Mathematics - 2012-04-15 (page 2 of 3)

t.b.

12:01

If a subsequence $f(x_{n_k})$ converges then it converges to $f(x)$. If the entire sequence didn't converge, there would be another limit point by compactness. Ooops

Benjamin Lim

@tb How can I prove that there is only one limit point?

I know that limits are unique but I don't thin they are the same

t.b.

It must be getting late down under... :)

Benjamin Lim

huhuh?

it's only 10pm

Matt N.

But limit points are unique means there can only be one per sequence. No?

Benjamin Lim

@MattN Yes because a metric space is hausdorff

t.b.

12:06

Once again: suppose $f(x_n)$ does not converge to $f(x)$ then there is a subsequence $f(x_{n_k})$ that stays at least $\varepsilon$ away from $f(x)$. This subsequence has a convergent subsequence $f(x_{n_{k_l}})$ by compactness, but then it must converge to $f(x)$, which is absurd.

user19161

@BenjaminLim It's 8pm here!

Matt N.

@BenjaminLim Assume that $x_n$ has two limits $a$ and $b$. Then there exists an $N_a$ such that for $n > N_a$ you have $d(a,x_n) < \varepsilon$ and for $n > N_b$ you have $d(x_n, b) < \varepsilon$. So you get $d(a,b) \leq d(a, x_n) + d(x_n, b) < \varepsilon $ which means $d(a,b) \to 0$ so $a = b$. I don't know where it needs Hausdorffity but it's all in $\mathbb R$ which is a metric space and therefore Hausdorff.

(in the third sentence you want to pick $N = \max (N_a, N_b)$)

Benjamin Lim

@tb Why must $f(x_{n_{k_i}})$ convergence $f(x)$, am I missing something here?

t.b.

that's what azarel shows.

@MattN Hausdorffity? :)

Matt N.

@tb : )

Benjamin Lim

12:12

@tb I am very confused now.....

t.b.

obviously...

Matt N.

: D

user19161

@BenjaminLim That means bed time.

Benjamin Lim

I think I need to take like a massive break from mathematics for a while.....

Matt N.

Why? Just sleep over it.

It'll all be "obvious" when you wake up.

Benjamin Lim

12:14

hmmmm ok

I have a driving test tomorrow too

Matt N.

Eww. Good luck.

user19161

@MattN Sleep = massive break.

Benjamin Lim

I hate it man sometimes I get caught up in knots

Matt N.

@JasperLoy Didn't know that. : )

user19161

@BenjaminLim Then don't do knot theory!

Benjamin Lim

12:15

Sometimes I feel like I get everything, can see many connections

sometimes feel like an idiot

@JasperLoy Now you're just being annoying.

Matt N.

@BenjaminLim Nothing : )

user19161

@BenjaminLim That is when you are a totally disconnected set!

user19161

@MattN I saw that before the edit!

Matt N.

:4233047 : )

user19161

I also saw tb's message before deletion!

user19161

12:17

Everyone's infected now...

t.b.

Ben: the thing to do in such a situation is to take a deep breath and state what you want to prove in full detail. Then you state what you know how to prove and then see if things still are confusing.

Benjamin Lim

@tb I want to prove that $x_n \rightarrow x$ implies that $f(x_n) \rightarrow f(x)$

Now let $z_n = (x_n,f(x_n)$ that lives in the graph of $f$

t.b.

hypotheses?

Matt N.

If $f$ is continuous?

t.b.

no.

Benjamin Lim

12:18

I want to show that $f$ is continuous

t.b.

Ben: you need to state the assumptions, too!

Benjamin Lim

hypotheses: $X$ is compact, so is the graph of $f$

@tb I have stated them.

Ok so now by compactness there is a subsequence $z_{n_k} = (x_{n_k}.f(x_{n_k})$ of $z_n$ that converges to say $(x,f(x))$.

t.b.

Yes, because the graph is closed the limit point $(x,y)$ is of the form $(x,f(x))$.

