Recent content by Gene Naden

@WWDG, what is d(x,K)? I think you have to have Hausdorff. Certainly, it is used in the proof I shared.

I found the proof on the Math Planet website at https://planetmath.org/PointAndACompactSetInAHausdorffSpaceHaveDisjointOpenNeighborhoods. Let y be a point in M-R. Then for all ##x\in R## there exist disjoint neighborhoods ##U_x## and ##V_x## around x and y. A finite number of the ##U_x## covers...

Thanks @fresh_42 . Some of these are the very same topics that have been giving me fits in my study of differential geometry. I think I will get a book. I got into d.g. because it is required for general relativity and didn't think topology was relevant to that subject, but perhaps it is.

Topology books are beyond me. I am at the very beginning of topology, in Chapter four of the differential geometry book.

As I stated, Hausdorff is assumed. Region is a subset of the surface. Not open, not a neighborhood. That is what I was trying to prove. I found proof at MathPlanet.com

This is problem 4.7.11 of O'Neill's *Elementary Differential Geometry*, second edition. The hint says to use the Hausdorff axiom ("Distinct points have distinct neighborhoods") and the results of fact that a finite intersection of neighborhoods of p is again a neighborhood of p. Here is my...

That doesn't sound like what it means for the inverse to be continuous. "A coordinate patch ##x:d\rightarrow E^3## is a one-to-one regular mapping of an open set D of ##E^2## into ##E^3##. Proper patches are those for which the inverse function ##x^{-1}x(D)\rightarrow D## is continuous. That...

Well yes my statement seems a little garbled... Let p be a point in the image of an arbitrary patch ##x:D\rightarrow M## in the surface. p is a point in the surface. There is a proper patch y such that a neighborhood of p is in the surface, by the definition of a surface. Every point in x(D)...

Apparently we don't need the corollary to show that the image of an arbitrary patch in the surface M is open. Since every point of M is has a neighborhood in the image of a proper patch, every point in the image of an arbitrary patch in the surface has a neighborhood in the surface. So the image...

With the aid of the textbook: ##F^*(\nabla_V W)=F^*(W(p+tV)^\prime(0)## Definition of covariant derivative ##(p+tV)=TC(p+tV)=C(p+tV)+a=F(p)+tC(V)## General property of isometry ##F^*(W(p+tV)^\prime(0)=(\overline{W}(F(p)+tC(v))^\prime(0)## Given in the problem statement ##F^*(v)=C(v)## Theorem...

O'Neill's Elementary Differential Geometry, problem 4.3.13 (Kindle edition), asks the student to show that the image of an open set, under a proper patch, is an open set. Here is my attempt at a solution. I do not know if it is complete as I have difficulty explaining the consequence of the...

Regarding post #30, yes I see that xy=0 gives you the union of the x and y axes, hardly a smooth curve.

I overlooked the condition on the function that implicitly defines the curve when I presented this problem. Perhaps if I had seen it much of this thread would have been simpler. It is still not clear to me what motivates the condition that the partials be nonzero. And it begs the question: "If...

Have not yet come to formulas like ##\Gamma \times S_1## in my studies. What is it, please?

So, @Orodruin, how can we patch this up? I think you agree that if you rotate a curve about an axis, that makes a surface. You said my argument ##f(x,y)=y-h(x)## ##g(x,y,z)=f(x,r)=y-h(r)## ##dg=dy-h^\prime dr\neq 0## is wrong. So how to prove the proposition?