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12:00 AM
@JonasTeuwen I just upgraded from 1.5Mbps, and I was usually only getting around 800kbps because I was so far from the DSL main office. Now I am on cable modem.
 
I believe they said they had a direct connection to the Amsterdam Internet Exchange.
@robjohn Oh right, that is a big issue in the US: everything is far apart :-).
Here they just put fibre glass in the streets.
You can get ADSL. They put some kind of modem on the fibre glass cable to change it into ADSL 8-).
 
@JonasTeuwen My old ISP did that, but they wouldn't give me the 5 static IPs that I wanted. Then I found out the cable company would, so I changed and got faster, too.
 
Oh, multiple static IPs!
For each computer or do you want to run DNS servers and stuff like that?
 
@JonasTeuwen got 5 :-)
@JonasTeuwen DNS, mail, web all that good stuff.
 
Mm, yes, nice :-). I "outsource" that to a server farm 8-).
 
12:03 AM
You need the static IPs to run a mail server. without reverse DNS, other mail servers won't talk to you.
 
Yes.
But if you put your computer in a data center, they just give you IPs without whining.
I just like to keep stuff out of my house as I basically almost have no house, square meter speaking (10).
 
@JonasTeuwen I don't know if there is such a thing here.
 
@robjohn UCLA probably has a data center? And... if you have UCLA there are probably many more data centers!
My university does not allow staff servers anymore :-(. They have gotten too many cease and desist request from scary things like Interpol apparently.
 
@JonasTeuwen I don't think that you can put your own computer in the datacenters at UCLA. Maybe there are programs that I don't know about, though.
@JonasTeuwen ow
 
@robjohn Oh well, you have your IPs, so you don't need it :-).
There are plentiful data centers all over this country. Small... big, smaller. Smallest.
I have been inside of one. Very cool stuff. Cooled floors, everything cooled. Generators!
And lots of cables.
 
12:11 AM
@JonasTeuwen A forest! :)
 
But neatly tucked away!
I am writing a paper, and for the moment it has the title "Beam me up, Scotty!".
It has too little $\lambda$.
8-).
 
12:38 AM
I have done some good work today. Good night guys!
 
See ya!
 
I'm off to have a walk with Lilly, too. Be back later!
 
@t.b. Done! I'm studying topology, Theo!
 
And I think this, this and this should be closed.
@PeterTamaroff Oh, cool! Are you having fun with it?
(and hi)
 
@t.b. Yes! It is mind twisting!
 
@PeterTamaroff indeed :) What are you doing right now?
 
@t.b. I am studying from Mendelson, don't know if it rings a bell to you.
 
1:26 AM
Yes, heard of it, but I never read it.
 
@t.b. I just finished the section on Neighborhood Spaces.
Now I want to make a general reading of my notes and move on.
 
Hey t.b.!
 
Hey, J.M.
 
The chapter basically explained the one one correspondence between all topological spaces $(X,\mathfrak I)$ and all nbhd spaces $(X,\mathfrak N)$, and how we can get one from the other.
 
@PeterTamaroff Does that mean topologies defined in terms of neighborhoods?
 
1:27 AM
@PeterTamaroff If it really is topology, your mind should be invariant with respect to the twisting...
 
@t.b. Yep.
@J.M. HAHAHHA =D
Anything out there worth downvoting?
 
you trying to get 11300 exact?
 
@J.M. I like how Dehn twists (by coincidence) aptly describe what they're doing: they stretch and twist. (Dehnen = German for "to stretch")
 
@anon Yes, Sherlock! =)
 
@t.b. Hah, I didn't notice. Very fortuitous!
 
1:31 AM
@PeterTamaroff Oh, wrong forum. I wanted to calculate that in Mathematica, I confused the stack's names. =D
 
@PeterTamaroff I hope you've got some good examples to work with. You should at least understand metric spaces pretty well before plowing into topology.
 
@GustavoBandeira You shouldn't have; one of the two is somewhat orange-y...
 
@t.b. I have the example of $(\Bbb N,\mathfrak I)$ with $\mathfrak I=\{\varnothing,O_1,\dots,O_n,\dots\}$ and $O_n=\{n,n+1,\dots\}$
We can define each $U\subset \Bbb N$ to be nbhd of a number $n$ if whenever $m\geq n$, then $m\in U$.
This induces a nbhd space, which is the corresponding nbhd space of the topological space $(\Bbb N,\mathfrak I)$
 
@PeterTamaroff Ah yes... This looks like the sequence $1/n$ in $\mathbb{R}$.
 
One b only!
 
1:35 AM
he takes his R's really black
 
yes, sorry, I'm too lazy to activate the script...
 