Benjamin Lim

@tb What has that got to do with the graph being closed?

forget it

the limit point must be in the set

so that $y$ must be of the form $f(x)$

t.b.

yes.

Benjamin Lim

12:22

ok

Right so now I know that $f(x_{n_k})$ converges to $f(x)$

t.b.

sheesh, Jasper's performing his dance again.

Now suppose $(x_n,f(x_n))$ does not converge to $(x,f(x))$.

Benjamin Lim

I think you mean suppose $f(x_n)$ does not converge to $f(x)$

t.b.

which is the same since $x_n \to x$.

Benjamin Lim

yes

so now suppose $f(x_n)$ does not converge to $f(x)$

then there is a sequence $f(x_{n_j})$ that stays within some distance $\epsilon$ away from $f(x)$

t.b.

within some distance away? :)

Benjamin Lim

12:26

this means that there is some ball of radius $\epsilon$ about $f(x)$ that contains only finitely many terms of the sequence $f(x_n)$

t.b.

not quite.

Benjamin Lim

why not?

t.b.

$d(f(x_{n_j}),f(x)) \geq \varepsilon$ for all $j$.

Benjamin Lim

that's what I said

t.b.

no.

Benjamin Lim

12:28

we say that $f(x_n)$ converges to $f(x)$ if any epsilon ball contains all but finitely many terms

t.b.

yes, the negation of that is not what you state.

@BenjaminLim here

Benjamin Lim

the negation of that is that there is an epsilon

such that

the ball of radius epsilon about $f(x)$ contains only finitely many terms of the sequence

@tb

t.b.

no.

Benjamin Lim

huhuhuhuh????

wait

t.b.

there is $\varepsilon \gt 0$ such that infinitely many terms of the sequence $f(x_n)$ are outside the $\varepsilon$-ball around $f(x)$.

Benjamin Lim

12:31

@tb that is the same as saying that all but finitely many are outside the ball right?

t.b.

no. Think of $f(x_n) = (-1)^n$ and $f(x) = 1$.

Benjamin Lim

yeah

take $\epsilon =0.5$

nothing is in the ball about of that radius about 1

ok

so now

t.b.

nothing? I count $\{2n\,:\,n \in \mathbb{N}\}$ as pretty infinite.

Benjamin Lim

ok

so now we have a subsequence $f(x_{n_j})$ such that given any epsilon, for all $j$

$d(f(x_{n_j}),f(x))\geq \epsilon$

user997704

Each of two players receives an
envelope containing money. The amount of money has been randomly
selected to be between 1 and 1000 rupees (inclusive), with each rupee
amount equally likely. The random amounts in the two envelopes are
drawn independently. After looking in their own envelope, the players
have a chance to trade envelopes. That is, they are simultaneously
asked if the would like to trade. If they both say yes, then the envelopes
are swapped and they each go home with the new envelope.

t.b.

12:36

^ would you mind not interrupting a conversation like this?

3

Benjamin Lim

@user997704 This is very rude

2

@tb so now

t.b.

by compactness of the graph, $z_{n_j} = (x_{n_j}, f(x_{n_j}))$ has a convergent subsequence.

Benjamin Lim

yes

let us call it $(x_{{n_j}_k},f(x_{{n_j}_k}))$

say the $f(x_{n_{j_k}})$ converges to some $z$

t.b.

Thus, $(x_{n_{j_k}}, f(x_{n_{j_k}})) \to (x,f(x))$ and in particular $d(f(x_{n_{j_k}}),f(x)) \to 0$, contradicting the choice of $f(x_{n_j})$.

Benjamin Lim

I think I get it now.

firstly we know that $x_{n_{j_k}}$ converges to $x$

so that forces $z$ to be $f(x)$

t.b.

12:42

@BenjaminLim exactly

Benjamin Lim

because the graph of $f$ is closed it must contain all the limit points

F'x

@user997704 I deleted your message above by mistake; I do apologize for that

badp

@user997704 math.stackexchange.com/questions/ask

Benjamin Lim

@tb Ok

let me say everyhing

I get it now

phwwwwwwwww

Antonio Vargas

Hello all.

