Let S CR be a subset. A collection {U| A E A} of open subsets UX...

 

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Let S CR be a subset. A collection {U\| A E A} of open subsets UX CR (with A any indexing set) is called an open cover of S if scUU. That is, for all x E S there exists X E A with x € U. a) Suppose {U XE 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 ACR is not bounded or not closed, then there exists an open cover {U XEA} 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 la, 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. 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 Jaround z with J C S. A subset A CRis closed if its complement S = R - A is open. Let S CR be a subset. A collection {U\| A E A} of open subsets UX CR (with A any indexing set) is called an open cover of S if scUU. That is, for all x E S there exists X E A with x € U. a) Suppose {U XE 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 ACR is not bounded or not closed, then there exists an open cover {U XEA} 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 la, 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. 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 Jaround z with J C S. A subset A CRis closed if its complement S = R - A is open.