@JasperLoy I never said I didn't like them. Other people said they didn't like them, so I took other opinions. Doesn't mean I wont read them.

One book was a bit more detailed, that is all. THey are good for different thigns

@anon Do you maybe have an idea about my question: math.stackexchange.com/questions/1562749/… ?

@robjohn lol, sry for the late (late!!) reply, but yes, integral of 0 to f'(x) of f'(t) dt.

hello

Hey, I have been trying to work on this on problem $\lim_{n \rightarrow \infty} n^{-2} \sum_{i=1}^n \sum{j=1}^n \frac{1}{\sqrt{n^2 + ni + j}$ I'm pretty sure someone has answered it before because it is from Stewart MultiVar Calculus, but I can't find, has anyone seen this problem before? I'm trying to avoid duplicate questions.

Whoops:

\lim_{n \rightarrow \infty} n^{-2} \sum_{i=1}^n \sum_{j=1}^n \frac{1}{\sqrt{n^2 + ni + j}

ughh: $\lim_{n \rightarrow \infty} n^{-2} \sum_{i=1}^n \sum_{j=1}^n \frac{1}{\sqrt{n^2 + ni + j}$

@Dair lemme check! :-)

$\lim_{n \rightarrow \infty} n^{-2} \sum_{i=1}^n \sum_{j=1}^n \frac{1}{\sqrt{n^2 + ni + j}}$ ?

There we go... @r9m: Thanks!

@Dair 'kay, can't find it atm but the idea is pretty simple .. just apply Stolz-Cesaro once, that should get you somewhere?

blegh .. we won't even need that, just need to convert it to riemann sum I guess ..

@Dair 'kay .. I think the limit is $0$.

@Ted am I right about the one above? ^

Oh... Um... I think there is a small error: it should be $\sum_{j=1}^{n^2}$ sorry about that.

for the second sum...

@Dair $$\sum\limits_{i=1}^{n}\frac{1}{\sqrt{n^2+ni+j}} = \frac{1}{n}\sum\limits_{i=1}^{n}\frac{1}{\sqrt{1+\frac{i}{n}+\frac{j}{n^2}}} = \int_0^1 \frac{1}{\sqrt{1+x}}\,dx + O(1)$$ again summing over $j$ and dividing by $n^2$ just gives a limiting value $0$

@Dair ah! in that case directly we have a reimann sum for a double integral: $\displaystyle \frac{1}{n^2}\sum\limits_{j=1}^{n^2}\sum\limits_{i=1}^{n}\frac{1}{\sqrt{n^2+ni+j‌​}} = \frac{1}{n^2}\sum\limits_{j=1}^{n^2}\frac{1}{n}\sum\limits_{i=1}^{n}\frac{1}{\sq‌​rt{1+\frac{i}{n}+\frac{j}{n^2}}} = \int_0^1 \int_0^1 \frac{1}{\sqrt{1+x+y}}$ :D

Ok. Lol, I was looking at the first like... 1 integral... this chapter is about double integrals though lol.

I see :-)

Oh nice. I see how it works now. That was pretty slick. Thanks.

@Dair welcome :)

@DanielFischer baba, I was thinking if we could use equidistribution of $\{n\theta\}$ (for an irrational $\theta$) to infer $f(x) = \sum\limits_{n=1}^{\infty} \frac{\{nx\}}{n^2}$ is continuous at $\theta$?

Something like maybe using $\sum\limits_{n=1}^{N} \frac{\{nx\}}{n^2} = \sum\limits_{n=1}^{N-1} \left(\frac{1}{n} - \frac{n}{(n+1)^2}\right) \cdot \frac{\sum\limits_{j=1}^{n} \{jx\}}{n} + \frac{1}{N}\frac{\sum\limits_{j=1}^{N} \{jx\}}{N}$ (form Abel summation)

Is there anyone around that can help me interpret this answer: math.stackexchange.com/a/1563380/187120 ?

@JasperLoy I'm so glad that see you once again here!

@JasperLoy Hello, I'm working on Riemannian Geometry now :-)

hiiiiii

@JasperLoy No, just study from Do Carmo, Sakai and Petersen!!!

hellooo??

