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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)
 
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
 
 
1 hour later…
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 $A\cap B=\{0\}$?
 
1:43 AM
Yes, same proof as for vector spaces
 
Alright, easy enough
 
 
1 hour later…
3:10 AM
When do we use Differential Under Integral Sign method for integrating?
 
 
1 hour later…
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"
 
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
 
5:59 AM
@bjorn93 What's the context?
 
 
2 hours later…
7:42 AM
@AlessandroCodenotti want a game?
 
Not now, sorry
 
9:25 AM
@adeshmishra then i agree with you if that's the case
 
10:05 AM
@StupidQuestionsInc What? I'm missing the context.
 
 
2 hours later…
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
 
12:18 PM
Grandmaster Finegold is very entertaining to watch
 
@Rithaniel yes he is
 
12:54 PM
I find him more annoying than entertaining, I don't like his humour
 
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
 
@Rithaniel wnana play?
@Rithaniel he tried and he failed miserably (at least according to A. Codenotti); the lesson is: never try
 
In class currently
 
uni?
 
Yeah, we're doing stuff with commutative diagrams, so my attention is primarily directed towards the professor
 
1:48 PM
@StupidQuestionsInc Thank you for agreeing with me. Please say something else on that topic.
 
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 $F \hookrightarrow F[x]/(p)$
 
@LucasHenrique so far so good
 
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. $F \subset E'$?
 
@LucasHenrique yes
 
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
 
2:50 PM
@LeakyNun isn't set theory like extremely relevant to algebra?
without Zorn's lemma, not all rings have maximal ideals
 
let's see
the only point where it goes wrong is if $F \cap E \ne \varnothing$
 
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
 
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 $E'$
 
@LeakyNun I'm not sure if this is possible
 
so you've evaded the set-theoretic issues
the set-theoretic issues is that their elements can happen to be equal as sets
 
3:01 PM
I mean: idk if there's a way to have $F\cap E \neq \emptyset$ unless $F \subset E$
 
let's just look at an example
let's look at the field F2 = {0,1}
 
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
 
that's ugly
 
that's why we don't care about set-theoretic issues
 
3:02 PM
but alright, you win. :p
 
the point is that the elements can be any set
we only care about things up to isomorphisms
we care about injective homomorphisms
 
@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.
 
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 $E$
 
@LeakyNun what does this mean?
 
as in, represent them by different sets
@loch could you phrase this better than I?
 
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
 
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?
 
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.
 
good enough
thanks @Leaky and @Thorgott
 
@Jacksoja wanna play?
 
4:21 PM
@Rithaniel?
if I go on I might actually need to create a chess chatroom lmao
 
In a little bit, sure. Lunch and decompressing after coming home
 
send me a challenge @Leaky
 
@AlessandroCodenotti sure
 
4:37 PM
I went to the analysis board and I can't see the messages anymore... @Leaky
 
@AlessandroCodenotti wow I didn't even see that move winning the bishop
 
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
 
completely didn't notice
 
Wow SF even says to give up the queen for two minor pieces in that position
 
lol
 
4:52 PM
@Rithaniel tell me when you're ready
 
Will do
 
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)
 
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?
 
5:08 PM
so, a generic 2-by-2 Hermitian matrix?
 
Yep
I haven’t solved a matrix in a while and I just can’t believe I can’t do it
 
Have you found the eigenvalues already?
 
They came out rather ugly
I was imagining you could just guess the eigenvectors for such a simple matrix
 
No way around that, to some extent.
 
That tends to be the way we always do it in part II
 
5:11 PM
So what did you get for the eigenvalues?
 
One way to make life a bit easier is to write this matrix as $H=\overline{E}I_2+H'$ where $\overline{E}=(E_1+E_2)/2$
 
Mhmm. How does that make things easier?
i feel like my brain has died.
 
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
 
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?
 
5:20 PM
Complex or real?
 
real
 
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
 
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.
 
