Recent content by tomkoolen

Yes!

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...

Two line segments of length 2, a "cross"

{(a,1), (1,b), (c,-1), (-1,d)} with a,b,c,d smaller or equal to 1? ABsolute value!

Oh I think I should think about the set of all elements of R2 that have fixed distance from (x0,y0) and draw that right?

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.

Thanks for replying! In R2 I could just visualise them as circles around a point (x,y) right?

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?