the continuity is what says that the inverse image of any closed set is closed, and dalle wrote in his last post. My comment on the union of two closed sets was responding to that.
Okay, I think I'm making progress now. If G_E \in (E,\mathbf{E}) is open, then we have G_E = G \cap E for some open set G in X.
This means that G_E^c=G^c \cap E where G_E^c is the complement of G_E in E and thus an arbitrary closed set in E. Since E is closed and the intersection of two...
Thanks, dalle. That seemed to be helpful, but I ran into a problem.
1. I have as a theorem I can use directly
I thought I could prove 2. easily, but the problem was it didn't seem to depend on E or F being closed.
Then I seem to be making some leap again to the final proof that doesn't depend...
Okay, we can forget my example. It was a bad example. But still don't see how to start on a positive proof. I was able to do it for E and F open, but I seem to run into problems for them both closed.
Obviously the continuity is the limitation on the topologies here and the problem must turn on that point. But I don't see how to work out a proof from that (even if I could maybe see how it applies to my specific bad example).
I'm having trouble with the third part of a three part problem (part of the problem is that I don't even see how what I'm trying to prove can be true).
The problem is:
Let X and Y be topological spaces with X=E u F. We have two functions: f: from E to Y, and g: from F to Y, with f=g on the...
Nevermind, I got it. I forgot about the possibility that int(A) and ext(A) could both be the empty set, and thus the closure would also be the empty set (which I guess meets my criteria anyway of int(A) = ext(A)).
Sorry, bad phrasing. I figured it should have been obvious what I meant from context. I believe my book calls them "interior points", which are points that are contained in some open set that is completely contained in the set in question.
int(A) = interior of A
ext(A) = exterior of A, or the interior of the complement of A
My thoughts are thus: int(A) and ext(A) are both open sets, so their union is an open set, and if we let B = union of int(A) and ext(A) then B = int(B). So the only way the closer is not equal to the entire...
I already posted the basis of my proof, the rest is trivial.
Pick two rational points (points with rational coordinates) on the unit circle:
e^{i \theta_1} and e^{i \theta_2}
Then it is really easy to show that the distance between the points with double the angles e^{2i \theta_1} and e^{2i...
The proof I have for the unit circle (and it's not my proof, I can't take credit for it) doesn't directly extend to circles with non-perfect square radii. However, there is another proof I've seen that might.
Why do you ask?
Yes, I mean points with rational coordinates on the unit circle. It isn't hard to prove that if you pick two points with rational coordinates on the unit circle the distance between the points with double the angles is also rational.
In case anyone is interested, I've now seen two proofs of this, both of which are in principle the same. It actually turns out to be pretty simple. And it's fine to work with the unit circle.
The basic idea is to make triangles with the center of the circle and two adjacent points. Then you...
This seems to mostly be a case of you not understanding the result.
The best way to think of it is that the MATLAB result is wrong, or rather incomplete (they are just taking the constant to be 0).
It should say:
\int \sin(x)\cos(x)dx = -\frac{1}{2}\cos^2(x) + C
Which is exactly the same as...
I understand that triangles are easy, I've never had a problem with that part, and that polygons can be divided into triangles. But I still don't see how that helps.
It's not enough to divide polygon into distinct triangles, you would have draw every possible triangle, all overlapping, and they...
Yes, that is correct. That was what I asked from the beginning, but I thought it would be easier to limit it to a unit circle when someone pointed out that it could depend on the radius, but I realize that might be too much of a limit.
I'd settle for an arbitrary circle. I just chose the unit circle since i figured if it would be true for any circle it should be true for the unit circle. However, it is entirely possible that it is only true for circles with irrational radii or something like that, which is fine too. I don't...
I don't see the answer. And I don't see how triangles alone help since you need each point to be a rational distance from EVERY other point.
But if you have something clever, PLEASE explain!
The exact position will be radius dependent, sure. But I'm only looking for a proof of possibility, not a constructive proof. So you can just as well assume the unit circle.
Does anyone have an idea how to prove the following (or prove that it is not true):
For any positive integer k, you can find k points on a circle such that each point is a rational distance from every other point.
Isn't each term looking for points mod p on the unit circle (you can think of a p lattice on the unit circle, and x maps to some point in the one of the the domains). You are in adding a bunch of number mod different primes in essence, which being all coprime might make it easier.
Anyway, it...