@JasperLoy But this is really strange, studying Riemannian geometry with ANT, CA and GT!!!

hi! does the proof here not make sense proofwiki.org/wiki/Finite_Field_Extension_is_Algebraic

shouldn't $n$ of them be linearly independent, so then that polynomial must have $a_n,a_{n-1},\cdots,a_0=0$? meaning that this is the zero polynomial?

@Katie No, there is no reason why $n$ elements should be linearly independent.

because $L$ is an n dimensional K-vectorspace, these are all basis elements, so linearly independent

oh

i see, there are $n+1$ elements in that

@Katie No, they need not be a basis just because there is the right number of them

could you explain it to me?

@Katie If you take $n$ elements in a vector space, they need not be linearly independent, even if the dimension is $n$.

oh i see, oops

okay thanks i get it :)

guys i need answer to question

i solved the problem but i need to check answer?

if i'm pulling a wagon with force = 120 newton in direction of vector v = <3,1,2>. How much work is done pulling along from p(0,0,0) to q(300,0,60)?

I got 122400 joules. correct??

If you pull in that direction, you will never get to the second point.

never mind. i think i got it. it's about 32712.7 j

thanks though

Just visiting to complain about [a, b) notation. How can one write that and live with themself.

@MickLH You prefer $[a,b [$?

I like $[a,b)$ :( I don't see the problem with it

They are both glorious endeavors of trolling

@MickLH What notation would you suggest?

@Axoren: You got the right answer from that guy. In $\Bbb R^3$, your "simple" closed form isn't right. If you choose the right coordinates (which is what's going on in the Jordan block answer), it's one single rotation about a single axis. If you want to do products of rotations about the coordinate axes, I believe you'll need all three.

So, as the guy said, the general form will be $PRP^{-1}$ for some orthogonal matrix $P$ and a block matrix $R$ (or, in the $\Bbb R^3$ case, a single rotation about the $z$-axis, say).

Hi @TedShifrin

Heya @Tobias

@TobiasKildetoft I am not suggesting this, but I'd like to say something like $a\le \mathbb{S}\lt b$ in advance, and then just refer to $\mathbb{S}$. Short of that though, I'd like an accented set of ( and ) or [ and ], where the accent inverts their respective functions.

@TedShifrin What are you up to these days?

Cooking a lot of dinners, recently :) A mob of 6 just left earlier after we saw/heard the Messiah. ... Trying to line up volunteer tutoring. And writing more recommendations for graduate school and fellowships than I think I've ever before done in one year. So much for retirement :P

And you?

Preparing for tomorrows exercise session, plus the talk I am going to give. And trying to figure out how to find a job for next autumn

I seem to be running into a lot of annoying technicalities and cost saving problems

Are you planning to change countries again?

Such as applying in Copenhagen only to be told that they actually have a hiring freeze at the moment, so they are only allowed to make hires using external funding

That sounds like a very not-permanent sort of job prospect.

I would prefer to go back to Denmark, but right now I would like to stay in Sweden for another year, as then we will be moving at the same time that my daughter would need to change from kinder garden to school anyway

Ah. My dad switched jobs as I was finishing 8th grade (many centuries ago). I left behind many, many friends, and it was one year off from the obvious change (in those days, high school typically started in 10th grade). But I did survive :)

Unfortunately, I canot get an extension on my current postdoc, even though there is money in the budget for it. Because I am technically not a postdoc but a "researcher", due to being offered the position a few weeks before defending my PhD (and a researcher position is fixed at 2 years and not allowed to be extended)

Agh ... one never thinks about such technicalities at the time.

Yeah, my postdoc mentor is not very happy about it. He found out when he asked if they could extend my stay using the extra money (before I even asked him if it was possible)

I wish you the best of luck with it all.

But at least they recently announced a postdoc position in Stockholm, which is about the same commuting time for me, so I will be applying for that.

thanks

Talk to you soon :)

Hmm, is there some clever way to quickly factorize $x^4 - x^2 - 6$ as a real polynomial? I could probably write up a bunch of equations and see it, but it seems more complicated than it ought to be (considering the rest of these exercises are fairly trivial).

@JasperLoy Sure, but how do you get to that quickly by hand?

@JasperLoy Ahh, of course. Now sure how I missed that

@JasperLoy It is late morning here

@TobiasKildetoft You're in Sweden?

