By contradiction maybe?

Isn't every finite set, countable anyway as you can form a bijection between the finite set and a subset of $\mathbb{N}$?

Hello. Off-topic: How can I prove that $\displaystyle\lim_{(x,y)\to(0,0)}\dfrac{x^4 y^4}{(x^2+y^3)^2}=0$ ?

Is that even possible, @Cristopher?

You could approach $(0,0)$ from many directions and I'm not sure they'll all give the same limit.

I thought this conundrum was why the directional derivative was introduced ...

@KhallilBenyattou The numerator has total degree 8, the numerator has total degree 6.

Numerator wins.

Isn't there the danger of dividing by zero, @Pedro?

I was going to consider $(x,y) \to (0,0)$ along lines of the form $y=mx$.

@KhallilBenyattou That strategy I think is mainly useful for disproving multivariate limits

Ah gotcha, @ᴇʏᴇs! What's a good course of action for proving multivariate limits instead?

I've never actually learned how to do that

I never learned epsilon-delta proofs for multivariables either

Me neither! I'm only vaguely familiar with single variable stuff. :-P

We only did problems where we showed that there was some direction where the limit wasn't the same, thus the function doesn't have a limit at that point..we never had to prove that the limit value was something

Oh I see. Yea, the latter seems like something that isn't taught until later on at institutions.

@Cristopher Recall that $(a+b)^2\geqslant 4ab$ (since $(a-b)^2\geqslant 0$), so that $$\frac{x^4y^4}{(x^2+y^3)^2}\leqslant \frac{x^4y^4}{4x^2y^3}=\frac 1 4x^2y$$

Oh, nice one @PedroTamaroff

@MikeMiller Prove that $\Bbb Q\otimes \prod_{p\geqslant 2}\Bbb Z/p\neq 0$.

Then you squeeze it between $0$ and $x^2 y/4$ ?

Yes.

Can I ask what the $\otimes$ means, @Pedro?

@KhallilBenyattou It's a tensor product, over the ring of integers.

Ah, gotcha! I've only heard of tensors and spinors from my tutor. It looks like interesting stuff! :-)

Oh. That sounds like physics.

@PedroTamaroff: I think we've done this before. In any case, no.

@MikeMiller We proved the $\hom$ is nonzero, IIRC.

OK, invoke duality.

Wait no.

The point is that the product has a decomposition as $D\oplus R$ where $D\neq 0$ is divisible.

Every $B_M(a;r)$ in a metric space is open in M. how come doesn't all of its points have to be interior points?

Can I ask what they mean by $f \equiv 0 \in \mathcal{B}[a,b]$? Specifically, the $f \equiv 0$ part.

(Circled in green)

Btw @KarimMansour, Idk if you've heard, but libgen is gone

Why is it gone @Clarinetist

wtf

wtf is it gone !!!!! omg

This is all I could find:

torrentfreak.com/…

@KhallilBenyattou Usually that means $f(x) = 0$ for all $x \in \mathbb{R}$, but I don't understand why you would need it in that proof

To show that $\mathcal{B}[a,b]$ is a vector space, it'd need to contain an additive identity I think, @Clarinetist. $f(x) = 0$ for all $x \in [a,b]$ is of course an additive identity of the set so we can say for sure that $\mathcal{B}[a,b]$ is a vector space.

@KhallilBenyattou I understand that, I guess it makes no sense to me that it's after a $\therefore$ symbol.

Haha yea, @Clarinetist. I know what you mean! It should've just been a short comment at the beginning. :-)

@KarimMansour @ᴇʏᴇs It looks like IFLS had something on this before that article: iflscience.com/editors-blog/…

that sucks

Should the partition in this case be $P = \{ a =: p_0, p_1, \dots, p_{k-1}, p_{k} := b \}$ as opposed to the one in the definition?

where can I download books now @Clarinetist

Lol @KarimMansour, Idk

@KhallilBenyattou I don't see why not

It seems like the notes are slightly inconsistent, @Clarinetist.

