Mathematics - 2020-01-17

12:00 AM

@MyWrathAcademia $\#\{k\in\Bbb Z:a/n<k<b/n\}$ @Thorgott

(Set size is sometimes written $\#A$ and sometimes $|A|$. For some reason I've usually seen the former with set-builder notation)

Thorgott

Yeah, that works too, of course

I think the reason for that is that |{....}| can be pretty ugly and harder to read

I remember once working with an expression of the form |{....|{...}|....}| and it wasn't pleasnt

Rithaniel

1:26 AM

So, if $A$ and $B$ are $R-$modules, then is it correct to say that $A\oplus B\cong A+B$ if and only if $A\cap B=\{0\}$?

Thorgott

1:43 AM

Yes, same proof as for vector spaces

Rithaniel

Alright, easy enough

Rafael Nadal

3:10 AM

When do we use Differential Under Integral Sign method for integrating?

bjorn93

4:10 AM

Does anyone have a definition of "non-trivial interval" ?

Is it a "non-empty interval" or "non-empty and contains more than 1 point"

Mike Miller

4:45 AM

@bjorn93 this is not something with a widely accepted definition, but I would guess it means [a,b] for a < b, aka your second statement.

user443836

5:05 AM

Hello

Akiva Weinberger

5:59 AM

@bjorn93 What's the context?

Leaky Nun

7:42 AM

@AlessandroCodenotti want a game?

Alessandro Codenotti

Not now, sorry

Leaky Nun

8:33 AM

GM Ben Finegold Breaks the Rules: Never play f3/f6, protect your pawns

►

@AlessandroCodenotti this is brilliant

Stupid Questions Inc

9:25 AM

@adeshmishra then i agree with you if that's the case

adesh mishra

10:05 AM

@StupidQuestionsInc What? I'm missing the context.

Stupid Questions Inc

11:45 AM

23 hours ago, by adesh mishra

@StupidQuestionsInc Exam wants me to solve 25 questions in 1 hour.

@adeshmishra you can click on the small arrow that appears before "@adeshmishra" to see what i was responding to

Rithaniel

12:18 PM

Grandmaster Finegold is very entertaining to watch

Leaky Nun

@Rithaniel yes he is

Alessandro Codenotti

12:54 PM

I find him more annoying than entertaining, I don't like his humour

Rithaniel

1:07 PM

I don't always laugh at his jokes, but it's more the fact that he's trying, and doing so with a degree of confidence.

Also, he's a good chess-piece-mover

Leaky Nun

@Rithaniel wnana play?

@Rithaniel he tried and he failed miserably (at least according to A. Codenotti); the lesson is: never try

Rithaniel

In class currently

Leaky Nun

uni?

Rithaniel

Yeah, we're doing stuff with commutative diagrams, so my attention is primarily directed towards the professor

adesh mishra

1:48 PM

@StupidQuestionsInc Thank you for agreeing with me. Please say something else on that topic.

Lucas Henrique

2:36 PM

Hey chat

Dumb question: let $p$ be an irreducible polynomial over a field $F$. We know that $F[x]/(p)$ is a field with a root of $p$; it's not hard to see that $F \hookrightarrow F[x]/(p)$

Leaky Nun

@LucasHenrique so far so good

Lucas Henrique

However, we usually want things to stay in the same set, even if we can describe them with homomorphisms. Can we always say that there's a field $E \cong F[x]/(p)$ s.t. $F \subset E$?

And more generally, if $F \hookrightarrow E$, is there an $E' \cong E$ s.t. $F \subset E'$?

Leaky Nun

@LucasHenrique yes

Lucas Henrique

how?

Leaky Nun

but it's an entirely set-theoretic argument and there's not much point pursuing it

generally when we do algebra we don't care about the set theory

Lucas Henrique

2:50 PM

@LeakyNun isn't set theory like extremely relevant to algebra?

without Zorn's lemma, not all rings have maximal ideals

Leaky Nun

let's see

the only point where it goes wrong is if $F \cap E \ne \varnothing$

ÍgjøgnumMeg

I mean, if you are talking about some containment often it's the case that what you reaaaally have is an isomorphic copy of the thing you're talking about inside the other object

but it's overly pedantic to talk about isomorphic copies rather than just talking about a containment

