Mathematics - 2020-06-15

user76284

00:02

Hmm nevermind, I guess it might end up being more complex.

Thorgott

if I have a function R^n->R^m that is totally differentiable at a point and partially differentiable in a neighborhood of that point, can the partial derivatives be discontinuous at that point?

this ought to be possible, but I've always been bad with these multivariate counter-examples

orientablesurface

00:57

What functors do I need to show that the canonical injection between $X_i\rightarrow \coprod_iX_i$ is natural onto the image in the disjoint union in $\textbf{SET}$? $id_{\textbf{SET}}$ and what else?

I was able to construct a commutative diagram which shows that the canonical injection could be a natural isomorphism

but I feel like I shouldn't even consider $\textbf{SET}$, but rather a subcategory, perhaps?

Thorgott

well, category theory gives you a lot of freedom in how you can phrase things, so there's many way to go about this

you could, for example, note that the inclusions in the $j$-th coordinate $X_j\hookrightarrow\coprod_iX_i$ give a natural transformation between the "projection on the $j$-the coordinate" and "taking disjoint union" functors from $\mathbf{Set}^I$ to $\mathbf{Set}$

orientablesurface

perhaps the category consisting of the objects $\{$ $X_i$ : $i\in I$ $\}$ $\cup$ $\{$ $X_i^*$ $:$ $i\in I$ $\}$ ? where $X_i^*$ denotes the image?

Thorgott

if you really only care about the image, then this comes down to saying that the identity functor on $\mathbf{Set}$ is naturally isomorphic to the "taking the cartesian product with $\{j\}$" functor, which is even more tautological

orientablesurface

what's $\textbf{Set}^I$? @Thorgott

Thorgott

the $I$-th power of $\mathbf{Set}$

the objects are tuples $(X_i)_{i\in I}$ where each $X_i$ is a set and the morphisms $(X_i)_{i\in I}\rightarrow(Y_i)_{i\in I}$ are tuples $(f_i)_{i\in I}$ where $f_i\colon X_i\rightarrow Y_i$ is a map

orientablesurface

01:11

ahh okay

Thorgott

the projection on the $j$-th coordinate functor takes $(X_i)_{i\in I}$ to $X_j$ and $(X_i)_{i\in I}\stackrel{(f_i)_{i\in I}}{\longrightarrow}(Y_i)_{i\in I}$ to $f_j\colon X_j\rightarrow Y_j$ (unsurprisingly)

the disjoint union functor takes $(X_i)_{i\in I}$ to $\coprod_{i\in I}X_i$ and $(X_i)_{i\in I}\stackrel{(f_i)_{i\in I}}{\longrightarrow}(Y_i)_{i\in I}$ to the map $f\colon\coprod_iX_i\rightarrow\coprod_iY_i$ that acts like $f_i$ on $X_i$ (which exists by the universal property of coproducts)

once those definitions are unraveled, my above remark translates into a rather uninteresting tautology

robjohn

01:59

@TedShifrin back from walking the dogs. I see you're no longer here.

ktm5124

02:27

Hi, could anyone please tell me what is the highest order of infinity?

Balarka Sen

Huh. So the Sobolev lemma states that if $f \in H^s(\Bbb R)$ and $s > k + 1/2$ then there is some $\phi \in C^k(\Bbb R)$ such that $f = \phi$ almost everywhere

Yuvraj

02:57

@rob John morning sir

Edward Evans

05:30

@Thorgott double cosets are made up

robjohn

@Yuvraj It is somewhere ;-p

Ted Shifrin

Howdy @Edward, and re @robjohn

Edward Evans

Hullo @Ted

Fargle

hello once more, @Ted

Ted Shifrin

Heya @Fargle

Alekos Robotis

05:45

how's it going

Ted Shifrin

re @Alekos

Pig

hi all

rarely so many people :o

Ted Shifrin

yo Piggy

Pig

hey ted

Alekos Robotis

finishing up preparing something for tmrw lol

about to sleep

figured I'd see what was going on in here

Fargle

05:47

mostly I'm just being stir-crazy and wondering what textbook I should delve back into

Pig

@AlekosRobotis what did you prepare?

a talk or something?