Benjamin Lim

12:48

hi antonio

@tb I should go now

man this tiredness is affecting my ability to think straight

thanks as always

t.b.

@BenjaminLim no problem. Sleep well!

Hi Antonio

user997704

please can I speak now?

Antonio Vargas

I was wondering about this question. Does anyone know if, in algorithms, is it more common for O(...) to mean "for all inputs" or "for inputs large enough"? The OP didn't say.

t.b.

@user997704 sure, nobody said you weren't allowed to speak. Just try to avoid bursting in like that in the future.

3

Matt N.

@AntonioVargas It means asymptotically, so: for large ones.

Antonio Vargas

12:54

@MattN Ok, thanks.

In that case the question (part 2) makes less sense to me. Why would you need to "assume" that f(n) = O(g(n)) if you're already given that f(n) = c.g(n) for all n large enough?

t.b.

@AntonioVargas without further specification the big Oh notation f(n) = O(g(n)) means "for n -> infinity". It doesn't really make sense to write f(n) = O(g(n)) without specifying what behavior of n is intended.

Matt N.

@AntonioVargas I haven't read the question but if you say your algorithm has running time $O(n)$ for example then you'd also have to specify whether this is worst case, average case or best case running time.

For example, worst case running time of quick sort is $O(n^2)$ but average case is $O(n \log n)$.

So it's a bit more complicated than just "for large inputs" : )

Saying $f(n) = c g(n)$ is the same as saying $f(n)$ is $O(g(n))$ for large $n$.

Antonio Vargas

@MattN While studying heaps I realized they could easily be used to sort things. My version of heapsort didn't end up running nearly as fast as the usual implementation but at least it was faster than insertionsort :)

That's the extent of my sorting complexity knowledge

t.b.

I always liked bubblesort. All the others are far too slick :)

Antonio Vargas

Me too. I like how intuitive it is---anyone could come up with it on their own.

Alright I guess I'll go ask that poster for some clarification...

Matt N.

13:19

@AntonioVargas My sorting knowledge is almost non-existent. So I can relate to that : )

@tb Yes, that's my favourite : )

Can someone explain this comment to me?

t.b.

@MattN It might be intended as a counterexample: while being a patent clerk he developed his great ideas while after pursuing physics and maths, he made strange remarks about dice and the like :)

Matt N.

It won't count then since while he was working at the patent office he used all his free time to work on his paper : )

t.b.

describe all possible convex domains?

user997704

13:35

hey guys i have question

I asked on stack exchange didn't get convincing answer

Rob

@MattN There is no way of "Putting aside the question of defining 'intelligence'" and then asking will doing math make you more intelligent than not doing math, in my opinion.

Matt N.

Isn't $\{x_0\} \subset S^1$ as well?

Matt N.

13:58

Someone else had mercy on this one : )

N3buchadnezzar

How does one open ps documents?

Matt N.

Double-click it and then Preview will convert it into a pdf. : )

Antonio Vargas

@N3buchadnezzar SumatraPDF will open them.

Assuming you're on windows.

N3buchadnezzar

thanks =)

Man i feel tired

Antonio Vargas

No problem.

Matt N.

14:04

Is $B(X,X)$ all bounded linear operators on $X$?

user997704

anyone interested in game theory/

Matt N.

Ah yes. Kreyszig page 396. : )

N3buchadnezzar

The game

Is there any courses teaching "advanced geometry"?

Eg Euclidian theory , Carlyle circles , etc

Antonio Vargas

@MattN Oh wow, that must have taken some effort.

Matt N.

@AntonioVargas Indeed : D

Matt N.

14:56

Can someone explain this comment to me?

J.M. is around : )

t.b.

@MattN OP seems to be misunderstanding the problem incorrectly. IOW does not make much sense.

Matt N.

What's IOW?

t.b.

in other words

Matt N.

Oh, Thanks! : )

Btw, I caught you making a particular (English) mistake twice. Would you like to hear?

t.b.

tell me.