It seems natural to include the endpoints of the closed interval in the partition, right?

@KhallilBenyattou Yeah, that should be the case

Obviously you can't be working outside $[a, b]$ for that definition

Otherwise our step function $\varphi$ wouldn't be defined there.

hi @SamuelYusim

how's things

things are going alright

my research project has been destroying me for a week or two but that's par for the course

how about you?

that's good to hear

better a week or two than a year or two, eh

truly

things are going alright on my end, keeping busy

oh yeah?

lots of reading to do and a couple projects to work on

fun

well, sometimes reading is fun. sometimes it's painful

yeah, a little too much sometimes, but i think 'a little too much' is just the right workload for me

keeps me on my toes

preach it, brother

what's your project?

it's in graph theory. we're dealing with a concept called the metric dimension of a graph, and looking at how it plays with certain classes of graphs

it's neat stuff, there's a very algorithmic flavour to it and I've never really dealt with that kind of thing so that's been fun

did this guy just try to snatch a bounty last-hour with a nonanswer?

that sounds like fun. can you be more specific about what you mean by "how it plays with certain classes of graphs" or are you not comfortable with that?

oh, sure. We're finding an algorithm to calculate the metric dimension of outerplanar graphs right now, for instance

haha

ah, i see

outerplanar graphs are those with a planar embedding that has all the vertices on the outer face

so the idea is something like that, even though the problem in general is absurdly hard, when you have extra structure you should be able to make a better algorithm?

yeah

e.g., we demolished the problem for what are called cographs earlier in the summer

very cool

we're actually working on a variant of the metric dimension w.r.t. outerplanar graphs right now though, which is called the strong dimension

@TedShifrin John Lee's textbooks all have exercises. And nearly all my graduate physics textbooks have exercises. And so do the economics books I have. Are mathematicians who write grad textbooks exercise-averse?

it's already been determined that the decision problem "$G$ has metric dimension $n$" is NP-hard for outerplanar graphs $G$, but we think we can get the strong dimension in polynomial time which would be pretty rad

yeah, since that should still be NP-hard in full generality, right?

yep

(not that I know anything about this, I just googled some definitions, and they don't look that far apart from each other)

the proof isn't so bad. you basically show the strong dimension problem is reducible to the minimum vertex cover problem

and this is doable by coming up with a graph that has the same vertices but different edges, and on which a cover will be a strong resolving set for the original

if that makes sense

is that how one usually shows things are NP-hard? you show that you can reduce to something already known to be, in polynomial time?

yep

but usually it's not vertex covering from what I can tell

i guess there's a lot of problems known to be NP-hard so you just pick the one that looks closest to yours

yep

although somehow the way they proved metric dimension was np-hard on outerplanar graphs was to reduce it to the boolean satisfiability problem, which you can look up if you haven't heard of it and immediately notice that it has nothing to do with metric dimension

that proof is long, terrifying, and miraculous

miracles like that are obnoxious

make it very difficult to understand why something is true, at least without having a deep understanding of their proof

(i say 'why something is true' in a moral sense here, not in the sense of 'how to prove it')

I know exactly what you mean and completely agree

i think this is a fairly common belief

this wasn't exactly a proof I had the time or willingness to get deeply acquainted with either

yeah, if only it was common among what I've seen of the world's algorithmic graph theorists

i think the way these things happen rather often is that people know how the proof 'should go', hit a snag, and make severe modifications so that the proof that should work really does work; but it might look way different now

they'll be a three line proof away from showing that some step of an algorithm can be done completely based on concrete observations and they'll relegate everything to case analysis anyway. this is so frustrating

also, that.

i also get the feeling that referees tend not to like philosophical diversions in papers, like sections about why something is true or why one would think to prove it this way

I dont still have any idea why is this interesting area closed

@Agawa001 Not enough activity

but it is so interesting subject

so many SE sections arent that worthy to be here more that AI and machine learning

and they are both closed

alright, I should go to bed. night, @MikeMiller

night!

good luck on your stuff

after 1 year inactivity in beta

btw, mathematics is still beta ?

