# Check my work (Spivak problem in Calculus on Manifolds)

1. Aug 11, 2008

### krcmd1

Problem: given compact set C and open set U with C $$\subset$$U, show there is a compact set D $$\subset$$ U with C $$\subset$$ interior of D.

My thinking:

Since C is compact it is closed, and U-C is open. Since U is an open cover of C there is a finite collection D of finite open subsets of U that cover C. Some of these subsets may be interior to C. Each of the other subsets contain a point x in (U-C), and since U-C is open, there is an open rectangle about x in U-C. Close these subsets at x. Then D is closed and bounded, D is a subset of U and C is interior to D.

Are these statements true? Working alone I don't have anyone to check my reasoning.

How does one write this with acceptable conventional notation?

Thanks.

2. Aug 13, 2008

### mathwonk

are you working in euclidean space? i presume.

if so there is a minimum distance of points of your compact set from the closed exterior of your open set, so cover your compact set with little closed balls of radius half that distance.

by the way your reasoning is flaky since you are taking a finite subcover of an open set instead of an open cover.

3. Aug 13, 2008

### krcmd1

Thank you.

4. Aug 14, 2008

### maze

How can you know that an infinite union of closed balls will be closed? For example, (0,1) may be thought of as an infinite union of individual points which are all closed, yet the result is open.

5. Aug 14, 2008

### krcmd1

I need a tie-breaker!

6. Aug 14, 2008

### morphism

C is compact, so we really need only finitely many closed balls.

7. Aug 14, 2008

### krcmd1

Thank you.

8. Aug 14, 2008

### maze

Ahh, so then you would cover it with open balls, find the finite subcover, and then take the closure of the subcover.

9. Aug 14, 2008

### krcmd1

That was my thinking. But see Mathwonk above.

10. Aug 14, 2008

### maze

Err, I the reasoning in the first post is wrong but for a different reason than mathwonk says. In there the "cover" used is {U}, so the finite subcover is just {U} again, but the closure of U is not necessairily contained in U so it cant be used to prove the theorem.

However, if you take small open balls for the cover such that the balls are well within U, then there should be no problem when you take the closure of their union.

11. Sep 1, 2008

### Castilla

But the question of Maze was left unanswered. It can not be granted that the collection of little closed balls will be a closed set, let alone a compact set.

12. Sep 1, 2008

### morphism

A finite union of closed/compact sets is closed/compact, respectively.

13. Sep 1, 2008

### Castilla

Let say you have a curve K (therefore a compact set). This K is inside an open region S.

We cover K with an infinite number of open discs, each centered in a point of K, and each with the same radius r that allows all the discs to be inside of S. I understood the existence of "r"; it follows from compactness of K and "closedness" of Compl.(S).

(Not very good English).

We now take closed discs, again centered in each point of K, but now with radius r/2.

Apostol (proof of his theorem 9.63, first edition of his M. Analysis) says that this collection of closed discs is closed.

I fail to see why. I know that an infinite union of closed sets no necesarily is a closed set.

14. Sep 1, 2008

### morphism

Have you tried visualizing what that set looks like? If I'm understanding what you're saying correctly, then it should be pretty clear why such a set is closed (and in fact compact).

15. Sep 1, 2008

### mathwonk

my solution was completely correct as stated. although it is true a union of closed sets may not be closed, nonetheless the union of the sets i described is closed. think about it.

in fact someone already proved it by taking finite sub covers.

16. Sep 2, 2008

### Castilla

We have the curve K inside an open region S. We cover K with open discs centered in each point z of K and all of them with the same fixed radius "r" which allows all of them to be inside of S.

Now we take the closed discs N(z, r/2). We have to prove that this last set of discs (which I call "T") is compact. I know it is bounded (it is inside S). To prove closedness I tried this:

A closed set contains all its accumulation points. Let's suppose T does not. So there it is a point "a" which is a.p. of T and does not belong to T. Therefore (remember T is inside S) "a" belongs to the interior of S - T.
So there is a disc N(a) in the interior of S - T.
But, being "a" an a.p. of T, there are points of T inside N(a)
But N(a) is inside of S - T. So we got points of T inside S - T. Contradiction.

But then I realized my work was flawed, because of this deduction: "So there it is a point "a" which is a.p. of T and does not belong to T. Therefore "a" belongs to the interior of S - T."

That deduction is not true, because "a" no necesarily is in the interior of S - T. The point "a" might be located in the boundary of T, so I can not take a disc N(a) to follow up my "proof".

How can I fix this?

17. Sep 2, 2008

### Castilla

I think I have seen something more.

In my last post I saw that my "proof" was flawed because "a" might be in the boundary of T. (I will denote it B(T) ).

But a boundary is always a closed set. Now let's take one of the compact (closed and bounded) discs of T, denoting it "D". We are supposing that B(T) and disc D do not intersect. (If they did it, the problem would be solutioned). Then there is a distance between B(t) and the disc D.

So if "a" is in B(T) , there is a fixed distance between it and its "accumulated points" which all belong to one or other closed disc. End.

Please tell me if this is ok.

18. Sep 5, 2008

### morphism

I don't really understand what you're saying here.

It could be correct, but here's another proof (the best I could think of at this hour). Suppose x isn't in T. Then x is a distance of >r/2 away from any point in K. For a contradiction, suppose that the open ball B(x, 1/n) intersects T for every n. Then for each n there exists a y_n in B(x, 1/n) and a z_n in K such that $d(y_n, z_n) \leq r/2$. Consequently, $d(x, z_n) \leq d(x, y_n) + d(y_n, z_n) < r/2 + 1/n$.

Since {z_n} is a sequence in the compact set K, it admits a subsequence {$z_{n_k}$} which converges to some point z in K. Thus we can find a $z_{n_k}$ such that $d(z_{n_k}, z) < 1/k$. But then,

$$d(x,z) \leq d(x, z_{n_k}) + d(z_{n_k}, z) < \frac{r}{2} + \frac{1}{n_k} + \frac{1}{k}.$$

Letting $k \to \infty$ will yield that $d(x,z) <= r/2$, our desired contradiction.

19. Sep 7, 2008