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COMPACT SETS IN METRIC SPACES NOTES FOR MATH 703 ANTON R. SCHEP In this note we shall present a proof that in a metric space ( X,d ) a subset A is compact if and only if it is sequentially compact, i.e., if every sequence in A has a convergent subsequence with limit in A . The usual proofs either use the Lebesgue number of an open cover or reduces an open cover first to a countable cover. For compact sub- sets of the real line with respect to the Euclidean topology this result has an easier proof that one can easily derive from the equivalence of compactness to being closed and bounded. This theorem is called the Heine-Borel theorem and is usually derived from the theorem that a closed bounded interval is compact. This latter theorem has at least two different proofs. The first one uses the ordering of the real line by considering the supremum of the set of x [ a,b ] such that [ a,x ] has a finite subcover and showing that this supremum equals b . The other proof is by contradiction. One assumes that one has an open cover without finite subcover and splits the interval into closed subintervals, at least one of which has no finite subcover. Repeating this one gets a sequence of nested closed interval without finite subcovers whose di- ameters tend to zero, so by the so-called Nested Interval theorem this sequence has an intersection consisting exactly of one point. As this point is covered by an open set of the covering, we can also capture the closed intervals from the nested sequence from some point on, which gives the desired contradiction. Our approach here uses the ideas of this second proof to prove the above mentioned equivalence of com- pactness to sequential compactness in a metric space. First we recall
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This note was uploaded on 02/05/2012 for the course MATH 703 taught by Professor Schep during the Fall '11 term at South Carolina.

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