@Agawa001 Math is WAYYYYYYYYY past beta, I'm pretty sure

@Clarinetist how long, i remember it was trial maybe atleast two years ago

Idk, I haven't been here that long

@Clarinetist i have been here as visitor, not regular member

MSE has been out of beta since 2010.

and it helped me often in my studies

AI need to be be reopened, people need to be comitted again

i mean wth? serious study field and scientific domain like this closed whereas something like bicycles and movies graduated ?

and beer ?

Where there is a supply and demand, a market will arise. In this case for knowledge.

Bicycles, beer, and movies have much larger markets than AI.

At least, when it comes to the number of people who consumer knowledge about them.

But I understand your point.

Is robotics insufficient?

no its insufficient

Was there a proposal for AI at one point?

i see there was three proposals

one marked as duplicate od data science

two closed

even machine learning is closed

That is annoying I agree. I myself would like to ask questions about ANN design, but am not sure where to ask that. AI would be a natural place for that, yes?

neuronal network ?

yeah sorry

Artificial Neural Network learning type stuff

e.g. Hand writing recognition

Artificial forms of perception

i hope people commit back in it, i only see facebook groups and google sites , but these communities arent so popular

SE can gather people all arround the world

how would I prove $(n!)^2 > n^n$?

assuming the latter, I get $((n+1)!)^2 = (n+1)^2(n!)^2 > (n+1)(n!)^2 > (n+1)n^n$

(after which I get stuck)

Hi

if its possible of someone could give me the intuition of the following

Let (S,d) be a metric subspace of (M,d) and let X be a subset of S. Then X is open in S iff X = A intersect S for some set A which is open in M.

I don't completely understand why it works

@VibhavPant There's a famous proof of $2\sum_{i=1}^ni=n(n+1)$. Adapt it to this setting.

@KarimMansour Do you see that it works for open balls in $S$?

yes @KarlKronenfeld

ok @Karim, for general $X$ is there a direction of the iff that is easy for you to prove?

yeah the converse is easy

but what about the direction going from right to left?

You mean, suppose $A$ is open in $M$, prove $A\cap S$ is open in $S$?

yh

Ok.

What does being open in $M$ mean for $A$?

I see I got it

:)

thanks karl I just needed first push

So quiet in here. I need a diversion from my writing, and nobody is speaking here :(

I see that the two users who had obscene avatars are gone.

Good.

Hello @TobiasKildetoft, I am here.

@JasperLoy Hi

France and Germany are my favourite countries. I am still deciding which of the two to be reborn in for my next life.

France :-)

I divided the six populated continents into three groups of two, based on quality of life. I am living in the worst group.

Any Ubuntu users here? What is your favourite desktop?

I really miss Windows. I will buy myself a Windows laptop when I am rich.

@JasperLoy I use Cinnamon, though it is not officially supported any longer

It allows me to customize a lot of things I want, which was not possible in the others

@TobiasKildetoft I know about the Cinnamon PPA and Cubuntu, which is not an official derivative. I hope they make one soon, to complete the family.

You may want to try Ubuntu Mate.

@JasperLoy It used to be supported by the developers, but as far as I recall, they stopped doing so

Mate and Cinnamon are similar, just like Gnome and Unity.

@JasperLoy I mainly need the possibility of adding menus to the top panel is a more customizable way than what unity or gnome allows

which is done through applets in Cinnamon (I had to do some modifications to the code of some of the applets though to get them precisely as I wanted)

I see. Mate certainly allows that.

@JasperLoy Might be worth looking into

though unless I can make things work so I can configure the menus using the same syntax, it might be too much work to make it all look like I have it now

@TobiasKildetoft You can ask me if you need help with Mate. In Mate, you can have one or two panels, and you can add and remove things to them and move them around easily.

Is KDE supported by Ubuntu?

@BoniTea You can use Kubuntu, which is an official derivative.

