Here is an alternate proof to the one given in Exercise 3.3.9 for the final implication in the Heine Borel Theorem. Consider the special case where K is a closed interval. Let {O_Lambda: Lambda} be an open cover for [a, b] and define S to be the set of all x [a, b] such that [a, x] has a finite subcover from {O_Lambda: Lambda}. Argue that S is nonempty and bounded, and thus s = sup S exists. Now show s = b, which implies [a, b] has a finite subcover. Finally, prove the theorem for an arbitrary closed and bounded set K.
Show transcribed image text Here is an alternate proof to the one given in Exercise 3.3.9 for the final implication in the Heine Borel Theorem. Consider the special case where K is a closed interval. Let {O_Lambda: Lambda} be an open cover for [a, b] and define S to be the set of all x [a, b] such that [a, x] has a finite subcover from {O_Lambda: Lambda}. Argue that S is nonempty and bounded, and thus s = sup S exists. Now show s = b, which implies [a, b] has a finite subcover. Finally, prove the theorem for an arbitrary closed and bounded set K.