Suppose A CR is closed and bounded, and f: A R is continuous. Using the results...

 

Transcribed Image Text:

Suppose A CR is closed and bounded, and f: A R is continuous. Using the results of Q4, and results proved in class, show: a) f is bounded, b) f takes on its maximum and minimum, b) f is uniformly continuous. The following notions are relevant for Q4,Q5: Recall that S C R is open if for all x E S, there exists an open interval J around x with JCS. A subset A CR is closed if its complement S = R – A is open. %3D Q4 (6 points) Definition: Let SCR be a subset. A collection {U\| A € A} of open subsets U CR (with A any indexing set) is called an open cover of S if scU U. XEA That is, for all x E S there exists A E A with x € U,. a) Suppose {Ulde A} is an open cover of a closed interval [a, b]. Prove that [a, b] is already covered by finitely many of the U,'s. b) Show that the statement from part (a) remains true if a, b is replaced by any bounded, closed subset A CR. c) Prove the converse: that if ACRIS not bounded or not closed, then there exists an open cover {U E A} of A such that A can not be covered by finitely many of the U,'s. Hints (a): Follow the strategy from the proof of the Intermediate Value Theorem, starting with a suitable SC [a, b]. (b): Reduce to part (a), by choosing a closed interval containing A. (c): If A is not closed, choose a eR - A such that every open interval around a meets S, and consider the open subsets Us = {x| |x – a| > 8}. Note: There is an approach to these results using sequences -- but we haven't covered the relevant facts, so you may not use sequences. Suppose A CR is closed and bounded, and f: A R is continuous. Using the results of Q4, and results proved in class, show: a) f is bounded, b) f takes on its maximum and minimum, b) f is uniformly continuous. The following notions are relevant for Q4,Q5: Recall that S C R is open if for all x E S, there exists an open interval J around x with JCS. A subset A CR is closed if its complement S = R – A is open. %3D Q4 (6 points) Definition: Let SCR be a subset. A collection {U\| A € A} of open subsets U CR (with A any indexing set) is called an open cover of S if scU U. XEA That is, for all x E S there exists A E A with x € U,. a) Suppose {Ulde A} is an open cover of a closed interval [a, b]. Prove that [a, b] is already covered by finitely many of the U,'s. b) Show that the statement from part (a) remains true if a, b is replaced by any bounded, closed subset A CR. c) Prove the converse: that if ACRIS not bounded or not closed, then there exists an open cover {U E A} of A such that A can not be covered by finitely many of the U,'s. Hints (a): Follow the strategy from the proof of the Intermediate Value Theorem, starting with a suitable SC [a, b]. (b): Reduce to part (a), by choosing a closed interval containing A. (c): If A is not closed, choose a eR - A such that every open interval around a meets S, and consider the open subsets Us = {x| |x – a| > 8}. Note: There is an approach to these results using sequences -- but we haven't covered the relevant facts, so you may not use sequences.