Alekos Robotis

yeah, I'm telling a friend of mine a few things about hodge theory tomorrow

we're doing a little reading course together

Pig

oh wow nice

are you going the analytic route or algebraic route?

Alekos Robotis

mmm more algebraic, but it seems I might actually need to learn the analysis eventually if I want to do more with this topic

Pig

nic nice

Alekos Robotis

05:56

I've been using it more as an excuse to learn some complex AG

Pig

i'm actually curious how complex geometry and algebraic geometry are different these days

i suppose working over C gives you more tools, but i don't really know what concrete problems/areas where you can say much more in complex AG than "general" AG

Alekos Robotis

some people definitely care about arithmetic things

and I guess the tools become quite different, but I don't know much about that

Pig

yea i'm curious too

Yuvraj

@robjohn AT LEAST AT MY YARD

sir have you noticed my post yesterday?

robjohn

@TedShifrin Hey there. I'm fixing an answer I wrote a long time ago, but someone just pointed out that I misread the question.

@Yuvraj It is an hour to midnight here. I think I may have to rethink whether the Earth is flat...

Yuvraj

06:07

lol

my question

i am in doubt whether p would be an exact integer?

robjohn

@Yuvraj yes, why post that microscopic text in a large blank page, and then post it again? No, $p\approx1.110720735$, but that was a numerical search. I didn't know an exact solution right off.

Yuvraj

sorry sir actually question was in the phone so i posted it as it is

robjohn

@Yuvraj ah... Let me show a better way to repost something like that.

@Yuvraj that will go to the previous post without a large image here.

You can also ask:

$\textbf{P55.}$ Find the smallest positive number $p$ for which the equation $\cos(p\sin x)=\sin(p\cos x)$ has a solution for $x\in[0,2\pi]$.

Yuvraj

06:25

@robjohn right sir but that time i was on phone(in my phone mathjax do not work,so i am habitual of posting picture).

till the time i am on my phone!

robjohn

yes, but what I showed above links to the previous question in the transcript so that it doesn't need to be posted twice.

right before you said "i am in doubt whether p would be an exact integer?"

Yuvraj

yes sir

robjohn

Just a suggestion to keep the clutter down

but do you think there is an exact solution? I got a numerical one above.

Yuvraj

it could be

because actually in last it is scaler which makes the equality feasible.

robjohn

$p=1.1107207345395$ is about as close as I can get

That is very close to $\frac\pi{2\sqrt2}$.

as in all digits match

And it seems that $x=\frac\pi4$

So I don't have a proof yet, but $p=\frac\pi{2\sqrt2}$ and $x=\frac\pi4$ might be the answer.

where $\sin(x)=\cos(x)=\frac1{\sqrt2}$

and $p\sin(x)=p\cos(x)=\frac\pi4=x$

I am sure that is the solution.

robjohn

07:05

here is the image

If you look at that image, you can see that when the curves first touch, the derivatives are equal at the points where the curves touch. When the curves intersect more steeply, the derivatives do not match.

@Yuvraj: so checking out $p=\frac\pi{2\sqrt2}$ and $x=\frac\pi4$, we see that those conditions are met.

so that is the solution

Yuvraj

yes i got it sir

but still, I have to search out the analytical solution for this

robjohn

@Yuvraj Why? For $p\lt\frac\pi{2\sqrt2}$, the curves don't intersect. For $p\gt\frac\pi{2\sqrt2}$ they do intersect. They are tangent for $p=\frac\pi{2\sqrt2}$.

is this for a class?

Yuvraj

no sir you are correct.

actually i have solved this question analytically can you check?

@robjohn

actually, you are correct we do not much analytical solution now because graph clear almost everything.

robjohn

07:35

@Yuvraj where?

