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18:05
anyone that can help me with some proof in rudin
twink
yes?
well
which proof?
it is theorem 3.7 in walter rudin
The subsequential limits of a sequence P_n in a metric space X form a closed subset of X
and which part don't you understand?
18:13
is it possible to write with latex code here $\{p_n\}$
no i guess not
well
he says that P_ni converges to q therefore q is in E^*
last sentence
are you with me?
yes
he constructed a subsequence that converges to $q$
so $q$ is a subsequential limit
yes..
yes
but
@PedroTamaroff: I'm so sorry
does this say that q is in E^*
yes because $E^*$ is the set of subsequential limits
and $q$ is a subsequential limit
that means $q$ is an element of $E^*$
18:18
hold on ..
@FernandoMartin why are you so sorry?
I got Pedro into this cookie game.
Yes, @Fernando, it seems you should be very sorry!
what happened?
I am, I am!
18:19
so he first assumed that q is a limit point of E^*
@Danny: You can type LaTeX, and it you click on the $\LaTeX$ in chat link on the right, you'll get instructions to see it appear all processed :P
@Fernando: It's almost as bad as @Pedro first dragging me into this chatroom months ago.
yes Danny
ok hold
Hi @Twink
a set is closed if and only if it contains all its limit points
hi @TedShifrin
so he took a limit point of $E^*$
and showed it's contained in $E^*$
18:21
yes but he first assumed that q was a limit point and then showed that it is in E^*
yes
thanks twink!
I don't have Rudin here with me, but it seems that you want a diagonal argument to construct a new subsequence with $q$ as its limit.
confused
hold on guys i could not find the latex button
do you see the link to the right of Etiquette guidelines on the right, @Danny?
18:24
but what did @FernandoMartin do to @PedroTamaroff ?
I don't understand
got him hooked on some intrepid game
and why should he be sorry?
Pedro can decide to play the game or not
because now @Pedro is addicted to something stupid :D
LOL
ah, well, addictions are sometimes out of one's control
but I have to go prepare my course, so I'm getting out of this place :P
18:26
$\epsilon$
yes, @Danny, that's an $\epsilon$ ... although I prefer $\varepsilon$ :D
@Twink: once you play, you can't just decide to stop playing
hehe
lol
$\epsilon$
but
why
18:27
So, @Fernando, we have to punish you by not answering 10 of your math questions.
poor Pedro
:(
It's ok, @Fernando. You didn't really like differential geometry, anyhow :D
render mathjax works but not start mathjax
$\theta$
Haha, I know nothing about differential geometry. Can't say I dislike it.
18:28
weiii
were you guys from?
Did you drag the link into your bookmark bar, @Danny?
but i think it works now
Ah, ok. It look me a long time to get it to work on my iPad, but I got it there finally, too.
Lots of folks from Europe and South America — oh yeah, and Australia — are around here, @Danny. There's a few of us from the US from time to time.
Iam from sweden Ted shifrin
Very cool, @Danny. That's a country I've always wanted to visit.
18:31
hehe yeah, nothing special to see execept for a lot of beautiful girls
many Phd:s here
LOL ... I had a very beautiful Danish girl who was an exchange student here and in my classes for a year back about 25 years ago. She was phenomenally brilliant, too. All the guys stayed in the class all year to "work" with her :D
haha
Does anyone know a nice text on CW-complexes?
And, come to think of it, a very brilliant, handsome Danish guy a few years before that. He's now world famous — and turned out to be gay :)
Ted are you a genious
18:34
Um, no, @Danny, I don't believe so. But I'm not stupid :P
i have a really hard time with basic math .. iam doing Real analysis now
iam working my ass off
I don't know of a text that focuses just on CW complexes, @Fernando. They're usually part of an algebraic topology or differential topology course/book.
Real analysis is hard, @Danny, and Rudin is not a book that is easily digested.
but i saw your profile you are a teacher?
But some profs insist on using it, because that's how they were raised. (For example, I was taught out of that book 40+ years ago.)
at the university
18:35
Yes, @Danny, I am.
in what subject Ted
well, math, obviously :D
yeah i mean
hehe
well, there are many branches in math
I teach all sorts of things, but recently every year I have been teaching a multivariable calculus/analysis/linear algebra combined course that I created. Also taught abstract algebra last year, differential geometry this year.
Twink by the way thank you!
wow
differential geometry sound tough
18:37
@Danny: Rudin is very tough. There are more accessible real analysis books you can certainly find ... for example, one with pictures :D
;)
@TedShifrin: Ok, thanks! Any suggestions? I have seen Spanier's book on algebraic topology on my uni's library.
NOOOOO @ Spanier. It's like Rudin. Unreadable.
Ted so you are a PHD student
Haha, ok. What would you suggest?
18:38
sorry
have been
Munkres wrote a much more readable book on homology/cohomology (no fundamental group stuff) that talks about it. Hatcher's topology book is all the rage now, and you can download a .pdf free from his webpage.
