I think $Y$ must be Hausdorff, because then compact sets in $Y$ are also closed

@leo I do think $\lambda$ is continuous without any extra conditions on $Y$.

@user1 how?

lets denote $V(A,B)=\{f\in C(X,Y) : f(A)\subseteq B\}$

@leo If $U$ is open in $Y$, note that $F^{-1}$ of $U$ is open in $X\times[0,1]$, where I use $F$ to mean the homotopy.

@leo You're already using the letter $F$. How about $V(A,B)$ instead (as seen on the wikipedia article on compact-open topology).

@PeterTamaroff hello. brb tho.

@leo If $f_t$ maps a compact set $K$ into $U$, then that means $K\times{t}\subseteq F^{-1}(U)$.

the set $\varphi=\{V(A,B):\text{$A$ is compact in $X$ and $B$ is open in $Y$}\}$. This set is a subbase for the compact open topology, so it's enough to show that the inverse image of any of those $V(A,B)\in \varphi$ by $\lambda$ is open on $[0,1]$. I'm stuck o that

@anon thanks

@leo Let $O\times V$ be an open box around $K\times{t}$ in the preimage of $U$.

okay

I think that the set of all $t$ in $V$ satisfy $f_t\in V(K,U)$

I did not use compactness of $K$, though, which worries me slightly.

@user1 I started that way. I as trying to proof that those $t$ are exactly $v$, which seemed weird to me as well

@user1 With Hausdorffness of $Y$, it follows easy that the complement of those $t$ is closed

I'll think

I hope

@leo I was wrong to assume that $K\times \{t\}$ fits neatly into an open box. What you have to do is construct a family of open boxes, one for each $\langle k,t\rangle$ and then use the compactness of $K\times \{t\}\subseteq K\times[0,1]$ to show that there is a finite subfamily.

@user1 On it. I'll let you know how it goes

@leo :)

@leo I actually have to go, but I would still like to know if this works.

@user1 I'll tell you later, thanks :-)

@anon Are you back?

yes

Yay

It is what I showed you before.

The best I could do is $$n|a_n|\leq 2\sum_{k=1}^n |a_k|$$

But that recurrence gives me a bigger bound, I think.

@anon

dunno

@anon Bahoolas.

I bet Brian Scott can figure this one out.

@user1 The key observation is $$\lambda^{-1}(V(K,U))=\{t\in I:K\times \{t\}\subseteq F^{-1}(U)\}$$

$$0<\sum_{n\leq x}\ln(\frac{\gcd(a,n)}{a})<\ln(a)$$

$\forall a$

$\forall a,b \mid \gcd(a,b)=1, 1\leq b\leq a$

This "emptying-the-unanswered-queue" business makes me reaching my daily-vote limit each day. Today 11h before the end of the day.

It is a noble cause crusade :-)

:-)

As Feynman said "Know how to solve every problem that has been solved."

There should be some more "crusaders". At the moment we are making some progress, but very slowly. Although sometimes it just needs checking a simple linear algebra answer and then upvote.

Hi @amWhy

@skullpatrol Hello, skull :)

@amWhy How are you?

@skullpatrol good...still half-asleep, but coming to... How are you?

@amWhy Fine thanks.

@leo I agree; it contains the problem in a space we know a lot more about (when comparing $K\times I\subseteq X\times I$ with $Y$).

@amWhy did you see anon's answer?

@skullpatrol where?

@amWhy here

It's saying pretty much what's already been said.

@amWhy why was it made into a community wiki?

if you exclude $x = y$, then the first method can be employed...but excluding $a = b$ means you omit a possible scenario. The second method doesn't exclude a = b.

anon chose to post as "CW".

@amWhy There is no y?

I meant if you exclude $a = b$...sorry.

np

@amWhy Thanks for all the help, I appreciate it :-)

@skullpatrol You're welcome! It's always a pleasure, especially when the "asker" is a "bud"...

