Mathematics

Peter Tamaroff

12:00 AM

@EdGorcenski Arg, happens every time.

Ed Gorcenski

Alright, beer and Rudin time. See you folks shortly.

Peter Tamaroff

12:20 AM

@J.M. Are you there?

Peter Tamaroff

12:37 AM

@ZhenLin Hello.

I have a simple question. Yes or no, basically.

Is the $\limsup$ and $\liminf$ of a set defined for bounded sets only?

(of course, you can tell me some more after the "Yes" or "No" =D)

@AymanHourieh

J. M.

@PeterTamaroff I'm here, but I'm about to leave in a few...

Peter Tamaroff

@J.M. Just that question up there.

@J.M. 'bout $\limsup$ and $\liminf$

J. M.

@PeterTamaroff Sure, if the sequence is unbounded from above, the limit superior is $\infty$...

If the sequence is unbounded from below, then the limit inferior is $-\infty$.

Peter Tamaroff

12:52 AM

@J.M. Hmmmm. I have this definition of $\limsup$: We call $x$ "almost an upper bound" of a set $A$ if for only a finite number of elements $y$ of $A$, it is $y\geq x$. Let $B$ be the set of all almost upper bounds of $A$. Then $\limsup A \; \mathop=\limits^{\text{def}}\;\inf B$

J. M.

@PeterTamaroff That seems to be a definition of the infimum, not limit superior...

Peter Tamaroff

@J.M. Sorry, the limsup of A is defined to be the infimum of B.

J. M.

Ah, you had it backwards.

Peter Tamaroff

@J.M. The thing is I have proven that $\inf B$ exists using the hypothesis that $A$ is bounded both from above and below.

Hence the question.

J. M.

Ah. The bit about unbounded sequences is by convention. If you've shown that the sequence under consideration is bounded on both sides, then you're golden, and you've nothing to worry...

Peter Tamaroff

12:56 AM

@J.M. But how is then the limsup of a set A defined if it is bounded from above only?

Gustavo Bandeira

@MeAndMath Acordei.

J. M.

@PeterTamaroff It's bounded from above, so you're still supposed to have a finite limit superior. (Unfortunately I can't be more elaborate, since I do not have a textbook on hand...)

Its limit inferior would be $-\infty$, tho.

Peter Tamaroff

@J.M. For example, if the set is $(-\infty,1)\cup \{1,2,3\}$ then $\limsup A=1$?

@anon

anon

yes

Peter Tamaroff

@anon OK. Seems strange. How do we use the $\limsup$?

anon

1:05 AM

dunno

I've only seen it used for sequences (which e.g. give statistical arithmetic results in analytic number theory that cannot otherwise be expressed with just limits).

J. M.

Well, gotta go. I'm sorry I can't be more helpful, @Pete.

Peter Tamaroff

@J.M. Hehehe, no problem. Thanks.

J. M.

Hi @anon.

anon

yo

J. M.

...and, see y'all later.

Peter Tamaroff

1:10 AM

@anon Dude.

Say I have a subset of the reals that is bounded from above but not from below.

anon

k\

Peter Tamaroff

@anon Let $B$ be the set of all almost upper bound of $A$.

$B$ is nonempty for it contains all upper bounds of $A$.

Now, I want to show $B$ is bounded from below.

But this might not be true, I think.

For example, let $A=\{\dots,-3,-2,-1,0\}$

anon

correct, like negative integers

Peter Tamaroff

@anon ;)

anon

or nonpositive, I'm not picky

Peter Tamaroff

1:14 AM

@anon Hhehe well.

Then what is $\limsup A$???

anon

Well B=A in that case; what's the inf of A? DNE.

Peter Tamaroff

Seems like $B=A$ in this case.

DNE?

anon

does not exist

Peter Tamaroff

@anon Oh, OK. So we "really" define the limsup and liminf of a set when it is bounded.

anon

actually B=$\Bbb R$, what am I thinking

Peter Tamaroff

1:16 AM

@anon Oh, right!

anon

not necessarily bounded.