Leaky Nun

so given a set $S$ we want a set $T$ that bijects with $S$ and such that $S \cap T = \varnothing$

that would be the source of our new elements

here we will let $S = F \cup E$

and then rename elements of $E \setminus F$

let's say $f:S \to T$ is our bijection

then set $E' := F \cup f(E \setminus \iota(F))$

where $\iota : F \to E$ is the embedding

@LucasHenrique are you convinced that we can have a field structure on $E'$ and that $E \cong E'$?

if so then we can begin constructing $T$

rather, start by establishing a bijection $E \to E'$ that makes the triangle commute and the induce the field structure on $E'$

Lucas Henrique

@LeakyNun I'm not sure if this is possible

Leaky Nun

so you've evaded the set-theoretic issues

the set-theoretic issues is that their elements can happen to be equal as sets

Lucas Henrique

3:01 PM

I mean: idk if there's a way to have $F\cap E \neq \emptyset$ unless $F \subset E$

Leaky Nun

let's just look at an example

let's look at the field F2 = {0,1}

Lucas Henrique

k

Leaky Nun

let's say F and E are both isomorphic to F2

the 0 in F is represented by the set {}
the 1 in F is represented by the set {{}}
the 0 in E is represented by the set {{}}
the 1 in E is represented by the set {{{}}}

so the 1 in F is equal to the 0 in E

Lucas Henrique

that's ugly

Leaky Nun

that's why we don't care about set-theoretic issues

Lucas Henrique

3:02 PM

but alright, you win. :p

Leaky Nun

the point is that the elements can be any set

we only care about things up to isomorphisms

we care about injective homomorphisms

Lucas Henrique

@LeakyNun it's just that sometimes (normally with infinite stuff) working with sets is easier than working with whole structures

for example: the construction of the algebraic closure of a set using transfinite induction.

Leaky Nun

if you want subsets, you can give up $F$ and look at $\iota(F)$ which is a subset of $E$

but it will be more tedious to give up $E$

Lucas Henrique

@LeakyNun what does this mean?

Leaky Nun

as in, represent them by different sets

@loch could you phrase this better than I?

Lucas Henrique

3:07 PM

I'm trying to formulate the algorithm (not the proof, because technical issues usually obfuscate the proof idea). Your example helped me a lot

Adam

3:40 PM

So do ordinal numbers only represent measures of the cardinality of finite subsets of $\mathbb N$, or can we refer the cardinality of n-tuple sets as ordinal numbers?

also what's with the users that appear with lamely self promoting user names that are complete descriptive sentences assumedly referring to the way they perceive themselves?

Thorgott

4:07 PM

If you have $F\hookrightarrow E$, then $F\cong\iota(F)\subseteq E$ is easy and gives you the same extension (algebraically) with a set containment. If you want to find $F\subset E^{\prime}\cong E$, this comes with the issues that Leaky mentioned. So we usually stick with the former to just phrase everything in terms of subsets (it ultimately matters little cause we only care about things up to isomorphism usually). Sometimes, we may already embed everything into an algebraic closure a priori.

Lucas Henrique

good enough

thanks @Leaky and @Thorgott

Leaky Nun

@Jacksoja wanna play?

Leaky Nun

4:21 PM

@Rithaniel?

if I go on I might actually need to create a chess chatroom lmao

Rithaniel

In a little bit, sure. Lunch and decompressing after coming home

Alessandro Codenotti

send me a challenge @Leaky

Leaky Nun

@AlessandroCodenotti sure

Alessandro Codenotti

4:37 PM

I went to the analysis board and I can't see the messages anymore... @Leaky

Leaky Nun

@AlessandroCodenotti wow I didn't even see that move winning the bishop

Alessandro Codenotti

I felt like there was something wrong with my bishop move, but I only realized what exactly after moving and hoped you wouldn't notice :P

Leaky Nun

completely didn't notice

Alessandro Codenotti

Wow SF even says to give up the queen for two minor pieces in that position

Leaky Nun

lol

Leaky Nun

4:52 PM

@Rithaniel tell me when you're ready

Rithaniel

Will do

Leaky Nun

Firouzja vs. Artemiev | Peter Svidler's Wijk aan Zee Round 3 Highlights

►

@AlessandroCodenotti I like his commentary

Alessandro Codenotti

Svidler is very nice

He always explains stuff for us mere mortals even though he's super strong

Other top GMs are super hard to follow while commentating (Grischuk just to name one, even though I really like watching his games)

Jake Rose

I'm beign very silly and can't solve this eigenvector problem. Is there something simple I'm missing here?