:D

@skullpatrol I like your $\ddot \smile$

@amWhy Thanks

How come I feel so inadequate when I look at math graduate programs?

Maybe because you're not a math graduate :P

:P

Just my opinion, please don't be offended...

Hi there!

Hi

@skullpatrol What is your major?

@PooyaM I don't have one.

@PooyaM What is yours?

Electrical Engineering, I was looking for a Mathematics major

@skullpatrol Electrical Engineering, I was looking for a Mathematics major

@skullpatrol no, I am nowhere near a math graduate - still in undergrad

hi

hello

@PooyaM you need one for the US or not country-specific?

@JulianKuelshammer Actually I want to talk to a Math major...

@JulianKuelshammer here

@JulianKuelshammer Seems like you are a Post-Doc in Math!!!

@PooyaM yep

@JulianKuelshammer Massive fail

@JulianKuelshammer Great...I'm looking for double majoring with Mathematics...What is your opinion?

@PooyaM That's always difficult to tell when you don't have that much information. That's why I asked which country you are in. It depends for example on how much the overlap is between mathematics and electrical engineering. And of course also on how you are currently doing with your math courses. From my perspective (teaching some of the engineers) what they do in their math courses is often quite different from what a mathematician does.

@JulianKuelshammer I agree with you about the difference of teaching mathematics in engineering schools and math schools...Actually, I've taken some online mathematics courses like Intro. of Algorithms(71/100), Linear and Discrete Optimization(Didn't finished it because of university midterms but till week 5/6 I completed 100/100 and I barely found complications throughout the course) Cryptography(which is little far from math ,97/100). I live in Iran, so I chose Electrical engineering to

@JulianKuelshammer get a good job. The other problem is if I wanted to pursue my interest, there were too many and too diverse. Even in Mathematics, I can't choose an area to do research in. Optimization, Number theory,Graph theory, Complexity theory and AI. I think that is the reason why I didn't pursue Math or Physics major at the first place!

@PooyaM What about talking to some math professor at your local university? They probably know better, what is possible, or maybe they also know people who already did what you are asking about.

Should I delete my response

@JulianKuelshammer I talked to three professors in out Math dept. They told me that if you don't want to be a professor, don't double major with math. They told me you should choose between a well paid job(which is earned by pursing Engineering) and a low paid stressful job a.k.a being a professor in math

@JulianKuelshammer I just don't know what to do. I find courses like Electrical Machines and Microelectronic circuits too naive. I see no deep idea in engineering courses but simple,linear,approximate yet working and robust

@PooyaM To be honest I know nothing about the job situation in Iran. In Europe there are plenty of jobs for mathematicians outside University, like working for insurances, for banks, as a consultant, or also in the physics/engineering direction. Here, I think most people would see it as an advantage to have a degree in math, but such things differ between countries. I can understand that you are lacking deep ideas (maybe also reasoning) in engineering courses from what I'm teaching engineers.

@JulianKuelshammer For example in Digital Logic Design we learned " mealy and moore algorithm " . But by learning I mean, I know how to do it just like the computer does! I don't know how that algorithm is designed. I think these have something to do with Markov Processes. And every time I read some articles in Wikipedia discussing these ideas, my heart starts beating and I think I'm wasting my time in linear time independent systems!

@JulianKuelshammer And here also there is job outside of univ. for math majors. But I don't think a bank clerk thinks about proving whether P=NP

@PooyaM Usually not, that's right. :-)

@JulianKuelshammer I think,the only way I have is choosing a single area and do research in it. And I should self study the prerequisites. But I loved to be like Von Neumann and Shannon

@JulianKuelshammer and Gauss and Poincare and Decartes and Leibniz :( :( :( :(

@PooyaM then you have to do a lot of research :-) Your professors might be right: Then it is a stressful and (at least in comparison) low-paid job. You have to make the decision. It's not obvious in the beginning how far one can get.

