@Ted yeah I really would like to understand the way Cartan thought better. I also need to get better at reading french. My deciphering of it from my knowledge of other romance languages makes it slow going.
Yeah, for someone who's knowledge of french is pretty basic (I should've payed more attention in school :( ) his french isn't super hard, but the way of thinking is very foreign to the modern mindset
Say, if $T$ is compact, it's not enough to have some polynomial $p$ so that $p(T)=0$ for one to conclude that $T$ is finite rank, right? We need self-adjoint too, I think. Also vice versa, we need compactness and self-adjoint and such a polynomial in order to conclude that, am I correct?
Great, thanks. i dont know why i made simple thing difficult :P
@EricSilva in general , if we found a sequence in $\overline{B(0,1)}$ that does converge , but to something outside the unit ball, it was enough for showing the unit ball is not compact, am i right? be we showed $l_2$ is not compact too.
finding somethign that converges to something outside $A$ shows that $A$ isn't closed, which is stronger than saying $A$ isn't compact, so yeah it does prove it but it proves something more too
@EricSilva im writing down now the proof that $l_2$ is not locally compact , and im not sure about one thing. we showed that each closed ball containing $x$ is not compact, but we need to show each $V$ nhbd of $x$ , $\overline V$ is not compact. we have that the closed ball is contained in $\overline V$ but what this implies $\overline V$ is not comapct?
Two dimensional random variable (X,Y) has a uniform distribution on the unit square [0,1]x[0,1]. Find the probability distribution of the random variable Z that is the distance of the point (X,Y) from the edge of the square.
Yeah, you take an open cover of the small set and tack on its complement, which covers the larger one, then a finite subcover (minus the complement if necessary) should do it
With a given $h(x)$ we want to solve $$xu_y-yu_x=u \\ u(x,0)=h(x)$$
I have solved it using two ways.
$$$$
First way:
For $x\neq 0$ we get $u_y-\frac{y}{x}u_x=\frac{u}{x}$.
We have that $$\frac{du}{ds}=\frac{du}{dx}\cdot \frac{dx}{ds}+\frac{du}{dy}\cdot \frac{dy}{ds}$$
Therefore, we hav...
I guess I don't care what professionals do in advanced courses and in research, as long as they're clear to their audience. But I think those of us saying/teaching things to beginners need to be clear and careful.
@TedShifrin I didn't understand the context here. Of course all notation is clarified in context; no topologist wants to confuse. I meant in that setting, not in a first course.
I currently study mathematics by myself and I'm going to create study group I'm going to go through these books : Introduction to formal logic by peter smith Naive set theory calculus spivak How to prove it
With a given $h(x)$ we want to solve $$xu_y-yu_x=u \\ u(x,0)=h(x)$$
I have solved it using two ways.
$$$$
First way:
For $x\neq 0$ we get $u_y-\frac{y}{x}u_x=\frac{u}{x}$.
We have that $$\frac{du}{ds}=\frac{du}{dx}\cdot \frac{dx}{ds}+\frac{du}{dy}\cdot \frac{dy}{ds}$$
Therefore, we hav...