Matt N.

15:00

It's "get on my nerves" not "go on my nerves".

It's a false friend : )

t.b.

Thanks for pointing it out. Wouldn't make that mistake if I actually thought about it, I believe.

@MattN maybe kuku wanted to say something like $|f(x) - f(0)| = x^2 \geq C |x|$ as soon as $|x| \geq C$.

Matt N.

@tb Yes, I believe you. : )

@tb He/She's deleted his/her comment.

t.b.

@MattN I was just about to point you to a relevant meta thread.

Matt N.

@tb Thanks. For not doing it. : )

(Resisted the temptation to write "knot doing it")

Sunny Marella

Hey guys,i posted a question once on mathoverflow:mathoverflow.net/questions/91462/thomas-clausens-puzzle-closed   Received very few replies,so i will re-post in a different manner here,hope someone can explain:I want to know what is the relation of branch cuts to the above puzzle.
On querying my teacher,he told me "Raising to a power essentially requires the logarithmic(Ln) function, which has many branches. So, it may not be true that (a^b)^c=a^(bc)."
I was explained that the Logarithmic(Ln) function is discontinuous

Matt N.

15:14

Hello JM : ) Now you've got balls?

J. M.

@MattN I always had balls. (Also, hi.)

Matt N.

: )

How are you doing? You're not back yet, are you.

t.b.

@SunnyMarella Did you read this?

J. M.

@MattN I think there might be positive developments this week... hopefully.

Matt N.

crosses fingers

t.b.

15:16

What Matt said. (Also, hi.)

J. M.

@tb Hi also. :)

@SunnyMarella if something won't fit in two lines, it's usually a better idea to post on main instead of in chat...

Antonio Vargas

Hello @JM.

J. M.

Hello @Antonio. How'd you do with that almost-Julia set you were working on?

Sunny Marella

@JM Yup.Thougt would try to caress it into chat for informal discussion.But will post on main anyway.Thks.

John Smith

anyone here good with graph theory

Matt N.

15:22

Finally. My headache is gone. : ) Had some (pleasant) visitors yesterday evening and far too much to drink. I wonder when I'll finally not drink any alcohol anymore.

Antonio Vargas

@JM Well, visualizing it allowed me to guess where a bunch of the zeros lived, but I still can't show it analytically. Essentially the problem now boils down to showing that the function $cos(cos(cos(sinh x))) - sin(sin(sin(cosh x)))$ with x real has a zero (or, preferably, infinitely-many).

J. M.

@AntonioVargas Interesting. Though I've a good feeling an analytical route would require a fair bit of trickery...

Antonio Vargas

Me too.

John Smith

anyone know about spanning trees?

or grid graphs

J. M.

@JohnSmith If it's a bit long, consider asking on main...

John Smith

15:28

it's not long

Nico Bellic

@JohnSmith ask the question.

John Smith

just trying to count the number of perf. mazes based on its dimensions

user997704

anyone interested in game theory

John Smith

which means at any given point there is only one path from that square to the start of the maze

J. M.

@JohnSmith Not sure the guys who can deal with that hang around here. Consider asking on main.

Antonio Vargas

15:50

@JM Thanks for prompting me to think about it more, I may almost have it now :)

t.b.

I truly dislike it when people have questions that are very interesting, extremely hard, utmost challenging, and so on.

J. M.

@AntonioVargas Heh; tell me if you manage to pull it off... :)

t.b.

Turns out that I almost always disagree...

J. M.

Well, apparently it's easier for people to just plaster the words "interesting" or "difficult" in their questions instead of formulating their questions such that their being interesting/difficult is already apparent, without having to say those words...

Matt N.

Google has messed up all the colours of my inbox :,(

But more annoying than that is the fact that I have to scroll in places where I didn't use to.

N3buchadnezzar

16:05

I do not like it when people choose not to answer my questions =(

J. M.

Bah, everybody these days is bent on turning GUIs into nuclear dashboards...

Matt N.