@JulianKuelshammer Thank you for talking with me :)

@nitrous2 I assumed that the OP was asking for a numerical method to get square roots, as one would get using a calculator, but not using a calculator. What you wrote is correct, just not what the OP was looking for (imo).

It certainly is quiet right now

@robjohn scream

@PeterTamaroff why?

@robjohn Sometimes there is no reason.

Or maybe there is, but one knows not why.

okay

@robjohn I have a problem.

yes?

@robjohn For a few days now.

Let $\langle a_n\rangle$ be a sequence of complex numbers, and $\langle p_n\rangle$ a sequence of non negative real numbers. Let $A(z)$ and $P(z)$ be their related powerseries. We write $A(z)\ll P(z)$ if $|a_n|\leq p_n$ for each $n$.

I have proven that $A\ll P$ and $B\ll Q$ implies $A+B\ll P+Q$ and $A\cdot B\ll P\cdot Q$.

Now, the problem is as follows:

Suppose $f(z)=z+a_2z^2+a_3z^3+\cdots$ (yes, $a_0=0,a_1=1$). Suppose further that $$\frac{zf'(z)}{f(z)}\ll \frac{1+z}{1-z}$$ Show that $|a_n|\leq n$ for each $n$.

Note that if we write $A(z)=zf'(z)/f(z)$, then $A(0)=1$. So we can write $$A(z)=1+\sum_{k\geq 1}A_kz^k$$ The above means that $|A_k|\leq 2$, since $$\frac{1+z}{1-z}=1+2z+2z^2+2z^3+\cdots$$

The best I could do is deduce $$n|a_n|\leq 2\sum_{k=1}^n |a_k|$$

@robjohn =)

Nothing comes to mind immediately. I will think about it.

@robjohn Thank you.

hello y'all

simple algebra and induction after that, it seems

@robjohn When I use the bookmark start Chatjax, it doesn't work. I have to click on render Chatjax every time instead. Do you know why this could be? (Does it matter where I keep the bookmark, say in my add-on bar for Firefox instead?)

@AlanH If the renderMathJax bookmark works, then if the start ChatJax bookmark is installed in the same place, it should work.

The only thing they do differently is start ChatJax starts a loop to rerender when needed

@robjohn You cannot have complex numbers on a series that is a majorant, can you? A majorant is supposed to be of positive real numbers. I mean, on the RHS of the $\ll$

I mean, the algebra of $\ll$ is not the same of $\leqslant$.

I'm reading Ahlfors and he says different regions may have the same closure. Can someone provide an example of this? I don't quite see it

@AlanH Take the open disk $D=\{|z|<1\}$

And then take $D\cup \{(1,0)\}$

Or $D\cup A$ where $A\subseteq \partial D$

More dramatically $\overline{\Bbb Q}=\overline{\Bbb R}=\Bbb R$ in the metric space $\Bbb R$ with the usual topology.

So $D$ union any piece of the unit circle has the same closure as D alone, is what you're saying?

@AlanH Yes.

Because in fact $\overline{D}=D\cup\partial D$.

@PeterTamaroff ah. thanks.

@robjohn Yes, that is exactly what I got, sire.

I think?

I must have slipped somewhere. You're getting $(n-1)|a_n|\leq 2\sum_{k=1}^{n-1}|a_k|$

@PeterTamaroff Then you use that $2(1+2+3+\dots+n)=(n+1)n$ and cancel

@robjohn Heh, I wasn't too far away. Damn me.

My induction hypothesis is $|a_k|\leq k$ for $k=1,\dots,n-1$. Done.

I will rewrite this to get it neat and tidy.

@robjohn Thank you dearly.

@PeterTamaroff no problem

Hi, @robjohn

Hi, @Peter

what's up?

@robjohn you asking me? Not much...just saying "hello"...

@amWhy okay. Hello

@robjohn What are you up to this Sunday?

@AlanH That is from Rudin

Yeah

@PeterTamaroff Are you able to view the whole thing? I copied it from texshop

@AlanH Let me try and explain.

