Week 1 Tutorial Solutions
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- Every closed interval is a Borel set. Each set is a finite union of closed intervals and therefore also a Borel set. Finally is a countable intersection of the and thus also a Borel set.
- We will prove that belongs to if and only if it can be written in the stated form with for all . First, suppose that belongs to . Define to be or according to which of the intervals in the points belongs. Thus define
and put . Inductively, define
and for all . By induction and
are in the same closed subinterval of for every so we must have
and has the desired representation.
Conversely, if
and every belongs to then each of the truncations
belongs to and therefore the limit must belong to .
- Define by
for every . By the previous question is surjective. It is injective because if in are not equal then there is minimal with . We must then have and is different sub-intervals of and is therefore injective.
- We have for all and therefore for all . As is a union of disjoint intervals of length we can say (using Problem 3 from Worksheet 1) that . Thus . Since was arbitrary we must have .