@FrankScience Yeah, at Uppsala University

Is there anything like tenure in U.S? @TobiasKildetoft

I heard that tenure is quite competitive.

@FrankScience I am not actually sure precisely how the system works here, though I know it has some differences to the American tenure system

I don't know whether I will go to U.S. for PhD or in Europe, yet.

@TobiasKildetoft Do they speak English in the University (esp. in class)? Or Swedish?

@FrankScience depends on the class. I teach in Swedish, but that is for undergraduates. Seminars and PhD courses are in English

@TobiasKildetoft Eh, postdoc can also teach classes?

@FrankScience Of course.

@TobiasKildetoft Which class are you teaching?

This is fun. When $X$ is a CW-complex, I can simply note that $X \hookrightarrow CX$ (inclusion into the bottom) is an isomorphism on $\pi_n$ for all $n$, and then apply Whitehead, right?

Right now it is an introduction to elementary algebra (I just head the exercise sessions)

OK, I see.

Is there always an exercise session for each class in mathematics? @TobiasKildetoft

@FrankScience Not sure about for the graduate courses and PhD courses

@TobiasKildetoft But for undergrads?

@FrankScience As far as I know yes (I don't know that much about how things work here)

@TobiasKildetoft Where did you do your PhD?

@FrankScience In Aarhus (Denmark)

Hey @BalarkaSen

Hi!

What's happening?

Very ill.

Ooff, that blows. Sorry to hear that.

I am answering topology questions in MSE because I can't do much math.

General topology or algebraic topology?

Algebraic :)

Which question are you answering?

The knot complement one?

Nah, people already answered that in the comments.

True.

It'd be cheating to post an answer without making it CW, but I am too lazy to do that. What's the point if I get no rep points? :P

Oh, you wrote like 4 answers today.

Yeah.

I am going to edit this answer to do it completely using long exact sequences and not cellular homology, because I think the current answer is not helpful for the OP.

Though he already accepted your answer

A circle with centre $A$ and radius $BC$ means 1) that the circle passes through $B$ and $C$ or that 2) the radius of the circle is equal to $d(A,B)\times d(A,C)$ where this is the euclidean metric?

You have an easy way to prove that $H_{n+1}(X) \cong 0$? I think I have to use a piece of the cellular homology diagram after all...

@iwriteonbananas Without an upvote. Anyway, it'd be fun to try to do that without cellular :D

@BalarkaSen Yeah, definitely.

If I use the LES for $(X, S^n)$, the relevant part is this.

@BalarkaSen It shouldn't be too hard, hold on.

$H_{n+1}(X) \to H_{n+1}(X, S^n) \stackrel{\partial}{\to} H_n(S^n) \to H_n(X) \to 0$

The $\partial$ is a map from $\Bbb Z$ to $\Bbb Z$, and the geometric interpretation can be used to prove that it sends $1$ to $m$.

So once I can prove that $H_{n+1}(X) = 0$, I have the short exact piece $0 \to \Bbb Z \stackrel{\times m}{\to} \Bbb Z \to H_n(X) \to 0$.

Thus, $H_n(X) \cong \Bbb Z/m\Bbb Z$, and all is over.

$H_{n+1}$ is being problematic. I don't know how can I avoid the cellular chain complex.

Ok, maybe i got an idea.

mhm?

The boundary map $H_{n+1}(D^{n+1}, S^n)\to H_n(S^n)$ is an isomorphism, right?

it's the suspension iso for $S^n$

yes.

now look at the LES of $(X, S^n)$:

$0\to H_{n+1}(X)\to H_{n+1}(X,S^n)\to H_n(S^n)$

so $H_{n+1}(X)\cong \ker (\partial)$

oh, yeah, fair enough.

but the boundary map is inj, so we're done.

identify $H_{n+1}(X,S^n)=H_{n+1}(D^{n+1},S^n)$ and see what happens to the boundary map under that identification.

I am being silly.

@iwriteonbananas Boundary map is multiplication by $m$.

@BalarkaSen oh, yeah.

that requires proof though.

$X$ is $D^{n+1}$ attached to $S^n$ by mult by $m$. $\partial$ sends relative cycles to boundaries.

