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00:00
@KannappanSampath Of what?
@AsafKaragila I see. There are not any questions on History while there are many on the constructions themselves.
I’m surprised that no one has pointed out that Christian Blatter’s paper is almost exactly what the OP seems to want.
Which OP?
Construction of the reals.
It’s not perfect, because one must either ignore finite sets or impose a simple equivalence relation, but I doubt that one can do much better.
Nice. The history question has one vote left for opening.
Well, either way. I am going to sleep now.
00:11
@Brian I'd like to discuss something about connected sets. Will you help me out?
(On Metric Spaces)
@KannappanSampath Connectedness is not my strong suit, but I can have a look.
It's not a problem, but some preliminary understanding of the definitions.
Rudin starts out this way:
With Seperated Sets:
Let $(X,d)$ be a metric space. Two subsets of $X$, $A$ and $B$ are seperated if closure of one does not intersect the other.
Right, that’s standard. (Note the correct spelling, though: sepArated.)
00:16
(i.e.) $A \cap \overline B=\varnothing=\overline A \cap B$
@BrianMScott Thank You for the pointer.
Now, my teacher defined connecteness through separatedness.
$X$ is connected iff it can’t be written as the union of two non-empty separated sets?
@BrianMScott Yes. But, I think that's bad (And, Mariano once pointed out). Since, connectedness is the property of a individual set, we must have a nicer definition.
Do we have one?
On the other hand, connectedness intuitively says that a set can’t be broken into pieces in any natural way, so it’s not surprising that the definition involves subsets.
I agree. But, this becomes messier sometimes while writing proofs.
For many purposes I prefer a similar but slightly different characterization: $X$ is connected iff it is not the union of two disjoint, non-empty clopen subsets.
Another characterization: $X$ is connected iff every continuous function from $X$ to the two-point discrete space is constant.
00:23
@BrianMScott I have proved this equivalence. But, there was an intermediate characterisation:
A space $X$ is connected iff the clopen sets in $X$ are $\varnothing$ and $X$.
In the interest of giving advanced notice: I'll need to be away for a few weeks, during which I have no way of accessing the Internet. I hope I can return at a better time for me. So, for now: sayonara, auf Wiedersehen, and whatnot.
12
That’s essentially the one that I gave here.
[Asaf unavailable to pin JM's notice this to notice board (:-(), hence a star from me]
@JM You’ve also been infected by the whatnot bug! Anyway, auf Wiedersehen!
@BrianMScott :) Farewell.
00:27
@BrianMScott This is a bit roundabout way of saying, no?
Perhaps, but it’s the form in which one often wants it.
I see. I remember Mariano telling me a open set version. Is there one or just that I misremember sth?
what do you mean you'll have no way to access the internet? going to Mars?
$X$ can’t be written as the disjoint union of two non-empty open sets. Follows immediately from the clopen version.
But they are not equivalent. Are they?
00:31
Sure they are.
Two disjoint open sets whose union is $X$ are necessarily also closed.
I see Clopen $\implies$ open implication. The other?
Each is the complement of an open set.
Suppose $X \neq A \cup B$ , $A$ and $B$ open. Then,
@BrianMScott I mean, I don't see what this means here.
Someone is messing with the events that the chat generates. MathJax is not catching the new $\LaTeX$ that is being entered.
I had noticed. :-(
00:36
Perhaps a message to meta is in order...
@KannappanSampath $X$ is the union of two disjoint non-empty open sets iff $X$ is the union of two disjoint non-empty clopen sets, because such open sets are necessarily clopen.
Haha Yes.
I am fine with the arguments now.
@BrianMScott I'll TeX them and show you. Will you be around for $\dfrac{1}{2}$ an hour.
Good. :-)
Most likely, though I’ll probably be checking in and out.
Alright. In any case, I'll leave a link to some file sharing site. Do you prefer TeX or pdf?
sharing TeX files is like sharing Word files...
00:42
PDF is probably a little easier.
@BrianMScott Fine, then. Thank you for your time, Brian.
