I think the main problem is that I just can't imagine an example of a convex, pointed cone that is not closed. Can someone give me some examples?
I am having a really hard time in a course I am not sure why I am taking. Most of my background is in number theory and probability so I really need help with this linear algebra stuff.
Anyway, these are the problems.
I'm not sure why it's so small but here's a link.
Anyway, this is what I have so far:
a) This is almost tautological. If this set is the empty set for all <v,w> in gamma, for all v in V there must be a w in gamma such that <v,w> is less than or equal to zero. In this case, either <v,w> = {0}, or if there are such points where <v,w> < 0, there must also be points where <v,w> = 0 because cones are continuous.
In the case of euclidean spaces, <v,w>=|v||w|cos(-), so anywhere where (-)=pi/2 means that <v,w>=0. If there exists (-) > pi/2, there must also exist some w for which (-) = pi/2.
b) In this case there must be some cases where for v, all <v,w> > 0.
In a 2D situation, there must exist a point for every v where the (-) of w = pi/2. This is only possible if the full angle of the cone is pi, but at least one edge cannot be included because then the intersection of gamma and negative gamma > {0}, and the cone would not be pointed. In higher dimensions this is more complicated.
For a cone to be convex, then it cannot take up more than one half of the vector space in one dimension (all of that dimension must either be negative or positive). In such a case as well, the cone must take up less than half of the entire vector space for gamma to only intersect gamma at zero. Cases such as a book-shaped cone are also not closed, convex cones because gamma and negative gamma intersect at only one point.
So, for example, [1,1] would need a [1,-1] vector as its w for <v,w> = 0. For every single v vector, we would require a transposed w vector ([1,3] to [3, -1], [5,4] [4,-5], I do not quite remember what this function was called), which is impossible unless the cone takes up half of the vector space. If the cone takes up half of the vector space, however, then gamma and -gamma would have more of an intersection than simply {0} if the cone is closed. It may turn out that the cone only approaches taking up half of the vector space, in which case the cone is not closed.
This is true for all dimensions. (I'm not sure how to prove this)
c) If gamma is pointed, there exists a hyperplane H such that H intersect gamma = {0}.
In one dimension for the cone to be pointed and closed, gamma must either be strictly positive or negative, and the cone cannot take up one half or more of the vector space, in which case the hyperplane can lie tangent to {0}. The hyperplane splits the vector space in half and can do so without splitting the cone because the cone takes up less than half the space in R contiguously.
d) This seems pretty obvious, so maybe it would be better to look at situations where this is not true. It is not true for cones that exceed 1/2 of the vector space. Essentially, non closed, non pointed cones. (this is all I have of this question)
I know these are all completely wrong, and today when I thought I had an answer right, I was laughed out of the classroom for being completely wrong. Please help me. I have no idea what I'm doing.