So it sends the relative cycle represented by $[(X, S^n)]$ to the boundary $S^n$, thus it sends $1$ to $m$.

Ok, i think i could show rigorously without geometric arguments that $\partial$ is mult by m

sure. use naturality.

and geometric arguments are rigorous.

:P

Alright, whatever. :P i think your argument is fine.

thanks for the heads up about $\ker \partial$. should have seen it, guess I am just ill.

Glad to be of assistance.

@DanielFischer Could you tell me if I got the following right? Let $Y,Z$ be Banach spaces. $K\subset Y$ be compact and let $T_n, T\in B(Y,Z)$ with $\lim_n T_ny=Ty$ for all $y\in Y$. I want to prove that $\lim_n \sup_{y\in K} \|T_ny-Ty\|=0$.

So, my argument goes as follows:

The sequence $\left( (T_n-T) y \right)_n$ convergees for all $y\in K$, therefore $\sup_n \|(T_n-T)y\|<\infty$.

@TedShifrin i got some free time during the break. Any suggestions on what to study? I feel adventurous and craving something geometrical.

By the Banach-Steinhaus theorem, one has $\sup_n \sup_{y\in K} \|(T_n-T)y\|<\infty$

So we can interchange taking limit and taking sup as follows: $$\lim_n \sup_{y\in K} \|(T_n-T)y\| = \sup_{y\in K} \underbrace{\lim_n \|(T_n-T)y\|}_{=0}=0$$

Actually, something is fishy I think. But I don't know what.

So, @iwriteonbananas, how's life?

@iwriteonbananas To what are you applying Banach-Steinhaus, and how do you justify the interchange of taking limit and $\sup$? You can do that in several correct and several incorrect ways. I'd go with "$\lim T_n y$ exists for all $y$, thus by Banach-Steinhaus $\sup \lVert T_n\rVert < +\infty$. By Ascoli-Bourbaki, on an equicontinuous family, the uniform structures of pointwise convergence and of uniform convergence on compact sets coincide."

@DanielFischer I'm not sure I grasp that last sentence. What is Ascoli-Bourbaki? Is that the same as Arzela-Ascoli?

@BalarkaSen So-so. I was forced to do a seminar for some "soft-skill" the entire weekend. The uni makes you do a couple of those throughout your studies nowadays. It felt like being back in high school.

Maybe full strength of Arzela-Ascoli isn't needed.

Suppose $\sup_n\lVert T_n-T\rVert\le M$. Given $\epsilon>0$, we can choose $\delta>0$ such that $M\delta<\epsilon$.

@iwriteonbananas It's the generalisation of Ascoli-Arzelà. We need only a part of it, that for equicontinuous families pointwise convergence implies uniform convergence on compact sets.

You can cover $K$ by finite number of $\delta$-balls, then consider the convergence at centers of these balls.

You can choose a big $N$ such that for each $n\ge N$, we have $\lVert(T_n-T)c\rVert<\epsilon$ for each center $c$ of these balls.

Thus $\sup_{x\in K}\lVert (T_n-T)x\rVert<2\epsilon$.

Arzela-Ascoli is nontrivial, however.

@iwriteonbananas done.

@DanielFischer Ok, uhm. I don't see right now why $\{T_n\}$ is equicontinuous. How difficult is it to prove this generalization of Arzela-Ascoli?

@iwriteonbananas Soft-skill?

@iwriteonbananas Is my proof okay?

@FrankScience Ah, hold on. You're proving that $\{T_n\}$ is equicontinuous?

@iwriteonbananas No, I directly proved that they converges uniformly on $K$.

Oh, sorry I got it now. Give me a second to understand it :)

any number theory people have ideas on math.stackexchange.com/questions/1563968/…

@iwriteonbananas $\sup \lVert T_n\rVert < +\infty$ means the family is even equilipschitz, which implies uniform equicontinuity, which implies equicontinuity. It's not hard to prove the generalisation, the hardest part is getting used to the involved concepts.

Eh, sorry, equicontinuous follows directly from Banach-Steinhaus @iwriteonbananas

@DanielFischer For the generalization, you mean one for compact Hausdorff spaces or compact metric spaces?

@FrankScience Yes, it looks good to me. Thanks for chiming in.

@FrankScience Locally compact domain and a Hausdorff uniform space as codomain.