@t.b. Can we create a topology in $\Bbb R$ with $O_n=(0,1/n)$?
 
Oh hullio TeeBee.
 
...or something of the sort?
 
@PeterTamaroff sure, every subset of $\mathbb{R}$ inherits a topology from $\mathbb{R}$ (but you'll get there). And yes, the neighborhoods of $0$ give you exactly the tails you described before.
@Gigili 'Ello, Gigili :)
 
1:38 AM
so I'm perusing some rep theory notes. it takes things like maximal chains and Zorn's completely for granted, but then it says Here i is a complex number such that i^2 = −1 and pi denotes the area of a circle of radius one). strange how views on which topics can be taken for granted and which can't differs between people.
2
 
heh :)
 
@t.b. Do you know any good book to get exercises (if not complementary theroy) from?
 
@t.b. neighborhoods of 0? huh?
 
@DavidWheeler $(0,1/n)$
 
oh, those....reading too far up. my bad.
 
1:41 AM
@PeterTamaroff Are there none in the book you use? Then burn it get 'nother one.
 
@t.b. Yes of course there are! But the more the merrier!
 
@PeterTamaroff Hm. I learned topology from a German book that was never translated (Boto von Querenburg) and the other books I read are a bit too advanced to recommend. People tend to like Munkres, though.
(I haven't read that one)
 
@t.b. For example, it had this exercise. Prof Scott gave an awesome extension of the exercise!
Wrong link before.
 
@PeterTamaroff Sorry it's a bit long for me to read right now. Do you mean his mentioning of the Cantor set?
 
@t.b. Well, I just mean to show you the exercise, not Scott's answer. =P
 
1:45 AM
I see.
 
The THEOREM I state was given to prove as an exercise too, for example.
 
I didn't intend to stay too long and the room looked quiet before I entered. woosh, half an hour gone... Dangerous place.
2
 
@t.b. Hehehe true.
 
Well, don't let us keep you from the more important stuff...
 
I usually talk about math here so I usually benefit from this =D
 
1:51 AM
@J.M. right, yes, I should actually be sleeping... So, see you guys, always nice to talk to you all. See you soon.
 
Meanwhile, I need a few executioners here...
 
@J.M. Done.
 
2:24 AM
@J.M. I probably should have read it first, but my impression is that you're trustworthy :)
 
@DylanMoreland In which case, thanks for the trust. :D
 
@anon I think that (at a different level) this is a nice idea. From experience, many people are told, "Oh, well, you're going to do algebraic something-or-other, so make sure to learn algebra well." So you do that, and then you start reading a paper on representation theory and go, "Whoaa look at all this analysis; I have been tricked."
Every time an author reminds me of basic complex/Fourier/functional analysis, I am wildly thankful.
But I can remember the definition of $i$. That's a weird one.
 
It's very opposite with me. I'd never studied any modern (abstract) algebra until a couple years ago, it was all just analysis. (That was all I was ever offered!)
 
Someone (I think Etingof) has rep. theory notes intended for high schoolers (good ones, I would think) online. Maybe then.
 
I'm looking at Murnaghan's.
 
2:32 AM
This is quite old!
 
eh?
 
Oh, maybe they're related. When I Googled the first result was a review of a book written in 1939.
 
is there a faithful representation of A4 of degree less than 4?
 
I think there's one of degree 3
 
2:46 AM
oh cool
what is it?
oh, i bet i know
 
A4 acts on K^4 (via permutation rep) and has an invariant subspace of codim 1, namely vectors whose components add to 0
 
well i was thinking just find the 3x3 matrices that preserve a regular octahderon (centered at the origin), and A4 will be a subgroup
now that i see how it can be done...doing it seems...tedious
and what do you mean by K?
 
base field
 
ah...any particular restrictions on the field?
 
the article says in char 2 it's reducible but not indecomposable. I think that means it should still be faithful.
 
3:04 AM
@anon Could I ask your help with an answer Arturo made (to a question of mine)?
It's about the AXIOM OF CHOICE
 
if it's interesting
 
@anon Well you tell me.
I'm only confused about the part in the DUAL THEOREM where it starts with the AXIOM OF CHOICE here
 
what about it?
 
Well, the part he defines $$g:B\to \bigcup_{b\in B}f^{-1}(b)$$
and then says it is the inverse of $f$.
 
He constructs a choice function.
 
3:10 AM
right inverse, yes. and?
 
It's not just what you wrote, the part that requires AC comes afterward
 
Well, I don't get why that is the desired inverse.
 
"the"? there is more than one. one for each choice function.
 