Sure thing!
00:59
@BrianMScott Question posted to meta
We’ll hope that it gets a useful response.
@Brian Corrected that blunder, Thank You. I was hoping sth would get balanced!
(Though, I was reasoning the odd way!)
01:15
I suspect that you were (possibly unconsciously) making the potential mistake that I mentioned at the end of my answer.
@BrianMScott Yes. Something of this kind.
01:43
@BrianMScott A pedagogical/notational doubt: Is it wise to define the notion of connectedness for every subset or for the space and then every subset when considered as a subspace?
Doesn’t really make much difference, I think.
The second might be a bit easier.
Oh, I'll then stick to it.
01:57
Dear all: You may find this relevant: math.ucr.edu/home/baez/elsevier_boycott_poster.pdf
(On the Elsevier and the movement initiated by Gowers)
02:28
@Brian The file is growing. I'm TeXing forever. I am sorry. I'll ping you here once I am done.
02:51
That’s okay. TeXas is a big place. :-)
A doubt:
@Brian A separation for a space $X$ is the union $V \cup W$ such that $V \cap W= \varnothing$ right?
And $V$ and $W$ are open and non-empty.
Is open required?
Sure: $\mathbb{R}$ is connected, but it’s also the disjoint union of $(\leftarrow,0)$ and $[0,\to)$.
Ah. Sure.
I don't seem to be thinking these days. :/
03:00
Some days are like that, unfortunately.
03:31
Jury needed: Here
04:09
Here and not in the previous link. That was the duplicate. :-)
04:44
Hi all. I'm studying for a manifolds exam, and I have a question - an open subset of a manifold is always a submanifold, but is it always regular (i.e. embedded)?
I feel like it should be, since it should just inherit both its topology and differentiable structure, so it'll have the subspace topology & the restricted smooth structure.
 
2 hours later…
06:30
Morning everyone : )
 
2 hours later…
08:01
@JM : (
At least he thought to give notice.
@JonasTeuwen Did you mean $f = \sum \langle e_n, f \rangle e_n$?
@BrianMScott Yes!!
08:35
One last paramedic is needed: math.stackexchange.com/questions/111043/…
@AsafKaragila Resuscitation successful.
takes off latex gloves and closes suitcase with defibrillator in it
LaTeX gloves?
Yes!
08:39
And a high-TeX defibrillator, no doubt.
Well. I have to prepare a lecture on Whitehead's problem and I have to finish the paper in algebraic topology, which I first have to get myself to start.
I'll see you folks later.
@AsafKaragila You could write it backwards ...
As long as I finish before next Thursday it's fine.
09:35
I thought that the symbol was the Fourier transform of the differential operator. But a simple example like $P := a(x) \frac{\partial}{\partial x}$ applied to $f$ and then transformed yields $\widehat{Pf(x)}= \xi \widehat{a(x) f(x)}$ whereas the symbol is supposed to be $\xi a(x)$. What am I missing?
Hi Rajesh. Long time no see.
yeah i had to attend two weddings
how are you doin ?
I see. I am doing quite OK. On the wall that separates a good day from a bad one.
Now don't ask me for a existential proof (for the wall).
How has your day been? :-)
just lazy...
contemplating which side to jump ???
09:50
@RajeshD Given a chance, I'd choose to jump on the Good day side. But, unfortunately, I don't have a choice.
Looks around if Asaf will pounce with some Set-Theoretic argument to support/disprove my claim
Anyway, hope to catch you later. I am TeXing on. Probably life story is a bit simpler to TeX up. ;-)
k cu
10:09
$$
\begin{align}
Pf(x)
&=\mathscr{F}^{-1}(a(x)\xi\mathscr{F}f(\xi))(x)\\
&=a(x)\mathscr{F}^{-1}(\xi\mathscr{F}f(\xi))(x)\\
&=a(x)\frac{1}{2\pi i}\frac{\partial}{\partial x}f(x)
\end{align}
$$
@KannappanSampath why are you on a wall? That seems like such a precarious place to be. :-)
@robjohn Thank you!! (I'll have a closer look later, got to run now to Japanese class) Hope I can catch you later : )
@MattN Probably
I just got a comment on an answer from @Srivatsan :-) It seems he might be getting homesick for MSE. Perhaps we will see him on chat again soon.