@DanielFischer Could you please tell me what equilipschitz means? :)

@BalarkaSen It's modern lingo referring to stuff like presentation skills, social skills yada yada. I'll have a look at your edited answer once I got this problem straight.

@DanielFischer Locally compact is enough?

@iwriteonbananas Ugh. And no worries, you can have a look whenever you want.

@iwriteonbananas All members of the family are Lipschitz-continuous and there is a Lipschitz constant that works for all members. (Equivalently, $\sup \operatorname{Lip} f < +\infty$.)

@DanielFischer Aha, that makes sense.

@FrankScience Yes. Of course one takes compact convergence and not uniform convergence then. (And, just in case, "locally compact" includes Hausdorff, I'm not sure what remains true if we take a "locally quasicompact" domain.)

Okay, I see. Maybe you're referring to compact convergence.

@DanielFischer Okay, you're using Bourbaki's term of compactness.

@FrankScience Sure. It's the correct one.

I cannot remember under what circumstance we can apply diagonal process to prove Ascoli's Theorem.

@FrankScience We need separability of the domain.

Okay, so for compact metric spaces, the proof works.

In general, one needs another proof.

@r9m If we let $f_n(x) = \frac{\{nx\}}{n^2}$, then $\lVert f_n\rVert_{\infty} = \frac{1}{n^2}$, so the series converges uniformly. For irrational $\theta$, every $f_n$ is continuous at $\theta$, and by uniform convergence, $f$ is continuous at $\theta$.

@DanielFischer right right! thanks :) of course $\frac{\sum\limits_{j=1}^{n} \{jx\}}{n}$ is also continuous at every irrational $x$, (if I were to think along that unnecessarily complicated expression with abel sum)

@DanielFischer so when we are dealing with a function that looks like $f(x) = \sum\limits_{n=1}^{\infty} \frac{a_n(x)}{n^2}$, does equidistribution of $a_n(x)$ say something about continuity of $f$ at $x$ in general?

@r9m I don't think so. You could always add a function with a jump at the given $x$ to $a_1$, that wouldn't influence equidistribution of $(a_n(x))$, but it would change the (dis)continuity at $x$.

@DanielFischer Thanks :-) got it!

hello someone reminds me what is the name of the formula which turns cos/sin in terms of exponontiel

i know its french name

please*

formula de ...

@Agawa001 De Moivre

ah , i forget that often

thank u tobias

@DeutschMathematiker I just realized you changed your name (Algean) .. seeing you after a long time in chat! :-)

@TobiasKildetoft can I ask you an analysis question?

@Paradox101 Sure, but I probably can't answer

Oh ok. I just need to confirm something. If we're supposed to prove the equivalence of the manhattan and the sup norm metric, then we assume that there exists a sequence $x_n$ that converges to $x$ and so the manhattan metric : $d_1(x,y)\rightarrow 0$. as $x_n$ converges to $x$ so the sup norm metric converges to zero and hence they are equivalent. Is this sufficient to prove the equivalence?

@Paradox101: Yes. BTW, a sequence $x_n$ converging to $x$ with respect to a metric $d_1$ is defined as $d_1(x_n, x) \to 0$.

If you can prove that any sequence converges in both metrics to the same limit, then you have proven that the metrics are equivalent

@Huy ok thanks. so this just has a relatively simple proof?

also if we have a metric composed of functions and have to find out for which functions it's a metric on the real numbers, we need to employ the 4 set of conditions that make a metric?

@skullpetrol welcom back from the dead

I don't understand your questions. Can you maybe give an example?

@Huy if we have a metric $d(x,y)=|f(x)-f(y)|$ and we need to prove for which functions $f(x)$ is a metric on $\mathbb R$ do we merely need to manipulate the conditions for which $d(x,y)$ is a metric? i.e. $d(x,y)>0$ etc

@Paradox101: If I understand you correctly, the answer is yes.

$d(x,y) \geq 0$ and $d(x,y) = 0 \iff x = y$.

@Huy if we go by this then it's a metric on $\mathbb R$ for all positive functions which are symmetric and follow the triangle inequality?

@Paradox101: I don't know what you mean by "a function follows the triangle inequality".

