MATH 202 A - Problem Set 11

نویسنده

  • Walid Krichene
چکیده

proof E is a Borel set: for each n ∈ N, On is open as the union of open sets. Therefore E is the countable intersection of open sets, thus is a Borel set. E is of the second category: we first prove that O n is nowhere dense. Indeed, we have for all qk ∈ Q, B(qk, 1 n2k ) ⊂ On, thus qk ∈ int(On). Therefore Q ⊆ int(On), therefore R = cl(Q) ⊆ cl(int(On)), i.e. cl(int(On)) = R. Taking complements, and using the fact that for any set X, int(X) = cl(X) and cl(X) = int(X), we have int(cl(O n)) = ∅ i.e. O n is nowhere dense. Now we have R = E ∪ E = E ∪ (∪n=1O n) therefore E is not of the first category, otherwise we would have R is of the first category (countable union of nowhere dense sets), which contradicts Baire’s theorem. m(E) = 0. We prove that for all > 0, m(E) ≤ . Let > 0, and let n0 ∈ N such that 4/n0 ≤ . Then we have

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تاریخ انتشار 2012