# 一題代數證明題 (和有限聯集覆蓋有關)

If D is any class of sets and R(D) is the ring generated by D, then every set in R(D) can be covered by (that is, is contained in) a finite union of sets of D. [Hint: The class K of sets that can be covered by finite unions of sets of D forms a ring.]

### 1 個解答

• 最佳解答

Claim: K is a ring.

Proof:

Let A and B be sets in K. There are collections {A1, A2..., An} and {B1, B2, ..., Bm} of sets in D such that

A ⊆ A1 ∪ A2 ∪ ... ∪ An

B ⊆ B1 ∪ B2 ∪ ... ∪ Bm

Then, clearly

A \ B ⊆ A1 ∪ A2 ∪ ... ∪ An (since A \ B ⊆ A)

A ∪ B ⊆ A1 ∪ A2 ∪ ... ∪ An ∪ B1 ∪ B2 ∪ ... ∪ Bm

so A \ B and A ∪ B are both in K.

Hence, K is a ring. ::

Claim: Every set in R(D) can be covered by a finite union of sets of D.

In other words, R(D) ⊆ K.

Proof: K is a ring containing D (since every set in D is covered by itself). R(D) is the smallest ring containing D; hence R(D) ⊆ K.

::

2006-07-18 03:14:28 補充：

（我的書裡沒有 "rings contain the empty set" 這一條）The empty set is in K, since it is (and hence is covered by) the union of 0 sets. (Note that D could be empty.)