$\begin{pmatrix}
E_1 & V_0 e^{i\omega t}\\
V_0 e^{-i \omega t}& E_2
\end{pmatrix}$

any ideas for finding the eigenvectors?

Semiclassical

5:08 PM

so, a generic 2-by-2 Hermitian matrix?

Jake Rose

Yep

I haven’t solved a matrix in a while and I just can’t believe I can’t do it

Semiclassical

Have you found the eigenvalues already?

Jake Rose

They came out rather ugly

I was imagining you could just guess the eigenvectors for such a simple matrix

Semiclassical

No way around that, to some extent.

Jake Rose

That tends to be the way we always do it in part II

Semiclassical

5:11 PM

So what did you get for the eigenvalues?

Jake Rose

Semiclassical

One way to make life a bit easier is to write this matrix as $H=\overline{E}I_2+H'$ where $\overline{E}=(E_1+E_2)/2$

Jake Rose

Mhmm. How does that make things easier?

i feel like my brain has died.

Semiclassical

if you write the energy difference between the unperturbed levels as $2 Delta = E_1-E_2$ (assuming $E_1>E_2$), then that becomes $$H' = \begin{pmatrix} \Delta & V_0 e^{i\omega t} \\ V_0 e^{-i\omega t} & -\Delta \end{pmatrix}$$

so that $H'$ is traceless

vesii

Hi guys, small question here if you may. I have the following equation: $x^7-\frac{7\alpha}{2}x^2+2=0$. I'm trying to find the number of solutions it has. Can someone provide some guidelines?

Thorgott

5:20 PM

Complex or real?

vesii

real

Semiclassical

What's cute to recognize now is this: $$\begin{pmatrix} \Delta & V_0 e^{i\omega t} \\ V_0 e^{-i\omega t} & -\Delta \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & e^{-i\omega t} \end{pmatrix}\begin{pmatrix} \Delta & V_0 \\ V_0 & -\Delta \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & e^{i\omega t}\end{pmatrix}$$

which means that the matrix of interest is conjugate to the case of $\omega=0$, so it suffices to find eigenvectors for that

and then apply the change of basis

It's still a bit tedious to find eigenvectors for $\begin{pmatrix} \Delta & V_0 \\ V_0 & -\Delta \end{pmatrix}$, but less so

Thorgott

A definite answer can be obtained by Sturm chains. A more elementary algorithmic way is to compute derivatives and their gcd with the polynomial using Euclid's algorithm.

Jake Rose

Mhmm. Seems like quite a tedious work around though.

Although that trick really is cute

Semiclassical

My usual approach for finding the eigenvectors of the final matrix there is to let $B:=\sqrt{\Delta^2+V_0^2}$ and write $\Delta=B \cos \theta,V_0=B\sin \theta$

Thorgott

5:26 PM

Once you know the eigenvalues you can just do row reduction

Semiclassical

So that, aside from an overall factor of $B$, one has the matrix $$\begin{pmatrix} \cos \theta & \sin \theta \\ \sin \theta & -\cos \theta\end{pmatrix}$$

which leads to nice expressions for the eigenvectors in terms of half-angle identities

Jake Rose

Where did you learn this type of method? @Semiclassical

Semiclassical

Also, cute thing: $\begin{pmatrix} \Delta & V_0 \\ V_0 & -\Delta \end{pmatrix}^2=B^2 I_2$

@JakeRose physics

Jake Rose

@Semiclassical I was hoping more so for a textbook ;) I’m currently doing physics myself

Time dependent perturbation theory to be exact

Semiclassical

yeah, I gathered

To get the motivation behind the steps, let $\vec{B}=(V_0\cos \omega t,V_0\sin \omega t,\Delta)$

Then one can write $H'=\sigma_x B_x+\sigma_y B_y+\sigma_z B_z$

Ted Shifrin

5:30 PM

@vesii Did you use the usual calculus techniques for graphing? It's going to depend on the value of $\alpha$.

Semiclassical

where the sigmas are Pauli matrices

Jake Rose

Also, in terms of the orgiinal Hamiltonian, does this really help? Because we still have to find the overall eigenvectors, not just of the second matrix once you split it?