The chat "frame" (aren't frames from the 90ies?) for example.

J. M.

SE, Google, ...

N3buchadnezzar

Then I do not know how to improve my question to obtain help

J. M.

@N3buchadnezzar which?

Matt N.

16:05

@JM Yes but the simpler (read: more blank space) the better.

I'm sure more than 50% of the users would prefer the old look.

J. M.

@MattN Yes, I'm complaining about the obsession with nuclear dashboards too.

I still don't like the current way profile pages in SE are set up, for instance...

Matt N.

Which bit?

N3buchadnezzar

@JM Here

J. M.

@MattN I mean, compared to the old look of the profile pages, the current one is quite overwhelming...

Matt N.

But the gtalk thing is really terrible. I used to be able to see the entire list of contacts that were online. Now I have to scroll.

J. M.

16:09

@N3buchadnezzar I don't know what Maple is capable of these days; is there nothing in the help file on restricting the results of plots with inequalities?

@MattN Ah, that; very vexing, I agree...

Matt N.

@JM I thought that when they changed it. Now I can't remember the old one.

@JM I think it calls for a facepalm.

J. M.

@MattN Maybe Google is counting on that happening also... :)

Matt N.

: )

(not sure why I'm posting a smiley, it makes me want to cry actually)

J. M.

@MattN I don't really know who to bitch to about things like these, sadly...

Matt N.

Well on SE, it's probably meta we can bitch to.

t.b.

16:13

$\color{red}{(\text{status-declined})}$

Matt N.

Yes, I had a feeling that would be the case.

J. M.

With Google, I've never seen them undo a new feature, ever...

t.b.

Or (status-ignored) seems more to the point.

Matt N.

It's not tagged feature-request.

N3buchadnezzar

16:29

hi

Kannappan Sampath

16:45

Can we generalize this post for a prime $p$ so that this shall be a good original for subsequent posts? — Kannappan Sampath 13 mins ago

Antonio Vargas

What's the etiquette for posting incomplete but lengthy ideas as answers? I could certainly write it as a few comments. Just wondering.

J. M.

@AntonioVargas If it's long enough, just say at the outset that it's incomplete, but a starting point nevertheless...

Antonio Vargas

I probably won't think about it more since I'm getting tired of the problem. Maybe I should just leave it be.

J. M.

You did say you made some headway; it'd be a mess to split it across multiple comments I think...

Antonio Vargas

Well I just don't have anything solid. Interested in hearing my thoughts?

J. M.

16:56

e.g. this is one of my "incomplete" answers, but it seems nobody complained.

@AntonioVargas Sure; what's on your mind?

Antonio Vargas

Alright, so I mentioned that I basically want to show that the function $f(x) = \cos(\cos(\cos(\sinh x))) - \sin(\sin(\sin(\cosh x)))$ has infinitely many zeros. The approach I was considering was to show that it alternates sign infinitely-many times. Numerically $f(1)$ is negative, and I wanted to try to find some periodicity in the function to conclude.

J. M.

Well, $\cosh$ and $\sinh$ are $2\pi i$-periodic...

Antonio Vargas

Yes, but I'm considering real x.

J. M.

...I can't plot, but I do believe it looks nasty on the real line...

Antonio Vargas

If $g(x) = \cosh^{-1}(\cosh x + 2\pi n) - x$ then $\cosh(x + g(x)) = \cosh x + 2\pi n$, so that $\sin(\sin(\sin(\cosh(x+g(x))))) = \sin(\sin(\sin(\cosh x)))$.

So replacing $x$ by $x+g(x)$ in the right term of $f$ leaves it unchanged. As for the left term...

It appears that replacing $x$ by $x+g(x)$ in the left term asymptotically minimizes that term. Here, I'll make you a picture...

J. M.

17:08

@AntonioVargas Okay, no trickery so far, I see...

Antonio Vargas

J. M.

@AntonioVargas o_O wiggly, that.

Antonio Vargas

Actually that's a terrible plot. I'm getting mixed up. One minute...

