How many copies of curly braces occur in the expanded notation for Ƥ⁵(Ø), in the representation where an empty set appears as Ø ?

(Hint: see Binary nature of power set)

Answer:

Let C(Ƥⁿ(Ø)) be the number of elements in Ƥⁿ(Ø)
Let B(Ƥⁿ(Ø)) be the number of curly braces occurrence in Ƥⁿ(Ø)

Based on the observation in the Binary nature of power set,

1. when n > 0, C(Ƥⁿ(Ø)) = 2^C(Ƥn-1(Ø))
   
2. when n > 1, every element in Ƥ^n-1(Ø) is included in exactly half of all the element sets in Ƥⁿ(Ø)
   
3. every element in a non-empty power set is a set and therefore has an outer pair of curly braces except the element Ø
   
4. a non-empty power set always has exactly one pair of outer most curly braces
   

This translates to:

  B(Ƥn(Ø)) = (B(Ƥn-1(Ø)) - 1) · C(Ƥn(Ø))/2 + C(Ƥn(Ø)) where n > 1

  Ƥ0(Ø) ≡ Ø
  Ƥ1(Ø) ≡ {Ø}
  Ƥ2(Ø) ≡ {Ø,{Ø}}
  Ƥ3(Ø) ≡ {Ø,{Ø},{{Ø}},{Ø,{Ø}}}
  ...

  C(Ƥ0(Ø)) = 0
  C(Ƥ1(Ø)) = 2C(Ƥ0(Ø)) = 1
  C(Ƥ2(Ø)) = 2C(Ƥ1(Ø)) = 2
  C(Ƥ3(Ø)) = 2C(Ƥ2(Ø)) = 4
  C(Ƥ4(Ø)) = 2C(Ƥ3(Ø)) = 16
  C(Ƥ5(Ø)) = 2C(Ƥ4(Ø)) = 65536

  B(Ƥ1(Ø)) = 1
  B(Ƥ2(Ø)) = (B(Ƥ1(Ø)) - 1) · C(Ƥ2(Ø))/2 + C(Ƥ2(Ø)) = 0 + 2 = 2
  B(Ƥ3(Ø)) = (B(Ƥ2(Ø)) - 1) · C(Ƥ3(Ø))/2 + C(Ƥ3(Ø)) = 1 · 4 /2 + 4 = 6
  B(Ƥ4(Ø)) = (B(Ƥ3(Ø)) - 1) · C(Ƥ4(Ø))/2 + C(Ƥ4(Ø)) = 5 · 16 /2 + 16 = 56
  B(Ƥ5(Ø)) = (B(Ƥ4(Ø)) - 1) · C(Ƥ5(Ø))/2 + C(Ƥ5(Ø)) = 55 · 65536 /2 + 65536 = 1867776

So there are 1,867,776 copies of curly braces occur in the expanded notation for Ƥ⁵(Ø)