Here is the required notation for the above two questions: Definition: Let SCR be a subset. A

 

Here is the required notation for the above two questions:

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Definition: Let SCR be a subset. A collection {U\| A E A} of open subsets U CR (with A any indexing set) is called an open cover of S if dEA That is, for all x E S there exists A E A with x € U. a) Suppose {UllE 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 [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 = {r| |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. Definition: Let SCR be a subset. A collection {U\| A E A} of open subsets U CR (with A any indexing set) is called an open cover of S if dEA That is, for all x E S there exists A E A with x € U. a) Suppose {UllE 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 [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 = {r| |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.