Semiclassical

So this is effectively a spin-1/2 particle subject to magnetic field $\vec{B}$, where $\vec{B}$ is rotating

Sure it does. It's a change of basis.

Once you find the eigenvectors (in what amounts to the rotating frame) you use the change of basis to find the eigenvectors in the original basis.

Ted Shifrin

Hi, a @Balarka.

Jake Rose

Also what’s the motivation in choosing B to be the signature like that? Couldn’t you arbitrarily choose it to be anything?

Semiclassical

5:33 PM

as in, why the letter B?

Jake Rose

No as in, why can you let $\Delta =B cos\theta$ etc

Semiclassical

That's...what I did?

oh

because $\Delta^2+V^2=B^2$

and $\Delta, V$ are real

so this is just polar coordinates for $(\Delta, V)$

Jake Rose

Ahhhh

Semiclassical

(To bear out my point above: If $H'=U^{-1}H''U$ and $H''\psi = \lambda \psi$, then $H'(U^{-1}\psi) = U^{-1} H''\psi = \lambda (U^{-1}\psi)$. So $\psi$ is an eigenvector of $H''$ iff $U^{-1} \psi$ is an eigenvector of $H'$, in each case corresponding to eigenvalue $\lambda$.)

Jake Rose

That is clever

Semiclassical

5:37 PM

Again, physically the motivation is to turn the problem into a spin-1/2 particle in a (rotating) magnetic field

Ted Shifrin

@Semiclassic: You were right that the OP was right about the determinant (aside from a typo). If he'd written $\epsilon^{ijk}_{i'j'k'}$ instead of the product of $\epsilon$s, I wouldn't have hesitated.

Semiclassical

and if you pick a reference frame which is rotating along with the magnetic field, you just have a constant magnetic field

vesii

@TedShifrin Hi thanks, I did try to use the basic techniques and I understand that it will be depend on $\alpha$. Not sure how to get it though.

Leaky Nun

hi @Ted

Jake Rose

ohhh and because H only differs from H’ by an identity factor, the eigenvectors are the same

Semiclassical

5:38 PM

right. additive shifts don't matter

@TedShifrin tbh, the main reason I recognized it was because of old flirtations with Penrose graphical notation

c.f. en.wikipedia.org/wiki/Penrose_graphical_notation#/media/…

Leaky Nun

@Rithaniel the initial position is +0.1 at depth 36

Ted Shifrin

@vesil: You should find that there are critical points at $x=0$ and $x=\alpha^{1/5}$. The value of the function at $\alpha^{1/5}$ will tell you the answer. Obviously there's at least one real root (because the polynomial has degree $7$) and it's negative.

Hi @Leaky

Rithaniel

Yeah. Also, that's 9 games in a row that you have demonstrated that you are better at chess than me

Ted Shifrin

@Semiclassic: How interesting. In my entire life I've never seen that before.

@Rithaniel: Only 9?

Rithaniel

Yeah, but 9 consecutive ones

Semiclassical

5:41 PM

@JakeRose what mostly makes the problem tedious imo is that, while the eigenvectors are nice enough in the polar $(B,\theta)$ parametrization, they're pretty nasty in the Cartesian ($\Delta,V_0$) parametrization

and even worse in the original $(E_1,E_2,V_0,\omega)$ setting

Rithaniel

I didn't even think about what would happen if you got a rook onto the back row

Jake Rose

Mhmm,

Semiclassical

@JakeRose is this leading to Berry phase stuff?

I could see a textbook following that path

Jake Rose

what’s the motivation for the initial splitting of the Hamiltonian into the additive and the difference bits?

Nope, this is just the first problem in an example sheet

in terms of the unitary transform, is there any motivation for that too?

Semiclassical

just to make it traceless, really

traceless operators having fewer parameters

for the unitary transformation, the motivation is to get it down from a complex Hermitian matrix to a real symmetric one

Jake Rose

5:46 PM

Why is trace less beneficial?