10:56
Finally! My girlfriend was signed in and I couldn't find how to signout of the chat 8-).
@MattN Yes, that is what I mean.
11:09
@JonasTeuwen did she sign in MSE?
@Ilya No, on History.SE.
ah, and you use the same computer so you had to sign out of the chat?
Yes.
Which was quite "hard" since it didn't have a signout button!
@Jonas: I met this problem when was changing my name
you have to sign out of the main page, I believe and also sign out of your openid on openid.com
at least that worked for me - I don't know though which login do you use to log in
Of MSE :-).
11:18
@JonasTeuwen the "leave" link above the gravatar bar? :-)
@robjohn That leaves the room! Does not logout!
@Jonas: do you have nice notebooks in your department?
I think that in ours they buy each time crappier that the previous one
@JonasTeuwen if you want to leave MSE, there is a "log out" option:
@JonasTeuwen That "log out" link is http://math.stackexchange.com/users/logout
Yes, but I was logged in on MSE as Jonas and my girlfriend on History.SE and when I wanted to go to this chatroom I logged on as her...
@JonasTeuwen seems that the chat is SE-wide
11:31
Are you saying that if you leave the chat room and log out of MSE that there are still some active cookies?
I logged her out of History.SE then I logged in as myself on this chat 8-).
@Ilya Wouldn't that log you out of any place that used OpenID?
@robjohn I don't think so - at least if you checked "remember me" on OpenID
@JonasTeuwen Did you have to use anything more than the "leave" and "log out" links?
11:42
@robjohn No, but I had to guess the website she was logged on to, sign out there, so I could chat here...
@JonasTeuwen Ah, I see. You didn't know the SE section to which she was logged in.
Any tips on $\sum_{n=2}^{\infty}\left( \frac{1}{\log n}\right)^3$ ? I know that it diverges, but I am having some difficulty proving it
@N3buchadnezzar compare with $\sum\frac1n$
@robjohn Thanks
$$
\lim_{x\to\infty}\frac{\log(x)}{x}=0
$$
implies that
$$
\lim_{x\to\infty}\frac{\log(x)}{x^\alpha}=\frac1\alpha\lim_{x\to\infty}\frac{\log(x^\alpha)}{x^\alpha}=0
$$
11:49
wth
$x$
@Rob: your script has stopped working for new formulas which appear
hm, worked now. weird
@Ilya It stopped for me for a while, it seems to be working for me again. I posted on meta about it.
@Ilya when you enter a comment or rerun the bookmark, it renders. Did you have a lot of the log visible at the time?
There might be a problem with something that was posted a while ago, and it might be scrolled off of my backlog.
If you still had it in your backlog/scrollback, that might explain your problem.
@robjohn hm, no - it happened with your last formulas here (message that I refer to)
I know that, but sometimes I have a lot in the backlog that I can scroll back to, and other times I can only scroll back a little bit. I was wondering if you still had the offending TeX in your backlog, whereas I didn't.
Ok, now I can't check it - since it lasted for 3 minutes with you new formulas and with my $x$ - but then it became normal, quite suddenly
there might be some TeX that holds up the processing of the Ajax "complete" events.
11:55
now it is ok, the next time I'll try to check it out
It is the "complete" events that trigger rendering
they are supposed to happen all the time, but perhaps MathJax gets tangled in some bad LaTeX and clogs the pipe.
@robjohn LaTeX troubles. Will you help me a bit?
@KannappanSampath whassup?
I'd like to use updown arrow in a proof environment.
@robjohn And, it should align nicely in the middle of the statements it connects. How do I do this?
(There are 4equivalences)
$\begin{array}{c}\text{like}\\\updownarrow\\\text{this}\end{array}$?