J. M.

@AntonioVargas The function itself is rather terrible, you know... :)

Antonio Vargas

Let me rename my $g(x)$ to $g_n(x)$, since it's the $n$ that's varying in that plot.

Indeed!

The curve on that plot is $\sin \sin \sin \cosh(1+g_x(1))$, and the dots are $\sin \sin \sin \cosh(1+g_n(1))$

for $n$ ranging over integers 0, 1, 2, ..., 20

No, I am mistaken.

The curve on that plot is $\cos \cos \cos \sinh(1+g_x(1))$, and likewise for $n$.

Now, there is an expression for $\sinh(1+g_n(1))$,

$$\sqrt{\frac{-1+2 \pi n+\cosh 1}{1+2 \pi n+\cosh 1}} (1+2 \pi n+\cosh 1)$$

J. M.

17:24

@AntonioVargas This is from $\sinh$'s addition formula, yes?

Antonio Vargas

To be honest it is from Wolfram Alpha ;)

Clearly this is near $1 + 2\pi n + \cosh 1$ for large $n$, so I suspected $\cos \sinh(1+g_n(x))$ would be near $\cos(1+\cosh 1)$, but this is not the case. The functions are simply too wild for an argument like that to work.

J. M.

@AntonioVargas "The functions are simply too wild" - hence what I said about needing trickery... so far it's all straightforward, which means the solution is still a bit far away... ;)

Antonio Vargas

Indeed. Well, if I could just figure out why this choice apparently almost minimizes the left-hand term I would be golden.

Maybe I'll take a look at the fixed point of cosine.

Meh. Maybe this is just the totally wrong approach.

J. M.

...well, my time's up for today. I hope you find a better route.

See you!

Antonio Vargas

Thanks for the ear! See you later, @JM.

Matt N.

17:39

@JM Bye!

robjohn

17:51

Ack. I just missed JM. I wanted to chat about a question we had both answered.

Matt N.

@Ilya This is the "encounter" I'd mentioned. : ) Couldn't resist undeleting it.

robjohn

18:07

@MattN I haven't seen Ilya for over a week. Have you seen him recently?

Matt N.

@robjohn No, I haven't seen him either. But he'd asked me about my Watman encounter so I thought I'd post it (since I undeleted it) : )

robjohn

@MattN that pair of comments with Didier?

Matt N.

Yes. : )

robjohn

Who is Watman? am I missing some reference?

Matt N.

I call him the Watman since his W answer.

It's a reference to this talk here.

robjohn

18:11

I see. That was a while ago

Matt N.

Yes.

robjohn

@MattN That was very humorous :-)

Matt N.

@robjohn Yes! It's my favourite talk (on software development) : D

robjohn

@MattN I see that Didier has 174 upvotes and 4 downvotes on the "W" answer.

Matt N.

@robjohn : (

robjohn

18:19

@MattN does that upset you?

Matt N.

@robjohn Yes. On a scale from 1 to 10 where 10 is super-helpful, how helpful would you judge his answer?

N3buchadnezzar

I am thinking about learning python

Matt N.

Go for it.

N3buchadnezzar

but I am also thinking about ruby, javascript or C++. But seeing as I am only going to use this for mathematics, python seems like the best choice, yes, no ?

Matt N.

Don't think it matters at all.

Antonio Vargas

18:23

If you know one programming language well then you can learn another very quickly.

Well, switching from an interpreted language to a compiled language can be a bit rocky.

N3buchadnezzar

Well I know the syntax of a few programs like Matlab, and Maple and so forth. and I am thinking that the transition to python would be rather fast (learning the new syntax.)

Python 2 or 3 ? Hmmm ponders

Zhen Lin

Programming languages are much more than just syntax. Sure, you'll be able to write programs, but you need to learn the culture if you want to write good programs...

Eric Gregor

what do you mean by good?

efficient? clever?

robjohn

19:00

@ZhenLin I'm not sure what the culture difference is between programming languages, but once one knows good programming strategies, is there a real difference between writing a good program in one language vs another?

user19161

@N3buchadnezzar Python is voted the best programming language and also best scripting language by the readers of linuxjournal.com last year.