Fix $\epsilon >0$, and $x_1\in X$ arbitrary. First, I want to show that for this given $\epsilon$ we can find a finite set of points $\{x_1,x_2,\dots,x_p\}$ such that the $p$ open balls $B(x_i,\epsilon)$ cover $X$.

@PeterTamaroff You're just showing the dense part right? Or what part are you explaining?

Now, if $X\subset B(x_1,\epsilon)$, we're done. If not, there exists $x_1\neq x_2$ such that $d(x_1,x_2)\geq \epsilon$. If $X\subset B(x_1,\epsilon)\cup B(x_2,\epsilon)$ we're done. If not, there exists $x_3\neq x_2,x_1$ such that $d(x_1,x_3),d(x_2,x_3)\geq \epsilon$.

Now, if this process didn't terminate, we would find a sequence $\{x_i\}$ in $X$ such that $d(x_i,x_j)\geq \epsilon$ if $i\neq j$, which is absurd, since this would have no convergent subsequence, hence no limit point.-

yes, exactly

Now, we have proven in particular that for each $n\in\Bbb N$ there exists an integer $p_n$ such that $$\{x_{n,1},\dots,x_{n,p_n}\}$$ is a $1/n$-net for $X$; that is, the set of open balls with radius $1/n$ and centers at the $x_{n,i}$ cover $X$.

Now consider the countable set $$\{x_{n,i}:n\in \Bbb N\; , 1\leq i\leq p_n\}$$

Show this is dense in $X$.

@AlanH I mean that $$\bigcup_{i=1}^{p_n} B(x_{n,i},1/n)\supset X$$

@PeterTamaroff right, so I thought it was just dense by the fact that we have an open cover. Every point in X is either a limit point or is one of those centers

@AlanH This last part shouldn't be much of a fuzz. Fix $\epsilon >0$, and take $x\in X$, and let $K=\{x_{n,i}:n\in \Bbb N\; , 1\leq i\leq p_n\}$. We want to show $B(x,\epsilon)\cap K\neq \varnothing$.

Now, take $N$ such that $1/N<\epsilon$.

okay okay stop

why is it necessary to do that?

that's where I was confused. isn't our construction already enough to say its dense by open covers?

@AlanH That is what dense means: $K$ is dense in $X$ iff for each $\epsilon >0$ and $x\in X$, $K\cap B(x,\epsilon)\neq \varnothing$, that is $\overline K=X$.

I was using the definition that every element of $X$ is either a member of the set (in this case the centers), or is a limit point of that set. equivalent, no?

Having taken such $N$, note that $$X\subseteq \bigcup_{i=1}^{p_N}B(x_{N,i},1/N)$$ But then $x$ must be in some of said balls, so $d(x_{N,j},x)<1/N<\epsilon$ for some $1\leq j\leq p_N$, so $x_{N,j}\in B(x,\epsilon)\cap K$.

@AlanH Sure.

@PeterTamaroff or maybe you should just finish explaining and i'll ask my questions afterwards

@AlanH I'm done.

You can read it now, think about it. It is a nice exercise.

@PeterTamaroff So are there other points in $X$ besides the centers?

@AlanH You mean in $K$; yes?

@PeterTamaroff just doing some things around the house. Probably go out for an early dinner.

@robjohn Do you have Father's Day there?

@PeterTamaroff I think it is a pretty universal holiday

@robjohn Heh, didn't know.

Mother's Day is on a different day there.

@PeterTamaroff really?

when is it there?

@robjohn October 20th

@PeterTamaroff Why do you say $K$? Earlier you stated to fix $\epsilon$ and take $x\in X$.

@PeterTamaroff That is quite a ways off. Is it always that date or is it always a Sunday?

@AlanH Could you repeat your question?

5am here in australia

@PeterTamaroff K is just the set of centers, right? Is there anything else in K?

@AlanH Nothing else. The centers.

They are the points that make the $1/n$-nets for $X$ for each $n=1,2,\dots$.