@Huy I mean for $d(x,y)$ to be a metric it has to follow the rule that : $d(x,y)<d(x,z)+d(z,y)$ where $x$,$y$,$z$ $\belong to$ $\mathbb R$

And how can a function $f(x)$ follow that rule?

Oh it can't. But then when we are checking whether it is a metric the triangle inequality is also checked. But here I can't determine much about the nature of the function from the inequality. So does the proof only contain the other conditions?

@Paradox101: No, it has to satisfy all conditions.

is anyone here from UNC?

@Huy it has to but if I just write out the inequality that won't really prove anything will it? I mean we need it to be a metric, but merely stating that it follows the triangle inequality won't be enough. Using the previous conditions we know that the function should be positive and symmetric but we don't have any other info on the function

@Paradox101 Clearly, $f$ has to be injective.

@MartinSleziak how so?

If $f(x)=f(y)$ for some $x\ne y$, then you have $d(x,y)=0$.

Which contradicts the definition of metric.

To prove that $d(x,y)=d(y,x)$ you simple need to check whether $|f(x)-f(y)|=|f(y)-f(x)|$.

To check the triangle inequality you only need to verify $|f(x)-f(z)|\le |f(x)-f(y)|+|f(y)-f(z)|$.

See also this question @Paradox101 Which properties must a function, $f$, fulfill for $d$ to be a metric

@I'mmostlyjustanidiot Are you sure you're not Alex Clark? :P

@MartinSleziak oh ok I didn't consider that. Thanks a lot!

@BalarkaSen Why do people think that?

You two have a lot of things in common, I guess.

@BalarkaSen Both learning geometry?

No.

Both delete messages :D?

both are mostly idiots?

lol!

@Huy Haha I'll have to meet him!

How can a person meet himself, I wonder.

I don't mind if people think I am @AlexClark I guess

I am checking out 'Four pillars'.

hello again, plz if u know any known method to simplify a linear equation $\sum_n\sqrt{1-x_n^2}=0$ where $\sum x_n=0$

Hartshorne is my recommendation.

haha

Or rather, Lurie's Higher Topos Theory

@I'mmostlyjustanidiot I'd be surprised that you're not Alex. Oh well. He (or his real account) seems to be gone for a while from this chat.

@BalarkaSen Did I join when he left?

Not quite.

hi all

@JessyCat is it complete?

@JessyCat can you write in the definitions please(uniform continuity) it's been 2 years since I've done it but I have an idea

hello, i need some exercises on compactness in a general topological space

not in a metric space

can someone help me please

@TedShifrin or @robjohn can you help me with a book or any thing else

@Vrouvrou You have some exercises which you want to solve? Or are you asking for some suggestions of a resource where you can find such exercises?

@Vrouvrou Google "compactness metric space exercise" or use the tagged questions which tagged the things you want

aresource where i can find such exercises

i don't need in metric space

@AlecTeal

@Vrouvrou I think Alec's suggestion was simply to look on questions tagged compactness+general-topology.

Then remove that tag from your search

What use are the tag nazis if you don't use them?

There are some posts on the main asking for books with problems from general topology. Like this one or this one.

Or just google "[thing I want questions in] assignments"

@AlecTeal, okay so no more compactness. I'll cut that part out of my argumen

From the suggestions given there, Schaum's Outline of General Topology is probably one of the most elementary. @Vrouvrou

Totally bounded: $\forall \epsilon > 0$, $\exists \cup_{k=1}^{n}B(x_{k},\epsilon)$ such that $X \subseteq \cup_{k=1}^{n}B(x_{k},\epsilon)$.

So, by the definition of uniform continuity, if a point $x \in X$ is contained inside a delta-ball, its image should be contained inside an epsilon-ball

So, since each point in f(X) is contained inside an epsilon-ball, the union of such points is contained in the union of those epsilon-balls.

@AlecTeal?

@Vrouvrou Or you can try Elementary Topology. Textbook in Problems which is freely available online (at least the abridged version): pdmi.ras.ru/~olegviro/topoman

@JessyCat it's the uniform continuity definition I wanted added

But yeah that sounds right @JessyCat IIRC one delta works for all x

@AlecTeal $\forall \epsilon > 0$, $\exists \delta > 0$ s.t. for $u,v \in X$, if $\rho(u,v)<\delta$, then $\sigma (f(u), f(v)) < \epsilon$, where $\rho$ is the metric for $X$ and $\sigma$ is the metric for $Y$.