Jonas Teuwen

Hi guys.

Matt N.

Long time no see : )

How are you?

Hey Jonas : )

Zhen Lin

19:16

@robjohn: Writing a good program in an object-oriented language is a quite different process compared to writing a good program in a functional programming language, say.

Matt N.

@JonasTeuwen Did you see?

N3buchadnezzar

Mission: Dinner
Status: 98%

Jonas Teuwen

@MattN Man, I'm wasted... Traveled >500km today by train and I hardly slept.

Matt N.

High five then! We drank 5 bottles of red wine last night. Worst headache in a long time : )

N3buchadnezzar

@JonasTeuwen Driving a train, drinking and not sleeping is often a bad combination.

Jonas Teuwen

19:22

@MattN Great!

Matt N.

No : )

N3buchadnezzar

Beer before wine is fine, but wine before beer makes you queer? (There was some Mnemonic for this, but I can not recall it )

Eric Gregor

19:36

beer before liquor, never been sicker. liquor then beer, you're in the clear.

user19161

@MattN It depends on how many people is "we".

Matt N.

Four.

user19161

That is quite a lot of wine then!

user19161

The most I ever took is half a bottle at one go.

Matt N.

Yes, 'tis.

user19161

19:38

After that my voice dropped an octave.

Matt N.

Didier has removed his downvote.

robjohn

20:09

@ZhenLin Okay, there is a big difference between object-oriented programs and pre-OOP functional programming, but if the latter is modularized, the differences are minimized and the modular programs benefit from a lot of the advantages of OOP.

run-on sentence?

Matt N.

20:22

The teddy is too fast.

Foool

20:34

I have a question,

\lim_ {x \to \infty} x = \infty can we say (here) that this limit does not exist?

I think it's better to say that the limit is infinity here.

and does not exists in cases where it is undefined.

Am I right?

Eric Gregor

people say both interchangeably. technically $\infty\notin\mathbb{R}$

so the limit doesn't exist. but of course you could adjoin $\pm\infty$ to $\mathbb{R}$ and usually it's fine to be casual about the context

azarel

You would have to rework the statements of a bunch of theorems about limits

Eric Gregor

so really all you're saying is that some sequence is unbounded in a specific direction

Foool

Aha Thanks!

Matt N.

Hey teddy. I'm trying to understand your answer here. I'm particularly slow today, so be merciful. Why is a closed set minus a closed set open?

t.b.

20:42

It's open in the relative topology on $B$.

@MattN Probably the easiest way of saving your example in the real case would be to take the balls of radius $\sqrt{2}$ around $\pm e_n$ where $e_n = (0,\ldots,0,1,0,\ldots)$.

Matt N.

@tb Ah yes, the sets need to be open in the subspace topology. Thanks : )

t.b.

(I think, I haven't checked thoroughly)

@MattN yes, and open sets $U$ in $B$ are of the form $V \cap B$, so if you insist on an open cover in the surrounding space, just replace the relatively open sets $U$ by sets $V$ such that $U = B \cap V$.

Eric Gregor

anyone want to take a crack at this finely posed question: math.stackexchange.com/questions/132169/…

Matt N.

@tb Yes, thank you.

I like this question : )

t.b.

@MattN seen the comments to my answer? Here's a good exercise: Prove that the unit ball in an infinite dimensional normed space is never compact using Riesz's lemma.

Matt N.

20:55

@tb Yes I have. Will do this exercise. But not today, I'm done in, can't think straight (I mean even less than normally)

t.b.

@MattN oh, you're not the only one. The painful days after too much alcohol will never be a lesson I'll be able to learn, apparently :)

Matt N.

: D I still have hope that I will!

John Smith

anyone good with graph theory?

t.b.

I heard rumours that Béla Bollobás knows a little bit about it... But I know no one among the regulars in chat who is especially acquainted with graph theory.

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$Mathematics$ Mathematics