@JasperLoy Right, but I need specifically to be able to add menus to the top panel, which consist of submenus where each item points to a file on the computer

@TobiasKildetoft I would expect most major desktops to support like.

@BoniTea Back when I decided this would be useful, it took me a long time to find one that allowed it

and even then (as I mentioned), I had to do a bit of code editing myself to make it work

Hello@iwriteonbananas

@Rememberme Hey, did you manage to find as example of what I asked yesterday?

I have been all into school work @Tobias .. Sorry I didn't try it yet

that's fine

@TobiasKildetoft My example works doesn't it?

The hausdorff one

Not sure what example you mean

The topological space$\{X,\tau\}$ with a toplogy $\tau =\{\emptyset , X\}$@Tobias

@Rememberme That is certainly non-Hausdorff, and it satisfies the first version of what I asked

Actually, I think my stronger version is not quite possible (I made it stronger than I intended)

Which one?

ohh, nevermind, it can be done. The stronger version os where the is some $x$ such that the constant sequence $a_n = x$ converges to everything, but for any $y\neq x$ the sequence $b_n = y$ converges to only $y$

(and the space should have more than one element)

Converges to everything ? It just converges to x doesn't it since it is a constant sequence@Tobias

@Rememberme Well, in your example any constant sequence will converge to everything (you should check this)

in fact, any sequence will converge to everything in that example

Can you please elaborate on what you mean by a constant sequence @TobiasKildetoft

@Rememberme The sequence $x,x,x,x,...$

Okay I thought it would be it

I can take any value for x and it will converge to any value in the the topological space @Tobias

@Rememberme right (in your example)

you should prove this of course

hmm....

@Rememberme anyway, I need to run an errand. Be back shortly

Okay

@StanShunpike Sorry, Stan: I was thinking (but didn't type) more advanced graduate texts. First-year material generally has exercises. And it's not a universal thing. As I've opined before, I think exercises in large part make or break a book. I'm most proud of the exercises in my books.

hi @Remember @Soham

hey, @Ted. How's things?

Just fine, and you?

Not too bad, had a little trouble at school because I forgot to get my lab coat for Chem. The usual :P

Well, they make you do that so you don't burn holes in your clothes :P

When writing a longer document such as a thesis, is it "okay" to use the same index (such as i) in different contexts, as long as it is made clear to the reader which values the index can take / what its meaning is?

Yeah, but we're just making pretty colors now. No fun concentrated stuff yet, haha. Anyway, it's Friday, which means that I can do a bunch of things I've been planning to.

@user2249626 Yeah!

Yes, @user2249626, but better not to do so in close proximity if possible

@Hippalectryon

@TedShifrin I have a nice example.

Thanks. I don't want to confuse the reader but I feel that introducing more letters might also be confusing.

@Chris'ssistheartist ?

When there are consistent different index ranges, I will sometimes establish rules for a section by saying: Here $1\le i,j,k\le n$, $n+1\le \mu,\nu\le N$, etc.

Salut @Hippa

@TedShifrin o/

@TedShifrin "Let $v, \nu$ be different integers?" :P

That's a different matter, @Soham.

I know, I know.

I should go, actually. Later :)

bubye

@Hippalectryon do you see Ovidiu's problem? 3.62.?

@Chris'ssistheartist yep

@Hippalectryon How about the variant with $\zeta(3)$? :-)

:D

:-)

@Rememberme Back

Hello again @Tobias

@TedShifrin Hi

@Rememberme Did you prove that all sequences converge to everything in the indiscrete topology yes?

@Rememberme Take any $y$ and show that the sequence converges to $y$ (recall what this means precisely)

actually, start by writing down the definition of "the sequence $a_n$ converges to $y$" here

Okay wait

(it is a good idea to try from memory first, before looking it up)

We say that that $a_n$ converges to y if for some$ \epsilon >0$ $d(a_n,y)<\epsilon$ @Tobias

@Rememberme No, this is not metric spaces, it is topology

so we need the definition that only uses the topology

Okay But I just know the metric space definition @Tobias

@Rememberme Ahh, then it is of course going to be tricky, since these topologies are not metrizable

I thought you were studying general topology

Yes I am

Why?