@AlecTeal, what's IIRC?

@JessyCat About your comment: Every function is surjective if you consider it as a function from $X$ to $f(X)$. So asking: "Is $f(X)$ totally bounded?" is the same as asking: "Is $Y$ totally bounded, assuming $f$ is surjective?"

But I did not notice that the other question is about uniform spaces, not metric spaces.

@MartinSleziak, thanks.

"if I recall correctly"

gotcha

Are there only finitely many epsilon-balls in the image, though?

@JessyCat to go with what Martin Sleziak said, maths.kisogo.com/… continuity of non surjective functions.

You really need to write down what you know, and draw a picture of some balls and what they map to

Fantastic. I'll do that. Would be great if eventually today somebody posted a solution so I'd have something to compare it against. Also, it seems to me it would work for regular continuity, too: $\forall \epsilon > 0$, $\exists \delta > 0$ s.t. $f(B(x,\delta)) \subseteq B(f(x), \epsilon)$.

@JessyCat Just use $\varepsilon$ and $\delta$ from this definition. Now if you are given $\varepsilon>0$ and you know that $X$ can be covered by finitely many balls of diameter at most $\delta$, what can you say about $f(X)$?

But these things RARELY work for regular continuity, so I'm second-guessing.

Remember an open set is the union of open balls

X is open in itself....

Regular as in pointwise

maths.kisogo.com/index.php?title=Continuous_map several equivalent definitions of continuity

But then, if we had different deltas for each x, we might have a problem covering all the y's in f(X)

Yeah that's where the uniform part comes in

IIRC it's one delta for all x rather than one delta for each x

So I'm correct then in assuming it doesn't work for pointwise convergence?

Oh come on!!!

Don't be a Socrates.

Not being deliberately unhelpful I just don't have any paper nor (for the 5th time) remember what uniform continuity is.

Argh.

@JessyCat You probably mean "continuity" and not "convergence". The function $f\colon(0,1]\to(0,\infty)$ is a counterexample showing that the claim is not true for continuous functions.

Yes, I mean continuity.

and thanks @MartinSleziak

@MartinSleziak what function is that?

Sorry, I wanted to write $f\colon x\mapsto \frac1x$.

But any continuous surjection would work.

$\frac{1}{x}$? As in 1/x?

I don't think that's quite right but I see what you mean

$\frac1x$ and $1/x$ are just two different notations for the same thing.

Yes, but usually, I write $\frac{1}{x}$.

I thought you needed the two sets of curly brackets

No I mean for (0,1) I'm sure 1/x goes from infinity to 1

use chatjax, for gods sake

@TedShifrin Are you still around? I don't see how my post is wrong and how the Jordan form is applicable to vectors in $V$. The matrix he's shown, if it truly is a Jordan matrix, would be similar, but he's presented one that is twice the length along the diagonal. Therefore, it's not a similar matrix. Am I missing something?

Mathjax

And all that jax

;P

I mean this.

Anyway @JessyCat write out your definitions

Yup, Imma do that

Whenever you want to start a=>b start writing out b, it just means you can bring in a at some point. So total-bounded-ness involves an epsilon right? Your first line should be "let epsilon>0 be given"

Then at some point you can go "but wait! a means blah blah blah"

@TedShifrin Also, it's rather unfortunate to have to describe a rotation matrix with a similar matrix. The reason as to why it's unfortunate is that the matrix $P$ in $R = PJP^{-1}$ ends up being a rotation matrix for which I'd also need the closed form for practical use.

However, this is definitely a step as I've described $R$ as a product of 3 matrices instead of $\dim(V) - 1$

Thanks, @AlecTeal! :) :) I'm going to get to that right now.

NP, let me know when you're done @JessyCat

@AlecTeal, probably won't be right away, I've got to make a call and go to the doctor's. But, I'll try to write down as much as I can now and then smooth out the loose ends later.

Let me know either way

@Balarka: Huy wasn't joking. Hartshorne has a Euclidean geometry book.

anyone good with combinatorics or recursions

@DanielFischer hi, in Cantor's intersection theorem for metric complete space, is there another proof other than choosing a sequence ? (to avoid axiom of choice)

@MikeMiller Oh, I didn't know that. Thanks.

