1873_solutions

Suppose that x n is a sequence in x and that x is the

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: B 1 , B 2 , , B n , such that n  Bj. X j1 For each j we choose a member U j of the family such that B j U j and, since n  Uj X j1 we have found finitely many members of that cover the space X. 3. Suppose that X is a compact metric space and that is a family of closed subsets of X. Prove that if  then there must exist a finite subfamily of  such that  We observe first that 182 .  X H H  X H H X  X. Therefore, since X is compact, it is possible to find finitely many members H 1 , H 2 , , H n of the family such that n X X Hj j1 and we see at once that n  Hj  . j1 4. In a theorem we have just studied we saw that if X is a compact metric space then every sequence in X that has no more than one partial limit must be convergent. Is it true that if every sequence in a given metric space having no more than one partial limit must be convergent, that X must be compact? Yes, it is true. If every sequence that has no more than one partial limit has to be convergent then every sequence must have at least one partial limit. Now let’s change the question. Suppose that we know that, in a given metric space, that every sequence with precisely one partial limit must be convergent. Must the space be compact. Again the answer is yes. To see why, suppose that a metric space X fails to be compact and choose a sequence x n in X such that x n has no partial limit. We now look at the sequence x 1 , x, x 2 , x, x 3 , x, . This sequence has precisely one partial limit but is not convergent. 5. Suppose that X is a compact metric space and that S X. Prove that the metric space S is compact if and only if S is closed in X. The result follows at once from Theorem 7.12.14 6. Prove that if A and B are compact subspaces of R then A  B is also compact. Hint: Show that every sequence in the set A  B has a subsequence that converges to a point of A  B. Suppose that x n is a sequence in the set A  B. For each n, choose numbers a n A and b n B such that x n  a n  b n . Using the fact that A is compact, choose a subsequence a n j of a n that converges to a point a of the set A. For each j we have x n j  a n j  b n j . Using the fact that B is compact, choose a subsequence b j i of b n j that converges to a point b of the set B. The subsequence x n j i of x n converges to the point a  b of the set A  B. 7. Given a metric space X, is it true that every bounded infinite subset of X has a limit point if and only if every closed bounded subset of X is compact as a subspace of X? Yes, this assertion is true. Suppose that every closed bounded subset of X is compact as a subspace of X and suppose that S is a bounded infinite subset of X. Since S is compact, every infinite subset of S must have a limit point. Therefore S must have a limit point. Suppose that every bounded infinite subset of X has a limit point and that Y is a closed bounded subset of X. Every infinite subset of Y, being bounded, must have a limit point and, since Y is closed, any such limit point mus...
View Full Document

This note was uploaded on 11/26/2012 for the course MATH 2313 taught by Professor Staff during the Fall '08 term at Texas El Paso.

Ask a homework question - tutors are online