No, no, @Danny. I finished my Ph.D. in 1979. I'm old.
@TedShifrin: Thanks, will look them up!
@Fernando: Hatcher requires maturity to read, but he has the best exercises I've ever seen in any algebraic topology text.
yes ok Ted
i really look up to people that are good at math
beacuse i know how tough it is
@TedShifrin do you like the joke in my profile? :D xD
18:45
by the way twink
when he writes choose $n_i$ such that $p_n{^i}$
Um, I'm not sure I get it entirely, @Twink.
such that..
$p_n^i \neq q$
in the parenthesis he writes that if no such $n_i$ exist then $E^*$ han only one point
they're just saying Robin is Batman's boyfriend Ted xD
does he mean that no subsequence converges
to any other point than $q$
he means that if there wasn't such $n_i$
18:50
yes...
the sequence would be $(q,q,q,...)$
so it would converge to $q$
See you guys later. I have work to do :) Have fun!
and $q$ would be in $E^*$
ok bye Ted
that's why there's nothing to prove
Ciao Ted
18:51
Thanks for your help @Ted. See you!
yes i got it twink !!!
See you, @Fernando.
twink can i bother you a little more?
i have some other question on another proof
sure
it is theorem 3.17
give me a second so i can formulate my question..
18:55
Looking at the stars, I see Ted emphatically disagreed with someone
with who?
I don't know
why do you think that?
its right over there ---------------------->
"NO" xD
@KevinDriscoll thanks for unignoring me :$
18:59
okey now, it is 3.17 (b) ,
will $s_n$ converge
to $y$
and why?
I frequently ignore/unignore you depending on how I'm feeling and whats being said
Hello
howdy Charlie
hi
@KevinDriscoll fine and you?
@Danny hi Danny, how.are you?
19:03
@Charlie doing well, just trying to wake up and get some work done while I watch football
iam fine thanks charlie!!
@KevinDriscoll ah, interesting :)
twink are you there
@Danny good :D
yes
19:04
do you know what i mean
not $s_n$
but a subsequence of $s_n$ would converge to $y$
ok yeah some subsequence of sn
ofcourse
but!
how do i know that a subsequence of $p_n$ converges?
sorry $s_n$
@alizter
@FernandoMartin LAWL @TedShifrin
@pedro Welcome back
19:08
Pedro nicolás
@Charlie Yiss?
Now that we somehow convinced him to change his name to Pedro, I think we should start calling him Pyotr
@PedroTamaroff how are you?
@KevinDriscoll Pyotr Nikolai.
@Pedro Tamarrof not Russian enough for you?
19:10
@KevinDriscoll Tamaroff.
or is that your middle name?
I think one can write it as Tamarov.
Nah, that's my surname.
Good to know you'ree fine
@Danny because there are infinetely many points to the right of $x$
@Charlie You so silly. I was about to answer.
19:10
points of the sequence
I had a good coffee and now feeling good.
I had a little headache because I had no coffee when I woke.
@PedroTamaroff I asked you first, but you answered.kevin firstly >:(
@Pedro Coffee in the afternoon? Someone mustve been sleepy.
@Charlie *first.
is there any theorem about this?
or is is clear..?
19:12
@PedroTamaroff I don't care
>:(
@KevinDriscoll Well, I didn't have my morning coffee!
@PedroTamaroff Ah okay. Sleep late or have another engagement? I'm not a coffee drinker myself. Quite odd for a grad student.
@KevinDriscoll Come againz?
every sequence must have a subsequence that converges to a real number or to $+\infty$ or $-\infty$
@Pedro Just curious why you missed your morning cup
19:17
@KevinDriscoll Oh, someone made tea, and I said: "Oh, why not?"
and since there are infinitely many points on the right of $x$
Aaaaah! Okay. Didn't quite do it for you? @Pedro
that would define a sequence
@KevinDriscoll Not at all. =D
@Pedro I'm a tea drinker, but not for the caffine
19:18
so there must be a subsequence of that sequence that converges
oh it was that simple
@KevinDriscoll I am a coffee junkie.
There, I admited it!
but does it matter if it is a subsequence or not
the possibilities $y=+\infty$ and $y=-\infty$ are not excluded
of course
we can just say that either
19:19
it matters
@Danny What are you discussing?
It's okay @Pedro you're among friends here. And quite possibly a number of fellow addicts
$s_n$ converges to $+\infty$ or to some point
we are discussing Theorem 3.17 of Rudin @PedroTamaroff
pedro some theorems of Rudin
19:20
no Danny
@Twink Checks...
$s_n$ may not converge to any real number nor to $+\infty$
oh
but there will always be a subsequence
that converges either to a real number, or diverges to $+\infty$ or $-\infty$
it's part (b) of Theorem 3.17 @PedroTamaroff
$(i)$ We have $s^\ast \in E$ and $(ii)$ If $x>s^\ast$ there is an integer $N$ such that $n\geqslant N$ implies $s_n<x$?