@JeSuis Well, if we had $\bigcap C_k = \varnothing$, then $\{ C_1 \setminus C_k : k \geqslant 2\}$ would be an open (in $C_1$, which we may assume to be the whole space) cover of $C_1$. There would then be a finite subcover, and that means $C_k = \varnothing$ for some $k$.

@DanielFischer I was thinking about the claim for closed subspaces whose diameters tend to zero.

Can $F(n) = \frac{1}{n} \sum_{i=1}^{k} F(n-i)$ be restated in terms of harmonic numbers?

@JeSuis Ah. Then it depends. Avoiding sequences there is no problem, the nested family of nonempty closed sets with diameters shrinking to $0$ is a base of a Cauchy filter. Being complete means every Cauchy filter converges, by closedness the limit lies in the intersection. But, if you call a metric space complete if every Cauchy sequence converges, you need to prove the equivalence "All Cauchy sequences converge $\iff$ All Cauchy filters converge".

The direction $\Leftarrow$ is straightforward, $\Rightarrow$ needs some choice, I think.

@DanielFischer: Do you know how I can bound $\langle Tv, w \rangle_{L^2(TM)}$ by $\|v\|_{L^2(M)}$? Think of $T$ as the gradient. I tried partial integration but I'm not very successful. Maybe also assume $v \in C_c^\infty(M)$ if that makes it a lot easier (for a start).

(this is to find the domain of its adjoint, so we're not really allowed to do any calculations with a divergence / Laplacian yet)

what does multiplicity mean when we are talking about the roots of polynomial equations?

@deostroll $(x-2)^3$, 2 has multiplicity 3

@Huy $M$ is presumably a Riemannian manifold and $TM$ its tangent bundle?

yes, @DanielFischer

I define $L^2(TM)$ inner product by integrating over the Riemannian metric

(is that "the" usual definition or is a different one more useful?)

@Huy I would expect that if you can do it for $C_c^{\infty}(\mathbb{R}^n)$, you can patch things together to lift it to your manifold. But at the moment, I'm not even sure about that case.

hm, I do not know wtha filter is but it seems interesting, I will learn it. Thank you

Woo. Beating cellular homology using long exact sequences was fun.

@DanielFischer where ca I learn about filter?

Not claiming that cellular homology is useless, but it's fun as a passtime work to use weaker tools to do that problem.

Obviously this won't apply in more general frameworks.

@JeSuis Every better book about point set topology should treat filters.

@DanielFischer Ok, great

@BalarkaSen: It seems to me that what you've written down is basically part of the proof that cellular homology works.

@MikeMiller Is it so? I haven't explicitly constructed the cellular chain complex, note that.

Ok, I guess the 2 terms in the long exact sequence essentially forms the chain complex as other terms are all 0.

I'm not paying close attention, but your exact sequence looks almost exactly like one of the zigzags in Hatcher's proof.

And I would be surprised if what you do doesn't show up in the same proof.

You're claiming that computing the snake map is essentially what we do in cellular homology? Interesting, I haven't thought about it that way.

I need to review the construction of cellular homology someday.

We've just handled cellular homolgy in my class this friday

It's a bit more than just computing the snake map though

Any chance we could get a few upvotes to the answers on this question? I've never campaigned for upvotes before but, in this case, the OP has repeatedly denied overwhelming evidence from multiple sources that his question has an error. It would be nice to get some back up.

@BalarkaSen There isn't much to it. $C_n^{CW}(X)=H_n(X^n,X^{n-1})$ and the boundary map $C_n^{CW}(X)\to C_{n-1}^{CW}(X)$ is the boundary map from the LES of the triple $X^{n-2}\subset X^{n-1}\subset X^n$.

Does anyone know where the notation $\Gamma(E)$ for the sections of a vector bundle $E$ comes from? Why not $\Sigma(E)$ or $S(E)$?

No, but nobody uses $\Gamma(-)$ for anything else, so there's no worry of conflicting notation.

hello

hello. going to write my first answer. can you help me as to how I can write math symbols, etc. in my answer?

@JasperLoy Oh. Ok. Will look into that. Thanks. :)