Mhmm. Seems like quite a tedious work around though.
Although that trick really is cute
 
My usual approach for finding the eigenvectors of the final matrix there is to let $B:=\sqrt{\Delta^2+V_0^2}$ and write $\Delta=B \cos \theta,V_0=B\sin \theta$
 
5:26 PM
Once you know the eigenvalues you can just do row reduction
 
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
 
Where did you learn this type of method? @Semiclassical
 
Also, cute thing: $\begin{pmatrix} \Delta & V_0 \\ V_0 & -\Delta \end{pmatrix}^2=B^2 I_2$
@JakeRose physics
 
@Semiclassical I was hoping more so for a textbook ;) I’m currently doing physics myself
Time dependent perturbation theory to be exact
 
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 $H'=\sigma_x B_x+\sigma_y B_y+\sigma_z B_z$
 
5:30 PM
@vesii Did you use the usual calculus techniques for graphing? It's going to depend on the value of $\alpha$.
 
where the sigmas are Pauli matrices
 
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?
 
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.
 
Hi, a @Balarka.
 
Also what’s the motivation in choosing B to be the signature like that? Couldn’t you arbitrarily choose it to be anything?
 
5:33 PM
as in, why the letter B?
 
No as in, why can you let $\Delta =B cos\theta$ etc
 
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 $(\Delta, V)$
 
Ahhhh
 
(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 $\lambda$.)
 
That is clever
 
5:37 PM
Again, physically the motivation is to turn the problem into a spin-1/2 particle in a (rotating) magnetic field
 
@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.
 
and if you pick a reference frame which is rotating along with the magnetic field, you just have a constant magnetic field
 
@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.
 
hi @Ted
 
ohhh and because H only differs from H’ by an identity factor, the eigenvectors are the same
 
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
 
@Rithaniel the initial position is +0.1 at depth 36
 
@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
 
Yeah. Also, that's 9 games in a row that you have demonstrated that you are better at chess than me
 
@Semiclassic: How interesting. In my entire life I've never seen that before.
@Rithaniel: Only 9?
 
Yeah, but 9 consecutive ones
 
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
 
I didn't even think about what would happen if you got a rook onto the back row
 
Mhmm,
 
@JakeRose is this leading to Berry phase stuff?
I could see a textbook following that path
 
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?
 
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
 
5:46 PM
Why is trace less beneficial?
 
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 $(\sigma_x B_x+\sigma_y B_y+\sigma_z B_z)^2=(B_x^2+B_y^2+B_z)^2I_2$
 
@AlessandroCodenotti it's also his colourful language that makes his commentaries interesting
 
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
 
traceless means at least one coefficient of the characteristic polynomial vanishes, which is nice
 
right
so it's analogous to completing the square
 
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”?
 
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?
 
Why is it the inverse?
 
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.)
 
Want to play another game? @Leaky
 
@Semiclassical one last q
is there a quick way to see the eigenvectors for the polar matrix?
 
5:59 PM
@TedShifrin Thanks. I got $f(a^{1/5})=2 - \frac{5 a^{7/5}}{2}$. What does it say to us?
 
There is, actually: Use the trigonometric double-angle identities on $\cos\theta$ and $\sin\theta$ (taking $\theta=2(\theta/2)$)
 
@AlessandroCodenotti sorry not now
 
You'll need to play with that to make it work, though
 
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.
 
6:02 PM
Hi @Ted
 
@Semiclassical Wdym? Can you then see them by inspection?
 
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.)
 
Mhm. I’ve got the eigenvalues but I’m struggling to find the egenvectors
 
just to check, eigenvalues are what?
 
0,1?
 
6:08 PM
no
 
One sec
 
trace is zero
so the sum of the eigenvalues is...?
 
Oops
 
wrong way around. With debt and trace
$\lambda = 1, -1$?
 
