电大数学分析专题研究形考一18题答案

证明：先证f (A∪B )−1⊃f

(A )−1⋃f (B )−1 事实上，因为A ∪B ⊃A ，故f (A∪B )−1⊃f

(A )−1 同理有f (A∪B )

−1⊃f (B )−1.故有f (A∪B )−1⊃f (A )−1f (B )−1 再证：f (A∪B )−1⊂f

(A )−1Uf (B )−1 任取x ∈f (A∪B )−1，则∃y ∈A ∪B 使得f (x )=y

若y ∈A 则x ∈f (A )

−1，否则y ∈B ，x ∈f (B )−1，故x ∈f (A )−1Uf (B )−1 从而f (A∪B )−1⊂f (A )−1⋃f (B )−1

于是有f (A∪B )−1=f

(A )−1⋃f (B )−1