He says since $g(b)\in f^{-1}(b)$ then $g(f(b))=b$
 
that implication should be obvious
 
3:11 AM
@anon Aren't inverses unique?
 
@PeterTamaroff not one-sided inverses, no
 
Oh, well. it is the first time I read about "one sided" inverses.
 
Consider $f=\{(0,1),(1,1),(2,2)\}$ where there are two right-inverses.
@PeterTamaroff You meant here f(g(b))=b.
 
@CLarue Right inverse should be $fg=\operatorname{id}$, correct?
 
yes
 
3:16 AM
Well, you have $f:A\to B$ $A=\{0,1,2\}$ and $B=\{1,2\}$
 
yes
 
and $f(0)=1$, $f(1)=1$, $f(2)=2$
Then I should seek $g:B\to A$
 
it's kinda like this
 
Now, let's follow Arturo's discussion (we don't need AC in finite cases though)
 
Wait.
 
3:18 AM
 
Because $g$ must lose a value.
$g(1)=0$ and $g(2)=2$
$g(1)=1$ and $g(2)=2$
or
Is that what you mean @CLarue ?
 
Exactly.
 
how do i write is an element of in math.se?
 
\in
gives $\in$
@CLarue I see
But that is because $f$ is not one-one.
@CLarue OK
 
In my image $f$ maps $A\to B$. The red ovals designate preimages aka inverse images of the elements of $B$ under $b$. We can find a choice function $g:B\to A$ that maps $b$ somewhere into $f^{-1}(b)$. Obviously applying $g$ and then $f$ will preserve all of $B$; it is the identity. Not so in reverse; applying $f$ and then $g$ will map each inverse image to the particular representative of it the choice function chose.
 
3:20 AM
the part that involves the axiom of choice is that we have to "choose" something in $f^{-1}(b)$, in order to have some "definite" function
 
how do i write natural number in latex
the bold n?
 
@ChuckFernández $\bf N$?
 
\Bbb N or \mathbb{N}
 
\bf gives boldface
 
since when is $\bf n$ naturals?
 
3:21 AM
by the way what is the name of the latex language we use?
 
latex
 
it isnt exactly latex is it?
 
@DavidWheeler OK.
 
\mathbb is the latex command \Bbb is jax, i think
 
because $a \mod b\$ doesnt work
 
3:22 AM
Close it
$a \mod b$
or $a\pmod b$
 
in normal latex it works
 
@anon Some people use bold instead of blackboard bold for their number sets; there was a discussion on this in meta...
 
@DavidWheeler But we choose exactly one from the $f^{-1}(b)$ to get one of the so many inverses, correct?
I think I'm getting it.
 
@DavidWheeler kile says \Bbb is obsolete
 
yeh i want blackboard bold
 
3:23 AM
@J.M. It was more the lowercase..
 
the one with white in the middle
 
@ChuckFernández $\bf N$ should work
@ChuckFernández $\mathbb N$ or $\Bbb N$
 
@anon Ah, that I haven't seen before...
 
@J.M. Why did that cause discussion?
 
what about logical and operator
"and"
 
3:25 AM
the axiom of choice is subtle....to have a "choice function" it's not enough to have the set on which it is defined...you must also specify the method of choosing
 
\wedge, \&, \cdot
 
@ChuckFernández \land
 
Everybody stop.
 
I used this religiously when I started with latex
 
(easier to remember than \wedge in this context...)
 
3:26 AM
NO ONE TELLS THE CHATROOM TO STOP
 
chatroom, stop!
 
@anon I'M NO ONE
 
your argument is invalid
 
$\not\exists\ne \emptyset$, HAM SANDWICH FALLACY
 
3:27 AM
I could freeze this room, but I don't want to abuse my abilities...
 
yes, but that is different than telling to chatroom to stop...that's actually stopping it
 
@J.M. Freeeeeeeeeeeeze it!
SYMBOLS Enjoy, everyone.
 
why do you want to stop it?
 
You may continue.
 
error 404
 
google drive is better btw
 
@ChuckFernández Says who?
 
the gold standard is meg.. nevermind
2
 
@anon LOL
It is coming back, however!
 
3:31 AM
what makes you say that?
 
@anon Dotcom is free. I heard.
 
@PeterTamaroff Since when?
(and all his assets were frozen IIRC)
 
yup
 
So, free, but somewhat poor.
 
they're letting some people have access to their file collections I heard, but in the process they will be scrutinized so this will only work for a select subset of the 25,000TB.
they = some kind of law enforcement
 
WW is an 'nym I take it?
 
Wouldn't know.
 
4:07 AM
 
Somewhat apt...
 