19:22
@Danny asked why there is such a $y$
ok!
part (b)
hello
Pedro will explain it better @Danny
no it was good explanation ! the only thing
19:24
@Twink Oh, I guess you put it out clearly.
does every sequence have a convergent subsequence?
$+ infty $ included
@Danny If you admit infinite limits, yes.
and $-\infty$ included too
is there a theorem about this...
@Danny Bolzano-Weiertrass.
19:26
sorry for asking so much . iam a beginner
take a look at theorem 3.6
Plus the fact that if $\langle x_n\rangle$ is unbounded then it has a sequence "converging" to $+\infty$ or $-\infty$.
it handles the case where the space $X$ is compact
that is Bolzano-Weierstrass theorem
yes
but is $\mathbb{R}^k$compact?
look at part (b)
a bounded sequence always has a convergent subsequence
convergent means it converges to a real number
19:28
hold
and as Pedro said
if it's unbounded
ok !! i think i get it
we can put it in two cases
there is a subsequence that converges to $+\infty$ or $-\infty$
either it is bounded then it converges
yes, boundedn, and unbounded
19:29
or it is unbounded then it converges to +- $\infty$
thanks pedro and twins! , were are you guys from?
Danny see part (a) of Theorem 3.17
yes?
they handle the case that $s_n$ is unbounded
Hello people of chat room
Howd @alizter
19:32
they say: "$s_n$ is not bounded above, and there is subsequence convergins to $+\infty$
yes. give me a moment
just that sentence
they're saying what we just told you
so, there are 2 subcases
Is {1,1,1,1,1,......} a subsequence of {1,-1,1,-1,1,-1,1,.......}?
if $s_n$ is not bounded above, there is a subsequence converging to $+\infty$
and if $s_n$ is not bounded below, there is a subsequence converging to $-\infty$
yes!!
19:37
if $s_n$ is not bounded above nor below, there are at least two subsequences one converging to $+\infty$ and one converging to $-\infty$
yes @KevinDriscoll
it's the subsequence $s_{2n-1}$
Twink what do u study?
math and you?
haha
yeah
i mean what courses
measure theory and linear algebra
okey i see
19:40
So in the sequence $s_n = (-1)^{n+1}$ the subsequence $s_{2n-1}$ 'converges' to $+\infty$ and the subsequence $s_{2n}$ 'converges' to $-\infty$........kind of odd
are you from US? twink
errrr sums I mean
@KevinDriscoll Well, no.
What you want is $0$ and $1$ as limits.
@Pedro Ya I meant the elements sum to those values, not that the sequence goes to those values
@KevinDriscoll Note the partial sums are $1,0,1,0,1,0,1,0,\ldots$.
No $\infty$ possible.
19:43
@Pedro are you not allowed to rearrange the terms?
@KevinDriscoll That is another story, dawg. We're talking subsequences here.
Ya, I realized
Oh okay. Can't change the order of the elements when going from a sequnece to its subsequence. Didn't know that
@KevinDriscoll Right. Subsequences $x_{n_i}$ are meant to satisfy $n_{i}<n_{i+1}$.
19:49
What about $1-\frac{1}{2}+\frac{2}{3}-\frac{1}{3}+\frac{2}{4}-\frac{1}{4}+\frac{2}{5}-\frac{1}{5}+\frac{2}{6}-\frac{1}{6}+\cdots$?
is it $\sum_{n=2}^\infty \frac{1}{n}$?
can I just associate the terms like that?
$\sum_{n=2}^{\infty} \frac{1}{n}$ I think you mean
What is the ZFC representation of a limit?
yes
Hello guys, just a quick question. Does the function $f(x) = exp(\frac{-1}{x})$ for $x < 0$ belong to Schwartz space?
thanks
can I do that @PedroTamaroff?
19:53
@Alizter What...?
@masfenix The definition of $S$ that I've seen requires the function to be defined on all of $\mathbb{R}$
@Twink What you're doing is looking at $a_1,a_3,a_5,\ldots,$ if I am not crazy.
@masfenix but I'm not an expert. If its reasonable to define it only on $x<0$ then it seems like it should be.
no I'm just trying to find the limit of that series
@KevinDriscoll Thanks
19:56
but I was wondering if I can associate the terms like that
@masfenix Actually no
and get the series with $1/n$
@masfenix I wasa wrong, just looked at a graph. It doesn't decrease as $x \to -\infty$ fast enough
or isn't it permitted?
I am actually a little bit confused about Shwartz spaces. Could you give me a little intuition on them?
19:57
@Twink You have found one subsequence converging to $+\infty$. That's a start.
So if the sum converges it must go to $+\infty$.
Can you show this?
is that a subsequence of the partial sums sequence? :S
@masfenix They're the space of all functions that are infinitely differentiable and both the function and the derivatives decrease faster than ANY power of $x$ as $x \to \pm \infty$
@Twink Yes.
If everything can be represented with ZFC what about $\lim_{x\to0}f(x)$

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