@AliCaglayan and quadratic forms I would like this semester to build very good understanding in commutative algebra because I want during the summer to study Category theory through Allufi chapter 0 and also algebraic geoemtry.
I am following Michael Atiyah commutative algebra. In one of the proposition he proved $nil(R) = \cap P$ where P is prime ideal. One can easily see the inclusion $nil(R) \subset P$. However, for the other inclusion we used Zorn lemma. I understand the proof, but I don't really get it conceptually...
A Steiner Triple System is a set $\mathcal{S}$ of $v \geq 3$ elements together with a set $\mathcal{B}$ of $3$-subsets (triples) of $\mathcal{S}$ such that every $2$-subset of $\mathcal{S}$ occurs in exactly one triple of $\mathcal{B}$. As an example, the following forms a Steiner Triple System o...
Can you give me an example where Borel Sigma algebra is properly contained in sigma algebra ? i know that Borel sigma algebra is the intersection of all sigma algebras of subsets of R that contains the open sets
@MikeMiller How did you immediately see that the one-point compactifications of (open) moebius strip x $\Bbb R$ have different $\pi_1$? If I am not wrong nbhd of the point at infinity looks like a cone over the Klein bottle (whereas for solid torus it's just cover over the torus), right?
I was dealing with a problem as follows, "Let f be a function of a real variable such that it satisfies f(r+s)=f(r)+f(s) for all r,s . Let m and n be integers, then f(m/n) is ?" The answer is (m/n)f(1) and what I assume the solution is that using f(x+y)-f(x)=f(y) we get that the derivative is constant and the function is of the form f(x)=kx for some k. But doing so would require the function to be differentiable wouldn’t it? How do I go ahead with proving that?
@DHMO I think you can find a very nice explanation that Cantor set has measure 0 ,uncountable ,closed in Royden Real Analysis , it has nice lucid explanations there
A pipe can empty a tank in 40 Minutes , A second Pipe with diameter twice as much as that of the first pipe is also attached to the tank to empty it .The two pipes can together empty the tank is
I'm trying induction on the rationals. My argument is, given a statement 'P' true on an interval '[a,b]' and two rationals 'x' and 'y' in 'I' . Would showing that 'x' and 'y' satisfying the statement implies that 'x+y' satisfies the statement show the statement to be valid on [a,2b] ? Or am I missing a few values in [b,2b]?
when 2 points of an arc form an angel at the center of the circle and at the circumference, the angle at the circumference is half the angle at the center
Well, that's what a punctured neighborhood should look like. Whereas a punctured neighborhood of the one-point compactification of the solid torus is $T \times (0, \infty)$, which are not topologically the same so they're not homeomorphic
@BalarkaSen Those two neighborhoods are not homeomorphic, but it's not obvious that there's not some neighborhood in the latter that is homeomorphic to $K \times (0,\infty)$.
Have to develop a fundamental group at infinity, which is irritating. You could also take local homology at the basepoints, which would show they're not the same. But all of this is kind of more work than I think this should need.
@DHMO When I tried generalizing that to the rationals I'm able to show that $f(x)=xf(1)$ for x=0 & $x= \lim_{n \to \infty} 2^{-n}$ and that $f(x)=xf(1)$ being true over the interval [0, $\lim_{n \to \infty} 2^{-n}$] implies that its true for all the rationals, but I'm stuck at proving that it is.
(I really like the idea of distinguishing spaces by using proper homotopy invariants, or finding local invariants of spaces by considering the homotopy-type-at-infinity of the complement of the point.)
I honestly don't know how to compute fundamental group at infinity effectively. Sometimes it's just obvious that they're weird (eg, Whitehead manifold, because the end is itself very weird), but I have no idea how to compute it in general.