414/514 – Compactness

Here are some extra credit problems dealing with the notion of compactness. The deadline to turn them in is Wednesday, December 14, at noon. I will not post hints for now, but feel free to stop by my office as you work on the problems, if you want to double check how your approach is going, and we may discuss suggestions then. These results are proved in different sources, try not to look these solutions up.

  • (Vitali’s covering lemma)

Suppose that (X,d) is a metric space and K\subseteq X is compact. Let \{B_{\epsilon_i}(x_i)\mid i\in I\} be an open covering of K. Show that there is a finite set J\subseteq I such that the balls B_{\epsilon_i}(x_i) for i\in J are pairwise disjoint, and

K\subseteq\bigcup_{i\in J}B_{5\epsilon_i}(x_i).

If X={\mathbb R}, can the constant 5 be improved? (I.e., can it be replaced by a smaller number?) If so, can you find the optimal constant?

The Vitali covering lemma has nice consequences. For example, it allows us to prove Lebesgue’s differentiation theorem.

  • (Antisocial coverings)

This is a result of Krantz and Parsons. Again, let (X,d) be a metric space. A self-centered covering of a non-empty subset S\subseteq X is a collection \{B_{r_x}(x)\mid x\in S\} of open balls, with open ball centered at each point of S. An antisocial family is a collection of balls with the property that if B_\alpha(x) and B_\beta(y) are distinct balls in the collection, then x\notin B_\beta(y) and y\notin B_\alpha(x). Prove the following:

Suppose that {\mathcal C}=\{B_{r_x}(x)\mid x\in K\} is a self-centered covering of the compact set K, and suppose that the function x\mapsto r_x is continuous on K. Then there is an antisocial family {\mathcal A}\subseteq {\mathcal C} that covers K.

Also, show that any such family {\mathcal A} must be finite. Can you find a “reasonable” assumption on the map x\mapsto r_x, not as restrictive as continuity, but sufficient to ensure the result?

(By the way, the topic of covering theorems is very interesting. Let me know if you think you may want to explore this further.)


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