Week 1 Tutorial Solutions

  1. C2=[09,19][29,39][69,79][89,99] C2=[027,127][227,327][627,727][827,927] [1827,1927][2027,2127][2427,2527][2627,2727]
  2. Every closed interval [a,b] is a Borel set. Each set Cn is a finite union of closed intervals and therefore also a Borel set. Finally C is a countable intersection of the Cn and thus also a Borel set.
  3. We will prove that x belongs to C if and only if it can be written in the stated form with x(n){0,2} for all nN. First, suppose that x belongs to C. Define d(1) to be 0 or 2 according to which of the intervals in C1 the points x belongs. Thus define d(1)={00x13223x1 and put y(2)=3xd(1). Inductively, define d(n+1)={00y(n+1)13223y(n+1)1 and y(n+1)=3y(n)d(n) for all nN. By induction x and n=1Nd(n)3n are in the same closed subinterval of CN for every NN so we must have x=n=1d(n)3n and x has the desired representation.

    Conversely, if x=n=1d(n)3n and every d(n) belongs to {0,2} then each of the truncations n=1Nd(n)3n belongs to CN and therefore the limit must belong to C.
  4. Define ϕ:{0,1}N by ϕ(α)=n=12α(n)3n for every α{0,1}N. By the previous question ϕ is surjective. It is injective because if α,β in {0,1}N are not equal then there is NN minimal with α(N)β(N). We must then have ϕ(α) and ϕ(β) is different sub-intervals of CN and ϕ is therefore injective.
  5. We have CCn for all nN and therefore Λ(C)Λ(Cn) for all nN. As Cn is a union of 2n disjoint intervals of length 3n we can say (using Problem 3 from Worksheet 1) that Λ(Cn)2n/3n. Thus Λ(C(2/3)n. Since nN was arbitrary we must have C=0.