@PeterTamaroff Then what is X comprised of?

@AlanH Isn't $X$ your space to begin with?

You said "Let $X$ be a metric space..." ¬¬

@PeterTamaroff yeah

@AlanH So...?

@BenjaLim 4pm here!

@PeterTamaroff I don't know, I"m getting really confused. I've been on this problem for two days now. bah.

@AlanH OK, let's start over.

@PeterTamaroff no no, its just the dense part I'm having trouble with

@AlanH OK. The two important things are: for each $\epsilon$, we can make $1/n<\epsilon$ for some $n$. For each $n$, $$\bigcup_{i=1}^{p_n}B(x_{n,i},1/n)$$ covers $X$.

So, pick $x\in X$ in your space, and $\epsilon >0$.

To show $K$ is dense, we ought to show $B(x,\epsilon)\cap K\neq \varnothing$ for each $x\in X$ and $\epsilon >0$. This is saying that $X\subseteq \overline{K}$. Since trivially $\overline K\subseteq X$, we get that $\overline K=X$, that is, $K$ is dense.

But given $\epsilon >0$; we can pick $n_0$ such that $1/n_0<\epsilon$. Since the balls corresponding to said $n_0$ cover $X$, $d(x,x_{n_0,j})<1/n_0<\epsilon$ for some $j$ (that is $x$ must be in some of the balls, here I'm saying it is in the ball with center in $x_{n_0,j}$), which means that $x_{n_0,j}\in B(x,\epsilon)\cap K$. Do you see?

Hi @MarianoSuárez-Alvarez

@MarianoSuárez-Alvarez how are you?

Hi @Argon

Hi @keri welcome to the chat.

hello

@keri how are you?

awesome, just spending the evening with graph theory :)

and u?

@keri Fine thanks :-)

good :D

:D

why is it so quiet?

Sometimes it gets that way, while other times you can't get a word in edge-wise...

...it really depends on who is talking.

So let's throw a math party :D

We have a "Party Zone" for that :D

I mean something more math than party ;)

Oh... icic

;-)

@keri Gotta run, see ya later.

bye

:D

@skullpatrol Hi Skull

@Argon Hi/bye

@skullpatrol Bye

:D

@anon

yes

Remember my problem the other day, about $\ll$?

the one starred on the side?

@anon Oh, yes. Didn't see it was starred.

I have my screen with too much zoom.,

Well, I should be getting that $$n|a_{n+1}|\leq 2\sum_{k=1}^n |a_k|$$ Assuming $|a_k|\leq k$ for $1,\dots,n$ one gets $$n|a_{n+1}|\leq 2\sum_{k=1}^n k\implies |a_{n+1}|\leq n+1$$ so the assertion is proven.

pretty sure you wrote $n|a_n|\le $ yesterday (otherwise I could have told you that!)

@anon Sorry?

Oh, well, I keep failing in my manipulations, yes.

I attempted the induction idea using what you wrote yesterday, which had $a_n$ on the left instead of $a_{n+1}$.

@anon But it fails, doesn't it?

does one know a category theoretical proof of sylow group theorems ?

@PeterTamaroff you just proved that $|a_k|\le k$ for $k=1,\cdots,n$ implies $|a_{n+1}|\le n+1$, did you not?

or by "I should be getting that [blah]" are you saying that you haven't got [blah] but you think you need to get there (which sounds reasonable)?

@anon Yes, that is what I'm saying. I should be getting that, but I am not. =)

I must be overlooking something in my manipulations. Gurrdarmn.

Does anyone know why movies look so much homemade in a "HD" television (compared to the more unrealistic one has in "older" TVs)? The image has a faster dynamic, for example, makes it look strange.

@anon Why are there so many groups of order $2^n$?

@PeterTamaroff framerate + motion interpolation

@anon So, how does one fix that? Watching newer movies? Even "The Dark Night Rises" looked silly, or "The Avengers", though this one not so much.