Not analysis

@Rememberme Well, then you probably want to look up the definition of convergence in general topological spaces (and try to see why this is the correct notion when compared to the metric space one)

Munkres hasn't yet introduced the definition of convergence in topological spaces @Tobias

@Rememberme Ahh, then it might be better to skip it until you get to that part

@Tobias Would the open set definition work??

@Rememberme what do you mean?

@Rememberme Not quite the right quantifiers

also, you should not require all the points of the sequence to be in the open set, as that is not what convergence is about (you should be able to ignore any initial segment)

I think now it is fine

Not quite (you still have a requirement of the entire set of points of the sequence which is way too strict)

Okay then what should be the requirement then...

just think of what that would mean for the sequence $y,x,x,x,...$ in a metric space

so you need the requirement to be that for any such open set $U$ there is some $N$ such that all $a_m$ with $m\geq N$ are in $U$

Okay .. but whats the reason for this definition@Tobias

@Rememberme As I said, otherwise the sequence $y,x,x,x,...$ would not converge in a metric space (if $y\neq x$)

which is obviously absurd

Yes it makes sense

convergence is all about what the tail of the sequence looks like

Not the head..

Also @Tobias The example of a non hausdorff space which I gave was quite boring.. Is there any interesting one?

Hey@Huy

@Rememberme Sure, that was what I was trying to move towards with my exercise. But I thought you were familiar with convergence in topological spaces

@Rememberme But if you take a finite set then how many Hausdorff topologies are there in that set?

There can be many

But for me infinite is more interesting

What about a infinite set .. Is there any non hausdorff topologies in it?

@Rememberme No, there cannot be many

and yes, the indiscrete topology is always non-Hausdorff. Another example is the cofinite topology.

co countable?

no, cofinite (if you want cocountable to be non-Hausdorff, then you need the set to be uncountable)

In munkres there is this notation saying about restricting a function what do we mean by that @Tobias

@Rememberme In general, it just means we have a function $X\to Y$ and a subset $Z\subseteq X$ and then define a function $Z\to Y$ in the "obvious" way

Okay and inclusion mapping means a mapping from a set A to its subspace right?

@Rememberme Other way around (from subspace to parent space)

Okay is it a homeomorphism?

not in general no. It is what is known as a topological embedding, meaning that it is a homeomorphism when you consider the codomain to be only the image

Okay you mean bijective...

not just bijective

any injective map is bijective when we restrict the codomain to be the image. But that might not make it a homeomorphism

Yes I did the exercise

Showing that not only being a bijection proves it to be a homeomorphism

Example this:

$f:[0,1)\to S^1 , f(x)=\{\cos(2\pi a), \sin(2\pi a)\}$

Hey, @Rememberme

Hi. I asked a question last night, but when it was answered I was absent, so I wanted to thank you @PedroTamaroff for helping me out. Also thanks, @KhallilBenyattou for chiming in too :)

No prob, @Cristopher! It was an interesting question! ^_^

Thanks :P

When it comes to two or more variables, we don't have L'hopital so the evaluation becomes more complicated sometimes, but also more challenging, in a good way

@SohamChowdhury nothing you'd find interesting. ps : wouldn't be able to go at Sarat tomorrow. am a bit ill.

I was wondering, is there any interest in a "product" integral defined by $\displaystyle\int_a^b f=\lim_{n\infty}\frac{b-a}{n}\prod_{k=1}^nf\left(a+(b-a)\frac{k}n\right)$ ? (for strictly positive functions)

Surely someone has already studied that before

@Hippalectryon just putting (b-a)/n in front of the integral isn't going to normalize it well enough to get a limit

Or, $\displaystyle\int_a^b f=\lim_{n\infty}(b-a)\sqrt[n]{\prod_{k=1}^nf\left(a+(b-a)\frac{k}n\right)}$

but yes, "product integral" is a thing (google it), and it's just $\exp \int_a^b \ln f(x)\,{\rm d}x$

ok, thanks :D

Lately I have been having back problems. To fix this I went to a physiotherapist who gave me some really good advice about how to fix my posture. She also said that I should avoid sitting. Therefore I have now rearranged my screen and my keyboard so that I can spend time by the computer standing instead of sitting.

