Fixed radius disjoint ball covering vs. open covering

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November 22nd 2010, 05:33 AM
SlipEternal

Fixed radius disjoint ball covering vs. open covering

My professor posted the following problem, and I'm at a loss for where to begin. This is an introduction Real Analysis course, so I'm expecting the answer to be blatantly obvious. In fact, for small cases, I can see that it is true, but I am not sure how to prove it for the general case.

Notation: M_r(p) means the open neighborhood of radius r about p in the metric space M (this is the notation used by the professor).

X7. Suppose M is a sequentially compact metric space and let $alpha$ be a positive number. Then any sequence of disjoint balls in M of radius $alpha$ must be finite.
Prove this by contradiction: Suppose there is an infinite sequence M(x_n) of such balls. Can this sequence x_n of the centers of these balls have a convergent subsequence?

I solved this question with little difficulty.

X8. Problem X7 was used to prove the equivalence of sequential compactness and covering compactness for metric spaces. Using covering compactness we can strengthen X7 as follows:
Suppose M is a compact metric space and let $alpha$ be a positive number. Then there is an integer N, depending only on $alpha$ , so that any sequence of disjoint balls in M of radius $alpha$ has at most N elements.

Prove this as follows:
(a) Start with the collection of all open balls in M of radius , and show that there is a finite collection of balls of radius $alpha$ which covers M. Choose such a collection: M_ $alpha$ (c_n) for 1 <= n <= N.
(b) Now suppose M_ $alpha$ (x_k), for 1 <= k <= K, is any collection of disjoint balls of radius $alpha$ . Show that, for each k, there is an index n(k) so that c_n(k) is an element of M_ $alpha$ (x_k). Why is n(k) not equal to n(j) if k not equal to j?
(c) Why is K less than or equal to N?

For part (a), I'm having trouble showing that if you know that any sequence of disjoint balls of fixed radius in M can have a maximum of N elements, then there exists a finite collection of N open balls of that fixed radius in M such that the collection is an open cover for M. At least, I think that is what the question is asking, and intuitively, it seems correct. Starting with the finite cases, it seems quite plausible, making me wonder where I can apply the triangle inequality and hoping that some induction can prove this easily. Unfortunately, such a proof seems elusive at the moment, so any guidance on where to begin would be helpful.
November 22nd 2010, 06:40 AM
SlipEternal

As an addendum, I am not looking for anyone to solve this problem for me. I am just looking for some guidance in the direction in which I should focus my thoughts. For instance, if looking into a potential Lebesgue number would help, I would be quite appreciative. On the other hand, if the problem can easily be solved with judicious use of the triangle inequality, or perhaps something equally ubiquitous, I would appreciate some advice for which distances to compare. For instance, I considered attempting to prove it by contradiction, suggesting that every collection of N balls of radius $alpha$ possesses points which are not covered, then I wanted to show for the finite case, where N=1, that this was a contradiction, then use induction to show that it is always a contradiction. So, I began with this rough sketch of a proof:

Because any sequence of disjoint balls of radius $alpha$ in M can have at most one element, the maximum distance between any two points in M must be less than 2 $alpha$ . Otherwise, let x,y be points in M such that d_M(x,y)>=2 $alpha$ . Because a ball of radius $alpha$ about x and y both would be disjoint, implying that there exists a sequence of two disjoint balls of radius $alpha$ . I get stuck here, as I am not sure if I need additional cases to show that the base case is true, or if I have enough here to demonstrate that it is. Either there exists points in between the furthest two points, or there are none. If there are points between, then centering the ball on them would suffice for an open cover. Alternatively, if the points x,y are disjoint, they must be within $alpha$ distance from each other, or within a distance of $alpha$ from a single point p in M, otherwise, again, an additional ball could be added to the sequence, but we know that 1 is the maximum number of disjoint balls allowed.

Then, for the induction step, I'm completely at a loss for how to proceed.
November 22nd 2010, 07:01 AM
Plato

Here are some hints on 8X.
Suppose that $\left\{ {M_\alpha (c_i )} \right\}_{i = 1}^n$ is the finite open cover in part (a).
If $M_\alpha (p )~\&~ M_\alpha (q )$ are disjoint $\alpha$ balls, let’s suppose that $\{p,q\}\subset M_\alpha (c_k )$ for some $1\le k\le n$ .
That leads to a contradiction: $2\alpha\le d(p,q)<2\alpha$ .

Can you finish?