高等微積分證明題
2009-05-18 5:21 pm
回答 (1)
2009-05-18 6:58 pm
✔ 最佳答案
x∈A-U(B_n)<=>x∈A and x ∈/ U(B_n)
<=>x∈(A-B_1),x∈(A-B_2),x∈(A-B_3),...
<=>x∈∩(A|B_n)
So A-U(B_n)=∩(A|B_n)
Let say x∈U(A_n)
Then x∈A_i for some i
Since A_i is open, there is an B_ε(x) such that B_ε(x)⊂A_i
That is B_ε(x)⊂U(A_n)
So U(A_n) is open
收錄日期: 2021-04-26 13:49:42
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20090518000051KK00299