(:

@BalarkaSen ach, okay.

hey, @Mike.

@BalarkaSen hi

hi soham

Back from dentist. Well, it was just horror, and very grateful I've survived. :-)

For almost 3 hours I did no integral, series and limit! Well, I couldn't even think of them. Bad.

hmmm, @robjohn is missing today as I can see.

morning

@Chris'ssistheartist didn't you eat icecream later?

@SohamChowdhury No, I didn't and I won't :-). I'm in a special fitness program these days, so I need to be careful about eating sweets.

There's a PhD in biology chat who insists P(U=a) for a uniform distribution on the reals[0,1] is zero, and that the integral of zero across [0,1] can be zero and the integral of the probability measure can be one across the same interval without causing a paradox. They are correct; probability measures are only sigma-additive and [0,1] is uncountable.

Why can't I define a uniform probability on integers and reconstitue the paradox that way?

That is, P(Z=a) is zero for integer a, but the integral across P(Z=a) for all integer a =1.

Does that make sense?

Because there are no paradoxes. What you're describing right there is precisely the reason there is no uniform probability on the integers - $\sigma$-additivity.

More generally then there are no uniform distributions on countably infinite sets?

Yeah, though I wouldn't say more generally

@MikeMiller I guess the integers are exactly the same size as other countably infinite sets. At the risk of asking to reformalize probability and measure theory, is there a simple reason why this is formalized this way? Strengthening the third axiom to uncountably infinite disjoint sets seems like a handy property.

If it's not easy to explain, I can just take decades of mathematicians word for it.

lol, I love to be in a special fatness program

@Resonating Well, because we don't want it. The whole point of measure theory is to make rigorous the probability theory studied on $[0,1]$ (or on $\mathbb R$) in a more general setting; if we couldn't use the uniform distribution on $[0,1]$ as a measure there wouldn't be much point.

@Chris'ssistheartist I'm here, but I am having problems with my ISP. I am not able to access ajax.googleapis.com and without access to that site, much of SE is inoperable. I didn't notice that I had been logged off of chat. I am now connecting via my iPhone.

@robjohn I was just asking since I didn't see you around. I also was off for some hours, but I'm back now. OK, hope things improve soon.

One reasonable argument would be to try to go in the other direction: well, why don't we reduce this to only having finite additivity? And I think the intuition here is just that you should be able to take the "limit" of the probabilities of the first $n$ events together. e.g. if I know what $\mu([1/n,1/(n+1)])$ is for all $n$, I ought to know what $\mu([0,1])$ is by just putting them all together - after all, I can count them all one by one.

Rob use vpns

@Resonating If we had uncountable additivity, we would need any subset of the reals to have either $0$ or infinite measure

@Resonating any infinite subset I mean

in fact its not isp problem, its DNS problem

@Agawa001 you mean VPN? I'd need to have an account somewhere. I have VPN through work, but I don't want to access MSE through work.

@robjohn yea cyberghost , or openvpn , some tools need to be purchased each month

@TobiasKildetoft I am now using Cinnamon on Debian Jessie.

@JasperLoy Neat

@TobiasKildetoft An uncountably infinite subset. Countably infinite subsets with measure 1/2^n for instance can have finite measure in total, like mike's example. Are there things that can be summed across uncountably infinite sets to produce finite numbers? If we did have such an entity what would be the point of it? You couldn't do much math to it.

I'm beginning to understand how this is all laid out now. Thanks.

The $...$ text.

Can I write smaller text in this?

Not just subscript, but lines of smaller maths code?