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?
 
well, $\pm B$. but $B$ is pretty much irrelevant here
so yeah
 
@topologicalmagician You need to assume that the union is countable (or do not require manifolds to be second countable)
 
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}
 
Yeah sorry
 
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.)
 
6:14 PM
@topologicalmagician This is not a sensible sentence now
 
@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?
 
Are you assuming the union to be manifold and you want to show that the index set is countable?
 
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
 
@JakeRose okay, I need to head out
 
6:17 PM
@AlessandroCodenotti I really apologize if I wasn't clear.
 
if you've got access to textbooks, maybe take a look at Shankar
later
 
Oh they do come out nice don’t they
 
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?
 
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
 
@Thorgott oops I forgot to mention that they must be non-empty
 
What if they're all equal?
 
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 $\frac bn - \frac an + 1$"
 
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.
 
6:45 PM
@Thorgott if they are empty then the index set is the empty set, right?
 
No?
 
@Thorgott if they are equal then $I$ is finite. If they're empty then what happens?
 
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.
 
@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 $k$
 
It doesn't. $1\not\ge7/6$
 
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
 
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.
 
Yes $ 1 \not\ge1.166666667$
So your'e saying that for example you round $1.166666667$ to $2$?
 
Yes. And $k=2$ is the right answer.
 
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 $k \ge \frac an$?
 
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.
 
7:07 PM
Wow that makes so much sense. What are those type of brackets called?
 
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...
 
@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
 
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
 
Demonstrating failure of second-countability in case of $J$ uncountable sounds like a good approach
 
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?
 
@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?
 
@topologicalmagician I can define an injective map $\Bbb N\to\Bbb R$, does that mean that $\Bbb R$ is countable?
 
@AlessandroCodenotti images of countable sets are countable
$X_i^*$ is countable
 
No, the image of $f$ is countable
Think about my $\Bbb N$ and $\Bbb R$ example
 
Does any one know how to view bookmarks that were created for conversions in a chat room?
 
8:05 PM
Click on info on the right, then "conversations"
 
@AlessandroCodenotti If $f:A \rightarrow B$ is injective and $B$ is countable then so is $A$.
 
@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.
 
Sure but why would $X_i^\ast$ be countable?
Just show that the space is not second countable
 
@AlessandroCodenotti $X_i$ is second countable, it has a countable cover.
 
$\Bbb R$ is second countable, it has a countable cover and it is also very uncountable
 
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?
 
@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.
 
@Thorgott definitely
 
Meaning a proof necessarily has to demonstrate the failure of secound-countability
 
Should be clear that you can't mess being $T_2$ or locally Euclidean with a disjoint union
 
Isn't paracompactness also required sometimes?
Though that should also behave well under disjoint union
 
8:20 PM
Hi chat
 
Hi Lucas
 
@Thorgott that's trivial I think
 
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
 
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
 
every second countable space is lindelof, thats a theorem in the notes I have.
 
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
 
ohhhhhhhhhh
dammit
 
You can bookmark them in your browser
 
@AlessandroCodenotti @Thorgott thanks guys so sos os sososoooooo much
 
Oh you mean a permalink is basically for you to copy and paste it into your browser's bookmarks?
 
Yes, it's just a link to the message in the archived transcript of this chat
 
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
 
Don't know if that's possible
 
Yeah, I'll have to find out. Great chat today @Thorgott. Signing out....
 
Bye
 
Yeah, see ya.
 
@MyWrathAcademia in fact, it does. But it's not intended for that.
 
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.
 
 
1 hour later…
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.
 
rehi, demonic @Alessandro
 
10:45 PM
Not a lot going on here tonight
 
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
 
Open sets of a subspace are not necessarily open in the big space
 
yes I know
 
They are the intersection of an open set of the big space with the subspace though, and those can be extended to a basis
 
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)
 
11:45 PM
what do you call a property that holds regardless of the sign of all operands?
 

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