一題代數證明 (是代數但不是西格馬代數的東西)

Question

Let D consist of those sets which are either finite or have a finite complement. Then D is an algebra. If X is not finite, then D is not a sigma-algebra.

忘了說，X is the universal set，不過你好像已經看出來了。:-)

Answer 1

Claim: Let D consist of those sets which are either finite or have a finite complement. Then D is an algebra.
Proof:
D is nonempty, since the empty set is a finite set and hence is in D.

Let A, B ∈ D.
(closed under complements)
If A is finite, then ￢A has finite complement (which is A), so ￢A ∈ D.
Otherwise, ￢A is finite, so ￢A ∈ D.

(closed under unions)
If both A and B are finite, then A ∪ B is also finite and hence is in D.
Otherwise, at least one of ￢A and ￢B is finite, so
￢(A∪B) = (￢A)∩(￢B) is finite, and hence is in D.

Hence, D is an algebra. ::


Claim: If X is not finite, then D is not a σ-algebra.
Proof: Assume X is not finite. There exists a countable sequence of elements of X:
x(1), x(2), x(3), ...
with x(n) ≠ x(m) whenever n ≠ m.

Let A(k) = {x(2k)} ∈ D for all k,
A = ∪A(k) = {x(n): n even}. Then A is not finite.
Its complement ￢A contains the countable set B = {x(n): n odd}, and hence is not finite either.
Therefore, A = ∪A(k) is not in D. Hence, D is not a σ-algebra. ::