Hey has someone an idea how ti find a $C$ such that $C||a+b||<=||\lambda/x+(1-\lambda)b/y||$ with $x,y>0$ and $\lambda \in [0,1]$? Where $a,b \in \mathbb{R}^n$
And the norm is the 2-s norm. Any help highly appreciated!
okay so i had to cut a wire in to two parts one for a circle and a triangle, i created an expression for the sum of both these shape's areas, and now told to find the minimum value by differentiating, but when i do this the only solution for x is 0 so i think i have gone about this all wrong =/
Hey guys, it is known that every neighbourhood is an open set in a topology. Does the converse hold in general, or it is possible to find a topology such that an open set V of the topology does not contain any open set containing a point x, hence V failed to be a neighbourhood of x?
@Secret: Although I agree that "neighborhood" means "open set," apparently not everyone agrees with that. But remember that no matter what topology you take on $X$, for any $x$, the entire space $X$ is always an open set containing $x$. :)
The following is my attempt at understanding why open sets are so named (without referring to open intervals in metric space). Given a topology $\tau$, a set is defined to be open if it does not contain all of its limit points, that is, there exists a net in the set such that it converges to a point not in the set. This thus generalise the intuition that open sets are missing points that sort of "bound" it
Given V vectorial space, be $W_1,W_2$ subspaces of V, give the basis and equations of $W_1$ intersection (don't remember the symbol) $W_2$ and $W_1+W_2$
I am given the equations of $W_1$ and basis of $W_2$
@Maks: So I told you how to give equations for $W_2$ (that was the rows stuff). Then you have equations for everything and you just find the nullspace of that big matrix to find $W_1\cap W_2$.
For $W_1+W_2$, it's probably easiest to find a basis for $W_1$ (standard algorithm), and then use the vectors and the vectors for $W_2$ as columns and find constraint equations (when is a vector in the column space of that giant matrix?).
OK, let's go slow, I find the equations by creating a matrix with vectors as rows and equal it to (x,y,z...) And then I reduced it the null rows are the equations
In mathematics (especially algebraic topology and abstract algebra), homology (in part from Greek ὁμός homos "identical") is a general way of associating a sequence of algebraic objects such as abelian groups or modules to other mathematical objects such as topological spaces. Homology groups were originally defined in algebraic topology. However, similar constructions are available in a wide variety of other contexts, such as groups, Lie algebras, Galois theory, and algebraic geometry.
The original motivation for defining homology groups was the observation that two shapes can be distinguished...
@TedShifrin that one is a neither set, since its complement is not open and itself is not labelled as open in the usual topology (where all open sets are open intervals). Hmm, I have no idea if open sets can be described by referencing to only limit points as e.g. an open interval in the reals clearly contains some limit points
One of my major intuition problem I am facing is I don't really understand why continuity of a function f in a topological space is defined to be "the image f(U) of an open set U is open. For example:
Is it true that the only continuous function that map from open sets to closed set and vise versa are the constant maps as shown in that MSE? NB I am learning point set topology, I don't think I really understand what a pullback is
I am also thinking about it! ok, pulls back is nothing but pre-image of $f$ , that is if we take open set in the co-domain then $f^{-1}$ must map it open set...
Why I cannot pullback from an open set to a closed set even if f is not injective. As far I know, I struggle to get a counterexample of this but I am not sure how to prove it is impossible for any topology (housedoff or not)
All the proofs I had found so far that continuity implies preimage of open set is open make use of metric arguments. But what about topological space where a metric is not defined?
@YashFarooqui Yeah, the point is that, volume is alternating. Volume of parallelogram spanned by vectors v and w is minus the volume spanned by w and v.
Ah, that's a good question. The point is signed volume carries more geometric data than volume alone. Eg, magnitude of cross product of two vectors v x w in R^2 is precisely the signed volume of the parallelogram spanned by them. And |v x w| = - |w x v|. That's sort of the thing which you are trying to generalize.
On the other hand, in the continuous picture, note that differential forms (continuous generalizations of alternating k-forms) can be integrated, and you want it to satisfy nice properties like $\int_0^1 fdx = -\int_1^0 fdx$. So there's always the story of "oriented volume" hovering around.
okay. That isn't completely satisfying because the cross product itself seemed to be pulled out of nowhere but the latter part seems to suffice for now
I guess I'll work through a few more exercises and maybe move on to the definition of differential forms (am working through Spivak) to understand
@YashFarooqui Suppose I take the usual topology in the reals thus all open intervals are open sets say (1,6) and (4,9). By referencing to these open intervals, 2,3,4,5 etc. and 5,6,7,8 etc. are close to each other. Now since the union of any open sets is open, by taking the union I get the open interval (1,9). Now 2,9 are close to each other. So if I have a map that maps (1,3) U (5,9) to (1,9), I have a weird scenario where 2 and 9 are close to each other and far away at the same time?
No problem, you do raise interesting questions. I guess volume with respect to some "right-hand rule"/orientation has become so natural to me that I can't give you an easy intuitive picture :)
But it's certainly more interesting than just raw unsigned volume, I can tell you that.