高等微積分證明題

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