Proof that Compact Subset of Metric Space is Bounded

In summary, Francis Su proves in his Real Analysis lecture that a compact subset of a metric space is bounded by showing that every open cover of the subset has a finite subcover. This is done by considering a set of open balls with a radius of r, centered on each point of the subset, and finding a maximum distance s between the center points of the subcover. By adding 2r to s, a larger open ball can be constructed that includes the subset. This explains the use of the number 2 in the proof. Furthermore, it is clarified that x represents an arbitrary point in the subset A, rather than the center of one of the open balls.
  • #1
Rasalhague
1,387
2
Compact --> bounded

In lecture 8 of Francis Su's Real Analysis online lecture series, he has a proof that a compact subset of a metric space is bounded: Given a metric space (X,d), if A is a compact subset of X, then every open cover of A has a finite subcover. Let B be a set of open balls of radius r, one centered on each point of A. This is an open cover of A, so it has a finite subcover {Bi}. Let C be the set of distances between pairs of center points of the elements of the subcover {Bi}. Let s = max C. Then B(x,s+2r) is an open ball that includes A.

My question: why the 2? Isn't B(x,s+r) also an open ball that includes A? If u is a point in A, then

d(x,u) <= d(x,w) + d(w,u) <= s + r

where w is the center of an open ball in {Bi} that contains u.
 
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  • #2


set up a horizonal row of spheres tangent to each other. then s=r*2*(k-1), k=number of spheres. the total diameter is s+2r=r*2*k=d. now draw B(x, d) and note: but where's x? is it in the center or somewhere else?
 
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  • #3


Ah, I see now. I made x be the center of one of the open balls in my proof, although I forgot to mention that. He must have been been taking x to be an arbitrary point of A. Thanks, xaos.
 

1. What is a compact subset of a metric space?

A compact subset of a metric space is a set that is both closed and bounded. This means that the set contains all of its limit points, and that all of its points are contained within a finite distance from each other.

2. Why is it important to prove that a compact subset of a metric space is bounded?

Proving that a compact subset of a metric space is bounded is important because it allows us to make conclusions about the behavior of the set. Boundedness ensures that the set is finite and contained within a certain range, making it easier to analyze and work with.

3. How is boundedness related to compactness in a metric space?

Boundedness is a necessary condition for compactness in a metric space. In order for a set to be compact, it must be both closed and bounded. If a set is not bounded, then it cannot be compact.

4. What does it mean for a subset of a metric space to be closed?

A subset of a metric space is closed if it contains all of its limit points. This means that if a sequence of points within the subset converges, the limit point must also be contained within the subset.

5. Can you provide an example of a compact subset of a metric space that is not bounded?

No, because boundedness is a necessary condition for compactness in a metric space. Therefore, by definition, all compact subsets of a metric space must also be bounded.

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