Peter Tamaroff

@anon I had the silly idea the a.u.b.'s had to be taken from $A$.

anon

$(-\infty,-1)\cup\{0\}$ ain't bounded but its limsup is -1.

Peter Tamaroff

@anon Right.

@anon OK. NOw I have to prove that $\liminf A\leq \limsup A$.

$\limsup A\leq \sup A$ and $\liminf A\leq \inf A$

anon

If limsupA<liminfA then pick x in between these two values. By limsupA<x we find that A and (x,inf) have finite intersection; symmetrically by x<liminfA we know that A and (-inf,x) have finite intersection. So A is finite...

Peter Tamaroff

1:26 AM

@anon I just found that $\limsup \{1/n:n\in\Bbb N\}=0$. Am I crazy?

anon

correct on both counts

Peter Tamaroff

@anon Seems like the set of "almost lower bounds" of $\{1/n:n\in\Bbb N\}$ is $\{x:x\leq 0\}$ so that $\liminf =0$, too.

anon

< not $\le$

oh never mind

Peter Tamaroff

@anon No. $0$ is also an "alb"

@anon Oh,OK.

@anon Now I gotta do the same for $n\in \Bbb Z-\{0\}$

@anon I got the same results... same sets of aub and alb, same limsup and liminf, 0.

anon

yes

Peter Tamaroff

1:39 AM

@anon Now I $\{1/n\}\cup\{0\}$ =P

Peter Tamaroff

1:51 AM

@anon This one is odd.

Let $A=\{x:0\leq x\leq \sqrt 2 ,x\in\Bbb Q\}$, then $\limsup A=\sqrt 2$ and $\liminf A=0$.

Well, not really odd.

Here $\limsup A=\sup A$

and $\liminf A =\inf A$, unlike the other cases.

@BrianMScott

Peter Tamaroff

2:31 AM

@anon Could you elaborate on that?

@anon Don't you mean something like $(x,\sup A)$ and $(\inf A,x)$?

@MarianoSuárez-Alvarez Estás?

anon

Suppose $\limsup A<x<\liminf A$. Then $\limsup A<x$ so $x$ is an almost bound, i.e. the number of elements of $A$ in $(x,\infty)$ is finite, ie the intersection of A with (x,inf) is finite. Symmetrically for $x<\liminf A$.

I mean, the intersection with (x,supA) and (infA,x) will also be finite, but who cares.

Peter Tamaroff

@anon OH! You meant "inf" as $\infty$...

anon

yes

Peter Tamaroff

@anon Use $\LaTeX$!!!

I was :confus:

@anon I just proved that $\inf A\leq \liminf A$ and that $\limsup A\leq \sup A$, so I'm missing the middle part.

@anon Tell me what you think of this argument:

anon

What do you mean by "middle part"?

Peter Tamaroff

2:41 AM

@anon The inequality between limsup and liminf.

anon

That's just the negation of what you want to prove.

Peter Tamaroff

@anon What?

anon

limsupA<liminfA is the negation of $\liminf A\le \limsup A$, the latter of which you are trying to prove. the idea is to use contradiction.

Peter Tamaroff

@anon I know that. Dunno what you understood I said.

@anon Given an infinite bounded set $A$; let $\overline A$ be the set of upper bounds,and $A^*$ the set of almost upper bounds. Then $\overline A\subseteq A^*$, and since $\inf A^*=\limsup A$ and $\inf \overline A=\sup A$, we must have $\limsup A\leq \sup A$

Similarily, for $\underline A$ and $A_*$, we have $\underline A\subseteq A_*$ and since the situation is with $\sup$s, we have the reversed ineq.

anon

Wait. Are you trying to prove $\liminf A\le \limsup A$ or are you tryint to prove limsup<=sup and liminf<=inf? You wrote the former above.

1 hour ago, by Peter Tamaroff

@anon OK. NOw I have to prove that $\liminf A\leq \limsup A$.

