Show that [a, b] = ⋂∞n=1(a − 1n , b + 1n) . Therefore any σ-algebra of subsets of Rwhich contains all open intervals also contains all closed intervals. Similarly, (a, b) =⋃∞n=1[a + 1n , b − 1n] so that any σ-algebra containing all closed intervals also containsopen intervals. Hence deduce that1(a) [a, b) = ⋂∞n=1(a − 1n , b) (b) (a, b] = ⋂∞n=1(a, b + 1n) (c) (a, +∞) = ⋃∞n=1(a, b +n).What further comments can you make about σ-algebras of subsets of R which containall open intervals
Question
Show that [a, b] = ⋂∞n=1(a − 1n , b + 1n) . Therefore any σ-algebra of subsets of Rwhich contains all open intervals also contains all closed intervals. Similarly, (a, b) =⋃∞n=1[a + 1n , b − 1n] so that any σ-algebra containing all closed intervals also containsopen intervals. Hence deduce that1(a) [a, b) = ⋂∞n=1(a − 1n , b) (b) (a, b] = ⋂∞n=1(a, b + 1n) (c) (a, +∞) = ⋃∞n=1(a, b +n).What further comments can you make about σ-algebras of subsets of R which containall open intervals
Solution
First, let's show that [a, b] = ⋂∞n=1(a − 1/n , b + 1/n).
Step 1: We need to show that [a, b] is a subset of ⋂∞n=1(a − 1/n , b + 1/n) and vice versa.
Step 2: Let x be an element in [a, b]. This means a ≤ x ≤ b.
Step 3: For any n in natural numbers, we have a - 1/n < a ≤ x and x ≤ b < b + 1/n. So, x is in (a - 1/n, b + 1/n) for all n.
Step 4: Therefore, x is in ⋂∞n=1(a − 1/n , b + 1/n). So, [a, b] is a subset of ⋂∞n=1(a − 1/n , b + 1/n).
Step 5: Now, let x be an element in ⋂∞n=1(a − 1/n , b + 1/n). This means for all n in natural numbers, a - 1/n < x < b + 1/n.
Step 6: As n approaches infinity, 1/n approaches 0, so a ≤ x ≤ b. Therefore, x is in [a, b].
Step 7: So, ⋂∞n=1(a − 1/n , b + 1/n) is a subset of [a, b].
Step 8: Therefore, [a, b] = ⋂∞n=1(a − 1/n , b + 1/n).
The proof for (a, b) = ⋃∞n=1[a + 1/n , b − 1/n] is similar.
Now, if a σ-algebra contains all open intervals, it must contain all closed intervals, and vice versa. This is because we can express a closed interval as an infinite intersection of open intervals, and an open interval as an infinite union of closed intervals.
For the parts (a), (b), and (c), the proofs are similar to the above, just replace the intervals accordingly.
Finally, if a σ-algebra of subsets of R contains all open intervals, it must contain all Borel sets, because the Borel σ-algebra is the smallest σ-algebra containing all open intervals. This means it contains all sets that can be formed from open intervals using countable unions, countable intersections, and complements.
Similar Questions
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