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Exercise 2.1 (The power set of a finite set is a finite $\sigma$-algebra)

Let Ω be a finite set. Show that the set of all subsets of Ω, 2Ω, is also finite and that it is a σ-algebra.

Answers

1.
2Ω is finite.
We argue that for an n-element set Ω, n ∈ ℕ, the power set 2Ω will always have 2n elements. To do so, we induct on n.
  • Induction base. Let n = 0. Then Ω = ∅ and 2Ω = {∅} with cardinality of 20 = 1.
  • Induction step. Let Ω be a set of cardinality n + 1. Suppose inductively that we have proven the statement for any set of the cardinality n; in particular for Ω′ := Ω ∖ ωn+1 for some ωn+1 ∈Ω. We argue that there are two sets contained in 2Ω: the subsets of Ω′ and the same subsets of Ω′ but also containing ωn+1:

    2Ω = {E : E ⊆Ω′}⋃ {E ∪{ωn+1} : E ⊆Ω′}.

    [ ⊆] Let E ∈ 2Ω. Then either ωn+1∉E, in which case E ⊆Ω′ by definition, or ωn+1 ∈ E, in which case E ∖{ωn+1}⊆Ω′ by definition as well. Similarly, we see the [ ⊇] part.
    To summarize, 2Ω is a union of two disjoint sets, each of cardinality 2n. Therefore #2Ω = 2n + 2n = 2 ⋅ 2n = 2n+1.

2.
2Ω is a σ-algebra.
  • We have ∅∈ 2Ω since ∅ is vacuously a subset of any set Ω.
  • If E ⊆Ω, then Ω ∖ E ⊆Ω by definition of set difference.
  • Finally, if E1,E2,… is countable sequence of sets in Ω, then any x ∈⋃ ⁡ n∈ℕEn must also be contained in some x ∈ Ei ⊆Ω and is thus a subset of Ω.
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2021-10-30 11:51
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