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# Solution abns c x r and c consists of all sets that

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Solution. ABNS (c) X = R and C consists of all sets that are either countable or have a countable complement. (Finite sets count as countable!) Solution. SA (d) X = R and C is the family of all open subsets of R . Solution. N (e) X = R and C is the family of all sets that are either open or closed. Solution. N (f) X = R and C is the family consisting of R , and all sets of one of the following forms: i. n [ k =1 ( a k ,b k ], ii. ( a 0 , ] n [ k =1 ( a k ,b k ], iii. ( -∞ ,b 0 ] n [ k =1 ( a k ,b k ]. (The number of sets n in the unions above varies, all n N are involved.) Solution. This was supposed to be ABNS, but is actually N. See Homework 7 for a corrected version. 3. The smallest σ -algebra containing all open subsets of a metric space X is known as the Borel σ -algebra of X , or the family of Borel subsets of X . We will write B ( X ) to denote the σ -algebra of Borel subsets of X . To repeat, B ( X ) is the σ -algebra generated by the family of open subsets of X . Prove that B ( X ) is also the σ -algebra generated by all closed sets of X . Solution. Let O be the family of all open subsets of X , C the family of all closed subsets of X . Let A ∈ O . Then A c is closed, hence A c σ ( C ), hence also A σ ( C ). This proves that σ ( C ) contains all open sets, being a σ -algebra containing all open sets, it must contain σ ( O ) = B ( X ). Similarly one sees that B ( X ) = σ

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Solution ABNS c X R and C consists of all sets that are...

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