12:01
$$
\begin{array}{c}
&A&\Leftrightarrow&B
\\
&\Updownarrow& &\Updownarrow
\\
&C&\Leftrightarrow&D
\end{array}
$$
What does $c$ in the code there stand for?
@Ilya Precisely, but not this convoluted.
@Ilya Oh, don't use align, it spaces weirdly
@KannappanSampath centering the items in their boxes
@KannappanSampath convoluted?
@Ilya better :-)
@robjohn thanks ) I forgot which one is better and when
12:04
@robjohn , @Ilya Thank you both.
I'll get back in a minute.
@Ilya Yeah, the align environment is good for equations, but array is better for regular placement.
Why do I never have any problems with latex in chat?
@N3buchadnezzar are you complaining?
No, but everyone else is, when my equations breaks the chat..
@N3buchadnezzar Was it your equation that broke things?
12:07
TeX died :/
@robjohn yes =(
ButIt appeared just fine for me!
@N3buchadnezzar if you know the offending $\LaTeX$, could you put it inside backquotes?
$$\begin{array}{c}
x \in cl_Y(E) \iff & \exists ~ \{x_n\} \subseteq E \subseteq Y \subseteq X ~\text{such that}~x_n \to x \in Y \subseteq X& \\&\Updownarrow&
\end{array}$$
@N3buchadnezzar so that it doesn't mess things up
Haha, it works here!
@robjohn I got it all. Thank you, Boss.
12:10
@KannappanSampath No charge... ;-)
" I=\int \frac{dx}{x(x^6-1)}=\int\frac{x^5}{x^6(x^6-1)}dx\\u=x^6-1\,=>\,du=6x^5\,dx\\I={1\over 6}\int\frac{du}{u(u+1)}={1\over 6}\int\left(\frac{1}{u}\,-\,\frac{1}{u+1}\right)\,du\\I={1\over 6}\ln|x^6-1|\,-\,\ln(x^6)\,+\,C={1\over 6}\ln\left|\frac{x^6-1}{x^6}\right|\,+\,C "
Thats what broke the chat latex thingy, me thinks.
@robjohn Why does the LaTeX complain "Extra alignment Tab has been changed" while I type in the same code I put in here?
Latex hates array
@N3buchadnezzar you have a lot of `\\` things in there
why?
Linebreaks
12:16
@robjohn There's just one!
@N3buchadnezzar those don't make line breaks inside an equation $...$
only inside special math environments
They do when you put $$
Double dollar signs
No, I get errors, even inside $$...$$
a\\b
12:19
Perhaps inside align then
$$ \stackrel{\large a}{b} $$
@N3buchadnezzar well, that would be one of the environments
Please take a look into what is wrong.
@KannappanSampath your page is too narrow for the line
@KannappanSampath line-wrapping is messing things up
@robjohn I see. Without those ampersands and blah, it works!
12:27
perhaps encapsulating the array in \small{...} would help
as in no complaints from the Compiler. Let me see what about another implications
@robjohn I'll try that as well.
No, it hates those ampersands.
(even with small.)
Try this $\small{\begin{array}{c} x \in cl_Y(E) &\iff & \exists ~ \{x_n\} \subseteq E \subseteq Y \subseteq X ~\text{such that}~x_n \to x \in Y \subseteq X \\&&\Updownarrow \end{array}}$
@robjohn The same errors. The alignment changed and...
@KannappanSampath hmm... let me try it in my LaTeX processor...
@robjohn I'd love that.
12:40
Why not just use tabular?
@N3buchadnezzar A sample,please. I know not what you mean!?
@KannappanSampath ?
@KannappanSampath what's that?
\begin{table}
\begin{tabular}{c c c}
$ x \in cl_Y(E) $ & $\iff$ & $\exists ~ \{x_n\} \subseteq E \subseteq Y \subseteq X ~\text{such that}~x_n \to x \in Y \subseteq X$ \\
& & $\uppdownarrow$
\end{tabular}
\end{table}
12:44
The array structure at the top of the code dispays oddly as in the accompanying pdf with some errors.