Yuvraj

@robjohn in my copy

actually i just too the inverse of the function and wrote $sinx=cos(\pi/2-x)$

and then $p(sinx+cosx)=2n\pi+pi/2$

then $\sqrt 2 psin(x+\pi/4)=2n\pi+\pi/2$

minimum of rhs will be at $x=\pi/2$ value

56 mins ago, by robjohn

and $p\sin(x)=p\cos(x)=\frac\pi4=x$

robjohn

@Yuvraj that was where the graphs first touch.

Yuvraj

yes

robjohn

07:51

The equation is equivalent to $\frac\pi{2p}=\sin(x)+\cos(x)=\sqrt2\sin\left(x+\frac\pi4\right)$

so the smallest that $p$ can be is $\frac\pi{2\sqrt2}$

because $\sin\left(x+\frac\pi4\right)\le1$

flowian

09:16

Anyone capable of posting an example of uncountable subset of a topological space with countable base?

Can't imagine how it is possible.

Fargle

09:39

@flowian $\Bbb R$ has countable base, and $[0,1]$ is an uncountable proper subset

(you can take open intervals which have rational endpoints as a countable base for the standard topology on $\Bbb R$, because you can make intervals which have irrational endpoints as arbitrary unions by doing limiting stuff; e.g. $(e,\pi)$ can be exhibited as the union of $(2.8,3)$, $(2.72, 3.1)$, $(2.719, 3.14)$, ...)

Thorgott

It's somewhat analogous to $\mathbb{N}$ being countable, but $\mathcal{P}(\mathbb{N})$ being uncountable. You may only have countably many sets in the base, but you can still be able to exhibit uncountably many open sets by taking unions of these sets in the base. So it's not all too surprising there are uncountable spaces (or rather, spaces with uncountably many open sets, the other case is trivial) with a countable base.

Fargle

yep---as it turns out, the fact that you're allowed arbitrary unions can do quite a lot of heavy lifting

you can also do this trick with any $\Bbb R^n$, taking as basis all rectangles $(a_1, b_1) \times \cdots \times (a_n, b_n)$ with the $a_i, b_i$ all rational

Thorgott

09:56

on the other hand, you of course also get that a space with a countable basis definitely has at most continuum many open sets, so there're limitations

I think this also forces the space itself to have cardinality at most continuum if you additionally assume some separation axiom

Fargle

that sounds right

I think Hausdorff might be enough, but that also might be overkill

Thorgott

$T_1$ is enough, according to Wikipedia

Fargle

okay yeah I buy that

I always forget which separation axiom is which

Madhuchhanda Mandal

Should a transitive Relation contain at least one example of transitive? Or if a relation does not have any counter example of transitive property, can it be considered to be transitive?

Fargle

if you never have a situation where $a \sim b$ and $b \sim c$ whatsoever, the relation is automatically transitive

Thorgott

10:01

I guess the way to prove this is by injecting the space into its topology by picking a suitable neighborhood of each point, but how do I do this

oh wait, using closed sets instead of open sets

$T_1$ implies singletons are closed

Fargle

in other words, all you need is that no counterexample exists---i.e. you can't have $a \sim b$ and $b \sim c$ but $a \not\sim c$

Madhuchhanda Mandal

Example : R = {(1,1),(2,2)} is it transitive? X={1,2} @Fargle

Okay. I got it

Can this be generalized for mathematical statements?

Fargle

well, here, there are situations where it comes up, they're just not particularly obvious or interesting---

1 is related to 1, and 1 is related to 1, so therefore...

Thorgott

more generally, if $(X,\tau)$ is a top space and $T_1$, then $|\tau|\ge|X|$

Fargle

but yes, in general, if you're given a property that says "if P then Q", and P is never true, then the property is automatically satisfied

this is known as being "vacuously true"

Thorgott

10:05

it's good to have observed that, I guess

Fargle

for example: the empty relation (nothing is related to anything) is always automatically transitive

Madhuchhanda Mandal

Okay. Thanks a lot

Fargle

no problem :)

Madhuchhanda Mandal

Yes. Although I should have chosen R={(1,2),(3,4)}, X={1,2,3,4} as the example to make it more obvious. But I got your point

flowian