Semiclassical

traceless means it has trace zero

and that's nice because the eigenvalues had better be $\pm \lambda$, since there's only two of them

also, in general, one has $(\sigma_x B_x+\sigma_y B_y+\sigma_z B_z)^2=(B_x^2+B_y^2+B_z)^2I_2$

Leaky Nun

@AlessandroCodenotti it's also his colourful language that makes his commentaries interesting

Semiclassical

whereas you originally had $H=\overline{E}I_2 +H' = \overline{E}I_2+\sigma_x B_x+\sigma_y B_y+\sigma_z B_z$

which doesn't square so nicely

Thorgott

traceless means at least one coefficient of the characteristic polynomial vanishes, which is nice

Semiclassical

right

so it's analogous to completing the square

Jake Rose

5:49 PM

So to find the eigenvectors we just have to apply the inverse of the transformation matrix on to the eigenvector of the H”?

Semiclassical

Right.

So you've got $H''=B\begin{pmatrix} \cos \theta & \sin\theta \\ \sin \theta & -\cos \theta\end{pmatrix}$

What are the eigenvalues of that?

Jake Rose

Why is it the inverse?

Semiclassical

because I wrote it as $H'=U^{-1} H'' U$ and not $H''=U^{-1} H' U$

(Note also that $U^\dagger U=I_2$ where $\dagger$ is Hermitian conjugate, to use the usual physics notation. So it's a unitary transformation as well as a similarity transformation.)

Alessandro Codenotti

Want to play another game? @Leaky

Jake Rose

@Semiclassical one last q

is there a quick way to see the eigenvectors for the polar matrix?

vesii

5:59 PM

@TedShifrin Thanks. I got $f(a^{1/5})=2 - \frac{5 a^{7/5}}{2}$. What does it say to us?

Semiclassical

There is, actually: Use the trigonometric double-angle identities on $\cos\theta$ and $\sin\theta$ (taking $\theta=2(\theta/2)$)

Leaky Nun

@AlessandroCodenotti sorry not now

Semiclassical

You'll need to play with that to make it work, though

Alessandro Codenotti

Ok

Ted Shifrin

Well, for example, if $\alpha$ is positive and big, you'll have $f(\alpha^{1/5})<0$, and so this tells you that you have $3$ real roots. If $\alpha$ is positive and small, you'll only have $1$ real root. If $\alpha$ is negative, $f(\alpha^{1/5})>0$ and you'll only have $1$ real root. You really want to sketch the graph (approximately).

Hi, demonic @Alessandro.

Alessandro Codenotti

6:02 PM

Hi @Ted

Jake Rose

@Semiclassical Wdym? Can you then see them by inspection?

Semiclassical

yep. though it's best to start by subtracting off the relevant eigenvalue and then figuring out which double-angle identity to use

(the reason it works out the way it does is because, just as you can rotate to make $B$-vector point in the $xy$-plane, you can rotate once more to align it with the $z$-axis. But rotations in spin space correspond to half-angles because it's half-integer spin. So there's an Euler angle story going on.)

Jake Rose

Mhm. I’ve got the eigenvalues but I’m struggling to find the egenvectors

Semiclassical

just to check, eigenvalues are what?

Jake Rose

0,1?

Semiclassical

6:08 PM

no

Jake Rose

One sec

Semiclassical

trace is zero

so the sum of the eigenvalues is...?

Jake Rose

Oops

topologicalmagician

Hello

Jake Rose

wrong way around. With debt and trace

$\lambda = 1, -1$?

topologicalmagician

6:10 PM

If each $X_i$ is a topological n $-manifold$ ,, how do I show that the index set of the disjoint union is countable?

Semiclassical

well, $\pm B$. but $B$ is pretty much irrelevant here

so yeah

Alessandro Codenotti

@topologicalmagician You need to assume that the union is countable (or do not require manifolds to be second countable)

Semiclassical

so, if you want eigenvectors, you need to find solutions for \begin{pmatrix} \cos\theta \pm 1 & \sin\theta\\ \sin\theta & \pm 1 -\cos\theta\end{pmatrix}

Jake Rose

Yeah sorry

Semiclassical

This is where double angle identities are handy: $\sin 2\phi=2\sin \phi\cos\phi$, $\cos 2\phi = 2\cos^2\phi -1=1-2\sin^2\phi$

So what I'd do at this point is consider $\pm 1$ separately and figure out which of those double-angle identities make life simplest :)

(If one knows how rotations are implemented in spin space, then this goes from being an algebraic trick to a sensible method. but that takes some getting used to.)

