@TedShifrin So I gave $S^n$ a cell structure which makes it into a CW complex using the definition of a CW Complex as a Hausdorff space with a cell decomposition (that satisfies some other conditions), but using that cell structure I can't build $S^n$ up inductively (which is the other defn of a CW Complex) since it's $n-1$ skeleton is empty
I see so many people working on start ups and getting really good money with alot less effort tahn doing pure math and sometimes i wonder if i regret spending hours lear ing math. I know i like it and enjoy it its just doesnt feel rewarding except in your head :p
I loved (and still love) teaching and that was my main motivator. Publishing some really cool papers was a (necessary) plus for me. Most people who go into research math are very driven by research.
I'm driven about not having anything else to do lmao
@LeakyNun the picture of $\Bbb Z_p$ is that you have $p$ circles (representing the reduction mod $\Bbb Z/p$), each with $p$ circles inside, etc; the actual elements of the p-adics correspond to your choice of infinite nesting sequence of circles, but it's useful to not think about that and instead just think of the nesting circles themselves
Then the picture I have of $\Bbb Q_p$ is a zooming out of this process of zooming in: you progressively add $p$ copies of the old bin of circles you had, and then draw a big circle around it
If the Cantor set looks like the number of ways to go down an infinite rooted tree (with "root" at y=0), then this looks like the number of ways to go down an infinite tree where the root is at unspecified nonpositive integer height
@Manolis: Here's a little story for you. I took 4 courses from Michael Artin (who's known in this room for his algebra textbook ... as well as his famous name). He commented to me once that he went into algebraic geometry because it was the hardest thing for him (by comparison with analysis, topology, etc.). Mathematicians seek challenges. That's what makes them thrive.
k, so if linear transformation $F$ has only eigenvalue $\lambda = 0$ and if $F(V) = \lambda V$ and $W$ is any lineary independent vector with $V$ and $F(W)=0$, thus i can conclude that $m(F) = 0$ because if we take a vector it can be either lineary dependent or independent with $V$, if its independent then $F(U)=0$ by assumption, otherwise $U = \alpha V$ thus $F(\alpha V) = \alpha F(V) = \alpha \lambda V = 0$ and so $m(F) = 0$, am i right?
well im just asking is valid conclusion that $m(F)=0$ if following conditions are satisfied: -$F$ is linear transformation in $\mathbb{R}^2$ -$F$ has only one eigenvalue $\lambda = 0$ -if $W$ is lineary independent with eigenvector $V$, then $F(W)=0$
im provin this little stuff: Let $\lambda$ be double zero of $\chi_A$, where $A \neq \lambda \mathbb{I}$, i. e. $\chi_A (x) = (x - \lambda)^2$. Then there exists two lineary independent vectors $U$ and $W$, such that $AU = \lambda U$ and $AW = U + \lambda W$
@Ted I bet, I was also interested in how hard it was. The questions were just like "Example: Divide 363 by 3", now design a question that is at the same difficulty level. The idea was (at least obviously in this case) that all the digits are divisible by $3$ so there's no carrying if you use the "bus stop method" (this is supposed to be for school kids)
of course, the examples got harder as you went on (what techniques are employed by the question and can you engineer a question using numbers that use only the techniques employed by the example)
@Ted that's cool, I actually don't plan on teaching primary or secondary education, but the ideas/experience I'll get (if I get it!) are useful nonetheless
@Ted yeah it definitely will; the position I'm going for is actually for GCSE standard (upper secondary school) so it's mostly teaching basic stuff, I think factorising and solving quadratics is on the more difficult side
@TedShifrin ok let me elaborate. I may have been giving uncessary riguor. Without clearly specifying the problem. I was doing a double integral $\int \int \frac{dxdz}{x^2+a^2+z^2}^(3/2)$ where both limits go to 0 to a. After doing the first integral i got that.
