Hello everyone,
I was wondering if someone could assist me with the following problem:
Let T be the topology on R generated by the topological basis B:
B = {{0}, (a,b], [c,d)}
a < b </ 0
0 </ c < d
Compute the interior and closure of the set A:
A = (−3, −2] ∪ (−1, 0) ∪ (0, 1) ∪ (2, 3)
I...
I really don't see it. Of course the metric maps R x R ---> R, for example the distance between (1,2) and (2,4) would be 2 but I don't know how to "draw" it. This is the part where I need a hint.
I am having some trouble visualising the following problem and I hope someone will be able to help me:
Let (X, dx) and (Y,dy) be metric spaces and consider their product topology X x Y (T1) and the topology T2 induced by the metric d((x1,y1),(x2,y2)) = max(dx(x1,x2),dy(y1,y2)) so the maximum of...
Sorry I didn't see I was writing in bold.
The question is: Why does the condition ∂f/∂z > 0 make g(x,y) exist on the level surface? I don't understand the theory behind it.
Hello everyone,
I have a theoretical calculus question. I am working on a exercise where you have to consider f(x,y,z) and express the variable z as a function of x and y on a certain level surface around a certain (x0,y0,z0).
I found out that the condition for this to be able is that the...
The question: Let f: G -> H be a homomorphism of groups with ker(f) finite, the number of elements being n. Show that the inverse image is either empty or has exactly n elements.
My work so far:
Let h be eH (identity on H). Then the inverse image is ker(f) so has n elements, which makes it...
f(g^p) = f(e) so that is e of Q (= 0). And you can conclude that order(f(x)) is finite as well and then you can conclude that because {0} is the only finite subgroup of Q all elements of G will be mapped to 0?
Thanks for the help!
Okay I understand that, but now I only know that there is an isomorphism to Z/nZ with n the number of elements. Then I know that f is one-to-one, so |ker(f)| = 1 and Im(f) would equal n, but that's not what I need. Can you give me one more hint?
Hello, I have to solve the following problem:
Show that a homomorphism from a finite group G to Q, the additive group of rational numbers is trivial, so for every g of G, f(g) = 0.
My work so far:
f(x+y) = f(x)+f(y)
I know that |G| = |ker(f)||Im(f)|
I think that somehow I have to find that...
I understand Cantor is not the way to go here but we were allowed to regard |N x N| = |N| as proven. I just want to know if there is any way to link the two?