Alessandro Codenotti

6:14 PM

@topologicalmagician This is not a sensible sentence now

topologicalmagician

@AlessandroCodenotti The disjoint union is second countable... I can cover it by a countable bases so I can find a countable sub cover? Perhaps thats the way?

Alessandro Codenotti

Are you assuming the union to be manifold and you want to show that the index set is countable?

topologicalmagician

yes

Alessandro Codenotti

I can't read your mind and guess the problem you're trying to solve if you don't write it

Then your idea works. If the union were uncountable, the space wouldn't be second countable

Semiclassical

@JakeRose okay, I need to head out

topologicalmagician

6:17 PM

@AlessandroCodenotti I really apologize if I wasn't clear.

Semiclassical

if you've got access to textbooks, maybe take a look at Shankar

later

Jake Rose

Oh they do come out nice don’t they

topologicalmagician

6:28 PM

let $(A_j)$ be a collection of countable sets.
If $\bigcup_{j\in J}A_j$ is countable then $J$ is countable.

right?

Thorgott

If $M$ is a differentiable submanifold of $\mathbb{R}^n$, we say a function $f\colon M\rightarrow\mathbb{R}$ vanishes a.e. if $f\circ\psi$ vanishes a.e. for every local parametrization $\psi\colon U\rightarrow M\cap V$. If $W\subseteq\mathbb{R}^n$, what does it mean to say $f$ vanishes a.e. on $M\setminus W$? That $f\circ\psi$ vanishes a.e. for every local parametrization $\psi\colon U\rightarrow M\cap V$ whose image is contained in $M\setminus W$?

@topologicalmagician what if they're all empty

topologicalmagician

@Thorgott oops I forgot to mention that they must be non-empty

Thorgott

What if they're all equal?

MyWrathAcademia

Hi @Thorgott, glad your'e here. About that count of numbers divisible by $n$ between $a$ to $b$ problem, I want to clear up some misunderstandings:

How is $k > = a/n$ the smallest $k$ when for $a = 7$ $b = 14$ $n = 6$, $k >= a/n$ would mean that the smallest $k$ is $1$ which is not true. Isn't it more correct to say that $k >= a/n$ is the lower bound for $k$ and that the actual smallest $k$ can only be found by plugging a value for $k$ starting from (i.e. at or near) this lower bound into $nk >= a$? Like wise for $k <= b/n$.

Can you also confirm that $\lceil\frac bn\rceil + 1$ doesn't literally mean $b/n - \frac an + 1$ since I think $\lceil\frac bn\rceil$ and $\lceil \frac an \rceil$ are inequalities. I was trying to write a program that can count the number of numbers from $a$ to $b$ divisible by $n$ without using any for loop.

Am I right in thinking that the formula $\lceil \frac bn \rceil - \lceil \frac an\rceil + 1$ would still require me to loop over values of $k$ in the range $\frac bn$ to $\frac an$ and compare each value to the condition $nk >= a$ and the condition $nk <= b$ in order to find the smallest $nk >= a$ and the largest $nk <= b$ in the range $b/n$ to $\frac an$?

Formatting mistake, I meant to say "Can you also confirm that $\lceil\frac bn\rceil - \lceil \frac an \rceil + 1$ doesn't literally mean $\frac bn - \frac an + 1$"

Thorgott

In your example, $k=1$ yields $6$, which does not lie in the range from $7$ to $14$. Yes, $\lfloor\frac{b}{n}\rfloor-\lceil\frac{a}{n}\rceil+1$ is not the same as $\frac{b}{n}-\frac{a}{n}+1$, the former expression is always integer, the latter usually not (try computing some examples). Using a loop to count expressions in unnecessary, because we've already counted them. Computing $\lfloor\frac{b}{n}\rfloor$ is the same as rounding $\frac{b}{n}$ down, which I'm sure any computer can do for you.

topologicalmagician

6:45 PM

@Thorgott if they are empty then the index set is the empty set, right?

Thorgott

No?

topologicalmagician

@Thorgott if they are equal then $I$ is finite. If they're empty then what happens?

Thorgott

Here's what I mean: Let $I$ be an uncountable set and $A$ a countable set. Define $A_i=A$ for all $i\in I$. Then $(A_i)_{i\in I}$ is a collection of countable sets. Furthermore, $\bigcup_{i\in I}A_i=A$ is countable, but $I$ is uncountable.