Peter Tamaroff

2:45 AM

@anon I'm not trying to do the latter, I succeded (I guess). I'm missing the former.

@anon Is the argument OK?

anon

Your argument is for limsup<=sup, right? Looks right.

Peter Tamaroff

@anon Yes. And for inf <= liminf

anon

So what's up now?

Peter Tamaroff

@anon I just got your proof. Very neat.

anon

cool

Peter Tamaroff

2:51 AM

@anon I think I'll call it a day!

@anon Bye byes.

anon

later

Gustavo Bandeira

3:18 AM

@WillHunting

BenjaLim

@anon Hey

Can I ask you something?

anon

uh okay

BenjaLim

Suppose $\Bbb{C}^3$ is the standard rep of the lie algebra $sl_3(C)$

Write $V = C^3$

In fulton and Harris they claim that $V^\ast$ is isomorphic to $\Lambda^2 V$

@anon Now I tried to write out how this is possible

Suppose the dual is spanned by the row vectors like $(1 0 0 ),(0 1 0)$ and $(0 0 1)$

anon

can you give a page #?

BenjaLim

I tried to send these respectively to $e_1 \wedge e_2, e_2 \wedge e_3$ and $e_1 \wedge e_3$

177

Let me just write out for the moment

that the Cartan Subalgebra is spanned by $H_1,H_2$

where $H_1 = diag(1,-1,0)$ and $H_2 = diag(0,1,-1)$

@anon Did you find it?

anon

3:35 AM

I don't even know what a cartan subalgebra is man.

BenjaLim

Ok but do you know why $V^\ast$ is isomorphic to $\wedge^2 V$?

anon

can't see why atm

BenjaLim

@anon Ok

math101

4:07 AM

Hey guys

Its dead

peoplepower

I killed it with my mind tricks.

math101

lol

I have a really stupid question that no one has been answering

idk it must be really annoying because I have posted it all over and everyone just skips over it

math101

4:25 AM

@peoplepower Its time to revive the dead

MJD

4:57 AM

@math101 For one thing , both versions have lousy titles.

math101

@MJD haha What shld I be titling them. I am awful at that

I think the answer that Milikan provided is incorrect. I am still working on it though

math101

5:20 AM

Hey @JayeshBadwaik

Jayesh Badwaik

@math101 Hello.

math101

How are things?

Jayesh Badwaik

Things are going okay. Not particularly great, but I think I will survive. :-)

math101

I hope they get better

Jayesh Badwaik

What about you?

math101

5:29 AM

Just trying to catch up in my math which isnt working

Jayesh Badwaik

My MAthJAX SVG renderer is not workin. I will try to restart my browser to see what the problem is. Will be back in a bit.

math101

cya

Jayesh Badwaik

MathJaX Test: $\lim\sup \{1/n : n\in \Bbb N\}=0$

math101

I think I am gonna head to bed. Its 1:30 am here

Jayesh Badwaik

@math101 good night.

MAthJAX : $\limsup \{1/n:n\in\Bbb N\}=0$

math101

5:37 AM

Good Night

Jayesh Badwaik

$\liminf x$

@robjohn I get this kind of rendering while using SVG renderer instead of HTML-CSS.
$\lim \sup$ is supported while $\limsup$ is not. Probably not important, just a heads up. If it works for you, then I will have to check if there is a problem with my browser.

Broseph

I'm grading homework for modern algebra 2. One of the problems is to show I+J is an ideal when I and J are ideals (and both are subrings of R). Most of the students are showing d*(a+b) is in I+J when d is in R and a+b is in I+J. Are they also supposed to show (a+b)d is in I + J? Why?

peoplepower

No, distributive law.

Broseph

5:53 AM

that's what I thought. For some reason the professor mentioned both in his solution.

thanks.

peoplepower

I'd say they should mention that it holds.

Broseph

Yea, that's what I'm doing for students who didn't mention it, but I'm not taking off points for that. I'll cut 'em some slack...this time.