Soemthing like that, I am at school, and can not test it.
@Kan: which editor do you use?
@N3buchadnezzar remove a p from uppdown arrow and it should be fine.
@Ilya TeXmaker.
@KannappanSampath thanks. let me try iy
@KannappanSampath try changing the {c} to {ccc} in mine.
12:47
Well I just gave you an idea=)
It works in my processor. They are more picky about the number of cells in the array
@robjohn Perfect fix! Thank you!
@KannappanSampath hah, interesting! I'll try it more detailed
@N3buchadnezzar Yours works just as fine!
@Ilya And will you please let me know?
@KannappanSampath Does it look good? I just typed it out of my ass!
^^
12:52
@Kan: you were talking about comparison of TeXworks and TeXmaker - could you briefly remind the main points?
@N3buchadnezzar The second line aligns haphazardly.
@Ilya TeXmaker and TeXworks are both cross platform.
TeXworks is plain without crowding your screen with menus.
TeXmaker shows you a lot of menus.
Both have Inbuilt-Viewer.
TeXmaker prompts for autocomplete while no such thing has been known in TeXworks.
This is all I remember. Is this enough or I should fish out that Chat thread?
@KannappanSampath Then you just need to add another & below, some \hspace{1cm} or something
Yeah, I have never used any of the menus in TexMaker, I just try to remember everything. Thinking about switching to some simpler less cluttered typewriting program.
@N3buchadnezzar Oh, I am in no mood to try it out right now. And, on this note, the table is at the top of the page preceding everything else before it. Strange!
I will go with Rob's code!
(for now atleast)
@KannappanSampath \begin{table}[!htbp]
Ofcourse you should go with robs code, because it uses a math enviroment. Whilst mine uses a table
13:01
I guess you could drop the table part entirely
It's tb's time now. But, I will get back in sometime. I am typing probably my life out!!!
Hi tb
Hi, Kannappan
@tb: good afternoon
@KannappanSampath it's enough for now - at least I have not mentioned any advantages of TeXworks in your comments. In fact, this is too plain for me
@N3buchadnezzar: each time your formula does not render you will delete it? )
@Ilya How can you mention sth in my comment? I have a little doubt.
@tb Oh, Ted!
13:07
@Asaf: Oh, As!
@KannappanSampath didn't get you
@Ilya One of these times I regret not writing my name "Assaf"... :-)
@AsafKaragila that was an idea :) but you can always change it
@robjohn I know that since $ lim_{n \to \infty} \frac{\log n}{n} = 0 $ it implies $ \frac{1}{a} \lim_{n \to \infty} \frac{\log\left( n^a \right)}{n^a}$. But how does the latter imply that $\lim_{n \to \infty} \frac{\log(n)^a}{x} = 0$ ?
@Ilya Nah. I usually dislike "Assaf".
Too many letters.
@AsafKaragila seen my comment on the AD vs reflexivity of $\ell^1$? I think that gives a better answer than using Vaeth's paper.
13:09
Indeed.
I'll research into that some more.
Which is good, since I was just gonna start the work on algebraic topology...
@N3buchadnezzar Just a minute. There is something breaking MathJax for me again.
(you'll find the link to the online version Fremlin's books in my answer you link to). The chapter on AD is pretty nice, I think.
How did you come up with my answer, by the way?
@robjohn I can not figure out the mistake :/
I saw that it was linked to my question, so I checked what the hell that was... If that's what you're asking.
13:11
Yeah, that's what I was asking. :-)
Is there a .pdf version or just .tex files?
@N3buchadnezzar raise things to the $\frac1a$ power
no, just the .tex files.
And Pincus's paper is online, at least I have a pdf of it...
Let me check for the link.
I'm on the relevant page.
I'll check the .ps file first. Since I'm only quoting results...
here we go (Pincus's paper)
(or do you mean Pincus-Solovay?)