MyWrathAcademia

@Thorgott I understand that $k = 1$ yields $6$. What I don't understand is for $k >= \frac an$ in the above example when $a = 7$ and $n = 6$ then the inequality $k >= 7/6$ (i.e. $k >= 1)$ means that $k$ is greater than or equal to $1$, but the minimum $k$ is $2$ for the example $a = 7$ $b = 14$ $n = 6$. Basically I'm asking why does the condition $k >= \frac an$ hold when $k = 1$ even though $k = 1$ is not the smallest $k$

Thorgott

It doesn't. $1\not\ge7/6$

MyWrathAcademia

6:58 PM

Oh crap, sorry @Thorgott I'm used to rounding down in programming that I forgot you don't round down in mathematics

Thorgott

Finding the smallest natural number $k\ge\frac{a}{n}$ is the same as rounding $\frac{a}{n}$ up, so that's what you want to do.

MyWrathAcademia

Yes $ 1 \not\ge1.166666667$

So your'e saying that for example you round $1.166666667$ to $2$?

Thorgott

Yes. And $k=2$ is the right answer.

MyWrathAcademia

If you round up for $k \ge \frac an$ does that mean that you round down for $k \le \frac bn$?

I'm guessing thats why you round down for $k \ge \frac an$?

Thorgott

Yes, you round $\frac{a}{n}$ up and $\frac{b}{n}$ down. That's exactly what $\lceil\frac{a}{n}\rceil$ and $\lfloor\frac{b}{n}\rfloor$ mean, respectively.

MyWrathAcademia

7:07 PM

Wow that makes so much sense. What are those type of brackets called?

Thorgott

Floor and ceiling functions

In mathematics and computer science, the floor function is the function that takes as input a real number x {\displaystyle x} and gives as output the greatest integer less than or equal to x {\displaystyle x} , denoted floor ⁡ ( x ) = ⌊ x ⌋ {\displaystyle \operatorname {floor} (x)=\lfloor x\rfloor } . Similarly, the ceiling function maps x {\displaystyle x} to the...

MyWrathAcademia

@Thank you, the celing function $\lceil \frac an \rceil$ and floor function $\lfloor \frac bn\rfloor$ are quite intuitive

@Thorgott now I understand $k \ge \frac an$ and $k \le \frac bn$ perfectly and that the former's celing function rounds up and the latter's floor function rounds down that answers my second question about what $\lceil \frac an\rceil - \lfloor \frac bn\rfloor + 1$ means

topologicalmagician

7:37 PM

If $(X_i)_{j \in J}$ is a collection of $n-manifolds$ and their disjoint union is an n-manifold then $J$ is countable.

I\m still quite stuck on this

Thorgott

Demonstrating failure of second-countability in case of $J$ uncountable sounds like a good approach

topologicalmagician

7:55 PM

@Thorgott. I can define a map $f$ $:$ $I \rightarrow X_i^*$ (where $X_i^*$ is the image of the canonical injection), the map is injective so $I$ is countable?

MyWrathAcademia

@Thorgott thank you. Your'e right that using a loop to count expressions is unnecessary, because we've already counted them. I have computed the expression $\lfloor \frac bn \rfloor - \lceil \frac bn \rceil + 1$ that allows me to observe that this expression has already counted the number of natural numbers divisible by $n$ between $a$ and $b$ and also that this expression is always integer.

@Thorgott your responses have been immensely helpful, thanks a lot! Is there a way to bookmark a conversation or series of comments in this chat room so that I could quickly refer to it?

Alessandro Codenotti

@topologicalmagician I can define an injective map $\Bbb N\to\Bbb R$, does that mean that $\Bbb R$ is countable?

topologicalmagician

@AlessandroCodenotti images of countable sets are countable

$X_i^*$ is countable

Alessandro Codenotti

No, the image of $f$ is countable

Think about my $\Bbb N$ and $\Bbb R$ example

MyWrathAcademia

Does any one know how to view bookmarks that were created for conversions in a chat room?

Alessandro Codenotti

8:05 PM

Click on info on the right, then "conversations"

topologicalmagician

@AlessandroCodenotti If $f:A \rightarrow B$ is injective and $B$ is countable then so is $A$.

Thorgott

@MyWrath Nice. I'm glad you understood. As for the bookmarking, you can hover over the message and click the little arrow on the left, then click on permalink, which gives you a permalink to that message in the transcript.

