Proof on boundedness of sets in n space

1. May 11, 2010

prettymidget

1. The problem statement, all variables and given/known data
Let S be a bounded set in n -space. Fix a d>0. Then it is possible to choose a finite set of points {pi....pm} in S such that every point p in S is within a distance d of at least one of the points p1, p2,....pm.

2. Relevant equations

None really.

3. The attempt at a solution

I've tried some methods but I have been stuck at some point of every attempt. A nudge (or two) in the right direction would be greatly appreciated.

Last edited: May 12, 2010
2. May 12, 2010

lanedance

is that exactly how the question is written & is there any additional info?

3. May 12, 2010

prettymidget

This is exactly how its written and thats all they gave.

4. May 12, 2010

Staff: Mentor

Show us what you've tried. At the very least you should be using the definition of a bounded set in Rn.

5. May 12, 2010

lanedance

as a start you could consider 2 cases, where the bound on S is less than d and when it is greater....

6. May 12, 2010

VKint

How much point-set topology do you know? I've got a proof, but it uses a property of compact spaces that you may not be expected to know.

7. May 12, 2010

Tedjn

What property of compactness did you need? Here is an idea that might nudge you in the right direction. Consider the special case of R, which is one-dimensional. Since S is bounded, it lies entirely between -M and M for some M. Certainly, if d >= 2M, we need just pick any point in S to be done. Every other point is within a distance 2M from that point. What if d = M? A first thought might be to choose the midpoint... but that might not be in S. How can we compensate?

8. May 13, 2010

VKint

Basically just the definition: Any open cover of a compact set has a finite subcover. To get a compact set, take the closure of S; to get an open cover, consider the union of all balls of radius d around every point in S...

9. May 14, 2010

Tedjn

That is pretty slick. Occasionally I have these bouts where I realize I'm in out of my depth and don't know enough in this case about metric spaces. I know that in Rn compactness isn't needed (at least not directly). But now I'm not so sure about arbitrary metric spaces.

10. May 14, 2010

Dick

It's not true for an arbitrary metric space. Think about the unit sphere in an infinite dimensional Hilbert space, or even simpler any space with an infinite number of elements with a discrete metric.

11. May 18, 2010

prettymidget

I don't know much topology, only the basic concepts and definitions such as limit point, open closed sets, boundary, closure, and such. The class is a multivariable analysis class, so only basic knowledge has been needed so far.

I actually asked my TA the question and the proof he gave was very long and something that didn't even look undergraduate level to me. If you've found a proof using basic concepts I would appreciate it greatly if you would post it. I haven't really been able to dent the problem.

I'm not sure. I thought about the problem a lot and I was completely convinced that the set S need be both bounded and closed for the rest to follow, so its limit points would be contained in the set.

12. May 18, 2010

Dick

If the set in contained in Euclidean n-space, then there is a way to prove it that doesn't require any topology at all. Practice by supposing n=2. Hint: for any bounded set you can pick a number S such that the set in contained in a square with side length S centered at the origin. Now pick a d>0 and describe a finite set of points that works.

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