Hey all! :-)
 
4:32 AM
@robjohn Hey
I got the badge recently =D
 
@PeterTamaroff How are things?
 
@robjohn Good.
 
@PeterTamaroff cool!
 
I need to learn Set Theory though.
I'm advancing in my Topology studies and I need to have good ground on ST.
 
@PeterTamaroff Set theory should be a prereq for topology, if it isn't.
 
4:34 AM
@robjohn I mean, I know Set Theory.
But I think I should be more proficient at it.
 
well, elementary-set-theory. not sure if you need cardinals and whatnot for your topology.
 
@anon Yeah. Elementary Set Theory I guess.
Halmos' book should suffice.
I'm looking now for Exercises in Set Theory by Sigler
It's supposed to be the complement to Halmos' lack of exercises.
And when I say looking I don't mean in a library.
 
you mean googling?
rapidshare? nope. dl4free? nope. let's see how abt scribd?
 
@DavidWheeler Yeah, yo.
 
you do no tneed a good ground on set theory for anything, really
 
4:41 AM
@MarianoSuárez-Alvarez Really?
 
topology needs nothing more than for you to know what unions, intersections and products are
 
@MarianoSuárez-Alvarez he wants to be better at it
 
@MarianoSuárez-Alvarez How so?
 
i think you need to be comfortable with complements, too
 
4:43 AM
so you can tell when something isn't not-not untrue
 
@MarianoSuárez-Alvarez Well, what about functions, families, relations?
 
to construct some exotic examples, one uses ordinals, and to prove things like very strong metrization theorems one uses well-orderings
that is mostly irrelevant to soeone starting
of yocourse tou need that, Peter
but you already know it
'Set theory" is nothing
a family is just a set, btw
you surely know what a function is
 
when people are first exposed to functions, they are of course introduced to things like the domain and image (often called, in my opinon a poor name, range).
 
@DavidWheeler Codomain! HAHAHAHA
 
and relations... you probably do no tneed them
 
4:46 AM
but often very little is said about the dual concepts of pre-image and co-domain
the image is not the codomain
 
I was just telling the guys here I'm finding it hard to get the part bout the AOC here
 
one way to state the axiom of choice is: every surjective function has right-inverse
 
@DavidWheeler That is what Arturo is proving.
@MarianoSuárez-Alvarez I'm really not trying to against the tide here, but my book says "The equivalence relation that is appropriate to topological spaces is called homeomorphism."
 
you can't "prove" it, at least not from other axioms
 
@DavidWheeler Well, showing equivalence, pardon me!
 
4:49 AM
beh nvm
 
Zorn's Lemma is $\equiv$ to the AOC yet it is called a Lemma.
 
for historical reasons
only
@PeterTamaroff, sure
relations are all over the place
but to study relations before starting topology is poinless
 
what?
 
@MarianoSuárez-Alvarez Equiv relations build up $\Bbb Z$ and thus build up $\Bbb Q$; which builds up $\Bbb R$! =P
 
4:51 AM
@MarianoSuárez-Alvarez I know you know, of course.
 
the fact that R is construcred that way is 100% irrelevant for all of math
except the existence of R
 
@MarianoSuárez-Alvarez hey
 
which is something that is irrelevant for anything
 
@anon I must confess I'm finding the symbol $X\setminus A$ much more useful than $A^c$ or $C(A)$
@MarianoSuárez-Alvarez Fair enough.
 
@PeterTamaroff I have found out what special topics in algebra will be on
 
4:53 AM
good.
 
@MarianoSuárez-Alvarez My special topics class is taught by jarod alper
 
@BenjaLim The pinata popped, ay?
 
Yes this is the course text @PeterTamaroff
@anon This semester will be brutal.....
 
hard-core
 
@robjohn Hey Rob. I've been meaning to add a telescopic view to that question you mentioned yesterday, and I finally found the time just now.
 
4:54 AM
I like our lecturer
he just arrived in australia!
 
@BenjaLim I like trains.
Wait! NOOOOOOOOOOOOOOOOO!
 
@anon only problem is that
the courses assumes some complex analysis/analysis on manifolsd
 
well, @PeterTamaroff, an awful lot of mathematics comes from replacing = with a suitable equivalence
 
4:58 AM
@DavidWheeler Yes...?
 
in topology such things pop up as "qutoient spaces", or in technical terms: sewing stuff together.
 
@BillDubuque Yeah, in essence, it is not too different than mine. Let's see if the mad downvoter strikes again :-)
 
except for the followers of Bourbaki, who insist on the term "gluing"
 
@BillDubuque If they really are hooked on that one method.
 

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