集合 - 証明

將要利用以下性質:
(1) X∪Y=(X-Y)∪Y
(2)∩(k=1→n)(XUYk)=XU[∩(k=1→n)Yk]



A1∪A2∪A3∪...∪An
=Ak∪A(k+1)∪A(k+2)∪...∪An∪A1∪A2∪...∪A(k-1), 其中 k=1, 2, ..., n
=Ak∪A(k+1)∪A(k+2)∪...∪An∪A1∪A2∪...∪A(k-1)∪Ak
=(Ak-A(k+1))∪A(k+1)∪A(k+2)∪...∪An∪A1∪A2∪...∪A(k-1)∪Ak (性質(1))
=(Ak-A(k+1))∪(A(k+1)-A(k+2))∪A(k+2)∪...∪An∪A1∪A2∪...∪A(k-1)∪Ak
=...
=(Ak-A(k+1))∪(A(k+1)-A(k+2))∪...∪(An-A1)∪(A1-A2)∪...∪(A(k-1)-Ak)∪Ak
=(A1-A2)∪...∪(A(k-1)-Ak)∪(Ak-A(k+1))∪(A(k+1)-A(k+2))∪...∪(An-A1)∪Ak

∩(k=1→n)[(A1∪A2∪A3∪...∪An)]
=∩(k=1→n)[(A1-A2)∪...∪(An-A1)∪Ak]
=(A1-A2)∪...∪(An-A1)∪[∩(k=1→n)Ak] (性質(2))

∴(A1∪A2∪A3∪...∪An)=(A1-A2)∪...∪(An-A1)∪[∩(k=1→n)Ak]



(A1 - A2) ∪( A2 - A3 ).∪...∪(An - A1) = ( A1∪A2...∪An ) - (A1∩A2∩...∩An).
是正確的。可以獨立証明，亦可以利用上面結果。

現利用上面結果証明
(A1 - A2) ∪( A2 - A3 )∪...∪(An - A1) = (A1∪A2...∪An ) - (A1∩A2∩...∩An).

將要利用以下性質:
(3) (A∪B)-C=(A-C)∪(B-C)
(4) 若C⊂B, 則 (A-B)-C=A-B


已証 (A1∪A2∪A3∪...∪An)=(A1-A2)∪...∪(An-A1)∪(A1∩A2∩...∩An)
兩邊減(A1∩A2∩...∩An)
∴(A1∪A2∪A3∪...∪An)- (A1∩A2∩...∩An)
=[(A1-A2)∪...∪(An-A1)]-(A1∩A2∩...∩An)
=[(A1-A2)- (A1∩A2∩...∩An)]∪...∪[(An-A1)-(A1∩A2∩...∩An)]∪[(A1∩A2∩...∩An)
- (A1∩A2∩...∩An)] (性質(3))
=[(A1-A2)]∪...∪[(An-A1)]∪ø (性質(4))
=[(A1-A2)]∪...∪[(An-A1)]

哇, 超感謝!
不過我想補充問一下, 我想的能再收歛一下, 變成更準確更小的等式嗎?

2012-08-05 15:49:49 補充：
另外, 性質2 可否証明一下? 這個雖然相當顯然易見...

2012-08-05 23:28:48 補充：
性質2 真的怪怪的...