Alessandro Codenotti

Sure but why would $X_i^\ast$ be countable?

Just show that the space is not second countable

topologicalmagician

@AlessandroCodenotti $X_i$ is second countable, it has a countable cover.

Alessandro Codenotti

$\Bbb R$ is second countable, it has a countable cover and it is also very uncountable

Thorgott

8:09 PM

If you don't require topological manifolds to be second-countable, even an uncountable disjoint union would be a topological manifold, no?

MyWrathAcademia

@Thorgott your'e a future professor in the making! Unless the lure of industry salaries lures you away from that path. I'll be back another time, it's late here and I have to go.

Alessandro Codenotti

@Thorgott definitely

Thorgott

Meaning a proof necessarily has to demonstrate the failure of secound-countability

Alessandro Codenotti

Should be clear that you can't mess being $T_2$ or locally Euclidean with a disjoint union

Thorgott

Isn't paracompactness also required sometimes?

Though that should also behave well under disjoint union

Lucas Henrique

8:20 PM

Hi chat

Thorgott

Hi Lucas

Alessandro Codenotti

@Thorgott that's trivial I think

topologicalmagician

Whats wrong with the following proof: Suppose $X$ is second countable, cover it by a basis $\mathbb{B}$. Since $X$ is lindelof, there exists a countable sub cover $\mathbb{B'}\subseteq \mathbb{B}$ such that $\mathbb{B'}$ is a countable sub cover. Hence $X$ $\subseteq$ $\bigcup_{B\in \mathbb{B'}}B$ since the countable union of countable sets is countable, it follows that $X$ is countable

Alessandro Codenotti

Why is $X$ Lindelof?

Also the basis has countably many elements, but they are not countable themselves. Think about the basis of $\Bbb R$ made of open intervals with rational endpoints

topologicalmagician

every second countable space is lindelof, thats a theorem in the notes I have.

MyWrathAcademia

8:25 PM

@Thorgott after permalinking multiple messages how do you locate those messages using the permalinks that were created? I expected a kind of bookmark feature that adds permalinks of messages somewhere for me to use to quickly refer to messages that I permalinked

topologicalmagician

ohhhhhhhhhh

dammit

Thorgott

You can bookmark them in your browser

topologicalmagician

@AlessandroCodenotti @Thorgott thanks guys so sos os sososoooooo much

MyWrathAcademia

Oh you mean a permalink is basically for you to copy and paste it into your browser's bookmarks?

Thorgott

Yes, it's just a link to the message in the archived transcript of this chat

MyWrathAcademia

8:27 PM

Apologies, I thought it would work like saving/ staring a post on any stack exchange website which adds it in a list somewhere in your account for you to view

Thorgott

Don't know if that's possible

MyWrathAcademia

Yeah, I'll have to find out. Great chat today @Thorgott. Signing out....

Thorgott

Bye

MyWrathAcademia

Yeah, see ya.

Lucas Henrique

@MyWrathAcademia in fact, it does. But it's not intended for that.

Dattier

8:57 PM

Hello, Do you know Schoof+ ?

A result which give the cardinal of an Elliptic curve in Wieirtrass form $y^2=x^3+bx+a$ when $b\neq 0$ and on the field $\mathbb Z_p$ where $p$ is a prime number ?

Bob

What do people think about the schaum series? I like it because there books are cheap.

Ted Shifrin

10:14 PM

@Bob: Very old-fashioned in some cases and a bit too formula-oriented for my taste. But I don't know many of the books.

Alessandro Codenotti

Hi @Ted

Ted Shifrin

rehi, demonic @Alessandro

Alessandro Codenotti

10:45 PM

Not a lot going on here tonight

topologicalmagician

10:56 PM

Can we always extend a basis for a non-empty subspace to a basis for the the space?

note: I mean topological basis

Alessandro Codenotti

Open sets of a subspace are not necessarily open in the big space

topologicalmagician

yes I know

Alessandro Codenotti

They are the intersection of an open set of the big space with the subspace though, and those can be extended to a basis

topologicalmagician

aha

Alessandro Codenotti

Note that there's nothing interesting going on: any family of open sets can be extended to a basis of the topology (take the whole topology for example)

Adam

11:45 PM

what do you call a property that holds regardless of the sign of all operands?

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$Mathematics$ Mathematics