I How to prove that compact regions in surfaces are closed?

[Gene Naden]

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 attempt at a solution. At a key step it makes an assertion without proof.

Consider a point p in M-R where M is the surface and R is the region in M. To show that p has a neighborhood outside of R, we need a lower bound on the distance of p from R.For every $\epsilon>0$ the set of all neighborhoods of points in R covers R, so the neighborhoods of a finite set S of points in R covers R. Let q be the point in S that is closest to p. The point in R that is closest to p might be in the $\epsilon$ neighborhood of q. If so, the distance of p from R is greater than or equal to $d(p,q)-\epsilon$ where d is distance. If we choose $\epsilon<d(p,q)$ then the distance of p from R is greater than zero, and there is a neighborhood of p in $M-R$ and so M-R is open and R is closed.

 

[member 587159]

How do you define region? Is it the same thing as neighborhood? In any Hausdorff space, compact subsets are closed.

 

[Gene Naden]

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

 

[member 587159]

You can find this in any topology book. Have a look here: https://math.stackexchange.com/ques...et-in-a-hausdorff-topological-space-is-closed. There is not even need for using the metric.

 

[Gene Naden]

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

 

[fresh_42]

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

I consider this as a case of wrong order. You need some topological basics before you learn differential geometry. Sure, you don't need to learn all the details, but the fundamental concepts are necessary. You can do this in parallel, but I recommend to read a book on topology at least simultaneously and at least the first 100 pages - a small paperback will do. What you should know are:

- topology
- closed, open, continuity, compact, bounded
- separation axioms
- coverings
- neighborhood, region (convex, connectedness, star shaped)
- metric spaces
- all varieties of connectedness (simple, path, homology)
- filtrations, refinements

You can also look them up on Wikipedia or nLab, however, a book is usually the better choice.

 

[Gene Naden]

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.

 

[WWGD]

I don't know if this is overkill, but it seems you could use the local homeos: Break down your compact set K into a finite union. Use chart maps to send them into $\mathbb R^2$ . Chart homeo H . will send K to $h(K) \subset \mathbb R^2$ , which is closed in $\mathbb R^2$ by Heine-Borel. Then use h again to map back, since homeos are closed maps. Use the Pasting Lemma to produce a global continuous func. Adapt if you need more than one chart map. Note that by compactness you will only need a finite number of chart maps. EDIT. Union of (finite)images of charts is compact , since homeos preserve compactness.

Or, maybe as a good exercise in point set, cover the region by (finitely many ) charts. Then consider the union of the images

 

[WWGD]

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 attempt at a solution. At a key step it makes an assertion without proof.

Consider a point p in M-R where M is the surface and R is the region in M. To show that p has a neighborhood outside of R, we need a lower bound on the distance of p from R.For every $\epsilon>0$ the set of all neighborhoods of points in R covers R, so the neighborhoods of a finite set S of points in R covers R. Let q be the point in S that is closest to p. The point in R that is closest to p might be in the $\epsilon$ neighborhood of q. If so, the distance of p from R is greater than or equal to $d(p,q)-\epsilon$ where d is distance. If we choose $\epsilon<d(p,q)$ then the distance of p from R is greater than zero, and there is a neighborhood of p in $M-R$ and so M-R is open and R is closed.

BTW, I think you need to rigorize: Hausdorff means that different points have _disjoint_ 'hoods, not different ones; different ones may overlap.

 

[Gene Naden]

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 R because R is compact. Let U be the union of these neighborhoods and V be the intersection of the corresponding neighborhoods around y. Then $R\subset U$ and $y\in V$.

U is open because it is the union of open sets. V is open because it is the intersection of a finite number of open sets.

U and V are disjoint because if z is in U then z is in one of the $U_x$ and so it is not in the corresponding $V_x$ and hence not in V. Since V is disjoint from U, and $R\subset U$, V is disjoint from R.

$\forall y \in M-R$, there is a neighborhood of y that is disjoint from R, so M-R is open and R is closed.

 

[WWGD]

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 R because R is compact. Let U be the union of these neighborhoods and V be the intersection of the corresponding neighborhoods around y. Then $R\subset U$ and $y\in V$.

U is open because it is the union of open sets. V is open because it is the intersection of a finite number of open sets.

U and V are disjoint because if z is in U then z is in one of the $U_x$ and so it is not in the corresponding $V_x$ and hence not in V. Since V is disjoint from U, and $R\subset U$, V is disjoint from R.

$\forall y \in M-R$, there is a neighborhood of y that is disjoint from R, so M-R is open and R is closed.

You can also , less generally, show that a compact subset of a metric space is closed. Consider K compact. . We show the complement is open. Since K is compact, it is bounded, so consider x in the complement. Since x is in the complement, d(x,K)>0 , so the open ball B(x, d(x,K)/2) is an open set containing x in the complement of K. Then the complement of K is open and so K is closed. But yours is more general, since metric implies Hausdorf.

 

[Gene Naden]

@WWDG, what is d(x,K)?

I think you have to have Hausdorff. Certainly, it is used in the proof I shared.

 

[WWGD]

@WWDG, what is d(x,K)?

I think you have to have Hausdorff. Certainly, it is used in the proof I shared.

Sorry, it is the distance between a point and a set, usually definied as the Inf of distances between x and points in K. And, yes, Metric spaces are Hausdorff, though not all Hausdorff spaces are metric/metrizable.