@PeterTamaroff that isn't entirely clear to me. it follows intuitively from the conjectre that the number if iso classes of groups of order p^n grows as p^O(f(n)), and the fact that f(n) is independent of p is the more fundamental fact in need of explanation

I do not have an hdtv, so I do not share your 1st world problems :-)

@anon What is $f(n)$?

in the context of my response it doesn't matter what f(n) is, but it happens to be more precisely put as (2/27)n^3+O(n^(8/3))

@anon I hadn't changed my TV set since 2007/8. =P

@PeterTamaroff @PeterTamaroff I always thought it is "The Dark Knight Rises"

@DominicMichaelis \MEFACEPALMS

I surely meant that.

Finally i found a mistake :D

OK, this is probably the tenth time I will do this calculation.

so the 10 th pairwise different answer ?

@DominicMichaelis No, I am doing something wrongly and I cannot see where.

It is to late for me to be a help

Finally, I can vote people down! (starting power trip) I will probably have to wait till school starts again before I get to put it to good use through.

FUCK YEAH. Did it right this time. @anon I got it.

I have to get me a treat now.

@keri define math party with ZFC

what is a math party ?

hey guys

@PeterTamaroff hi

@robjohn hi

@anon hi

@anon Let $B_n$ denote the possible ways of writing $n$ as a sum of $1,2,3,4$. Then $$\sum_{j\geq 0}B_j\zeta^j=\prod_{k=1}^4(1-\zeta^k)^{-1}$$ Yes?

Note that order counts.

@mick On what grounds?

@PeterTamaroff yes, but order does not count

@anon Oh, then I'm wrong?

@PeterTamaroff because most sieve methods assume some structure such as multiplicative functions or are probabilistic in nature. Further Arxiv is full of false proofs of the twin prime conjecture based on sieves.

@anon "Two sums consisting of the same terms but in different order are regarded as different."

@PeterTamaroff @PeterTamaroff So we know that there exists $x$ such that $d(x,x_{n_0,j}) < 1/n_0 < \epsilon$ because the union of open balls with radius $1/n_0$ covers $X$. Is this correct? I think the whole time I was struggling with why such an $x$ existed.

@anon greetings

@AlanH Wait!

waiting...

We picked an $x\in X$ arbitrarilty. What we want to show is that $x_{n_0,j}$ exists and is in $B(x,\epsilon)\cap K$!

@PeterTamaroff okay, so the question is flipped, how do we know there exists such an $x_{n_0,j}$ that is within an epsilon distance?

Maybe I need lessons in sieve theory @PeterTamaroff

:9918755 Because we chose $n_0$ so that $1/n_0<\epsilon$

And because for any $n$, $$X\subseteq \bigcup_{i=1}^{p_n}B(x_{n,i},1/n)$$

@PeterTamaroff Okay, now i believe it. Thanks for taking the time to help.

Shall I repay you in starred messages or by voting up some answers? The choice is yours

@AlanH I don't care about starring.

@mick hey there... what's up?

well, off to get Father's Day dinner. BBL

BBL ?

Bon appetit mon ami !

@mick "Big Baked Lasagna"

aha

im looking for introductions into sieve theory for dummies like me :)

@PeterTamaroff @robjohn @anon

@mick No clue, Mick.

something easier than Terry tao's writings plz

sieve theory is so different from other math or it seems ...

:/

@anon I am doing some combinatorics.

@PeterTamaroff I am eating some caramel delights.

@anon I need an "obviousmeter".

One can see that if you multiply by $1-x$ on the LHS, all the terms of the form $(1+x^{2^k})$ die.

Then, upon multiplication by $1-x^3$, all terms of the form $1+x^{3\cdot 2^k}$ die.

And so forth.

mhmm

So, this is saying "every integer can be written in the form $\text{odd}\cdot \text{pureblood even}$" @anon

yes

@anon What kind of treat are you having?

these mofos

@anon I am speechless.

@anon cf. 0:30

@anon What does that have?