維正 發問時間: 科學數學 · 1 0 年前

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.]

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請問第一個claim的證明需不需要加證the empty set is in K? 因為我的書上對ring的定義有這一條。

1 個解答

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  • Eric
    Lv 6
    1 0 年前
    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.)

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