Yeah, I had that.
My university has no access to this paper, though.
And I'm sure as hell ain't gonna pay 25 EUR just for an answer on m.SE :-)
Which is why I preferred the MR link...
13:14
I think I lost your email address...
got it.
That link doesn't work for me. However, I have the books compiled. I use them every day :)
In case you haven't received mail, you should check your spam folder...
13:20
I'll be right back, I need to restart my browser. Something's fishy.
Might be that tuna salad you ate for lunch...
What's a weak order unit?
@AsafKaragila Given a vector lattice, $e$ is called a weak order unit if $x = \sup{\{ne \wedge x\,:\,n \in \mathbb{N}\}}$ for every $x \geq 0$.
In $\ell^1$ you can simply take a non-negative sequence that is never zero, like $(2^{-(n+1)})_{n=0}^\infty$.
I see.
Also, you should check the waiting deletion votes in case you haven't done that. :-)
Is this guy the same as Diophantine Vassili?
(Gerry's friend)
Do you mean that guy on MO?
13:30
cool. first office hours, first students. done in 5 minutes
Of course, it's the zeta guy on MO, but there was this guy asking a ton of diophantine questions this fall
Yeah, the same guy.
He asked Diophantine stuff on MO as well. Eventually Scott Morrison asked him to ask on meta.MO first, and eventually he told him to stop asking at all.
Oh yes, he is. I just didn't notice that he changed his username.
So, for $\ell^1$ $AC_\omega(\mathbb R)$ is enough, instead of $AC_\omega$ in general?
It's not quite clear to me... Since I don't know whether the identification of $\ell^{\infty}*$ with the bounded additive set functions on $\mathcal{P}(\mathbb{N})$ works without any choice.
13:35
Well, what is the cardinality of $(\ell^\infty)^\ast$?
It's not clear to me without choice :)
What is the cardinality of $\ell^\infty$?
well, it's contained in $\mathbb{R}^\omega$ and contains $[0,1]^\omega$...
So continuum it is.
Yes, but I can't say anything about the dual. With choice it's the power set of the continuum, without it but dependent choice it can be continuum, without any choice, I have no idea. At least continuum.
@AsafKaragila By the way, what question specifically do you want me to take care of?
13:44
Nothing in particular. :-)
So... what is $\mathbb RP^n$?
real projective space...
The space of $1$-dimensional subspaces of $\mathbb{R}^{n+1}$.
Yes, but what is it?
@AsafKaragila it is homeomorphic to a circle for $n=1$
What do you mean, what is it?
A beautiful $n$-manifold? Orientable if $n$ is odd, non-orientable if $n$ is even... $SO(3)$ when $n=3$?
I mean, what is the exact definition? What are the homology groups? etc etc.
13:52
(Hello teddy. You're early today : ))
You can define it as $(\mathbb{R}^{n+1} \smallsetminus 0) / \sim$ where $v \sim w$ whenever $v = \lambda w$ with $\lambda \neq 0$.
@AsafKaragila but all that should be on wikipedia...
But Wikipedia is so far away!
It just came closer to you: here it is.
It's so long and arduous to read...
Not to mention dreadfully boring!
@AsafKaragila You don't have a choice.
13:54
Sheeesh, lazy SOAB
@MattN Tell me about it. It's either that or algebraic geometry next semester... :\
@AsafKaragila Why not both?
I wanted neither!
Hi, Matt! Yes, I'm early...
@AsafKaragila Life's a bitch.
13:56
@MattN I know. :-\
Just when I was planning to do actual work my favourite distraction pops up... Oh well.
@Matt After I started typing out a small note on dense sets, it has ever since grown into a repository of some results on connected sets. Would you mind looking at it?
@KannappanSampath I wouldn't but not today (or at least not right now). I really need to prepare a lecture about PDOs. Sorry.
@MattN I understand. I never saw your last message. I am sorry about that. I'll type it further and later a weekend, may be?!
@KannappanSampath Or Friday afternoon... whatever : )

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