) \rightarrow (D \rightarrow \sim J)]$

The only way it could be false is if $\sim D\vee (A\wedge B)$ were true and $(J\to\sim A)\to(D\to\sim J)$ were false. The only way the latter is false is if $J\to\sim A$ is true and $D\to \sim J$ is false. The only way $D\to\sim J$ is false is if $D$ is true and $J$ is true. The only way $J\to\sim A$ is true when $J$ is true is if $A$ is false. But then $\sim D\vee (A\wedge B)$ must be false instead of true! Therefore a counterexample is impossible.

nevermind... i retract that! :D

Does $(U\to V)\to(U\to W)$ simplify according to some rule?

probably, but the textbook with the rules in it is 35 miles away :(.

there is a rule called material implication which reduces conditionals to logic operators.

material implication is $P \rightarrow \sim Q$ becomes $\sim P \vee Q$

@HenryTHorton I have a doubt about a simple theorem on subsequences

Basically about the proof of it.

Doubt is good

Challenge your intuition, Peter-chan

@PeterTamaroff ???

@HenryTHorton Oh!

I was composing an answer, sorry.

@Bill Hey, Bill.

@HenryTHorton It is very silly, indeed: let $\{a_n\}$ converge to $a$ in a metric space $(X,d)$. Let $\{a_{\sigma(n)}\}$ be a subsequence of $\{a_n\}$, then it also converges to $a$.

Hey Peter. Usually I'm not really here when my avatar floats by, that's just my machine refreshing windows when it wakes up....

@BillDubuque Oh, well. OK.

Proof $\{a_{\sigma(n)}\}\subset \{a_n\}$ so that $a_{\sigma(N)}\in V$ whenever $a_N \in V$, $V$ a nbhd of $a$, which means $\lim a_{\sigma(n)}=a$.

@anon @anon i managed to get that into symbolic logic form by assuming $\sim ((J \rightarrow\sim A) \rightarrow (D \rightarrow\sim J))$. thanks for help.

@PeterTamaroff To clarify, I mean I may be at my machine, but the chat window may be buried, so I may not be watching. How did you make out with Truesdell?

@BillDubuque I read a little of it, but maybe I'm not ready to really read it yet.

@PeterTamaroff It's better browsed anyhow...

@BillDubuque Hehehe true.

What I mean is that there are some crazy functions and operators I can't manage yet.

@PeterTamaroff I would write the proof more carefully

@HenryTHorton That is my doubt =P

@HenryTHorton But I guess you understand my "point" in the proof.

You know that for every neighborhood $V$ of $a$, there is a positive integer $N_V$ such that $a_n \in V$ for all $n \geq N_V$

And $\sigma: \Bbb Z_+ \longrightarrow \Bbb Z_+$ satisfies $\sigma(1) < \sigma(2) < \sigma(3) < \cdots$

So in particular, $n \leq \sigma(n)$ for all $n$

So $\sigma(n) \geq n \geq N_V$ and therefore $a_{\sigma(n)} \in V$ whenever $n \geq N_V$ as well

@HenryTHorton Right. Perfect.

@HenryTHorton The next exercise is to prove the Monotone Convergence Theorem, and the next one is to prove the Weiertrass-Bolzano Theorem.

Oh, no. It is only the case for $\Bbb R$, not $\Bbb R^n$ in general.

Should the general case be more challenging?

@PeterTamaroff Yes

@HenryTHorton Much more challenging?

=D

I'm trying to remember the argument for the general case

@HenryTHorton Here I'm told to use $\limsup$'s and $\liminf$'s, apparently.

For Monotone Convergence Theorem?

@HenryTHorton No, no, for Bolzano Weierstrass.

I would use Monotone Convergence Theorem to prove Bolzano-Weierstrass

@HenryTHorton Oh, yes, that is suggested.

So start by proving MCT

If the sequence is bounded and increasing, when what do you think the limit should be?

@HenryTHorton I'm asked to prove it is the supremum of $\{a_1,a_2,\dots,a_n,\dots\}$

Yes

So first, how do you know the supremum exists?

@HenryTHorton The supremum exists because the sequence is bounded.

@HenryTHorton See this

$\sup \{a_n\}$ exists because $\{a_n\}$ is bounded above and $\Bbb R$ has the least upper bound property

@HenryTHorton Yes. $\Bbb R$ is complete.

Now you can basically prove that $\sup \{a_n\}$ is the limit of $\{a_n\}$ from the definition of the supremum

@HenryTHorton Yes, that I was expecting.

So, I'll just write down the proof for the MCT and then move on to BWT

@PeterTamaroff I have to go home now, so I won't be around if you do Bolzano-Weierstrass tonight.

Adios

@HenryTHorton Adios!

@anon I want to write some articles in there if I can, topospaces

I am the bear. Hear me roar.

@JasperLoy Yea bro...

@JasperLoy Do you love repositories as much as I do?

Kickass commands like: [jonasteuwen@jonas-arch puzzles]$ hg push.

But emacs handles that Godly and transparantly.

Almost like Buddha, so clean and slick.

I love it.

@JasperLoy Thanks!

Relax! Think: what would Buddha do?

Do you guys push your config files to a repository?

@JasperLoy I know! :-). But! You know what I mean?

@JasperLoy Why? My config files are like.... 2k SLOC in total?

@JasperLoy Excellent.

My precious config is repositoried.

@JasperLoy Yes.